heat 4e sm chap05

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

5-1

Solutions Manual for

Heat and Mass Transfer: Fundamentals & Applications Fourth Edition

Yunus A. Cengel & Afshin J. Ghajar McGraw-Hill, 2011

Chapter 5 NUMERICAL METHODS IN HEAT

CONDUCTION

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5-2

Why Numerical Methods?

5-1C Analytical solutions provide insight to the problems, and allows us to observe the degree of dependence of solutions on certain parameters. They also enable us to obtain quick solution, and to verify numerical codes. Therefore, analytical solutions are not likely to disappear from engineering curricula.

5-2C Analytical solution methods are limited to highly simplified problems in simple geometries. The geometry must be such that its entire surface can be described mathematically in a coordinate system by setting the variables equal to constants. Also, heat transfer problems can not be solved analytically if the thermal conditions are not sufficiently simple. For example, the consideration of the variation of thermal conductivity with temperature, the variation of the heat transfer coefficient over the surface, or the radiation heat transfer on the surfaces can make it impossible to obtain an analytical solution. Therefore, analytical solutions are limited to problems that are simple or can be simplified with reasonable approximations.

5-3C In practice, we are most likely to use a software package to solve heat transfer problems even when analytical solutions are available since we can do parametric studies very easily and present the results graphically by the press of a button. Besides, once a person is used to solving problems numerically, it is very difficult to go back to solving differential equations by hand.

5-4C The energy balance method is based on subdividing the medium into a sufficient number of volume elements, and then applying an energy balance on each element. The formal finite difference method is based on replacing derivatives by their finite difference approximations. For a specified nodal network, these two methods will result in the same set of equations.

5-5C The analytical solutions are based on (1) driving the governing differential equation by performing an energy balance on a differential volume element, (2) expressing the boundary conditions in the proper mathematical form, and (3) solving the differential equation and applying the boundary conditions to determine the integration constants. The numerical solution methods are based on replacing the differential equations by algebraic equations. In the case of the popular finite difference method, this is done by replacing the derivatives by differences. The analytical methods are simple and they provide solution functions applicable to the entire medium, but they are limited to simple problems in simple geometries. The numerical methods are usually more involved and the solutions are obtained at a number of points, but they are applicable to any geometry subjected to any kind of thermal conditions.

5-6C The experiments will most likely prove engineer B right since an approximate solution of a more realistic model is more accurate than the exact solution of a crude model of an actual problem.


5-3

Finite Difference Formulation of Differential Equations

5-7C A point at which the finite difference formulation of a problem is obtained is called a node, and all the nodes for a problem constitute the nodal network. The region about a node whose properties are represented by the property values at the nodal point is called the volume element. The distance between two consecutive nodes is called the nodal spacing, and a differential equation whose derivatives are replaced by differences is called a difference equation.

5-8 The finite difference formulation of steady two-dimensional heat conduction in a medium with heat generation and constant thermal conductivity is given by

022 ,

22 ∆1,,1,,1,,1 =+

+−+

+− +−+ eTTTTTT nmnmnmnmnmnmn &

ase by simply adding another index j to the temperature in the z direction, and another difference term for the z direction as

−m

∆ kyx

in rectangular coordinates. This relation can be modified for the three-dimensional c

0222 ,,

21,,,,1,,

2,1,,,,1,

2,,1,,,,1 =+

∆

+−+

∆

+−+

∆

+− +−+−+−

ke

z

TTT

y

TTT

x

TTT jnmjnmjnmjnmjnmjnmjnmjnmjnmjnm &

5-9 A plane wall with variable heat generation and constant thermal conductivity is subjected to uniform heat flux 0q& at theleft (node 0) and convection at the right boundary (node 4). Using th

e finite difference form of the 1st derivative, the finite

mal conductivity is constant and there is nonuniform

Analysis ft and right boundaries can be expressed analytically as

at x = 0:

difference formulation of the boundary nodes is to be determined.

Assumptions 1 Heat transfer through the wall is steady since there is no indication of change with time. 2 Heat transfer is one-dimensional since the plate is large relative to its thickness. 3 Therheat generation in the medium. 4 Radiation heat transfer is negligible.

The boundary conditions at the le

0)0( q

dxdTk =−

q0

∆x

e(x)

1

h, T∞

• •• ••0 2 3 4

at x = L : ])([)(

∞−=− TLThdx

LdTk

Replacing derivatives by differences using values at the closest nodes, the of the 1st derivative of temperature at the

ndaries (nodes 0 and 4) can be expressed as finite difference formbou

xTT

dxdT

xTT

dxdT

∆−

≅∆−

≅ 3401 and =4 m right,

Substituting, the finite difference formulation of the boundary nodes become

at x = 0:

= 0 m left,

001 q

xTT

k =∆−

−

at x = L : ][ 434

∞−=∆−

− TThxTT

k


5-4

e

ime. 2 Heat transfer is e-dimen e plate is large relative to its thickness. 3 Thermal conductivity is constant and there is nonuniform t gen di m. 4 Convection heat transfer is negligible.

alysis nditions at the left and right boundaries can be expressed analytically as

5-10 A plane wall with variable heat generation and constant thermal conductivity is subjected to insulation at the left (nod0) and radiation at the right boundary (node 5). Using the finite difference form of the 1st derivative, the finite difference formulation of the boundary nodes is to be determined.

Assumptions 1 Heat transfer through the wall is steady since there is no indication of change with ton sional since thhea eration in the me u

An The boundary co

At x = 0: 0or 0 ==−dxdx

k

At x = L :

)0()0( dTdT

])([)( 4 LT

LdTk =− εσ 4

surrTdx

−

boundari

Replacing derivatives by differences using values at the closest nodes, the finite difference form of the 1st derivative of temperature at the

es (nodes 0 and 5) can be expressed as

xTT

dxdT

xdx ∆= 0 m left,

TTdT∆−

≅−

≅=

45

5 m right,

01 and

Substituting, the finite difference formulation of the boundary nodes become

At x = 0:

Insulated

∆x

(x)

1

ε

•

e

• • ••0 2 3 4•

5

Tsurr

adiation R

0101 or 0 TT

xTT

k ==∆−

−

][ 445

45surrTT

xTT

k −=∆−

− εσ At x = L :


5-5

One-Dimensional Steady Heat Conduction

5-11C The finite difference form of a heat conduction problem by the energy balance method is obtained by subdividing the medium into a sufficient number of volume elements, and then applying an energy balance on each element. This is done by first selecting the nodal points (or nodes) at which the temperatures are to be determined, and then forming elements (or control volumes) over the nodes by drawing lines through the midpoints between the nodes. The properties at the node such as the temperature and the rate of heat generation represent the average properties of the element. The temperature is assumed to vary linearly between the nodes, especially when expressing heat conduction between the elements using Fourier’s law.

5-12C The basic steps involved in the iterative Gauss-Seidel method are: (1) Writing the equations explicitly for each unknown (the unknown on the left-hand side and all other terms on the right-hand side of the equation), (2) making a reasonable initial guess for each unknown, (3) calculating new values for each unknown, always using the most recent values, and (4) repeating the process until desired convergence is achieved.

5-13C In a medium in which the finite difference formulation of a general interior node is given in its simplest form as

022

11 =+∆

+− +−

ke

xTTT mmmm &

(a) heat transfer in this medium is steady, (b) it is one-dimensional, (c) there is heat generation, (d) the nodal spacing is constant, and (e) the thermal conductivity is constant.

5-14C In the finite difference formulation of a problem, an insulated boundary is best handled by replacing the insulation by a mirror, and treating the node on the boundary as an interior node. Also, a thermal symmetry line and an insulated boundary are treated the same way in the finite difference formulation.

5-15C A node on an insulated boundary can be treated as an interior node in the finite difference formulation of a plane wall by replacing the insulation on the boundary by a mirror, and considering the reflection of the medium as its extension. This way the node next to the boundary node appears on both sides of the boundary node because of symmetry, converting it into an interior node.

5-16C In the energy balance formulation of the finite difference method, it is recommended that all heat transfer at the boundaries of the volume element be assumed to be into the volume element even for steady heat conduction. This is a valid recommendation even though it seems to violate the conservation of energy principle since the assumed direction of heat conduction at the surfaces of the volume elements has no effect on the formulation, and some heat conduction terms turn out to be negative.


5-6

5-17 A circular fin of uniform cross section is attached to a wall. The finite difference equations for all nodes are to be obtained, the nodal temperatures along the fin and the heat transfer rate are to be determined and compared with analytical solutions. Assumptions 1 Heat transfer along the fin is steady and one-dimensional. 2 Thermal properties are constant. 3 Heat transfer by radiation is negligible. Properties The thermal conductivity of the fin is given as 240 W/m·K. Analysis (a) The nodal spacing is given to be ∆x = 10 mm. Then the number of nodes M becomes

61mm 10∆mm 501 =+=+=

xLM

5

es, and we can use the general finite difference relation expressed as

The base temperature at node 0 is given to be T0 = 350°C. There areunknown nodal temperatures, thus we need to have 5 equations to determine them uniquely. Nodes 1, 2, 3, and 4 are interior nod

0))((11 =−∆+∆−

+∆−

∞+−

mmmmm TTxph

xTT

kAx

TTkA

0)(22

11 =−∆

++− ∞+− mmmm TTkA

xhpTTT

where

04167.0 W/m240(

)m 01.0)(K W/m250(44 2222=

⋅=

∆=

∆kD

xhkA

xhp )m 01.0)(K⋅

he finite d equation for node 5 at the fin tip (convection boundary) is obtained by applying an energy balance on ent about that node:

T ifferencethe half volume elem

0)(2 5

54 =−⎟⎠⎞

⎜⎝⎛ +

∆+

∆−

∞ TTAxphxTT

kA → 0)(2 554 =−⎟

⎠⎞

⎜⎝⎛ +

∆∆+− ∞ TTAxp

kAxhTT

where

03125.0122

=⎟⎠⎞

⎜⎝⎛ +

∆∆=⎟

⎠⎞

⎜⎛∆ pxh ⎝

+∆

Dx

kxhAx

kA

Then, m = 1: 0)(04167.02 1210 =−++− ∞ TTTTT

m = 2: 0)(04167.02 2321 =−++ ∞ TTTTT −

m = 3: 0)(04167.02 3432 =−++− TTTTT ∞

m = 4: 0)(04167.02 4543 =−++− ∞ TTTTT

m = 5: 054 +−TT 0)(03125. 5 =−∞ TT

rmined by solving the 5 equations above simultaneously with an nk EES screen to solve the above equations:

67*(25-T_3)=0

T 2*T_4+T_5+ 7*(25-T_4)=

Solving by EES software, we get , ,

(b) The nodal temperatures under steady conditions are deteequation solver. Copy the following lines and paste on a bla T_0=350

T_0-2*T_1+T_2+0.04167*(25-T_1)=0

T_1-2*T_2+T_3+0.04167*(25-T_2)=0

T_2-2*T_3+T_4+0.041

_3- 0.0416 0

T_4-T_5+0.03125*(25-T_5)=0

C 304.1 °=1T C 269.9 °=2T C 245.9 °=3T , C 231.0 °= C 224.8 °=5T 4T ,

From Chapter 3, the analytical solution for the temperature variation along the fin (for convection from fin tip) is given as

mLmkhmL

xLmmkhxLmTTTxT

b sinh)/(cosh)(sinh)/()(cosh)(

+−+−

=−−

∞

∞


5-7

The nodal temperatures for analytical and numerical solutions are tabulated in the following table:

T(x),°C x, m

Analytical Numerical 0 350.0 350.0 0.01 304.0 304.1 0.02 269.7 269.9 0.03 245.6 245.9 0.04 230.7 231.0 0.05 224.5 224.8

The comparison of the analytical and numerical solutions is shown in the following figure:

x, m

0.00 0.01 0.02 0.03 0.04 0.05

200

250

300

350

Analytical

T, °

C

Numerical

(c) The rate of heat transfer from a single fin is simply the sum of the heat transfer from the nodal elements,

W99.2=

−⎟⎠⎞

⎜⎝⎛ +

∆+−+++∆+−

∆=

−==

∞∞∞

=∞

=∑∑

)(2

)4()(2

)(

543210

5

0 surface,

5

0 element,fin

TTAxphTTTTTxhpTTxhp

TThAQQm

mmm

m&&

From Chapter 3, the analytical solution for the heat transfer rate of fin with convection from the tip is,

W98.8=

°−°°=++

−= ∞

)7901.0)(C 25C 350)(C W/3848.0(sinh)/(coshcosh)/(sinh)( tipconv mLmkhmL

mLmkhmLTThpkAQ bc&

where

1m 41.20 −==ckA

hpm , m 03142.0== Dp π , 252

m 10854.74

−×==DAcπ

Discussion For part (b), the comparison between the analytical and numerical solutions is excellent, with agreement within ±0.15%. For part (c), the comparison between the analytical and numerical solutions is within ±0.5%.


5-8

5-18 A circular aluminum fin of uniform cross section with adiabatic tip is attached to a wall. The finite difference equations for all nodes are to be obtained and solved using Gauss-Seidel iterative method, and the nodal temperatures along the fin are to be determined and compared with analytical solution.

Assumptions 1 Heat transfer along the fin is steady and one-dimensional. 2 Thermal properties are constant. 3 Heat transfer by radiation is negligible.

Properties The thermal conductivity of the fin is given as 237 W/m·K.

Analysis (a) The nodal spacing is given to be ∆x = 10 mm. Then the number of nodes M becomes

61cm 1∆cm 51 =+=+=

xLM

2,

finite difference relation expressed in explicit form as

The base temperature at node 0 is given to be T0 = 300°C. There are 5 unknown nodal temperatures, thus we need to have 5 equations to determine them uniquely. Nodes 1, 3, and 4 are interior nodes, and we can use the general

0))((11 =−∆+∆−

+∆−

∞+−

mmmmm TTxph

xTT

kAx

TTkA

⎟⎟ ⎠⎝⎠⎝

The finite difference equation for

⎞⎜⎜⎛ ∆

++⎟⎟⎞

⎜⎜⎛ ∆

+= ∞+−

−

TkA

xhpTTkA

xhpT mmm

2

11

122

node 5 at the fin tip (adiabatic) is obtained by applying an energy balance on the half volume element about that node:

0))((2 554 − =−∆+

∆ ∞ TTxphxTT

kA → ⎟⎟⎠

⎞⎜⎜⎝

⎛ ∆+⎟

⎟⎠

⎞⎜⎜⎝

⎛ ∆+= ∞

−

TkA

xhpTkA

xhpT2

4

12

5 22

Then,

m = 1: 1875.04938.04938.0 201 ++= TTT

m = 2: 1875.04938.04938.0 312 ++= TTT

m = 3: 1875.04938.04938.0 423 ++= TTT

m = 4: 4 1875.04938.04938.0 53 ++= TTT

guesses as , the results obtained from successive iterations are listed in the following table:

dal temp ture,°C

m = 5: 09876.0 45 += TT

(b) By letting the initial

1875.

C 25054321 °===== TTTTT

No eraIteration T1 T2 T3 T4 T5

1 271.8 257.8 251.0 247.6 244.7 2 275.6 260.2 250.9 244.9 242.1 3 276.8 260.8 249.9 243.1 240.3 4 277.1 260.4 248.8 241.7 238.9 5 276.9 259.8 247.8 240.5 237.7 6 276.6 259.2 246.9 239.5 236.7 7 276.3 258.6 246.1 238.6 235.9 8 276.0 .0 .4 .9 .1 258 245 237 235··· ··· ··· ··· ··· ··· 52 273.7 253.9 240.1 232.0 229.3


5-9

1T , 2T , 240.1

Hence, the converged nodal temperatures are

= = °C 273.7 ° C 253.9 ° C C 232.0 °=4T , C 229.3 °=5T =3T ,

From Chapter 3, the analytical solution for the temperature variation along the fin (for adiabatic tip) is given as

mL

xLmTTTxT

b cosh)(cosh)( −

=−−

∞

∞


T(x),°C x, m

Analytical Numerical

0 300.0 300.0

0.01 273.5 273.7

0.02 253.5 253.9

0.03 239.6 240.1

0.04 231.4 232.0

0.05 228.7 229.3


x, m

0.00 0.01 0.02 0.03 0.04 0.05

T, °

C

200

220

240

260

280

300

320

AnalyticalNumerical

Discussion The comparison between the analytical and numerical solutions is excellent, with agreement within ±0.3%.


5-10

5-19 A plane wall with no heat generation is subjected to specified temperature at the left (node 0) and heat flux at the right boundary (node 8). The finite difference formulation of the boundary nodes and the finite difference formulation for the rate of heat transfer at the left boundary are to be determined.

Assumptions 1 Heat transfer through the wall is given to be steady, and the thermal conductivity to be constant. 2 Heat transfer is one-dimensional since the plate is large relative to its thickness. 3 There is no heat generation in the medium.

Analysis Using the energy balance approach and taking the direction of all heat transfers to be towards the node under consideration, the finite difference formulations become

40°C∆x

No heat generation

•

3000 W/m2

• • •• •• ••0 1 2 3 4 5 6 7 8

Left boundary node: 400 =T

Right boundary node: 03000or 0 870

87 =+∆−

=+∆−

xTT

kAqxTT

kA &

Heat transfer at left surface:

001surfaceleft =

∆−

+xTT

kAQ&

5-20 A plane wall with variable heat generation and constant thermal conductivity is subjected to uniform heat flux 0q& at theleft (node 0) an

d convection at the right boundary (node 4). The finite difference formulation of the boundary nodes is to be

heat the node under consideration, the finite difference

formulations become

Left boundary node:

determined.

Assumptions 1 Heat transfer through the wall is given to be steady, and the thermal conductivity to be constant. 2 Heat transfer is one-dimensional since

∆x

)(xe&

1

h, T∞

•

0q&

•• ••0 2 3 4

the plate is large relative to its thickness. 3 Radiation heat transfer is negligible.

Analysis Using the energy balance approach and taking the direction of alltransfers to be towards

0)2/(001

0 =∆+∆−

+ xAexTT

kAAq &&

0)2/()( 4443 =∆+−+

∆−

∞ xAeTThAxTT

kA & Right boundary node:


5-11

5-21 A plane wall with variable heat generation and constant thermal conductivity is subjected to insulation at the left (node 0) and radiation at the right boundary (node 5). The finite difference formulation of the boundary nodes is to be determined.

Assumptions 1 Heat transfer through the wall is given to be steady and one-dimensional, and the thermal conductivity to be constant. 2 Convection heat transfer is negligible.


Left boundary node: 0)2/(001 =∆+

∆−

xAexTT

kA &

Right boundary node: 0)2/()( 5544

54

surr =∆+∆−

+− xAexTT

kATTAσ &ε

ll is diation at the right boundary (node 2). The complete finite difference

nsional, and the thermal conductivity to be eneration.

ode under consideration, the

finite difference formulations become

Node 0 (at left boundary):

Insulated

∆x

1

ε•

)(xe& Radiation

Tsurr

54•0 • ••

3•2

5-22 A composite plane wall consists of two layers A and B in perfect contact at the interface where node 1 is. The wainsulated at the left (node 0) and subjected to raformulation of this problem is to be obtained.

Assumptions 1 Heat transfer through the wall is given to be steady and one-dimeconstant. 2 Convection heat transfer is negligible. 3 There is no heat g

Analysis Using the energy balance approach and taking the directionof all heat transfers to be towards the n

Insulated

∆x

1

ε

• ••0 2

A

Tsurr

RadiationB

0101 0 TT

xTT

Ak A =→=∆−

01210 =∆−

+∆−

xTT

AkxTT

Ak BA Node 1 (at the interface):

0)( 2142

4surr =

∆−

+−xTT

AkTTA Bεσ Node 2 (at right boundary):


5-12

5-23 A pin fin with negligible heat transfer from its tip is considered. The complete finite difference formulation for the determination of nodal temperatures is to be obtained.

Assumptions 1 Heat transfer through the pin fin is given to be steady and one-dimensional, and the thermal conductivity to be constant. 2 Convection heat transfer coefficient is constant and uniform. 3 Heat loss from the fin tip is given to be negligible.

ConvectioT0

h, T∞

εRadiation

Tsurr

∆x D 2•1 ••0Analysis The nodal network consists of 3 nodes, and the base temperature T0 at node 0 is specified. Therefore, there are two unknowns T1 and T2, and we need two equations to determine them. Using the energy balance approach and taking the direction of all heat transfers to be towards the node under consideration, the finite difference formulations become

Node 1 (at midpoint):

[ ] 0)273()273()())(( 41

4surr1

1210 =+−+∆+−∆+∆−

+∆x−

∞ TTxpTTxphxTT

kATT

k εσ

Node 2 (at fin tip):

A

[ ] 0)273()273()2/())(2/( 42

4surr2

21 =+−+∆+−∆+∆x−

∞ TTxpTTxphTT

kA εσ

here is the cross-sectional area and 4/2DA π= Dp π=w is the perimeter of the fin.


5-13

5-24 A uranium plate is subjected to insulation on one side and convection on the other. The finite difference formulation of this problem is to be obtained, and the nodal temperatures under steady conditions are to be determined.

Assumptions 1 Heat transfer through the wall is steady since there is no indication of change with time. 2 Heat transfer is one-dimensional since the plate is large relative to its thickness. 3 Thermal conductivity is constant. 4 Radiation heat transfer is negligible.

Properties The thermal conductivity is given to be k = 34 W/m⋅°C.

Analysis The number of nodes is specified to be M = 6. Then the nodal spacing ∆x becomes

m 01.01-61−M

This problem involves 6 unknown nodal temperatures, and thus we need to have 6 equations to determine them uniquely. Node 0 is oninsulated boundary, and thus we can treat it as an interior note by using the mirror image concept. Nodes 1, 2, 3, and 4 are interior nodes, and thus for the

m 05.0===∆

L

m we can use the general finite difference relation expressed as

x

022∆ kx

Finally, the finite difference equation for node 5 on the right surface subjected to convection is obtained by applying an energy balance on the

11 =++− +− eTTT mmmm &

, for m = 0, 1, 2, 3, and 4

half volume element about node 5 and taking the direction of all heat transfers to be towards the node er cons eration:

∆x

1•

e&

54••0 ••

3•2

Insulatedh, T∞

und id

0)2/()( :)convection -surface(right 5 Node

02

:(interior) 2 Node2

321 =++−

keTTT &

02

:(interior) 4 Node2

543 =++−

∆

keTTT

x&

02

:(interior) 3 Node2

432 =++−

keTTT &

02

:(interior) 1 Node

02

:insulated)-surface(Left 0 Node

545

210

2101

=∆+∆−

+−

∆

∆

=++−

=+∆

+−

∞ xexTT

kTTh

x

x

eTTTke

xTTT

&

&

&

where

This system of 6 equations with six unknown temperatures constitute the finite difference formulation of the problem.

(b) The 6 nodal temperatures under steady conditions are determined by solving the 6 equations above simultaneously with an equation solver to be

T0 = 552.1°C, T1 = 551.2°C, T2 = 548.5°C, T3 = 544.1°C, T4 = 537.9°C, and T5 = 530.0°C

Discussion This problem can be solved analytically by solving the differential equation as described in Chap. 2, and the analytical (exact) solution can be used to check the accuracy of the numerical solution above.

2∆ kx

C.30 and C, W/m60 C, W/m34 , W/m106 m, 01.0 235 °=°⋅=°⋅=×==∆ ∞Thkex &


5-14

Prob. 5-24 is reconsidered. The nodal temperatures under steady5-25 conditions are to be determined.

nalysis is solved using EES, and the solution is given below.

nt"

peratures can be determined"

2=0 "for node 5"

The nodT0 = 552°C, T1 = 551°C, T2 = 549°C, T3 = 544°C, T4 = 538°C, and T5 = 530°C

A The problem "GIVEN" e_gen=6e5 [W/m^3] "heat generation" dx=0.01 [m] "mesh size" h=60 [W/m^2-K] "convection coefficiek=34 [W/m-K] "thermal conductivity" T_inf=30 [C] "ambient temperature" "ANALYSIS" "Using the finite difference method, the nodal tem(T_1-T_0)/dx^2+e_gen/(2*k)=0 "for node 0" (T_0-2*T_1+T_2)/dx^2+e_gen/k=0 "for node 1" (T_1-2*T_2+T_3)/dx^2+e_gen/k=0 "for node 2" (T_2-2*T_3+T_4)/dx^2+e_gen/k=0 "for node 3" (T_3-2*T_4+T_5)/dx^2+e_gen/k=0 "for node 4" h*(T_inf-T_5)+k*(T_4-T_5)/dx+e_gen*dx/

al temperatures are determined to be


5-15

5-26 A long triangular fin attached to a surface is considered. The nodal temperatures, the rate of heat transfer, and the fin efficiency are to be determined numerically using 6 equally spaced nodes.

Assumptions 1 Heat transfer along the fin is given to be steady, and the temperature along the fin to vary in the x direction only so that T = T(x). 2 Thermal conductivity is constant.

Properties The thermal conductivity is given to be k = 180 W/m⋅°C. The emissivity of the fin surface is 0.9.

Analysis The fin length is given to be L = 5 cm, and the number of nodes is specified to be M = 6. Therefore, the nodal spacing ∆x is

m 01.01-6m 05.0

1==

−=∆

MLx

T0 h, T∞

θ ∆x • •• • ••0 1 2 3 4 5

Tsurr

The temperature at node 0 is given to be T0 = 180°C, and the temperatures at the remaining 5 nodes are to be determined. Therefore, we need to have 5 equations to determine them uniquely. Nodes 1, 2, 3, and 4 are interior nodes, and the finite difference formulation for a general interior node m is obtained by applying an energy balance on the volume element of this node. Noting that heat transfer is steady and there is no heat generation in the fin and assuming heat transfer to be into the medium from all sides, the energy balance can be expressed as

0})273([)( 0 44surrsurfaceconvrightleft

sides all ∆∆ xx

Note that heat transfer areas are different for each node in

11 =+−+−+−

+−

→= ∞+−

mmmmmm TTATThA

TTkA

TTkAQ εσ&

this case, and using geometrical relations, they can be expressed as

∑

( )[ ]( )[ ]

)cos/(2widthLength2

tan2/12width)Height(

tan2/12width)Height(

surfA ace

2/1@right

2/1@left

θ

θ

θ

xw

xmLwA

xmLwA

m

m

∆=××=

∆+−=×=

∆−−=×=

+

−

Substituting,

0]})273([)(){cos/(2

tan])5.0([2tan])5.0([2 1 −−− − mm TTmLkw θ

44surr

1

=+−+−∆+∆−

∆+−+∆

∆

∞

+

mm

mm

TTTThxwx

TTxmLkwx

x

εσθ

θ

Dividing each term by ∆x gives θtan2kwL /

( ) ( ) 0])273([sin

)()(sin

)()(2/11)(2/11 44surr

22

11 =+−∆

+−∆

+−⎥⎦⎤

⎢⎣⎡ ∆

+−+−⎥⎦⎤

⎢⎣⎡ ∆

−− ∞+− mmmmmm TTkL

xTTkL

xhTTLxmTT

Lxm

θεσ

θ

Substituting,

0])273([sin

)()(sin

)()(5.11)(5.01 41

4surr

2

1

2

1210 =+−∆

+−∆

+−⎥⎦⎤

⎢⎣⎡ ∆−+−⎥⎦

⎤⎢⎣⎡ ∆− ∞ TT

kLxTT

kLxhTT

LxTT

Lx

θεσ

θ m = 1:

0])273([sin

)()(sin

)()(5.21)(5.11 2321 +−⎥⎦⎤

⎢⎣⎡ −+−⎥⎦

⎤⎢⎣⎡ −

kTT

LTT

L4

24

surr

2

2

2=+−

∆+−

∆∆∆∞ TT

kLxTT

Lxhxx

θεσ

θ m = 2:

0])273([sin

)()(sin

)()(5.31)(5.21 43

4⎡m = 3: surr

2

3

2

3432 =+−∆

+−∆

+−⎥⎦⎤

⎢⎣⎡ ∆−+−⎥⎦

⎤⎢⎣

∆− ∞ TT

kLxTT

kLxhTT

LxTT

Lx

θεσ

θ

m = 4: 0])273([sin

)()(sin

)()(5.41)(5.31 44

4surr

2

4

2

4543 =+−∆

+−∆

+−⎥⎦⎤

⎢⎣⎡ ∆−+−⎥⎦

⎤⎢⎣⎡ ∆− ∞ TT

kLxTT

kLxhTT

LxTT

Lx

θεσ

θ

An energy balance on the 5th node gives the 5th equation,

m = 5: 0])273([cos

2/2)(cos

2/2tan2

2 45

4surr5

54 =+−∆

+−∆

+∆−∆

∞ TTxTTxhxTTxk

θεσ

θθ


5-16

4555

TTATThAQQ mm −++−== ∑∑∑ ∞ εσ&&

for the boundary nodes 0 and 5, and twice as large for the interior es 1, 2, , and 4, we have

Solving the 5 equations above simultaneously for the 5 unknown nodal temperatures gives

T1 = 177.0°C, T2 = 174.1°C, T3 = 171.2°C, T4 = 168.4°C, and T5 = 165.5°C

(b) The total rate of heat transfer from the fin is simply the sum of the heat transfer from each volume element to the ambient, and for w = 1 m it is determined from

])273[()( 4surr

0 surface,

0 surface,

0 element,fin m

mm

mmm

===

Noting that the heat transfer surface area is θcos/xw∆nod 3

[ ]

W537=]})273[(])273[(2

])273[(2])273[(2])273[(2])273{[(cos

)()(2)(2)(2)(2)(cos

4surr

45

4surr

44

4surr

43

4surr

42

4surr

41

4surr

40

543210fin

TTTT

TTTTTTTTxw

TTTTTTTTTTTTxwhQ

−++−++

−++−++−++−+∆

+

−+−+−+−+−+−∆

= ∞∞∞∞∞∞

θεσ

θ&


5-17

5-27 Prob. 5-26 is reconsidered. The effect of the fin base temperature on the fin tip temperature and the rate of heat transfer from the fin is to be investigated.

Analysis The problem is solved using EES, and the solution is given below.

"GIVEN" k=180 [W/m-C] L=0.05 [m] b=0.01 [m] w=1 [m] T_0=180 [C] T_infinity=25 [C] h=25 [W/m^2-C] T_surr=290 [K] M=6 epsilon=0.9 tan(theta)=(0.5*b)/L sigma=5.67E-8 [W/m^2-K^4] “Stefan-Boltzmann constant" "ANALYSIS" "(a)" DELTAx=L/(M-1) "Using the finite difference method, the five equations for the temperatures at 5 nodes are determined to be" (1-0.5*DELTAx/L)*(T_0-T_1)+(1-1.5*DELTAx/L)*(T_2-T_1)+(h*DELTAx^2)/(k*L*sin(theta))*(T_infinity-T_1)+(epsilon*sigma*DELTAX^2)/(k*L*sin(theta))*(T_surr^4-(T_1+273)^4)=0 "for mode 1" (1-1.5*DELTAx/L)*(T_1-T_2)+(1-2.5*DELTAx/L)*(T_3-T_2)+(h*DELTAx^2)/(k*L*sin(theta))*(T_infinity-T_2)+(epsilon*sigma*DELTAX^2)/(k*L*sin(theta))*(T_surr^4-(T_2+273)^4)=0 "for mode 2" (1-2.5*DELTAx/L)*(T_2-T_3)+(1-3.5*DELTAx/L)*(T_4-T_3)+(h*DELTAx^2)/(k*L*sin(theta))*(T_infinity-T_3)+(epsilon*sigma*DELTAX^2)/(k*L*sin(theta))*(T_surr^4-(T_3+273)^4)=0 "for mode 3" (1-3.5*DELTAx/L)*(T_3-T_4)+(1-4.5*DELTAx/L)*(T_5-T_4)+(h*DELTAx^2)/(k*L*sin(theta))*(T_infinity-T_4)+(epsilon*sigma*DELTAX^2)/(k*L*sin(theta))*(T_surr^4-(T_4+273)^4)=0 "for mode 4" 2*k*DELTAx/2*tan(theta)*(T_4-T_5)/DELTAx+2*h*(0.5*DELTAx)/cos(theta)*(T_infinity-T_5)+2*epsilon*sigma*(0.5*DELTAx)/cos(theta)*(T_surr^4-(T_5+273)^4)=0 "for mode 5" T_tip=T_5 "(b)" Q_dot_fin=C+D "where" C=h*(w*DELTAx)/cos(theta)*((T_0-T_infinity)+2*(T_1-T_infinity)+2*(T_2-T_infinity)+2*(T_3-T_infinity)+2*(T_4-T_infinity)+(T_5-T_infinity)) D=epsilon*sigma*(w*DELTAx)/cos(theta)*(((T_0+273)^4-T_surr^4)+2*((T_1+273)^4-T_surr^4)+2*((T_2+273)^4-T_surr^4)+2*((T_3+273)^4-T_surr^4)+2*((T_4+273)^4-T_surr^4)+((T_5+273)^4-T_surr^4))


5-18

T0 [C]

Ttip [C]

finQ& [W]

100 93.51 239.8 105 98.05 256.8 110 102.6 274 115 107.1 291.4 120 111.6 309 125 116.2 326.8 130 120.7 344.8 135 125.2 363.1 140 129.7 381.5 145 134.2 400.1 150 138.7 419 155 143.2 438.1 160 147.7 457.5 165 152.1 477.1 170 156.6 496.9 175 161.1 517 180 165.5 537.3 185 170 557.9 190 174.4 578.7 195 178.9 599.9 200 183.3 621.2

100 120 140 160 180 20090

110

130

150

170

190

T0 [C]T t

ip [

C]

100 120 140 160 180 200200

250

300

350

400

450

500

550

600

650

T0 [C]

Qfin

[W

]


5-19

5-28 A plane wall is subjected to specified temperature on one side and convection on the other. The finite difference formulation of this problem is to be obtained, and the nodal temperatures under steady conditions as well as the rate of heat transfer through the wall are to be determined.

Assumptions 1 Heat transfer through the wall is given to be steady and one-dimensional. 2 Thermal conductivity is constant. 3 There is no heat generation. 4 Radiation heat transfer is negligible.

Properties The thermal conductivity is given to be k = 2.3 W/m⋅°C.

Analysis The nodal spacing is given to be ∆x=0.1 m. Then the number of nodes M becomes

T0

∆x

1

h, T∞

•

e&

40 • ••3

•2

51m 1.0m 4.01 =+=+

∆=

xLM

The left surface temperature is given to be T0 = 95°C. This problem involves 4 unknown nodal temperatures, and thus we need to have 4 equations to determine them uniquely. Nodes 1, 2, and 3 are interior nodes, and thus for them we can use the general finite difference relation expressed as

)0 (since 02 02 11 =++− +− eTT mmmm &

11 ==+−→ +− eTTTTmmm & , for m = 0, 1, 2, and 3

ut node 4

irection of all heat transfers to be towards the node under consideration:

2∆ kx

The finite difference equation for node 4 on the right surface subjected to convectionis obtained by applying an energy balance on the half volume element aboand taking the d

0)( :)convection -surface(right 4 Node

02 :(interior) 3 Node02 :(interior) 2 Node02 :(interior) 1 Node

434

432

321

210

=∆−

+−

=+−=+−=+−

∞ xTT

kTTh

TTTTTTTTT

where

tures under steady conditions are determined by solving the 4 equations above simultaneously with an equation

(c) The rate of heat transfer through the wall is simply convection heat transfer at the right surface,

ribed in Chap. 2, and the analytical (exact) solution can be used to check the accuracy of the numerical solution above.

C.15 and C95 C, W/m18 C, W/m3.2 m, 1.0 02 °=°=°⋅=°⋅==∆ ∞TThkx

The system of 4 equations with 4 unknown temperatures constitute the finite difference formulation of the problem.

(b) The nodal tempera solver to be

T1 = 79.8°C, T2 = 64.7°C, T3 = 49.5°C, and T4 = 34.4°C

W6970=°°=−== ∞ C15)-)(34.37m C)(20. W/m18()( 224convwall TThAQQ &&

Discussion This problem can be solved analytically by solving the differential equation as desc


5-20

5-29 Prob. 5-28 is reconsidered. The nodal temperatures under steady conditions as well as the rate of heat transfer through the wall are to be determined.

Analysis The problem is solved using SS-T-CONDUCT, and the solution is given below. On the SS-T-CONDUCT Input window for 1-Dimensional Steady State Problem, the problem parameters and the boundary conditions are entered into the appropriate text boxes. Note that with a uniform nodal spacing of 10 cm, there are 5 nodes in the x direction.

By clicking on the Calculate Temperature button, the computed results are as follows.

The rate of heat transfer through the wall is simply convection heat transfer at the right surface,

W6970=°°=−== ∞ C15)-)(34.37m C)(20. W/m18()( 224convwall TThAQQ &&


5-21

5-30 A plate is subjected to specified heat flux on one side and specified temperature on the other. The finite difference formulation of this problem is to be obtained, and the unknown surface temperature under steady conditions is to be determined.

Assumptions 1 Heat transfer through the base plate is given to be steady. 2 Heat transfer is one-dimensional since the plate is large relative to its thickness. 3 There is no heat generation in the plate. 4 Radiation heat transfer is negligible. 5 The entire heat generated by the resistance heaters is transferred through the plate.


Resistance heater, 800 W

∆x

Base plate

1

85°C

•0 • •3

•2

Analysis The nodal spacing is given to be ∆x=0.2 cm. Then the number of nodes M becomes

41cm 2.0cm 6.01 =+=+

∆=

xLM

The right surface temperature is given to be T3 =85°C. This problem involves 3 unknown nodal temperatures, and thus we need to have 3 equations to determine them uniquely. Nodes 1 and 2 are interior nodes, and thus for them we can use the general finite difference relation expressed as

)0 (since 02 02 11 ++− +− eTT mmmm &

112∆ +−kx mmm

The finite difference equation for node 0 on the left surface subjected to uniform heat flux is obtained by applying an energybalance on the h

==+−→= eTTTT& , for m = 1 and 2

alf volume element about node 0 and taking the direction of all heat transfers to be towards the node under

consideration:

02 :(interior) 2 Node02 :(interior) 1 Node

0 :flux)heat -surface(left 0 Node

321

210

010

=+−=+−

=∆−

+

TTTTTT

xTT

kq&

where

tures under steady conditions are determined by solving the 3 equations above simultaneously with an equation

cribed in Chap. 2, and the analytical (exact) solution can be used to check the accuracy of the numerical solution above.

. W/m000,50)m 0160.0/()W800(/ and C,85 C, W/m20 cm, 2.0 22003 ===°=°⋅==∆ AQqTkx &&

The system of 3 equations with 3 unknown temperatures constitute the finite difference formulation of the problem.

(b) The nodal tempera solver to be

T0 = 100°C, T1 =95°C, and T2 =90°C

Discussion This problem can be solved analytically by solving the differential equation as des


5-22

5-31 A plane wall is subjected to specified heat flux and specified temperature on one side, and no conditions on the other. The finite difference formulation of this problem is to be obtained, and the temperature of the other side under steady conditions is to be determined.

Assumptions 1 Heat transfer through the plate is given to be steady and one-dimensional. 2 There is no heat generation in the plate.

∆x

1•

0q&

T0

54•0 • ••

3•2

Properties The thermal conductivity is given to be k = 1.8 W/m⋅°C.

Analysis The nodal spacing is given to be ∆x=0.06 m.

Then the number of nodes M becomes

61m 06.0

m 3.01 =+=+∆

=x

LM

Nodes 1, 2, 3, and 4 are interior nodes, and thus for them we can use the general finite difference relation expressed as

)0 (since 02 02 11 =++− +− eTT mmmm &

112∆ kx mmm

The finite difference equation for node 0 on the left surface is obtained by applying an energy balance on the ha

==+−→ −+ eTTTT& , for m = 1, 2, 3, and 4

lf volume element about node 0 and taking the direction of all heat transfers to be towards the node under consideration,

C3.48 0 m0.060 ∆

C60C) W/m8.1( W/m350 0 1

1201 °=⎯→⎯=°−

°⋅+⎯→⎯=−

+ TT

xTT

kq&

ther noda temperatures are determined from the general interior node relation as follows:

°=°=

9.242.1322 :4

C6.36

345 TTTm

Therefore, the temperature of the other surface will be 1.5°C

Discussion This problem can be solved analytically by solving the differential equation as described in Chap. 2, and the analytical (exact) solution can be used to check the accuracy of the numerical solution above.

O l

°=−×=−==

=C2.136.369.2422 :3C9.243.486.3622 :2

234

123

TTTmTTTm

−×=−=−×=−== 603.4822 :1 012 TTTm

C1.5°=−×=−==


5-23

5-32E A large plate lying on the ground is subjected to convection and radiation. Finite difference formulation is to be obtained, and the top and bottom surface temperatures under steady conditions are to be determined.

Assumptions 1 Heat transfer through the plate is given to be steady and one-dimensional. 2 There is no heat generation in the plate and the soil. 3 Thermal contact resistance at plate-soil interface is negligible.

Properties The thermal conductivity of the plate and the soil are given to be kplate = 7.2 Btu/h⋅ft⋅°F and ksoil = 0.49 Btu/h⋅ft⋅°F.

Analysis The nodal spacing is given to be ∆x1=1 in. in the plate, and be ∆x2=0.6 ft in the soil. Then the number of nodes becomes

Tsky

111ft 0.6

ft 3in 1in 51

soilplate=++=+⎟

⎠⎞

⎜⎝⎛∆

+⎟⎠⎞

⎜⎝⎛∆

=x

Lx

LM

The temperature at node 10 (bottom of thee soil) is given to be T10 =50°F. Nodes 1, 2, 3, and 4 in the plate and 6, 7, 8, and 9 in the soil are interior nodes, and thus for them we can use the general finite difference relation expressed as

)0 (since 02 02112

11 ==+−→=+∆ kx

The finite difference equation for node 0 on the left surface and node 5 at the interface are obtained by applying an energy balance onrespective volume elements and taking the direction of

+−+−

+ eTTTeTTTmmm

mmm &&

their all heat

transfers to be towards the node under consideration:

−m

0 :)(interface 5 Node

02 :(interior) 4 Node02 :(interior) 3 Node02 :(interior) 2 Node02 :(interior) 1 Node

0])460([)( :surface) (top 0 Node

2

56soil

1

54plate

543

432

321

210

1

01plate

40

40

=∆−

+∆−

=+−=+−=+−=+−

=∆−

++−+−∞

xTT

kx

TTk

TTTTTTTTTTTT

xTT

kTTTTh skyεσ

Convection

εh, T∞

0.6 ftSoil

Radiation

• • • • • •10

• • • • •

0 1 2 3 4 5 6 7 8 9

Plate1 in

0 :(interior) 9 Node 8

7

=T02 :(interior) 8 Node02 :(interior) 7 Node 876

=+−=−

TTTTTT

2

02 :(interior) 6 Node

109

98

765

+−

+=+−

TT

TTT

h⋅ft⋅°F,

where

∆x1=1/12 ft, ∆x2=0.6 ft, kplate = 7.2 Btu/h⋅ft⋅°F, ksoil = 0.49 Btu/

h = 3.5 Btu/h⋅ft2⋅°F, Tsky =510 R, ε = 0.6, F80°=∞T , and T10 =50°F.

This system of 10 equations with 10 unknowns constitute the finite difference formulation of the problem.

T0 = 74.71°F, T1 =74.67°F, T2 =74.62°F, T3 =74.58°F, T4 =74.53°F,

T5 = 74.48°F, T6 =69.6°F, T7 =64.7°F, T8 =59.8°F, T9 =54.9°F

Discussion Note that the plate is essentially isothermal at about 74.6°F. Also, the temperature in each layer varies linearly and thus we could solve this problem by considering 3 nodes only (one at the interface and two at the boundaries).

(b) The temperatures are determined by solving equations above to be


5-24

5-33E A large plate lying on the ground is subjected to convection from its exposed surface. The finite difference formulation of this problem is to be obtained, and the top and bottom surface temperatures under steady conditions are to be determined.

Assumptions 1 Heat transfer through the plate is given to be steady and one-dimensional. 2 There is no heat generation in the plate and the soil. 3 The thermal contact resistance at the plate-soil interface is negligible. 4 Radiation heat transfer is negligible.

Properties The thermal conductivity of the plate and the soil are given to be kplate = 7.2 Btu/h⋅ft⋅°F and ksoil = 0.49 Btu/h⋅ft⋅°F.

Analysis The nodal spacing is given to be ∆x1=1 in. in the plate, and be ∆x2=0.6 ft in the soil. Then the number of nodes becomes

111ft 0.6in 1soilplate ⎠⎝ ∆⎠⎝ ∆ xx

The temperature at node 10 (bottom of thee soil) is given to be T

ft 3in 51 =++=+⎟⎞

⎜⎛+⎟

⎞⎜⎛=

LLM

hem we can use the general finite ence relation expressed as

10 =50°F. Nodes 1, 2, 3, and 4 in the plate and 6, 7, 8, and 9 in the soil are interior nodes, and thus for tdiffer

)0 (since 02 0 112 ==+−→=+∆ +− eTTT

kx mmm &

The finite difference equation for node 0 on the left surface and node 5 at the interface are obtained by applying an energy balance onrespective volume elements and taking the direction of al

2 11 +− +− eTTT mmmm &

their l heat

transfers to e towards the node under consideration: b

0 :)(interface 5 Node platek

02 :(interior) 4 Node02 :(interior) 3 Node02 :(interior) 2 Node02 :(interior) 1 Node

0)( :surface) (top 0 Node

2

56soil

1

54

432

321

210

1

01plate0

=∆

+∆

=+−=+−=+−=+−

=∆−

+−∞

xk

x

TTTTTTTTTTTT

xTT

kTTh

h = 3.5 Btu/h⋅ft2⋅°F, , and T10 =50°F.


(b) The temperatures are determined by solving equations above to be

T0 = 78.67°F, T1 =78.62°F, T2 =78.57°F, T3 =78.51°F, T4 =78.46°F,

T5 = 78.41°F, T6 =72.7°F, T7 =67.0°F, T8 =61.4°F, T9 =55.7°F

Discussion Note that the plate is essentially isothermal at about 78.6°F. Also, the temperature in each layer varies linearly and thus we could solve this problem by considering 3 nodes only (one at the interface and two at the boundaries).

Convectionh, T∞

0.6 ftSoil

• • • • • •

•••••

0 1 2 3 4 5 6 7 8 9 10

1 inPlate

543

−− TTTT

02 :(interior) 6 Node 765 =+− TTT

02 :(interior) 8 Node 987 =+− TTT02 :(interior) 7 Node 876 =+− TTT

02 :(interior) 9 Node 1098 =+− TTT

where

∆x1=1/12 ft, ∆x2=0.6 ft, kplate = 7.2 Btu/h⋅ft⋅°F, ksoil = 0.49 Btu/h⋅ft⋅°F,

F80°=∞T


5-25

5-34 A plane wall with variable heat generation and variable thermal conductivity is subjected to specified heat flux and convection at the left boundary (node 0) and radiation at the right boundary (node 5). The complete finite difference formulation of this problem is to be obtained.

0q&

Assumptions 1 Heat transfer through the wall is given to be steady and one-dimensional, and the thermal conductivity and heat generation to be variable. 2 Convection heat transfer at the right surface is negligible.

Convectio

∆x

1

ε

k(T)

)(xe& h, T∞

Radiation

Tsurr

q0

20 •••



0)2/()( 001

000 =∆+∆−

+−+ ∞ xAexTT

AkTThAAq &&

0)(112

110

1 =∆+∆−

+∆−

xAexTT

AkxTT

Ak & Node 1 (at the mid plane):

0)2/()( 221

24

24

surr =∆+∆−

+− xAexTT

AkTTA &εσ Node 2 (at right boundary):

5-35 A pin fin with negligible heat transfer from its tip is considered. The complete finite difference formulation for the determination of nodal temperatures is to be obtained.

Assumptions 1 Heat transfer through the pin fin is given to be steady and one-dimensional, and the thermal conductivity to diation heat transfer is negligible. 4 Heat

the energy balance approach and taking the direction of all heat transfers to be towards the node under consideration, the

lations become

be constant. 2 Convection heat transfer coefficient is constant and uniform. 3 Raloss from the fin tip is given to be negligible.

Analysis The nodal network consists of 3 nodes, and the base temperature T0 at node 0 is specified. Therefore, there are two unknowns T1 and T2, and we need two equations to determine them. Using Convectio

T0

, Th ∞

• ••0 1 2D∆xfinite difference formu

0)( 11210 =−∆+

∆−

+∆−

∞ TTxhpxTT

kAxTT

kNode 1 (at midpoint): A

Node 2 (at fin tip): 0))(2/( 221 =−∆+

∆−TT

∞ TTxphx

kA

where is the cross-sectional area and 4/2DA π= Dp π= is the perimeter of the fin.


5-26

5-36 The handle of a stainless steel spoon partially immersed in boiling water loses heat by convection and radiation. The finite difference formulation of the problem is to be obtained, and the tip temperature of the spoon as well as the rate of heat transfer from the exposed surfaces are to be determined.

Assumptions 1 Heat transfer through the handle of the spoon is given to be steady and one-dimensional. 2 Thermal conductivity and emissivity are constant. 3 Convection heat transfer coefficient is constant and uniform.

Properties The thermal conductivity and emissivity are given to be k = 15.1 W/m⋅°C and ε = 0.6.

Analysis The nodal spacing is given to be ∆x=3 cm. Then the number of nodes M becomes

71cm 3cm 181 =+=+

∆=

xLM

The base temperature at node 0 is given to be T0 = 100°C. This problem involves 6 unknown nodal temperatures, and thus we need to have 6 equations to determine them uniquely. Nodes 1, 2, 3, 4, and 5 are interior nodes, and thus for them we can use the general finite difference relation expressed as

0])273()[())(( 44surr

11 =+−∆+−∆+∆−

+∆−

∞+−

mmmmmm TTxpTTxph

xTT

kAx

TTkA εσ

22

equation for node 6 at the fin tip is obtained by applying an energy balance on the half volume element about node 6. Then,

εσ

∞ TTkAxpTTkAxphTTT εσ

εσ

=+−∆+−∆++ ∞ TTkAxpTTkAxphTTT εσ

=+−∆+−∆++− ∞ TTkAxpTTkAxphTT εσ

Node 6:

h, T∞

Tsurr 6543210

3 cm

•••••••

or 0])273()[/())(/(2 surr11 =+−∆+−∆++− ∞+− mmmmm TTkAxpTTkAxphTTT εσ , m = 1,2,3,4,5

The finite difference

44

m= 1: T 0])273()[/())(/(2 41

4surr

21

2210 =+−∆+−∆++− ∞ TTkAxpTTkAxphTT

m= 2: − 0])273()[/())(/(2 42

4surr

22

2321 =+−∆+−∆++

m= 3: T 0])273()[/())(/(2 43

4surr

23

2432 =+−∆+−∆++− ∞ TTkAxpTTkAxphTT

0])273()[/())(/(2 44

4surr

24

254m= 4: 3 −

0])273()[/())(/(2 45

4surr

25

265m= 5: 4T

0])273()[2/())(2/( 46

4surr6

65 =+−+∆+−+∆+∆−

∞ TTAxpTTAxphxTT

kA εσ

where

The n ultaneously with an ion sol

2 3 4 5 6

(c) The total rate of heat transfer from the spoon handle is simply the sum of the heat transfer from each nodal element, and is determin d from

0surface,

0surface,

0 element,fin TTATThAQQ m

mm

mmm

mm εσ&&

where Asurface, m =p∆x/2 for node 0, Asurface, m =p∆x/2+A for node 6, and Asurface, m =p∆x for other nodes.

C W/m13 K, 295 ,C100 C,32 0.6, C, W/m1.15 m, 03.0 20 °⋅==°=°==°⋅==∆ ∞ hTTTkx surrε

and m 0.024cm 4.2)cm 2.01(2 and m 102.0 cm 0.2cm) cm)(0.2 1( 242 ==+=×=== − pA

The system of 6 equations with 6 unknowns constitute the finite difference formulation of the problem.

(b) odal temperatures under steady conditions are determined by solving the 6 equations above simequat ver to be

T1 = 54.1°C, T = 38.3°C, T = 32.8°C, T = 30.9°C, T = 30.2°C, and T = 30.1°C,

e

W0.92=−++−== ∑∑∑==

∞=

])273[()( 4surr

4666


5-27

5-37 One side of a hot vertical plate is to be cooled by attaching aluminum fins of rectangular profile. The finite difference formulation of the problem for all nodes is to be obtained, and the nodal temperatures, the rate of heat transfer from a single fin and from the entire surface of the plate are to be determined.

Assumptions 1 Heat transfer along the fin is given to be steady and one-dimensional. 2 The thermal conductivity is constant. 3 Combined convection and radiation heat transfer coefficient is constant and uniform.

Properties The thermal conductivity is given to be k = 237 W/m⋅°C. T0 h, T∞

∆x• • • • •

0 1 2 3 4

Analysis (a) The nodal spacing is given to be ∆x=0.5 cm. Then the number of nodes M becomes

51cm 5.0

cm 21 =+=+∆

=x

LM

The base temperature at node 0 is given to be T0 = 80°C. This problem involves 4 unknown nodal temperatures, and thus we need to have 4 equations to determine them uniquely. Nodes 1, 2, and 3 are interior nodes, and thus for them we can use the general finite difference relation expressed as

0))((11 =−∆+∆∆ xx

The finite difference equation for node 4 at the fin tip is obt

−+

−∞

+m

mmm TTxphTT

kAT

k

ained by applying an energy balance on t volume element about that node. Then,

m= 3: =−∆++− ∞ TTkAxphTTT

Node 4:

−mTA → 0))(/(2 2

11 =−∆++− ∞+− mmmm TTkAxphTTT

he half

m= 1: 0))(/(2 12

210 =−∆++− ∞ TTkAxphTTT

0))(/(2 22

321 =−∆++− ∞ TTkAxphTTT m= 2:

0))(/(2 32

432

0))(2/( 443 =−+∆+

∆−

∞ TTAxphxTT

kA

where

ined bove simultaneously with an equation

(c) The rate of heat transfer from a single fin is simply the sum of the heat transfer from each nodal element,

0surface,

0 element,fin

TTAxphTTTTxhpTTxhp

TThAQQm

mmm

m

C W/m30 C,80 C,35 C, W/m237 m, 005.0 20 °⋅=°=°=°⋅==∆ ∞ hTTkx

and m 006.6)m 003.03(2 and m 0.009m) m)(0.003 3( 2 =+=== pA .


(b) The nodal temperatures under steady conditions are determ by solving the 4 equations a solver to be

T1 = 79.64°C, T2 = 79.38°C, T3 = 79.21°C, T4 = 79.14°C

W172=−+∆+−++∆+−∆=

−==

∞∞∞

=∞

=∑∑

))(2/()3())(2/(

)(

43210

44&&

(d) The number of fins on the surface is

fins 286m 0.004) (0.003

m 2spacingfin essFin thickn

height Plate fins of No. =+

=+

=

Then the rate of heat tranfer from the fins, the unfinned portion, and the entire finned surface become

kW 53.8≅=+=+=

=°−××°⋅=−=

===

∞

W53,8254633192,49

W4633C35)m)(80 0.004 m 3C)(286 W/m30()(

W,19249 W)172(286)fins of No.(

unfinned totalfin,total

20unfinned`unfinned

fin totalfin,

QQQ

TThAQ

QQ

&&&

&

&&


5-28

5-38 One side of a hot vertical plate is to be cooled by attaching aluminum pin fins. The finite difference formulation of the problem for all nodes is to be obtained, and the nodal temperatures, the rate of heat transfer from a single fin and from the entire surface of the plate are to be determined. Assumptions 1 Heat transfer along the fin is given to be steady and one-dimensional. 2 The thermal conductivity is constant. 3 Combined convection and radiation heat transfer coefficient is constant and uniform. Properties The thermal conductivity is given to be k = 237 W/m⋅°C. Analysis (a) The nodal spacing is given to be ∆x=0.5 cm. Then the number of nodes M becomes

T0 h, T∞

∆x• • • • • • • 0 1 2 3 4 5 6

71cm 5.0

cm 31 =+=+∆

=x

LM


0))((11−mTA =−∆+

−+

−∞

+m

mmm TTxphTT

kAT

k →

uation for node 6 at the fin tip is obtained by applying an energy balance on the half volume element about tha T en,

=−∆++− ∞ TTkAxphTT

Node 6:

0))(/(2 211 =−∆++− ∞+− mmmm TTkAxphTTT

∆∆ xxThe finite difference eq

t node. h

m= 1: 0))(/(2 12


0))(/(2 22

321 =−∆++− ∞ TTkAxphTTT m= 2:

m= 3: 0))(/(2 32


0))(/(2 42

543 =−∆++− ∞ TTkAxphTTT m= 4:

m= 5: 4T 0))(/(2 52

65

0))(2/( 665 =−+∆+

∆−

∞ TTAxphxTT

kA

e 5 C,100 C,30 C, W/m237 m, 005.0 20 °⋅=°=°=°⋅==∆ ∞ hTTkx

×==== ππDA

simultaneously with an

2 3 4 5 6

he ra of heat transfer from a single fin is simply the sum of the heat transfer from the nodal elements,

=

∞=

∑∑))(2/()5()(2/

)(0

surface,0

element,fin

TTAxphTTTTTTxhpTTxhp

TThAQQm

mmm

m&&

wher 3 C W/m

and m 00785.0)m 0025.0( === ππDp

m 100.0491cm 0491.0/4cm) 25.0(4/ 2-4222

(b) The nodal temperatures under steady conditions are determined by solving the 6 equations aboveequation solver to be 1 T = 97.9°C, T = 96.1°C, T = 94.7°C, T = 93.8°C, T = 93.1°C, T = 92.9°C (c) T te

W0.5496=−+∆+−++++∆+−∆=

−==66

∞∞∞ 6543210

ber of fins on the surface is (d) The num

fins 778,27m) m)(0.006 (0.006

m 1 fins of No.2

==


kW 17.4 W17,383 ≅=+=+=

=°××°⋅=−=

===−

∞

2116267,15

W2116C30)-)(100m 10 0491.027,778-C)(1 W/m35()(

W15,267 W)5496.0(778,27)fins of No.(


2420unfinned`unfinned

fin totalfin,

QQQ

TThAQ

QQ

&&&

&

&&


5-29

5-39 One side of a hot vertical plate is to be cooled by attaching copper pin fins. The finite difference formulation of the problem for all nodes is to be obtained, and the nodal temperatures, the rate of heat transfer from a single fin and from the entire surface of the plate are to be determined. Assumptions 1 Heat transfer along the fin is given to be steady and one-dimensional. 2 The thermal conductivity is constant. 3 Combined convection and radiation heat transfer coefficient is constant and uniform. Properties The thermal conductivity is given to be k = 386 W/m⋅°C. Analysis (a) The nodal spacing is given to be ∆x=0.5 cm. Then the number of nodes M becomes

T0 h, T∞

∆x• • • • • • • 0 1 2 3 4 5 6

71cm 5.0

cm 31 =+=+∆

=x

LM


0))((11−mTA =−∆+

−+

−∞

+m

mmm TTxphTT

kAT

k →

uation for node 6 at the fin tip is obtained by applying an energy balance on the half volume element about tha T en,

=−∆++− ∞ TTkAxphTT

Node 6:

0))(/(2 211 =−∆++− ∞+− mmmm TTkAxphTTT

∆∆ xxThe finite difference eq

t node. h

m= 1: 0))(/(2 12


0))(/(2 22

321 =−∆++− ∞ TTkAxphTTT m= 2:

m= 3: 0))(/(2 32


0))(/(2 42

543 =−∆++− ∞ TTkAxphTTT m= 4:

m= 5: 4T 0))(/(2 52

65

0))(2/( 665 =−+∆+

∆−

∞ TTAxphxTT

kA

5 C,100 C,30 C, W/m386 m, 005.0 20 °⋅=°=°=°⋅==∆ ∞ hTTkx

×==== ππDA

(b) The n e simultaneously with an

1 2 3 4 5 6

he ra of heat transfer from a single fin is simply the sum of the heat transfer from the nodal elements,

=

∞=

∑∑))(2/()5()(2/

)(0

surface,0

element,fin

TTAxphTTTTTTxhpTTxhp

TThAQQm

mmm

m&&

where 3 C W/m

and m 00785.0)m 0025.0( === ππDp

m 100.0491cm 0491.0/4cm) 25.0(4/ 2-4222

odal temperatures under steady conditions are determined by solving the 6 equations abov solver to be equationT 98.6°C, T = 97.5°C, T = 96.7°C, T = 96.0°C, T = 95.7°C, T = 95.5°C =

(c) T te

W0.5641=−+∆+−++++∆+−∆=

−==66

∞∞∞ 6543210

ber of fins on the surface is (d) The num

fins 778,27m) m)(0.006 (0.006

m 1 fins of No.2

==


kW 17.8 W17,786 ≅=+=+=

=°××°⋅=−=

===−

∞

2116670,15

W2116C30)-)(100m 10 0491.027,778-C)(1 W/m35()(

W15,670 W)5641.0(778,27)fins of No.(


2420unfinned`unfinned

fin totalfin,

QQQ

TThAQ

QQ

&&&

&

&&


5-30

5-40 A circular fin of uniform cross section is attached to a wall with the fin tip temperature specified as 250°C. The finite difference equations for all nodes are to be obtained and the nodal temperatures along the fin are to be determined and compared with analytical solution.


Properties The thermal conductivity of the fin is given as 240 W/m·K.

Analysis (a) The nodal spacing is given to be ∆x = 10 mm. Then the number of nodes M becomes

61mm 10mm 501 =+=+

∆=

xLM

The base temperature at node 0 is given to be T0 =350°C and the tip temperature at node 5 is given as T5 = 200°C. There are 4 unknown nodal temperatures, thus we need to have 4 equations to determine them uniquely. Nodes 1, 2, 3, and 4 are interior nodes, and we can use the general finite difference relation expressed as

0))((11 =−∆+∆−

+∆−

∞+−

mmmmm TTxph

xTT

kAx

TTkA

0)(22

11 =−∆

++− ∞+− mmmm TTkA

xhpTTT

where

04167.0)m 01.0)(K W/m240()m 01.0)(K W/m250(44 2222

=⋅⋅

=∆

=∆

kDxh

kAxhp

m = 1:

Then,

0)(04167.02 1210 =−++− ∞ TTTTT

m = 2: 0)(04167.02 2321 =−++− ∞ TTTTT

m = 3: 0)(04167.02 3432 =−++− ∞ TTTTT

m = 4: 0)(04167.02 4543 =−++− ∞ TTTTT

(b) The nodal temperatures under steady conditions are determined by solving the 4 equations above simultaneously with an llowing lines and paste on a blank EES screen to solve the above equations:

67*(25-T_3)=0

T_3-2*T_4+T + 04167*(25-T )=

, ,

equation solver. Copy the fo

T_0=350

T_5=200

T_0-2*T_1+T_2+0.04167*(25-T_1)=0

T_1-2*T_2+T_3+0.04167*(25-T_2)=0

T_2-2*T_3+T_4+0.041

_5 0. _4 0

Solving by EES software, we get

C 299.9 °=1T C 261.3 °=2T C 232.5 °=3T , C 212.3 °=4T

From Chapter 3, the analytical solution for the temperature variation along the fin (for specified tip temperature) is given as

mL

xLmmxTTTTTTTxT bL

b sinh)(sinhsinh)/()()( −+−−

=−− ∞∞

∞

∞


5-31


T(x),°C x, m

Analytical Numerical

0 350.0 350.0

0.01 299.8 299.9

0.02 261.2 261.3

0.03 232.4 232.5

0.04 212.3 212.3

0.05 200.0 200.0


x, m

0.00 0.01 0.02 0.03 0.04 0.05

T, °

C

200

250

300

350

AnalyticalNumerical

Discussion The comparison between the analytical and numerical solutions is excellent, with agreement within ±0.05%.


5-32

5-41 A DC motor delivers mechanical power to a rotating stainless steel shaft. With a uniform nodal spacing of 5 cm along shaft, the finite difference equations and the nodal temperatures are to be determined. Assumptions 1 Heat transfer along the shaft is steady and one-dimensional. 2 Thermal properties are constant. 3 Heat transfer by radiation is negligible. Properties The thermal conductivity of the shaft is given as 15.1 W/m·K. Analysis (a) The nodal spacing is given to be ∆x = 5 cm. Then the number of nodes M becomes

61cm 5cm 251 =+=+

∆=

xLM

The base temperature at node 0 is given to be T0 = 90°C. There are 5 unknown nodal temperatures, thus we need to have 5 equations to determine them uniquely. Nodes 1, 2, 3, and 4 are interior nodes, and we can use the general finite difference relation expressed as

0))((11 =−∆+∆−

+∆−

∞+−

mmmmm TTxph

xTT

kAx

TTkA

022

1

2

1 =∆

++⎟⎟⎠

⎞⎜⎜⎝

⎛ ∆+− ∞+− T

kAxhpTT

kAxhpT mmm

where 6452.0 W/m5.15(

)m 05.0)(K W/m25(44 2222=

⋅⋅

=∆

=∆

kDxh

kAxhp

)m 025.0)(K

he finite difference equation for node 5 at the fi tip (convection boundary) is obtained by applying an energy balance on the half volume element about that node: T n

0)(2 5

54 =−⎟⎠⎞

⎜⎝⎛ +

∆+

∆−

∞ TTAxphxTT

kA

022

1 54 =⎟⎠⎞

⎜⎝⎛ +

∆∆+⎥

⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ +

∆∆+− ∞TAxp

kAxhTAxp

kAxhT

4032.0122

=⎟⎠⎞

⎜⎝⎛ +

∆∆=⎟

⎠⎞

⎜⎝kA

⎛ +∆ pxhwhere ∆

Dx

kxhAx

hen, T m = 1: 06452.06452.2 210 =++− ∞TTTT

m = 2: − 06452.06452.2 321 =++ ∞TTTT

m = 3: 06452.06452.2 432 =++− ∞TTTT

m = 4: 06452.06452.23 54 =++− TTTT ∞

m = 5: 04032.04032.1 54 =+− ∞TTT The nodal temperatures under steady conditions are determined by solving the 5 equations above simultaneously with an

lank EES screen to solve the above equations:

0.6452*20=0 T_2-2.6452* 3 _4+0.6452*2 0

1T , ,

equation solver. Copy the following lines and paste on a b T_0=90 T_0-2.6452*T_1+T_2+0.6452*20=0 T_1-2.6452*T_2+T_3+ T_ +T 0= T_3-2.6452*T_4+T_5+0.6452*20=0 T_4-1.4032*T_5+0.4032*20=0

ving by EES software, we get Sol = C 52.03 ° C 34.72 °=2T C 26.92 °=3T , C 23.58 °=4T , C 22.55 °=5T

Discussion The nodal temperatures along the motor shaft can be compared with the analytical solution from Chapter 3 for fin with convection fin tip boundary condition:

mLmkhmL

xLmmkhxLmTTTxT

b sinh)/(cosh)(sinh)/()(cosh)(

+−+−

=−−

∞

∞


5-33

5-42 Straight rectangular fins are attached to a plane wall. For a single fin, (a) the finite difference equations, (b) the nodal temperatures, and (c) heat transfer rate are to be determined. The heat transfer rate is also to be compared with analytical solution.


Properties The thermal conductivity is given as 235 W/m·K.

Analysis (a) The nodal spacing is given to be ∆x = 10 cm. Then the number of nodes M becomes

61mm 01mm 051 =+=+

∆=

xLM

The base temperature at node 0 is given to be T0 = 350°C. There are 5 unknown nodal temperatures, thus we need to have 5 equations to determine them uniquely. Nodes 1, 2, 3, and 4 are interior nodes, and we can use the general finite difference relation expressed as

0))((11 =−∆+∆−

+∆−

∞+−

mmmmm TTxph

xTT

kAx

TTkA

022

1

2

1 =∆

++⎟⎟⎠

⎞⎜⎜⎝

⎛ ∆+− ∞+− T

kAxhpTT

kAxhpT mmm

where

0275.0)m 01.0)(m 1.0m 005.0(2)K W/m154()()22( 2222

=+⋅

=∆+

=∆

wtkxwth

kAxhp

)m 1.0)(m 005.0)(K W/m235( ⋅

he finite difference equation for node 5 at the fi tip (convection boundary) is obtained by applying an energy balance on e half volume element about that node:

T nth

0)(2 5

54 =−⎟⎠⎞

⎜⎝⎛ +

∆+

∆−

∞ TTAxphxTT

kA

022

1 54 =⎟⎠⎞

⎜⎝⎛ +

∆∆+⎥

⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ +

∆∆+− ∞TAxp

kAxhTAxp

kAxhT

where

0203.01)(2

=⎥⎦⎤

⎢⎣⎡ +

∆+∆=⎟

⎠⎞

⎜⎛ +

∆ pkA

xh ⎝

∆wt

xwtk

xhAx

Then,

m = 1: 00275.00275.2 210 =++− ∞TTTT

m = 2: 00275.00275.2 321 =++− ∞TTTT

m = 3: 00275.00275.2 432 =++− ∞TTTT

m = 4: 00275.00275.2 543 =++− ∞TTTT

m = 5: 00203.00203.1 54 =+− ∞TTT


5-34

T ,

(b) The nodal temperatures under steady conditions are determined by solving the 5 equations above simultaneously with an equation solver. Copy the following lines and paste on a blank EES screen to solve the above equations:

T_0=350

T_0-2.0275*T_1+T_2+0.0275*25=0

T_1-2.0275*T_2+T_3+0.0275*25=0

T_2-2.0275*T_3+T_4+0.0275*25=0

T_3-2.0275*T_4+T_5+0.0275*25=0

T_4-1.0203*T_5+0.0203*25=0


, =2C 316.6 °=1T C 291.2 ° C 273.2 °=3T , C 261.9 °=4T , C 257.2 °=5T

(c) The rate of heat transfer from a single fin is simply the sum of the heat transfer from the nodal elements,

W445=

−⎟⎠⎞

⎜⎝⎛ +

∆+−+++∆+−

∆=

−==

∞∞∞

=∞

= 0 surface,

0 element,num ,fin

mmm

mm

∑∑

)(2

)4()(2

)(

543210

55

TTAxphTTTTTxhpTTxhp

TThAQQ &&

or straight angular fins, the analytical solution from Chapter 3 for the heat transfer rate is,

ThAQ bη&

here

F rect

W427=°−⋅=−= ∞ C )25350)(m 0105.0)(K W/m154)(813.0()( 22finfinexact fin, T

w

1m 19.162 -

kthm ==

m 0525.02/ =+= tLLc

2fin m 0105.02 == cwLA

8130tanh

fin .mL

mL

c

c ==η

Discussion The comparison between the analytical and numerical solutions is within ±4.3% agreement. One way to increase the accuracy of the numerical solution is by reducing the nodal spacing, thereby increasing the number of nodes.


5-35

5-43 A stainless steel plane wall experiencing a uniform heat generation is subjected to constant temperature on one side and convection on the other. The finite difference equations and the nodal temperatures are to be determined. Assumptions 1 Heat transfer through the wall is steady and one-dimensional. 2 Thermal properties are constant. 3 Heat transfer by radiation is negligible. Properties The thermal conductivity is given as 15.1 W/m·K. Analysis (a) The nodal spacing is given to be ∆x = 2 cm. Then the number of nodes M becomes

61m 2.0∆

m 11 =+=+=x

LM

n

or nodes, and we can use the general finite difference relation expressed as

The left surface temperature is given to be T0 = 70°C. There are 5 unknownodal temperatures, thus we need to have 5 equations to determine them uniquely. Nodes 1, 2, 3, and 4 are interi

02

2∆ kx11 =+

+− +− eTTT mmmm & → 02 2

11 =∆++− +− xk

The finite difference equation for node 5 on the right

eTTT m

mmm&

surface subjected to convection is obtained by applying an energy balance on the half volume element about that node:

0)(2 55

54 =−+∆

+∆−

∞ TThxexTT

k & → 02

1 5

2

54 =∆+∆

+⎟⎠⎞

⎜⎝⎛ ∆+− ∞xT

khe

kxTx

khT &

Then

=∆++− xkeTTT &

=∆++− xkeTTT &

r steady conditions are determined by solving the 5 equations above simultaneously with an ng lines and paste on a blank EES screen to solve the above equations:

x^2*e_gen)/(2*k)+(h*Dx/k)*T_inf=0

olving by S software, ge

m2·K), the right surface temperature would become approximately the same as the ambient fluid temperature (T5 ≈ T∞).

0)/(2 2121 m = 1: 0

m = 2: 0)/(2 22321 =∆++− xkeTTT &

0)/(2 2343 m = 3: 2

0)/(2 24543 =∆++− xkeTTT & m = 4:

m = 5: 0)/()2/()()/1( 52

54 =∆+∆+∆+− ∞TkxhkexTkxhT &

(b) The nodal temperatures undeequation solver. Copy the followi e_gen=1000

h=250

k=15.1

Dx=0.2

T_inf=0

T_0=70

T_0-2*T_1+T_2+(e_gen/k)*Dx^2=0

T_1-2*T_2+T_3+(e_gen/k)*Dx^2=0

T_2-2*T_3+T_4+(e_gen/k)*Dx^2=0

T_3-2*T_4+T_5+(e_gen/k)*Dx^2=0

T_4-(1+h*Dx/k)*T_5+(D

S EE we t C 62.5 °=1T , C 52.3 °=2T , C 39.5 °=3T , C 24.0 °=4T , C 5.87 °=5T

Discussion For a very large value of convection heat transfer coefficient (e.g. 20000 W/


5-36

5-44 Two cast iron steam pipes are connected to each other through two 1-cm thick flanges, and heat is lost from the flanges by convection and radiation. The finite difference formulation of the problem for all nodes is to be obtained, and the temperature of the tip of the flange as well as the rate of heat transfer from the exposed surfaces of the flange are to be determined. Assumptions 1 Heat transfer through the flange is stated to be steady and one-dimensional. 2 The thermal conductivity and emissivity are constants. 3 Convection heat transfer coefficient is constant and uniform. Properties The thermal conductivity and emissivity are given to be k = 52 W/m⋅°C and ε = 0.8.

hi Ti

∆x• • • • • • •

0 1 2 3 4 5 6

ho, T∞

Tsurr

Analysis (a) The distance between nodes 0 and 1 is the thickness of the pipe, ∆x1=0.4 cm=0.004 m. The nodal spacing along the flange is given to be ∆x2=1 cm = 0.01 m. Then the number of nodes M becomes

72cm 1cm 52 =+=+

∆=

xLM

This problem involves 7 unknown nodal temperatures, and thus we need to have 7 equations to determine them uniquely. Noting that the total thickness of the flange is t = 0.02 m, the heat conduction area at any location along the flange is

rtA π2cond = where the values of radii at the nodes and between the nodes (the mid points) are

r0 = 0.046 m, r1=0.05 m, r2=0.06 m, r3=0.07 m, r4=0.08 m, r5=0.09 m, r6=0.10 m r01=0.048 m, r12=0.055 m, r23=0.065 m, r34=0.075 m, r45=0.085 m, r56=0.095 m Then the finite difference equations for each node are obtained from the energy balance to be as follows:

Node 0: 0)2())(2(1

010100 =

∆−

+−xTT

trkTTtrh ii ππ

Node 1: 0]})273([)(){2/)](2/)(2[2)2()2( 41

4surr12121

2

1212

1

1001 =+−+−∆++

∆−

+∆−

∞ TTTThxrrtx

TTtrk

xTT

trk εσπππ

Node 2: 0]})273([)(){2(2)2()2( 42

4surr222

2

2323

2

2112 =+−+−∆+

∆−

+∆−

∞ TTTThxtrx

TTtrk

xTT

trk εσπππ

0]})273([)(){2(2)2()2( 43

4surr323

2

3434

2

3223 =+−+−∆+

∆−

+∆−

∞ TTTThxtrx

TTtrk

xTT

trk εσπππNode 3:

Node 4: 0]})273([)(){2(2)2()2( 44

4surr424

2

4545

2

4334 =+−+−∆+

∆−

+∆−

∞ TTTThxtrx

TTtrk

xTT

trk εσπππ

Node 5: 0]})273([)(){2(2)2()2( 45

4surr525

2

5656

2

5445 =+−+−∆+

∆−

+∆−

∞ TTTThxtrx

TTtrk

xTT

trk εσπππ

Node 6: 0]})273([)(]{t22/))(2/(2[2)2( 46

4surr666562

6556 =+−+−++∆+

−∞ TTTThrrrxt

TTtrk εσπππ

2∆x

where K 290 ,C250 C,12 0.8, C, W/m52 m, 01.0 m, 004.0 21 =°=°==°⋅==∆=∆ ∞ surrin TTTkxx ε

and W/m25 2=h .C, ×i K W/m105.67 C, W/m180 42-82 ⋅=°⋅=°⋅ σh

0 14 °C, T4 = 139.5°C, T5 = 137.7°C, and T6 = 136.0°C transfer from the flange under steady conditions is simply the rate

of heat transfer from the steam to the pipe at flange section

where Asurface, m are as given above for different nodes.

The system of 7 equations with 7 unknowns constitutes the finite difference formulation of the problem. (b) The nodal temperatures under steady conditions are determined by solving the 7 equations above simultaneously with an equation solver to be T 8.4°C, T = 1 = 147.0°C, T2 = 144.1°C, T3 = 141.6(c) Knowing the inner surface temperature, the rate of heat

W105.7=−++−== ∑∑∑==

∞=

])273[()( 4surr

46

1surface,

6

1surface,

6

1 element,fin TTATThAQQ m

mm

mmm

mm εσ&&


5-37

5-45 Prob. 5-44 is reconsidered. The effects of the steam temperature and the outer heat transfer coefficient on the flange tip temperature and the rate of heat transfer from the exposed surfaces of the flange are to be investigated.


"GIVEN" t_pipe=0.004 [m] k=52 [W/m-C] epsilon=0.8 D_o_pipe=0.10 [m] t_flange=0.01 [m] D_o_flange=0.20 [m] T_steam=250 [C] h_i=180 [W/m^2-C] T_infinity=12 [C] h=25 [W/m^2-C] T_surr=290 [K] DELTAx=0.01 [m] sigma=5.67E-8 [W/m^2-K^4] “Stefan-Boltzmann constant" "ANALYSIS" "(b)" DELTAx_1=t_pipe "the distance between nodes 0 and 1" DELTAx_2=t_flange "nodal spacing along the flange" L=(D_o_flange-D_o_pipe)/2 M=L/DELTAx_2+2 "Number of nodes" t=2*t_flange "total thixkness of the flange" "The values of radii at the nodes and between the nodes /-(the midpoints) are" r_0=0.046 "[m]" r_1=0.05 "[m]" r_2=0.06 "[m]" r_3=0.07 "[m]" r_4=0.08 "[m]" r_5=0.09 "[m]" r_6=0.10 "[m]" r_01=0.048 "[m]" r_12=0.055 "[m]" r_23=0.065 "[m]" r_34=0.075 "[m]" r_45=0.085 "[m]" r_56=0.095 "[m]" "Using the finite difference method, the five equations for the unknown temperatures at 7 nodes are determined to be" h_i*(2*pi*t*r_0)*(T_steam-T_0)+k*(2*pi*t*r_01)*(T_1-T_0)/DELTAx_1=0 "node 0" k*(2*pi*t*r_01)*(T_0-T_1)/DELTAx_1+k*(2*pi*t*r_12)*(T_2-T_1)/DELTAx_2+2*2*pi*t*(r_1+r_12)/2*(DELTAx_2/2)*(h*(T_infinity-T_1)+epsilon*sigma*(T_surr^4-(T_1+273)^4))=0 "node 1" k*(2*pi*t*r_12)*(T_1-T_2)/DELTAx_2+k*(2*pi*t*r_23)*(T_3-T_2)/DELTAx_2+2*2*pi*t*r_2*DELTAx_2*(h*(T_infinity-T_2)+epsilon*sigma*(T_surr^4-(T_2+273)^4))=0 "node 2" k*(2*pi*t*r_23)*(T_2-T_3)/DELTAx_2+k*(2*pi*t*r_34)*(T_4-T_3)/DELTAx_2+2*2*pi*t*r_3*DELTAx_2*(h*(T_infinity-T_3)+epsilon*sigma*(T_surr^4-(T_3+273)^4))=0 "node 3" k*(2*pi*t*r_34)*(T_3-T_4)/DELTAx_2+k*(2*pi*t*r_45)*(T_5-T_4)/DELTAx_2+2*2*pi*t*r_4*DELTAx_2*(h*(T_infinity-T_4)+epsilon*sigma*(T_surr^4-(T_4+273)^4))=0 "node 4" k*(2*pi*t*r_45)*(T_4-T_5)/DELTAx_2+k*(2*pi*t*r_56)*(T_6-T_5)/DELTAx_2+2*2*pi*t*r_5*DELTAx_2*(h*(T_infinity-T_5)+epsilon*sigma*(T_surr^4-(T_5+273)^4))=0 "node 5" k*(2*pi*t*r_56)*(T_5-T_6)/DELTAx_2+2*(2*pi*t*(r_56+r_6)/2*(DELTAx_2/2)+2*pi*t*r_6)*(h*(T_infinity-T_6)+epsilon*sigma*(T_surr^4-(T_6+273)^4))=0 "node 6" T_tip=T_6


5-38

"(c)" Q_dot=Q_dot_1+Q_dot_2+Q_dot_3+Q_dot_4+Q_dot_5+Q_dot_6 "where" Q_dot_1=h*2*2*pi*t*(r_1+r_12)/2*DELTAx_2/2*(T_1-T_infinity)+epsilon*sigma*2*2*pi*t*(r_1+r_12)/2*DELTAx_2/2*((T_1+273)^4-T_surr^4) Q_dot_2=h*2*2*pi*t*r_2*DELTAx_2*(T_2-T_infinity)+epsilon*sigma*2*2*pi*t*r_2*DELTAx_2*((T_2+273)^4-T_surr^4) Q_dot_3=h*2*2*pi*t*r_3*DELTAx_2*(T_3-T_infinity)+epsilon*sigma*2*2*pi*t*r_3*DELTAx_2*((T_3+273)^4-T_surr^4) Q_dot_4=h*2*2*pi*t*r_4*DELTAx_2*(T_4-T_infinity)+epsilon*sigma*2*2*pi*t*r_4*DELTAx_2*((T_4+273)^4-T_surr^4) Q_dot_5=h*2*2*pi*t*r_5*DELTAx_2*(T_5-T_infinity)+epsilon*sigma*2*2*pi*t*r_5*DELTAx_2*((T_5+273)^4-T_surr^4) Q_dot_6=h*2*(2*pi*t*(r_56+r_6)/2*(DELTAx_2/2)+2*pi*t*r_6)*(T_6-T_infinity)+epsilon*sigma*2*(2*pi*t*(r_56+r_6)/2*(DELTAx_2/2)+2*pi*t*r_6)*((T_6+273)^4-T_surr^4)

Tsteam [C]

Ttip [C]

Q& [W]

150 160 170 180 190 200 210 220 230 240 250 260 270 280 290 300

85.87 91.01 96.12 101.2 106.3 111.3 116.3 121.3 126.2 131.1 136 140.9 145.7 150.5 155.2 160

59.48 63.99 68.52 73.08 77.66 82.27 86.91 91.57 96.26 101 105.7 110.5 115.3 120.1 125 129.9

160 180 200 220 240 260 280 30080

90

100

110

120

130

140

150

160

60

70

80

90

100

110

120

130

140

Tsteam [C]

T tip

[C

]

Q [

W]

temperature

heat

h [W/m2.C]

T tip[C]

Q& [W]

15 20 25 30 35 40 45 50 55 60

155.4 145.1 136 128 120.9 114.6 108.9 103.9 99.26 95.08

87.7 97.31 105.7 113.2 119.7 125.6 130.8 135.6 139.8 143.7

15 20 25 30 35 40 45 50 55 6090

100

110

120

130

140

150

160

80

90

100

110

120

130

140

150

h [W/m2-C]

T tip

[C

]

Q [

W]

temperature

heat


5-39

5-46 Using EES, the solutions of the systems of algebraic equations are determined to be as follows:

olution: x1 = 2, x2 = 3, x3 = −1

2

11.964

olution: x1 = 2.33, x2 = 2.29, x3 = −1.62

"(a)"

3*x_1-x_2+3*x_3=0

-x_1+2*x_2+x_3=3

2*x_1-x_2-x_3=2

S

"(b)"

4*x_1-2*x_2^2+0.5*x_3=-

x_1^3-x_2+-x_3=

x_1+x_2+x_3=3

S


_4=6

4=2

olution: x1 = 13, x2 = −9, x3 = 13, x4 = −2

93

olution: x1 = 2.825, x2 = 1.791, x3 = −1.841

"(a)"

3*x_1+2*x_2-x_3+x

x_1+2*x_2-x_4=-3

-2*x_1+x_2+3*x_3+x_

3*x_2+x_3-4*x_4=-6

S

"(b)"

3*x_1+x_2^2+2*x_3=8

-x_1^2+3*x_2+2*x_3=-6.2

2*x_1-x_2^4+4*x_3=-12

S


5-40


=-1

olution: x1 = −2, x2 = −1, x3 = 0, x4 = 1

*x_4=-3

Solution: x1 = 0.263, x2 = −1.15, x3 = 1.70, x4 = 2.14

"(a)"

4*x_1-x_2+2*x_3+x_4=-6

x_1+3*x_2-x_3+4*x_4

-x_1+2*x_2+5*x_4=5

2*x_2-4*x_3-3*x_4=-5

S

"(b)"

2*x_1+x_2^4-2*x_3+x_4=1

x_1^2+4*x_2+2*x_3^2-2

-x_1+x_2^4+5*x_3=10

3*x_1-x_3^2+8*x_4=15


5-41

Two-Dimensional Steady Heat Conduction

5-49C A region that cannot be filled with simple volume elements such as strips for a plane wall, and rectangular elements for two-dimensional conduction is said to have irregular boundaries. A practical way of dealing with such geometries in the finite difference method is to replace the elements bordering the irregular geometry by a series of simple volume elements.

5-50C For a medium in which the finite difference formulation of a general interior node is given in its simplest form as : 4/)( bottomrighttopleftnode TTTTT +++=

(a) Heat transfer is steady, (b) heat transfer is two-dimensional, (c) there is no heat generation in the medium, (d) the nodal spacing is constant, and (e) the thermal conductivity of the medium is constant.

5-51C For a medium in which the finite difference formulation of a general interior node is given in its simplest form as

042

nodenodebottomrighttopleft k

(a) Heat transfer is steady, (b) heat transfer is two-dimensional, (c) there is heat generat

=+−+++le

TTTTT&

:

ion in the medium, (d) the nodal acing is constant, and (e) the thermal conductivity of the medium is constant.

sp


5-42

5-52 Two dimensional ridges are machined on the cold side of a heat exchanger. The smallest section of the wall is to be identified. A two-dimensional grid is to be constructed and the unknown temperatures in the grid are to be determined.

Assumptions 1 Heat transfer through the body is given to be steady and two-dimensional. 2 Thermal conductivity is constant. 3 There is no heat generation.

Analysis (a) From symmetry, the smallest domain is between the top and the base of one ridge.

5 mm

10 mm TA

TB

M

10 mm

(b) The unknown temperatures at nodes 1, 2, and 3 are to be determined from finite difference formulations

Node 1:

3010334022

022

21

1121

1121

=×==−=−+−+−

=∆

∆−

+∆

∆−

+∆∆−

B

BB

BB

TTTTTTTTT

xxTT

kxxTT

kxxTT

k

Node 2:

9024022

022

3T21

22321

22321

==−+−=−+−+−

=∆

∆−

+∆∆−

+∆

∆−

A

A

A

TTTTTTTTT

xxTT

kxxTT

kxxT

k

Node 3:

32

23

=+×=+=+−+++

AB

BBA

TTTTTT

241014

3

2

1

TTT

(c) The temperature T2 is 46.9ºC. Then the temperatures T1 and T3 are determined from equations 1 and 3.

• •

• • •

• • •

∆x

1

•

2 3

TA TA TA

TB

TB

TB

∆x

T

4 = TTT 1109010224

The matrix equation is

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−

−

1109030

410

C19.2°=⎯→⎯=−

=−

11

21

309.464

304

TT

TT

C39.2°=⎯→⎯=+−

=+−

33

32

11049.46

1104

TT

TT


5-43

5-53 A rectangular cross section is undergoing a steady two-dimensional heat transfer. The finite difference equations and the nodal temperatures are to be determined. Assumptions 1 Steady heat conduction is two-dimensional. 2 Thermal properties are constant. 3 There is no heat generation in the body. Analysis (a) There are 10 unknown nodal temperatures, thus we need to have 10 equations to determine them uniquely. For nodes 1 to 10, we can use the general finite difference relation expressed as

022

21 , ,1 ,

2 ,1 , ,1 =

∆

+−+

∆

+− +−+−

y

TTT

x

TTT nmnmnmnmnmnm

Since yx ∆= , we∆ have

)(25.0 ,11 , ,11 , , nmnmnmnmnm TTTTT +−−+ +++=

or )(25.0 rightbottomlefttopnode TTTTT +++=

Then Node 1: [ ]261 0)6/sin(10025.0 TTT +++= π

Node 2: [ ]3712 )6/2sin(10025.0 TTTT +++= π

Node 3: [ ]4823 )6/3sin(10025.0 TTTT +++= π

Node 4: [ ]5934 )6/4sin(10025.0 TTTT +++= π

Node 5: [ ]0)6/5sin(10025.0 1045 +++= TTT π

Node 6: [ ]716 0025.0 TTT +++=

Node 7: T [ ]8627 025.0 TTT +++=

+++=

Node 8: [ ]9738 025.0 TTTT +++=

Node 9: [ ]10849 025.0 TTTT +++=

Node 10: [ ]0025.0 9510 TTT

(b) The nodal temperatures under steady conditions are determined by solving the 10 equations above simultaneously with an ion solver. Copy the following lines and paste on a blank EES screen to solve the above equations: equat

T_1=0.25*(100*sin(pi/6)+0+T_6+T_2) T_2=0.25*(100*sin(2*pi/6)+T_1+T_7+T_3) T_3=0.25*(100*sin(3*pi/6)+T_2+T_8+T_4) T_4=0.25*(100*sin(4*pi/6)+T_3+T_9+T_5)

10+0) T_5=0.25*(100*sin(5*pi/6)+T_4+T_ T_6=0.25*(T_1+0+0+T_7) T_7=0.25*(T_2+T_6+0+T_8) T_8=0.25*(T_3+T_7+0+T_9) T_9=0.25*(T_4+T_8+0+T_10) T_10=0.25*(T_5+T_9+0+0)

olving S by EES software, we get T1 = 27.4°C, T2 = 47.4°C, T3 = 54.7°C, T4 = 47.4°C, T5 = 27.4°C T6 = 12.1°C, T7 = 20.9°C, T8 = 24.1°C, T9 = 20.9°C, T10 = 12.1°C Discussion The numerical solution can be verified using the following analytical solution:

( ))60/30sinh(

)60/sin()60/sinh(100,

πππ xy

yxT =

For example, at x = 30 cm and y = 20 cm, we have

( ) C 3.54)60/30sinh(

)60/30sin()60/20sinh(100cm 20 ,cm 30 °==π

ππT

When compared with the numerical solution, T3 = 54.7°C, the difference is within 0.8%.


5-44

5-54 A square cross section is undergoing a steady two-dimensional heat transfer. The finite difference equations and the nodal temperatures are to be determined.

Assumptions 1 Steady heat conduction is two-dimensional. 2 Thermal properties are constant. 3 There is no heat generation in the body.

Analysis (a) There are 4 unknown nodal temperatures, thus we need to have 4 equations to determine them uniquely. For nodes 1 to 4, we can use the general finite difference relation expressed as

022

21 , ,1 ,

2 ,1 , ,1 =

∆

+−+

∆

+− +−+−

y

TTT

x

TTT nmnmnmnmnmnm

Since yx ∆= , we∆ have

)(25.0 ,11 , ,11 , , nmnmnmnmnm TTTTT −−++ +++=

or )(25.0 leftbottomrighttopnode TTTTT +++=

Then

Node 1: )500100(25.0 321 +++= TTT

Node 2: )200100(25.0 142 TTT +++=

Node 3: )500300(25.0 413 +++= TTT

Node 4: )300200(25.0 324 TTT +++=

(b) By letting the initial guesses as T1 = 300°C, T2 = 150°C, T3 = 400°C, and T4 = 250°C the results obtained from successive iterations are listed in the following table:

Nodal temperature,°C Iteration

T1 T2 T3 T4

1 287.5 209.4 334.4 260.9

2 285.9 211.7 336.7 262.1

3 287.1 212.3 337.3 262.4

4 287.4 212.5 337.5 262.5

5 287.5 212.5 337.5 262.5

6 287.5 212.5 337.5 262.5


T1 = 287.5°C, T2 = 212.5°C, T3 = 337.5°C, T4 = 262.5°C

Discussion The finite difference equations can also be calculated using the EES. Copy the following lines and paste on a blank EES screen to solve the above equations:

T_1=0.25*(100+T_2+T_3+500)

T_2=0.25*(100+200+T_4+T_1)

T_3=0.25*(T_1+T_4+300+500)

T_4=0.25*(T_2+200+300+T_3)

Solving by EES software, we get the same results:

T1 = 287.5°C, T2 = 212.5°C, T3 = 337.5°C, T4 = 262.5°C


5-45

5-55 Two long solid bodies are subjected to steady two-dimensional heat transfer. The unknown nodal temperatures are to be determined.

Assumptions 1 Heat transfer through the bodies are given to be steady and two-dimensional. 2 There is no heat generation in the body.


Analysis The nodal spacing is given to be ∆x=∆x=l=0.01m, and the general finite difference form of an interior node for steady two-dimensional heat conduction for the case of no heat generation is expressed as

0404 nodebottomrighttopleft

2node

nodebottomrighttopleft =−+++⎯→⎯=+−+++ TTTTTk

leTTTTT

&

(a) There is symmetry about a vertical line passing through the nodes 1 and 3. Therefore, = 46 T= and

5421 and ,,, TTTT are the only 4 unknown nodal temperatures, and thus we need only 4 equations to determine them uniquely. Also, we can replace the symmetry lines by insulation and utilize the mirror-imageconcept when writing the finite d

23 TT , ,

ifference equations for the

interior nodes.

422

2451

12

=−+++=−+++

=−++

TTTTTTT

TT

olving t s above simultaneously gives

°C

T100

Insulated

•

• •

• • •

1

2 3

4 5 6

150

200

250

300

150

200

250

300

0402550 2 :(interior) 4 Node04200 :(interior) 2 Node

04201550 1 :(interior) 1 Node

044 :(interior) 5 Node 52 =− TT

S he 4 equation

T1 = 175°C

T2 = T3 = 200°C

T4 = T6 = 225

T5 = 200°C

(b) There is symmetry about a vertical line passing through the middle. Therefore, 23 TT = and 14 TT = . Replacing the symmetry lines by insulation and utilizing the mirror-image concept, the finite difference equations for the interior nodes 1 and 2 are determined to be

2

T3 = 85.7°C

iscussion Note that taking advantage of symmetry mplified the problem greatly.

• • • • 50 50

31 2 4• • • • •

Insulated

50 50

150 150 150 150 150

Insulated 5040150550 :(interior) 2 Node

04201550 :(interior) 1 Node 12

=−+++=−++TT

TT01

Solving the 2 equations above simultaneously gives

T1 = T4 = 92.9°C, T2 =

Dsi


5-46

5-56 A long solid body is subjected to steady two-dimensional heat transfer. The unknown nodal temperatures are to be determined.

Assumptions 1 Heat transfer through the body is given to be steady and two-dimensional. 2 There is no heat generation in the body.


Analysis The nodal spacing is given to be ∆x = ∆x = l = 0.02 m, and the general finite difference form of an interior node for steady two-dimensional heat conduction for the case of no heat generation is expressed as

4/)( 04 bottomrighttopleftnodenodebottomrighttopleft TTTTTTTTTT +++=→=−+++

There is symmetry about the horizontal, vertical, and diagonal lines passing through the midpoint, and thus we need to consider only 1/8th of the region. Then,

8642 TTTT ===

Therefore, there are there are only 3 unknown nodal temperatures, 521 and , , TTT , and thus we need only 3 equations to determine them

uniquely. Also, we can replace the symmetry lines by insulation andutilize the mirror-image concept w

9731 TTTT ===

hen writing the finite difference

equations for the interior nodes.

4/4 :(interior) 5 Node4/)2200( :(interior) 2 Node4/)

TTTTTT

==++=

225

152

21 2180180( :(interior) 1 Node T

T

++=

Discussion Note that taking advantage of symmetry simplified the problem greatly.

• • • • •

• • • • •

180 200 180

• • • • •

• • • • •

• • • • •

150 180 200 180 150

150 180 200 180 150

180 200 180

1 2 3

4 5 6

7 8 9

Solving the equations above simultaneously gives

C190C185

°=====°====

86542

9731

TTTTTTTTT


5-47

5-57 A long solid body is subjected to steady two-dimensional heat transfer. The unknown nodal temperatures are to be determined.

Assumptions 1 Heat transfer through the body is given to be steady and two-dimensional. 2 There is no heat generation in the body.


Analysis The nodal spacing is given to be ∆x = ∆x = l = 0.01 m, and the general finite difference form of an interior node for steady two-dimensional heat conduction for the case of no heat generation is expressed as

4/)( 04 bottomrighttopleftnodenodebottomrighttopleft TTTTTTTTTT +++=→=−+++

(a) There is symmetry about the insulated surfaces as well as about the diagonal line. Therefore T , and

421 and ,, TTT are the only 3 unknown nodal temperatures. Thus we need only 3 equations to determine them uniquely. Alwe can replace the symmetry lines by insulation and utilize the mirror-image concept when writing the finite differenceequations for th

23 T=so,

e interior nodes.

4/)22( :(interior) 4 Node

4/)2200( :(interior) 2 Node4/)180180( :(interior) 1 Node

324

142

321

TTTTTT

TTT

+=++=

+++=

Insulated

• • •

180 200

150 180 200

3

1 2

4

• • •

• • •

Insulated

Also, 23 TT =


C185C190

°=°===

1

432

TTTT

(b) There is symmetry about the insulated surface as well as the diagonal line. Replacing the symmetry lines by insulation, and utilizing the mirror-image concept, the finite difference equations for the interior nodes can be written as

4/)21402( :(interior) 4 Node4/)12140( :(interior) 3 Node

4/)4/)120120( :(interior) 1 Node

324

243

14

321

TTTTTTT

TTTTT

++=

=++=++++=

Discussion Note that taking advantage of symmetry simplified the problem greatly.

120120( :(interior) 2 Node 2T ++= • • • •100 120 140

120 120

3

1 2

4

• • • •

• • • •

Insulated

100 120 140


C128.6C122.9°==°==

43

21

TTTT


5-48

5-58 Starting with an energy balance on a volume element, the steady two-dimensional finite difference equation for a general interior node in rectangular coordinates for T(x, y) for the case of variable thermal conductivity and uniform heat generation is to be obtained.

Analysis We consider a volume element of size 1×∆×∆ yx centered about a general interior node (m, n) in a region in which heat is generated at a constant rate of e& and the thermal conductivity k is variable (see Fig. 5-24 in the text). Assuming the direction of heat conduction to be towards the node under consideration at all surfaces, the energy balance on the volume element can be expressed as

0elementelementbottom cond,right cond, topcond,left cond, =

∆t

for the steady case. Again assuming the temp ratures between the adjacent nodes to vary linearly and noting t

∆=++++

EGQQQQ &&&&&

e hat the heat transfer area is in the x direction and 1×∆y 1×∆x in the y direction, the energy balance relation above becomes

0)1()1(+

)1()1(+)1(

0,1,

,

,,1,

,1,,

,,1,

=×∆×∆+∆

−×∆

∆

−×∆+

∆

−×∆

∆

−×∆

−

++−

yxey

TTxk

xTT

yky

TTxk

xTT

yk

nmnmnm

nmnmnm

nmnmnm

nmnmnm

&

Dividing each term by and simplifying gives 1×∆×∆ yx

022

,∆0

21,,1,

2,1,,1 =+

+−+

+− +−+− nmnmnmnmnmnm

keTTTTTT &

For a square mesh with ∆x = ∆y = l, and the relation above simplifies to

∆ nmyx

042

0,1,1,,1,1 =+−+++ +−+− nmnmnmnmnm

leTTTTT

&

,nmk

It can also be expressed in the following easy-to-remem ber form

04node

20

nodebottomrighttopleft =+−+++k

leTTTTT

&


5-49

5-59 A long solid body is subjected to steady two-dimensional heat transfer. The unknown nodal temperatures and the rate of heat loss from the top surface are to be determined.

Assumptions 1 Heat transfer through the body is given to be steady and two-dimensional. 2 Heat is generated uniformly in the body.


Analysis (a) The nodal spacing is given to be ∆x=∆x=l=0.1 m, and the general finite difference form of an interior node equation for steady two-dimensional heat conduction for the case of constant heat generation is expressed as

042

nodenodebottomrighttopleft k

There is symmetry

=+−+++le

TTTTT&

about a vertical line passing through the middle of the region, and thus we need to consider only half of the region. Then,

lines by insulation and utilize the mirror-image concept when writing the fi te difference equations for the interior nodes.

4321 and TTTT ==

Therefore, there are there are only 2 unknown nodal temperatures, T1 and T3, and thus we need only 2 equations to determine them uniquely. Also, we can replace the symmetry

ni

04200150 :(interior) 3 Node ++

01100 :ior)

2

341

2

=+−+

=+

kleTTT

kle

&

&

Noting that 21 and TTTT = and substituting,

420(inter 1 Node 132 +−++ TTT

43 =

0C W/m150

m) )(0.1 W/m103(

0C W/m150

3220

237

13

×

=°⋅

+−+ TT

3350

m) )(0.1 W/m103(

31

237

=°⋅

+−+

×

TT

The solution of the above system is

(b) The total rate of heat transfer from the top surface can be determined from an energy balance on a volume element at the top surface whose height is l/2, length 0.3 m, and depth 1 m:

• • • •

• • • •

• • • •

• • • •100 100 100 100

120 3

1 2

0.1 m

200 200 200 200

120 1504

e&

150

C1159C1126°==°==

43

21

TTTT

topQ&

depth) (per

C100)-m)(1126 1(C)100120(2m 1)C W/m150(2m)2/1.03.0)( W/m103(

0100

121001202

12)2/13.0(

337top

10top

m

Q

lT

kll

lkleQ

W760,900=

⎟⎠⎞

⎜⎝⎛ °+°−°⋅−××−=

=⎟⎠

⎞⎜⎝

⎛ −×+

−×+××+

&

&&


5-50

5-60 Prob. 5-59 is reconsidered. The unknown nodal temperatures and the rate of heat loss from the top surface are to

nalysis The problem is solved using EES, and the solution is given below.

conductivity"

atures can be determined"

_3+e_gen*L^2/k=0 "for node 3"

_dot=e_gen*(0.3*1*L/2)+(2*k*L/2*(120-100)/L+2*k*L*(T_1-100)/L)

(a) The n°C

) The rate of heat loss from the top surface is

be determined.

A "GIVEN" e_gen=3e7 [W/m^3] "heat generation"k=150 [W/m-K] "thermal L=0.10 [m] "mesh size" "ANALYSIS" "(a) Using the finite difference method, the nodal temper100+120+T_2+T_3-4*T_1+e_gen*L^2/k=0 "for node 1" T_2=T_1 "for node 2" 150+200+T_1+T_4-4*TT_4=T_3 "for node 4" "(b) The rate of heat loss from the top surface is calculated using" Q

odal temperatures are determined to be T1 = T2 = 1126°C and T3 = T4 = 1159

W760900=Q& . (b


5-51

5-61 Prob. 5-59 is reconsidered. The effects of the thermal conductivity and the heat generation rate on the


]

s at the selected nodes are also given in the figure"

"

equations for the two unknown temperatures are determined to be"

_dot_top+e_dot*(3*l*d*l/2)+2*(k*(l*d)/2*(120-100)/l+k*l*d*(T_1-100)/l)=0

.C]

temperatures at nodes 1 and 3, and the rate of heat loss from the top surface are to be investigated.

A

"GIVEN" k=150 [W/m-C] e_dot=3E7 [W/m^3DELTAx=0.10 [m] DELTAy=0.10 [m] d=1 [m] “depth" "Temperature "ANALYSIS"(a)" l=DELTAx T_1=T_2 "due to symmetry" T_3=T_4 "due to symmetry" "Using the finite difference method, the two 100+120+T_2+T_3-4*T_1+(e_dot*l^2)/k=0 150+200+T_1+T_4-4*T_3+(e_dot*l^2)/k=0 "(b)" "The rate of heat loss from the top surface can be determined from an energy balance on a volume element whose height is l/2, length 3*l, and depth d=1 m" -Q

k [W/m

T1 [C]

T3 [C]

topQ& [W]

10 30.53 51.05 71.58 92.11 112.6 133.2 153.7 174.2 194.7 215.3 235.8 256.3 276.8 297.4 317.9 338.4 358.9 379.5 400

501.2

533.7 779000

151265040 3064 2222 1755 1458 1253 1102 987.3 896.5 823.1 762.4 711.5 668.1 630.7 598.1 569.5 544.1 521.5

151595073 3097 2254 1787 1491 1285 1135 1020 929 855.6 794.9 744 700.6 663.2 630.6 602 576.6 554

750725 752213 753701 755189 756678 758166 759654 761142 762630 764118 765607 767095 768583 770071 771559 773047 774536 776024 777512

0 50 100 150 200 250 300 350 4000

2000

4000

6000

8000

10000

12000

14000

16000

750000

755000

760000

765000

770000

775000

780000

T 1 a

nd T

3 [C

]

Qto

p [W

]k [W/m-C]


5-52

e& [W/m3]

T1 [C]

T3 [C]

topQ& [W]

100000 5.358E+06 1.061E+07 1.587E+07 2.113E+07 2.639E+07 3.165E+07 3.691E+07 4.216E+07 4.742E+07 5.268E+07 5.794E+07 6.319E+07 6.845E+07 7.371E+07 7.897E+07 8.423E+07 8.948E+07 9.474E+07 1.000E+08

129.6 304.8 480.1 655.4 830.6 1006 1181 1356 1532 1707 1882 2057 2233 2408 2583 2759 2934 3109 3284 3460

162.1 337.3 512.6 687.9 863.1 1038 1214 1389 1564 1739 1915 2090 2265 2441 2616 2791 2966 3142 3317 3492

13375 144822 276270 407717 539164 670612 802059 933507 1.065E+06 1.196E+06 1.328E+06 1.459E+06 1.591E+06 1.722E+06 1.854E+06 1.985E+06 2.117E+06 2.248E+06 2.379E+06 2.511E+06

0.0x100 2.2x107 4.4x107 6.6x107 8.8x107 1.1x1080

500

1000

1500

2000

2500

3000

3500

0.0x100

5.0x105

1.0x106

1.5x106

2.0x106

2.5x106

3.0x106

e [W/m3]

Qto

p [W

]

T 1 a

nd T

3 [C

]


5-53

5-62 A long solid body is subjected to steady two-dimensional heat transfer. The unknown nodal temperatures and the rate of heat loss from the bottom surface through a 1-m long section are to be determined.

Assumptions 1 Heat transfer through the body is given to be steady and two-dimensional. 2 Heat is generated uniformly in the body. 3 Radiation heat transfer is negligible.


Analysis The nodal spacing is given to be ∆x = ∆x = l = 0.04 m, and the general finite difference form of an interior node for steady two-dimensional heat conduction is expressed as

042

nodenodebottomrighttopleft =+−+++

kle

TTTTT&

where

C2.142C W/m45 °⋅kk

The finite difference equations for boundary nodes are obtained by applying an energy balance on the volume elements and t

)m 04.0)( W/m104( 23620

2de °=

×==

lele &&

aking the direction of all heatto be towards the node under consideration:

no

transfers

04-200240290260 :(interior) 3 Node

0 4-290325290350 :(interior) 2 Node

02

)(325

2290240

2 :)convection ( 1 Node

20

3

20

2

20

1111

=++++

=++++

=+−+−

+−

+−

∞

kle

T

kle

T

kle

TThll

Tlkl

Tkl

lTlk

&

&

&

where

5 362 °=×=°=° ∞Teh &

Substitut

(b) The rate of heat loss from the bottom surface through a 1-m long section is

h, T∞

Insulated

• • • •

• • • •

• • • •

• • • •

240

200°C350260 305

2903

1

2

4 cm 325

Convection

e&

4=k C20 , W/m104 C,. W/m50 C, W/m.

ing,

T1 = 281.2°C, T2 = 349.3°C, T3 = 283.1°C,

W1447=°+++×°⋅=

−+−+−+−=

−==

∞∞∞∞

∞∑∑

C20)/2]-(32520)-(281.220)-(24020)/2-m)[(200 1 m C)(0.04 W/m50(

)325)(2/()()240()200)(2/(

)(

21

surface, element,

TlhTThlThlTlh

TThAQQm

mmm

m&&


5-54

5-63 Prob. 5-62 is reconsidered. The unknown nodal temperatures and the rate of heat loss from the bottom surface


nt"

*L*(T_inf-T_1)+e_gen*L^2/(2*k)=0 "for node 1"

e heat loss from each node" _dot=h*L/2*(200-T_inf)+h*L*(240-T_inf)+h*L*(T_1-T_inf)+h*L/2*(325-T_inf)

(a) The nC

) The rate of heat loss from the bottom surface is

through a 1-m long section are to be determined.

A "GIVEN" e_gen=4e6 [W/m^3] "heat generation" h=50 [W/m^2-K] "convection coefficiek=45 [W/m-K] "thermal conductivity"L=0.04 [m] "mesh size" T_inf=20 [C] "ambient temperature" "ANALYSIS" "(a) Using the finite difference method, the 3 equations for the 3 nodal temperatures can be determined" k*L/2*(240-T_1)/L+k*L*(290-T_1)/L+k*L/2*(325-T_1)/L+h350+290+325+290-4*T_2+e_gen*L^2/k=0 "for node 2" 260+290+240+200-4*T_3+e_gen*L^2/k=0 "for node 3" "(b) The rate of heat loss from the bottom surface is calculated by summing thQ

odal temperatures are determined to be T1 = 281°C, T2 = 349°C, and T3 = 283°

W1447=Q& . (b


5-55

5-64 A rectangular block is subjected to uniform heat flux at the top, and iced water at 0°C at the sides. The steady finite difference formulation of the problem is to be obtained, and the unknown nodal temperatures as well as the rate of heat transfer to the iced water are to be determined.

Assumptions 1 Heat transfer through the body is given to be steady and two-dimensional. 2 There is no heat generation within the block. 3 The heat transfer coefficient is very high so that the temperatures on both sides of the block can be taken to be 0°C. 4 Heat transfer through the bottom surface is negligible.


0°C62 6 10

Insulated

3 7 7

0°C

8 kW heater Insulated

• • •

• • •

• • •

• • •4 8 8

1 5 5

Symmetry

Analysis The nodal spacing is given to be ∆x=∆x=l=0.1 m, and the general finite difference form of an interior node equation for steady 2-D heat conduction is expressed as

04 nodebottomrighttopleft

nodebottomrighttopleft

=−+++ TTTTTk

There is symmetry about a vertical line passing through thmiddle of the region, and we need to consider only half ofthe region. Note that all side surfaces are at T

042

node =+−+++le

TTTTT&

e

with unknown temperatures. Replacing the symmetry lines by insulation and utilizing the mirror-image co rence equations are obtained to be as fo

0 = 0°C, and there are 8 nodes

ncept, the finite diffellows:

022

1215100 =

−+

−+

−+

lTT

kll

TTlkl

TTlkNode 1 (heat flux): lq&

Node 2 (interior): 04 26310 =−+++ TTTTT


042 4830 =−++ TTTT Node 4 (insulation):

002

56510 =+

−+

−+

lTT

kll

TTlklq& Node 5 (heat flux):

Node 6 (interior): 2 04 6765 =−+++ TTTTT


Node 8 (insulation): 2 874 ++ TTT 04 8 =− T

s the finite difference formulation of the problem.

) The sly with

= 5.4°C, T6 = 15.0°C, T7 = 9.9°C, T8 = 8.3°C

eat transferred from the block. Therefore,

Discussion The rate of heat transfer can also be determined by calculating the heat loss from the side surfaces using the heat conduction relation.

where

l = 0.1 m, k = 23 W/m⋅°C, T0 =0°C, and 2200 W/m3200)m 5.0 W)/(58000(/ =×== AQq &&

This system of 8 equations with 8 unknowns constitute

(b 8 nodal temperatures under steady conditions are determined by solving the 8 equations above simultaneouan equation solver to be

T1 = 18.2°C, T2 = 9.9°C, T3 = 6.2°C, T4 = 5.2°C, T5 2

(c) The rate of heat transfer from the block to the iced water is 6 kW since all the heat supplied to the block from the top must be equal to the h kW 8=Q& .


5-56

5-65 Prob. 5-64 is reconsidered. The unknown nodal temperatures as well as the rate of heat transfer to the iced water are to be determined.

Analysis The problem is solved using SS-T-CONDUCT, and the solution is given below. On the SS-T-CONDUCT Input window for 2-Dimensional Steady State Problem, the problem parameters and the boundary conditions are entered into the appropriate text boxes. With a uniform nodal spacing of 10 cm, there are 6 nodes in the x direction and 4 nodes in the y direction. Note that on the top boundary the heat flux is

2200 W/m3200)m 5.0 W)/(58000(/ =×== AQq && .

By clicking on the Calculate Temperature button, the computed results are as follows. The rate of heat transfer from the block to the iced water is 8 kW since all the heat supplied to the block from the top must be equal to the heat transferred from the block. Therefore, kW 8=Q& .


5-57

The temperature contour for this problem can be plotted by selecting the Graphical Output tab, as follows.


5-58

5-66 A square cross section with uniform heat generation is undergoing a steady two-dimensional heat transfer. The finite difference equations and the nodal temperatures are to be determined. Assumptions 1 Steady heat conduction is two-dimensional. 2 Thermal properties are constant. 3 The heat generation in the body is uniform. Properties The conductivity is given to be k = 25 W/m·K . Analysis (a) There are 4 unknown nodal temperatures, thus we need to have 4 equations to determine them uniquely. For nodes 1 to 4, we can use the general finite difference relation expressed as

022 ,

21 , ,1 ,

2 ,1 , ,1 =+

∆

+−+

∆

+− +−+−

ke

y

TTT

x

TTT nmnmnmnmnmnmnm &

25.0 ,T nm = )/( 2 ,1 ,1 , ,1 ,1 kxeTTTT nmnmnmnmnm ∆++++ +−+− &

ince , we have

hen

Node 4: kxeTTT ∆++++= &

(b) By letting th 300°C, T2 = 150°C, T3 = 400°C, and T4 = 250°C the results obtained from successive iterations are listed in the following table:

temp °C

yx ∆=∆S

)/(25.0 2 , ,11 , ,11 , , kxeTTTTT nmnmnmnmnmnm ∆++++= −−++ &

)/(25.0 2nodeleftbottomrighttopnode kxeTTTTT ∆++++= & or

T

)/500100(25.0 2node321 kxeTTT ∆++++= & Node 1:

Node 2: )/200100(25.0 2node142 kxeTTT ∆++++= &

)/500300(25.0 2node413 kxeTTT ∆++++= & Node 3:

)/300200(25.0 2 node324

where C 20/2node °=∆ kxe& .

e initial guesses as T1 =

Nodal erature,Iteration

T1 T2 T3 T4

1 292.5 215.6 340.6 269.1 2 294.1 220.8 345.8 271.6 3 296.6 222.1 347.1 272.3 4 297.3 222.4 347.4 272.4 5 297.4 222.5 347.5 272.5 6 297.5 222.5 347.5 272.5 7 297.5 222.5 347.5 272.5

Hence, the converged nodal temperatures are T1 = 297.5°C, T2 = 222.5°C, T3 = 347.5°C, T4 = 272.5°C

also be calculated using the EES. Copy the following lines and paste on a :

Solving by EES software, we get the same results: T1 = 297.5°C, T2 = 222.5°C, T3 = 347.5°C, T4 = 272.5°C

Discussion The finite difference equations canblank EES screen to solve the above equations T_1=0.25*(100+T_2+T_3+500+20) T_2=0.25*(100+200+T_4+T_1+20) T_3=0.25*(T_1+T_4+300+500+20) T_4=0.25*(T_2+200+300+T_3+20)


5-59

5-67 Prob. 5-66 is repeated. Using SS-T-Conduct (or other) software, the nodal temperatures are to be solved.

Assumptions 1 Steady heat conduction is two-dimensional. 2 Thermal properties are constant. 3 The heat generation in the body is uniform.

Properties The thermal conductivity is given to be k = 25 W/m·K.

Analysis On the SS-T-Conduct Input window for 2-Dimensional Steady State Problem, the problem parameters and the boundary conditions are entered into the appropriate text boxes. Note that with a uniform nodal spacing of 1 cm, there are 4 nodes in each the x and y directions.



5-60


T1 = 297.5°C, T2 = 222.5°C, T3 = 347.5°C, T4 = 272.5°C

Discussion The temperature contour for this problem can be plotted by selecting the Graphical Output tab, as follows.


5-61

5-68E A long solid bar is subjected to steady two-dimensional heat transfer. The unknown nodal temperatures and the rate of heat loss from the bar through a 1-ft long section are to be determined.

Assumptions 1 Heat transfer through the body is given to be steady and two-dimensional. 2 Heat is generated uniformly in the body. 3 The heat transfer coefficient also includes the radiation effects.

Properties The thermal conductivity is given to be k = 16 Btu/h.ft⋅°C.

Analysis The nodal spacing is given to be ∆x=∆y=l=0.25 ft, and the general finite difference form of an interior node for steady two-dimensional heat conduction is expressed as

042


kle

TTTTT&

• • •

4 5 6

• •

• • •h, T∞

1 2 3

7 8 9

h, T∞ h, T∞

h, T∞

e&

(a) There is symmetry about the vertical, horizontal, and diagonal lines passing through the center. Therefore, 3T 971 TTT == and T= 864 TTT2 === , and

521 and ,, TTT are the only 3 unknown nodal temperatures, and thus we need only3 equations to determine them uniquely. Also, we can replace the symmetry lines by insulation and utilize the mirror-image c

oncept for the interior nodes.

The finite difference equations for boundary nodes are obtained by applying an energy balance on the volume elements and taking the direction of all heat transfers to be towards the node under consideration:

044 :(interior) 5 Node

02

)( 2


04

)(2

22


20

52

20

22521

20

112

=+−

=+−+−

+−

=+−+−

∞

∞

kle

TT

leTThl

lTT

kll

TTlk

leTTlh

lTTlk

&

&

&

where 350 ftBtu/h 1019.0 ⋅×= , l = 0.25 ft, k = 16 Btu/h.ft⋅°F, h =7.9 Btu/h.fte& 2⋅°F, and T∞ =70°F. The 3 nodal temperatures

under steady conditions are determined by solving the 3 equations above simultaneously with an equation solver to be

T T5 = 397.93°F

9731 TTTT === = 361.89°F,

864 TTT === = 379.37°F, 2

(b) The rate of heat loss from the bar through a 1-ft long section is determined from an energy balance on one-eight section of the bar, and multiplying the result by 8:

[ ]

flength)ft per (F70]2-37.379ft)[361.89 ft)(1 F)(0.25/2ftBtu/h 9.7(8

ft) 1()22

8ft) 1()(2

)(2

88

2

2121conv section,eight one

Btu/h 4750=°×+°⋅⋅=

−+×=⎥⎦⎤

⎢⎣⎡ −+−×=×= ∞∞∞− TTTlhTTlhTTlhQQ &&

Discussion Under steady conditions, the rate of heat loss from the bar is equal to the rate of heat generation within the bar per unit length, and is determined to be

length)ft per (Btu/h 4750ft) 1 ft 0.5 ft 5.0)(Btu/h.ft 1019.0( 350gen =×××== VeEQ &&&

which confirms the results obtained by the finite difference method.

=


5-62

5-69 Heat transfer through a square chimney is considered. The nodal temperatures and the rate of heat loss per unit length are to be determined with the finite difference method.

Assumptions 1 Heat transfer is given to be steady and two-dimensional since the height of the chimney is large relative to its cross-section, and thus heat conduction through the chimney in the axial direction is negligible. It is tempting to simplify the problem further by considering heat transfer in each wall to be one dimensional which would be the case if the walls were thin and thus the corner effects were negligible. This assumption cannot be justified in this case since the walls are very thick and the corner sections constitute a considerable portion of the chimney structure. 2 There is no heat generation in the chimney. 3 Thermal conductivity is constant.

• • • • 1 2 3 4

Tskyho, To

• • • •

• •

5 6 7 8

9 10hi, Ti

Insulated

Insulated

Hot gases

Properties The thermal conductivity and emissivity are given to be k = 1.4 W/m⋅°C and ε = 0.9.

Analysis (a) The most striking aspect of this problem is the apparent symmetry about the horizontal and vertical lines passing through the midpoint of the chimney. Therefore, we need to consider only one-fourth of the geometry in the solution whose nodal network consists of 10 equally spaced nodes. No heat can cross a symmetry line, and thus symmetry lines can be treated as insulated surfaces and thus “mirrors” in the finite-difference formulation. Considering a unit depth and using the energy balance approach for the boundary nodes (again assuming all heat transfer to be into the volume element for convenience), the finite difference formulation is obtained to be

Node 1: 0])273([222

)(2

41

4151210 =+−+

−+

−+− TTl

lTTlk

lTTlkTTlh skyo εσ

Node 2: 0])273([22

)( 42

426232120 =+−+

−+

−+

−+− TTl

lTT

kll

TTlkl

TTlkTTlh skyo εσ

Node 3: 0])273([22

)( 43

437343230 =+−+

−+

−+

−+− TTl

lTT

kll

TTlkl

TTlkTTlh skyo εσ

Node 4: 0])273([l22

)( 44

4484340 =+−+

−+

−+− TT

lTTlk

lTTlkTTlh skyo εσ

022

)(2

51565 =

−+

−+−

lTTlk

lTTlkTTlh ii Node 5:

022

)( 6267656 =

−+

−+

−+−

lTT

kll

TTlkl

TTlkTTlh ii Node 6:

022

)( 787379767 =

−+

−+

−+

−+−

lTT

kll

TTkl

lTTlk


(TlhNode 8: 0])273([22

) 48

48781084800 =+−+

−+

−+

−+− TTl

lTT

kll

TTlkl

TTlkT skyεσ

Node 9: 022

)(2

910979 =

−+

−+−

lTTlk

lTTlkTTlh ii

0])273([222

)(2

410

4109108100 =+−+

−+

−+− TTl

lTTlk

lTTlkTTlh skyo εσ Node 10:

where l = 0.1 m, k = 1.4 W/m⋅°C, hi = 75 W/m2⋅°C, Ti =280°C, ho = 18 W/m2⋅°C, T0 =15°C, Tsurr =250 K, ε = 0.9, and σ = 5.67×10-8 W/m2.K4. This system of 10 equations with 10 unknowns constitutes the finite difference formulation of the problem.


5-63


T1 = 94.5°C, T2 = 93.0°C, T3 = 82.1°C, T4 = 36.1°C, T5 = 250.6°C,

T6 = 249.2°C, T7 = 229.7°C, T8 = 82.3°C, T9 = 261.5°C, T10 = 94.6°C

(c) The rate of heat loss through a 1-m long section of the chimney is determined from

W3153=°+++×°⋅=

−+−+−+−=

−=== ∑∑∑

C261.5)/2]-(280229.7)-(280249.2)-(280250.6)/2-m)[(280 1 m C)(0.1 W/m75(4

)])(2/()()())(2/([4

)(444

29765

surface,surfaceinner element,chimney offourth -one

TTlhTTlhTTlhTTlh

TTAhQQQ

iiiiiiii

mmimi

&&&

Discussion The rate of heat transfer can also be determined by calculating the heat loss from the outer surface by convection and radiation.


5-64

5-70 Heat transfer through a square chimney is considered. The nodal temperatures and the rate of heat loss per unit length are to be determined with the finite difference method.

Assumptions 1 Heat transfer is given to be steady and two-dimensional since the height of the chimney is large relative to its cross-section, and thus heat conduction through the chimney in the axial direction is negligible. It is tempting to simplify the problem further by considering heat transfer in each wall to be one dimensional which would be the case if the walls were thin and thus the corner effects were negligible. This assumption cannot be justified in this case since the walls are very thick and the corner sections constitute a considerable portion of the chimney structure. 2 There is no heat generation in the chimney. 3 Thermal conductivity is constant. 4 Radiation heat transfer is negligible.

• • • • 1 2 3 4

ho, To

• • • •

• •

5 6 7 8

9 10hi, Ti

Insulated

Insulated

Hot gases

Properties The thermal conductivity of chimney is given to be k = 1.4 W/m⋅°C.

Analysis (a) The most striking aspect of this problem is the apparent symmetry about the horizontal and vertical lines passing through the midpoint of the chimney. Therefore, we need to consider only one-fourth of the geometry in the solution whose nodal network consists of 10 equally spaced nodes. No heat can cross a symmetry line, and thus symmetry lines can be treated as insulated surfaces and thus “mirrors” in the finite-difference formulation. Considering a unit depth and using the energy balance approach for the boundary nodes (again assuming all heat transfer to be into the volume element for convenience), the finite difference formulation is obtained to be

Node 1: 022

)(2

151210 =

−+

−+−

lTTlk

lTTlkTTlho

022

)( 26232120 =

−+

−+

−+−

lTT

kll

TTlkl

TTlkTTlho Node 2:

022

)( 37343230 =

−+

−+

−+−

lTT

kll

TTlkl

TTlkTTlho Node 3:

022

)( 484340 =

−+

−+−

lTTlk

lTTlkTTloh Node 4:

022

)(2

51565 =

−+

−+−

lTTlk


022

)( 6267656 =

−+

−+

−+−

lTT

kll

TTlkl

TTlkTTlh ii Node 6:

(lNode 7: 022

) 787379767 =

−+

−+

−+

−+−

lTT

kll

TTkl

lTTlk

lTTlkTTh ii

022

)( 87 −h 8108480 =+

−+

−+−

lTT

kll

TTlkl

TTlkTTlo Node 8:

Node 9: 022

)(2

910979 =

−+

−+−

lTTlk

lTTlkTTlh ii

022

)(2

109108100 =

−+

−+−

lTTlk

lTTlkTTlho Node 10:

where l = 0.1 m, k = 1.4 W/m⋅°C, hi = 75 W/m2⋅°C, Ti =280°C, ho = 18 W/m2⋅°C, T0 =15°C, and σ = 5.67×10-8 W/m2.K4. This system of 10 equations with 10 unknowns constitutes the finite difference formulation of the problem.


5-65

−== ∑∑∑

C263.7)/2]-(280235.2)-(280253.0)-(280254.4)/2-m)[(280 1 m C)(0.1 W/m75(4

)])(2/()()())(2/(

)(444

29765 TTlhTTlhTTlhTTl

TTAhQQQ

iiiiiii

&&&

Discussion The rate of heat transfer can also be determined by calculating the heat loss from the outer surface by convection.


T1 = 118.8°C, T2 = 116.7°C, T3 = 103.4°C, T4 = 53.7°C, T5 = 254.4°C,

T6 = 253.0°C, T7 = 235.2°C, T8 = 103.5°C, T9 = 263.7°C, T10 = 117.6°C

(c) The rate of heat loss through a 1-m long section of the chimney is determined from

= [4

surface,surfaceinner element,chimney offourth -one

hi

mmimi

W2783=°+++×°⋅=

−+−+−+−

=


5-66

5-71 Prob. 5-69 is reconsidered. The effects of hot-gas temperature and the outer surface emissivity on the temperatures at the outer corner of the wall and the middle of the inner surface of the right wall, and the rate of heat loss are to be investigated.


"GIVEN" k=1.4 [W/m-C] A_flow=0.20*0.40 [m^2] t=0.10 [m] T_i=280 [C] h_i=75 [W/m^2-C] T_o=15 [C] h_o=18 [W/m^2-C] epsilon=0.9 T_sky=250 [K] DELTAx=0.10 [m] DELTAy=0.10 [m] d=1 [m] “unit depth is considered" sigma=5.67E-8 [W/m^2-K^4] “Stefan-Boltzmann constant" "ANALYSIS" "(b)" l=DELTAx "We consider only one-fourth of the geometry whose nodal network consists of 10 nodes. Using the finite difference method, 10 equations for 10 unknown temperatures are determined to be" h_o*l/2*(T_o-T_1)+k*l/2*(T_2-T_1)/l+k*l/2*(T_5-T_1)/l+epsilon*sigma*l/2*(T_sky^4-(T_1+273)^4)=0 "Node 1" h_o*l*(T_o-T_2)+k*l/2*(T_1-T_2)/l+k*l/2*(T_3-T_2)/l+k*l*(T_6-T_2)/l+epsilon*sigma*l*(T_sky^4-(T_2+273)^4)=0 "Node 2" h_o*l*(T_o-T_3)+k*l/2*(T_2-T_3)/l+k*l/2*(T_4-T_3)/l+k*l*(T_7-T_3)/l+epsilon*sigma*l*(T_sky^4-(T_3+273)^4)=0 "Node 3" h_o*l*(T_o-T_4)+k*l/2*(T_3-T_4)/l+k*l/2*(T_8-T_4)/l+epsilon*sigma*l*(T_sky^4-(T_4+273)^4)=0 "Node 4" h_i*l/2*(T_i-T_5)+k*l/2*(T_6-T_5)/l+k*l/2*(T_1-T_5)/l=0 "Node 5" h_i*l*(T_i-T_6)+k*l/2*(T_5-T_6)/l+k*l/2*(T_7-T_6)/l+k*l*(T_2-T_6)/l=0 "Node 6" h_i*l*(T_i-T_7)+k*l/2*(T_6-T_7)/l+k*l/2*(T_9-T_7)/l+k*l*(T_3-T_7)/l+k*l*(T_8-T_7)/l=0 "Node 7" h_o*l*(T_o-T_8)+k*l/2*(T_4-T_8)/l+k*l/2*(T_10-T_8)/l+k*l*(T_7-T_8)/l+epsilon*sigma*l*(T_sky^4-(T_8+273)^4)=0 "Node 8" h_i*l*(T_i-T_9)+k*l/2*(T_7-T_9)/l+k*l/2*(T_10-T_9)/l=0 "Node 9" h_o*l/2*(T_o-T_10)+k*l/2*(T_8-T_10)/l+k*l/2*(T_9-T_10)/l+epsilon*sigma*l/2*(T_sky^4-(T_10+273)^4)=0 "Node 10" "Right top corner is considered. The locations of nodes are as follows:" "Node 1: Middle of top surface Node 2: At the right side of node 1 Node 3: At the right side of node 2 Node 4: Corner node Node 5: The node below node 1, at the middle of inner top surface Node 6: The node below node 2 Node 7: The node below node 3, at the inner corner Node 8: The node below node 4 Node 9: The node below node 7,at the middle of inner right surface Node 10: The node below node 8, at the middle of outer right surface" T_corner=T_4 T_inner_middle=T_9 "(c)" "The rate of heat loss through a unit depth d=1 m of the chimney is" Q_dot=4*(h_i*l/2*d*(T_i-T_5)+h_i*l*d*(T_i-T_6)+h_i*l*d*(T_i-T_7)+h_i*l/2*d*(T_i-T_9))


5-67

Ti [C]

Tcorner [C]

Tinner, middle [C]

Q& [W]

200 220 240 260 280 300 320 340 360 380 400 420 440 460 480 500

28.38 30.35 32.28 34.2 36.08 37.95 39.79 41.6 43.39 45.16 46.91 48.63 50.33 52.01 53.66 55.3

205.7 224.3 242.9 261.5 280.1 298.6 317.2 335.8 354.4 372.9 391.5 410 428.6 447.1 465.6 43.39

2441 2677 2914 3153 3392 3632 3873 4115 4358 4602 4847 5093 5340 5588 5836 335.8

200 250 300 350 400 450 50025

30

35

40

45

50

55

2000

2500

3000

3500

4000

4500

5000

5500

6000

Ti [C]

T cor

ner

[C]

Q [

W]

200 250 300 350 400 450 500150

200

250

300

350

400

450

500

Ti [C]

T inn

er,m

iddl

e [C

]


5-68

ε Tcorner

[C] Tinner, middle [C]

Q& [W]

0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1

51.09 49.87 48.7 47.58 46.5 45.46 44.46 43.5 42.56 41.66 40.79 39.94 39.12 38.33 37.56 36.81 36.08 35.38 34.69

263.4 263.2 263.1 262.9 262.8 262.7 262.5 262.4 262.3 262.2 262.1 262 261.9 261.8 261.7 261.6 261.5 261.4 261.3

2836 2862 2886 2909 2932 2953 2974 2995 3014 3033 3052 3070 3087 3104 3121 3137 3153 3168 3183

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 132.5

36.5

40.5

44.5

48.5

52.5

2800

2850

2900

2950

3000

3050

3100

3150

3200

εT c

orne

r [C

]

Q [

W]

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

261.5

262

262.5

263

263.5

ε

T inn

er,m

iddl

e [C

]


5-69

5-72 The exposed surface of a long concrete damn of triangular cross-section is subjected to solar heat flux aconvection and radiation heat transfer. The vertical section of the damn is subjected to convection with

nd water. The

damn. 3 Heat transfer through the base is negligible. 4 Thermal properties and heat transfer coefficients are

m, and all nodes are boundary nodes. N

temperatures at the top, middle, and bottom of the exposed surface of the damn are to be determined.

Assumptions 1 Heat transfer through the damn is given to be steady and two-dimensional. 2 There is no heat generationwithin theconstant.

Properties The thermal conductivity and solar absorptivity are given to be k = 0.6 W/m⋅°C and αs = 0.7.

Analysis The nodal spacing is given to be ∆x=∆x=l=1 ode 5 on the insulated boundary can be treated as an interior node for which 04 nodebottomrighttopleft =−+++ TTTTT . Using the energy balance approach and taking the direction of all h finite difference equations for the nodes are eat transfer to be towards the node, the obtained to be as follows:

[ ] 0)(45sin2/

2)(

2 10012

1 =−++−

+− TThqll

TTlkTTlh ssii &α Node 1:

Node 2: 022

)( 2324211 =

−+

−+

−+−

lTT

kll

TTlkl

TTlkTTlh ii

[ ] 0)(45sin 300

3532 =

Insulated

Water

hi, Ti

ho, To

• •

• • •

•

4 5 6

2 3

1

&qs

Node 3: −++−

+−

TThqll

TTkl

lTT

kl ss &α

Node 4: 022

)(2

45424 =

−+

−+−

lTTlk

lTTlkTTlh ii

042 5634 =−++ TTTT Node 5:

Node 6: [ ] 0)(45sin2 600

65 =−++ TThql

k ss &α

where

2/− lTTl

C, h T 2⋅°C, T0 =25°C, αs = 0.7, and The syst init formulation of the problem. The 6 nodal temperatures under steady conditions are determ ultaneously with an equation

T1 = Ttop =21.3°C, T2 =15.1°C, T3 = Tmiddle =43.2°C

T4 =15.1°C, T5 =36.3°C, T6 = Tbottom =43.6°C

Discussion Note that the highest temperature occurs at a location furthest away from the water, as expected.

2 W/m800=sq& . l = 1 m, k = 0.6 W/m⋅° i =150 W/m2⋅°C, i =15°C, ho = 30 W/mem of 6 equations with 6 unknowns constitutes the f e difference

ined by solving the 6 equations above simsolver to be


5-70

5-73 The top and bottom surfaces of an L-shaped long solid bar are maintained at specified temperatures while the left surface is insulated and the remaining 3 surfaces are subjected to convection. The finite difference formulation of the problem is to be obtained, and the unknown nodal temperatures are to be determined.

Assumptions 1 Heat transfer through the bar is given to be steady and two-dimensional. 2 There is no heat generation within the bar. 3 Thermal properties and heat transfer coefficients are constant. 4 Radiation heat transfer is negligible.


Analysis (a) The nodal spacing is given to be ∆x=∆x=l=0.1 m, and all nodes are boundary nodes. Node 1 on the insulated boundary can be treated as an interior node for which . Using the energy balance approach and taking the direction of all heat transfer to be towards the node, the finite difference equations for the nodes are obtained to be as follows:

04 nodebottomrighttopleft =−+++ TTTTT

Node 1: 04212050 12 =−++ TT

Node 2: 0120

250

2)( 221232

2 =−

+−

+−

+−

+−∞ lT

kll

TTkl

lTTlk

lTlkTThl

120°C

1 2 3• • •

h, T∞

Insulated

50°C

Node 3: 0120

22)( 332

3 =−

+−

+−∞ lTlk

lTTlkTThl

where

res under steady conditions are determined by solving the 3 equations above simultaneously with an equat

T1 = 78.8°C, T2 = 72.7°C, T3 = 64.6°C

l = 0.1 m, k = 5 W/m⋅°C, h = 40 W/m2⋅°C, and T∞ =25°C.


(b) The 3 nodal temperatuion solver to be


5-71

5-74 Heat conduction through a long L-shaped solid bar with specified boundary conditions is considered. The unknown nodal temperatures are to be determined with the finite difference method.

Assumptions 1 Heat transfer through the body is given to be steady and two-dimensional. 2 Thermal conductivity is constant. 3 Heat generation is uniform.


Analysis (a) The nodal spacing is given to be ∆x=∆x=l=0.015 m, and the general finite difference form of an interior node for steady two-dimensional heat conduction for the case of constant heat generation is expressed as

• • •

180

31 2

4 5 6 7 8• • • • •

h, T∞

InsulatedLq&

042


kleTTTTT

&

We observe that all nodes are boundary nodes except node 5 that is an interior node. Therefore, we will have to rely on energy balances to obtain the finite difference equations. Using energy balances, the finite difference equations for each of the 8 nodes are obtained as follows:

Node 1: 0422

)(22

2

01412

1 =+−

+−

+−+ ∞le

lTTlk

lTTlkTTlhlqL &&

0222

)(2

0252321

2 =+−

+−

+−

+−∞le

lTT

kll

TTlkl

TTlkTThl & Node 2:

0422

)(2

03632

3 =+−

+−

+−∞le

lTTlk

lTTlkTThl & Node 3:

02

18022

2

045441 =+

−+

−+

−+

lel

TTkl

lTlk

lTTlkNode 4: lqL &&

Node 5: 041802

05624 =+−+++

kle

TTTT&

04

32

1802

)(2

06766563

6 =+−

+−

+−

+−

+−∞le

lTTlkNode 6:

lT

kll

TTkl

lTTlkTThl &

Node 7: 02

18022

)(2

077876

7 =+−

+−

+−

+−∞le

lT

kll

TTlkl

TTlkTThl &

Node 8: 04

18022

)(2 0

8878 =+

−+

−+−∞

lel

Tlkl

TTlkTTlh & 2

.

This sys wns is the finite difference formulation of the problem.

ations above simultaneously with an equation solver to be

T1 = 221.1°C, T2 = 217.9°C, T3 = 213.3°C, T4 = 212.7°C, T5 = 209.6°C, T6 = 202.8°C,

T7 = 193.3°C, T8 = 191.4°C

Discussion The accuracy of the solution can be improved by using more nodal points.

where

, W/m8000 , W/m105 2360 =×= Lqe && l = 0.015 m, k = 45 W/m⋅°C, h = 55 W/m2⋅°C, and T∞ =30°C

tem of 8 equations with 8 unkno

(b) The 8 nodal temperatures under steady conditions are determined by solving the 8 equ


5-72

Transient Heat Conduction

5-75C The formulation of a transient heat conduction problem differs from that of a steady heat conduction problem in that the transient problem involves an additional term that represents the change in the energy content of the medium with time. This additional term represent the change in the internal energy content during ∆t in the transient finite difference formulation.

tTTxcA im

imp ∆−∆ + /)( 1ρ

5-76C The two basic methods of solution of transient problems based on finite differencing are the explicit and the implicit methods. The heat transfer terms are expressed at time step i in the explicit method, and at the future time step i + 1 in the implicit method as

Explicit method: t

TTcEQ

im

im

pi

∆−

=++

∑1

elementi

element gen,sides All

Vρ&&

tTT

cEQi

mi

mp

i

∆−

=++

+∑1

element1+i

element gen,sides All

1 Vρ&& Implicit method:

5-77C The explicit finite difference formulation of a general interior node for transient heat conduction in a plane wall is

given by τ

im

im

imi

mi

mi

mTT

kxe

TTT−

=∆

++−+

+−

12

11 2&

. The finite difference formulation for the steady case is obtained by

mply setting and disregarding the time index i. It yields im

im TT =+1si

022

11 =++− +− kTTT mmm

∆xem&

5-78C For transient one-dimensional heat conduction in a plane wall with both sides of the wall at specified temperatures, e stability criteria for the explicit method can be expressed in its simplest form as

th

2)( 2∆x

1≤

∆=

tατ

licit method can be expressed in its simplest form as

5-79C For transient one-dimensional heat conduction in a plane wall with specified heat flux on both sides, the stability criteria for the exp

21

)( 2≤

∆

∆=

xtατ

which is identical to the one for the interior nodes. This is because the heat flux boundary conditions have no effect on the stability criteria.


5-73

5-80C The explicit finite difference formulation of a general interior node for transient two-dimensional heat conduction is

given by k

leTTTTTT i

2inodei

nodei

bottomi

righti

topi

left1

node )41()(&

τττ +−++++=+ . The finite difference formulation for the steady

case is obtained by simply setting im

im T=+1 and disregarding the time index i. ItT yields

042


kle

TTTTT&

5-81C There is a limitation on the size of the time step ∆t in the solution of transient heat conduction problems using the explicit method, but there is no such limitation in the implicit method.

5-82C The general stability criteria for the explicit method of solution of transient heat conduction problems is expressed as follows: The coefficients of all i

m in the 1 expressions (called the primary coefficient) in the simplified expressions must be greater than or equal to zero for all nodes m.

T +imT

5-83C For transient two-dimensional heat conduction in a rectangular region with insulation or specified temperature boundary conditions, the stability criteria for the explicit method can be expressed in its simplest form as

4)( 2∆x

interior nodes. This is because the in

1≤

∆tατ

sulation or specified temperature boundary onditions have no effect on the stability criteria.

we should still use the smallest time step practical to minimize the numerical error.

=

which is identical to the one for the c

5-84C The implicit method is unconditionally stable and thus any value of time step ∆t can be used in the solution of transient heat conduction problems since there is no danger of unstability. However, using a very large value of ∆t is equivalent to replacing the time derivative by a very large difference, and thus the solution will not be accurate. Therefore,


5-74

5-85 A plane wall with no heat generation is subjected to specified temperature at the left (node 0) and heat flux at the right boundary (node 6). The explicit transient finite difference formulation of the boundary nodes and the finite difference formulation for the total amount of heat transfer at the left boundary during the first 3 time steps are to be determined.

Assumptions 1 Heat transfer through the wall is given to be transient, and the thermal conductivity to be constant. 2 Heat transfer is one-dimensional since the plate is large relative to its thickness. 3 There is no heat generation in the medium.

Analysis Using the energy balance approach and taking the direction of all heat transfers to be towards the node under consideration, the explicit finite difference formulations become

Left boundary node: C8000 °== TT i

Right boundary node: t

TTcxq

xTT

kii

p

ii

∆−∆

=+∆− +

61

60

65

2ρ&

Heat transfer at left surface: t

TTcxA

x∆surfaceleft TT

kAQii

p

ii

∆−∆

=−

++

61

601

2ρ&

Noting that , the total amount of heat transfer becomes ∑ ∆=∆=i

i tQtQQ &&

∆x T0

• • • • • • • 0 1 2 3 4 5 6

0q&

∑∑=

+

=

∆⎟⎟⎠

⎞⎜⎜⎝

⎛

∆−∆

+∆−

=∆=3

1

01

0103

1surfaceleft surfaceleft 2i

ii

p

i

i

i tt

TTcxA

xTT

kAtQQ ρ&

5-86 A plane wall with variable heat generation and constant thermal conductivity is subjected to uniform heat flux 0q& at left (node 0) and convection at the right boundary (node 4). The explicit transient finite difference formulation of the boundary nodes is to be determined.

the

e thermal conductivity to be constant. 2 Heat Radiation heat transfer is negligible.

y lance approach and taking the direction of all explicit finite

rence formulations become

Left boundary node:

Assumptions 1 Heat transfer through the wall is given to be transient, and thtransfer is one-dimensional since the wall is large relative to its thickness. 3

Analysis Using the energ baheat transfers to be towards the node under consideration, the

h, T∞∆x

• • • • • 0 1 2 3 4

0q&

),( txe&diffe

tTT

cxAxAeAqxTT ii − 01 kA

ii

pi

∆−∆

=∆++∆

+0

10

00 2)2/( ρ&&

Right boundary node:

t

TTcxAxAeTThA

xTT

kAii

piii

ii

∆−∆

=∆+−+∆− +

∞4

14

4443

2)2/()( ρ&


5-75

5-87 A plane wall with variable heat generation and constant thermal conductivity is subjected to uniform heat flux at the left (node 0) and convection at the right boundary (node 4). The explicit transient finite difference formulation of the boundary nodes is to be determined.

0q&

Assumptions 1 Heat transfer through the wall is given to be transient, and the thermal conductivity to be constant. 2 Heat transfer is one-dimensional since the wall is large relative to its thickness. 3 Radiation heat transfer is negligible.

Analysis Using the energy balance approach and taking the direction of all heat transfers to be towards the node under consideration, the implicit finite difference formulations become

Left boundary node:

t

TTcxAxAeAq

x∆TT

kAii

pi

ii

∆−∆

=∆++− +

+++

01

0100

10

11

2)2/( ρ&&


t

TTcxAxAeTThA

xTT

kAii

piii

ii

∆−∆

=∆+−+∆− +

+++∞

++4

141

41

41

14

13

2)2/()( ρ&

h, T∞∆x

• • • • • 0 1 2 3 4

),( txe&0q&


5-76

5-88 A hot brass plate is having its upper surface cooled by impinging jet while its lower surface is insulated. The implicit finite difference equations and the nodal temperatures of the brass plate after 10 seconds of cooling are to be determined. Assumptions 1 Transient heat conduction is one-dimensional. 2 Thermal properties are constant. 3 Convection heat transfer coefficient is uniform. 4 Heat transfer by radiation is negligible. 5 There is no heat generation. Properties The properties of the brass plate are given as ρ = 8530 kg/m3, cp = 380 J/kg·K, k = 110 W/m·K, and α = 33.9 × 10−6 m2/s. Analysis The nodal spacing is given to be ∆x = 2.5 cm. Then the number of nodes becomes M = L/∆x +1 = 10/2.5 + 1 = 5. This problem involves 5 unknown nodal temperatures, and thus we need to have 5 equations. The finite difference equation for node 0 on the top surface subjected to convection is obtained by applying an energy balance on the half volume element about node 0 and taking the direction of all heat transfers to be towards the node under consideration:

t

TTcx

xTT

kTThii

p

iii

∆−∆

=∆−

+−+++

+∞

01

01

01

110 2

)( ρ

00 =q&

022221or 010⎠⎝

∞kk

Node 4 is on insulated boundary, and thus we can treat it as an interior node by using the mirror image concept. Nodes 1, 2, and 3 are inodes, and thus for them we can

11 =∆

+++⎟⎞

⎜⎛ ∆

++− ++ TxhTTTxh iii ττττ

nterior use the general explicit finite

ifference lation expressed as d re

τ11 +++ iii

im

imi

mi

mi

mTT

TTT−

=+−+

++

++−

111

111 2

r 1 =++ +o 1(11 +− 0)2− mmm TTT ττ

hus, the nite difference equations are

imTτ

T explicit fi

022221 01

11

0 =∆

+++⎟⎠⎞

⎜⎝⎛ Node 0: ∆

++− ∞++ T

kxhTTT

kxh iii ττττ

111 =+++− +++ iii TTT τττ

34 =+ TTτ

Node 4: 13

14

13 ++− +++ iii TTT τττ

where ∆x = 2.5 cm, k = 110 W/m·K, h = 220 W/m2·K, T∞ = 15°C, α = 33.9 × 10−6 m2/s, and h∆x/k = 0.05. For time step of ∆t

Node 1: 0)21( 11

21

11

0 =+++− +++ iiii TTTT τττ

iNode 2: 2321 T

Node 3: )21( 113

12 ++− +++ iiii TT ττ

0)21(

0

04 =+ iT )21(

= 10 s. Then the mesh Fourier number becomes

5424.0)m 025.0(∆x

)s 10)(s/m 109.33( 26×∆ −tατ 22 === (for ∆t = 10 s)

n be determined by solving the 5 equations above

05*0.5424*15=0

*T_3+650=0

T0 = 631.2°C, T1 = 644.7°C, T2 = 648.5°C, T3 = 649.6°C, T4 = 649.8°C Discussion Unlike the explicit method, the implicit method does not require any stability criterion, and the solution will converge with large values of time step. However, the large time step tends to give less accurate the results.

(b) The nodal temperatures of the brass plate after 10 seconds of cooling casimultaneously with an equation solver. Copy the following lines and paste on a blank EES screen to solve the above equations: -(1+2*0.5424+2*0.05*0.5424)*T_0+2*0.5424*T_1+650+2*0.

0.5424*T_0-(1+2*0.5424)*T_1+0.5424*T_2+650=0

0.5424*T_1-(1+2*0.5424)*T_2+0.5424

0.5424*T_2-(1+2*0.5424)*T_3+0.5424*T_4+650=0

0.5424*T_3-(1+2*0.5424)*T_4+0.5424*T_3+650=0

Solving by EES software, we get the same results:


5-77

5-89 Prob. 5-88 is repeated. Using SS-T-Conduct (or other) software with explicit method, the temperature at the surface that is being cooled by the impinging jet as a function of time varying from 0 to 60 minutes is to be plotted. The duration for the surface to be cooled to 100°C is to be determined.

Assumptions 1 Transient heat conduction is one-dimensional. 2 Thermal properties are constant. 3 Convection heat transfer coefficient is uniform. 4 Heat transfer by radiation is negligible. 5 There is no heat generation.

Properties The properties of the brass plate are given as k = 110 W/m·K and α = 33.9 × 10−6 m2/s.

Analysis On the SS-T-Conduct Input window for 1-Dimensional Transient Problem, the problem parameters and the boundary conditions are entered into the appropriate text boxes.


5-78


The temperature at the surface as a function of time for 0 to 60 minutes is plotted as follows.

t, sec

0 1000 2000 3000

T 0, °C

0

100

200

300

400

500

600

700

From the results computed by the SS-T-Conduct software, the surface temperature reached 100°C at t = 3055 s.

Discussion When computing with explicit method, the time step should be chosen such that the stability criterion is satisfied. In this problem, the proper time step is ∆t ≤ 8.779 s.


5-79

5-90 A plane wall with variable heat generation and constant thermal conductivity is subjected to insulation at the left (node 0) and radiation at the right boundary (node 5). The explicit transient finite difference formulation of the boundary nodes is to be determined.

Assumptions 1 Heat transfer through the wall is given to be transient and one-dimensional, and the thermal conductivity to be constant. 2 Convection heat transfer is negligible.

Analysis Using the energy balance approach and taking the direction of all heat transfers to be towards the node under consideration, the explicit transient finite difference formulations become

Left boundary node:

t

TTcxAxAe

xTT

kAii

pi

ii

∆−∆

=∆

+∆− +

01

00

01

22ρ&


t

TTcxAxAe

xTT

kATTAii

pi

iii

∆−∆

=∆

+∆−

+−+

51

55

5445

4isurr 22

])()[( ρεσ &

ary and the finite difference formulation for the total amount of heat transfer at the

1 Heat transfer through the wall is given to be transient and one-dimensional, and the thermal conductivity to

heat transfers to be towards the node under it transient finite difference formulations become

Left boundary node:

∆x

)(xe&

• • • • • •0 1 2 3 4 5

Insulated ε Tsurr

5-91 A plane wall with variable heat generation and constant thermal conductivity is subjected to combined convection, radiation, and heat flux at the left (node 0) and specified temperature at the right boundary (node 4). The explicit finite difference formulation of the left boundright boundary are to be determined.

Assumptions be constant.

Analysis Using the energy balance approach and taking the direction of allconsideration, the explic

tTT

cxAxAexTT

kATThATTAAqii

pi

iiiii

∆−∆

=∆

+∆−

+−+−++

∞0

10

001

04

0surr04

22)(])([ ρεσ &&

t transf at right surface: Hea er

h, T∞

∆x

• • • • • 0 1 2 3 4

Tsurr TL),( txe&

0q&t

TTcxAxAe

xTT

kAQiright &

ii

pi

ii

∆−∆

=∆

+∆−

++

41

44

43surface 22

ρ&

Noting that

the total amount of heat transfer becomes

∑ ∆=∆= i tQtQQ && i

∑

∑

=

+

=

∆⎟⎟⎠

⎞⎜⎜⎝

⎛

∆−∆

+∆

−∆−

=

∆=

20

1

41

44

34

20

1surfaceright surfaceright

22i

ii

pi

iii

i

tt

TTcxAxAe

xTT

kA

tQQ

ρ&

&


5-80

x and (m, e

5-92 Starting with an energy balance on a volume element, the two-dimensional transient explicit finite difference equation for a general interior node in rectangular coordinates for for the case of constant thermal conductivity and no heat generation is to be obtained.

T x y t( , , )

Analysis (See Figure 5-49 in the text). We consider a rectangular region in which heat conduction is significant in the x and y directions, and consider a unit depth of ∆z = 1 in the z direction. There is no heat generation in the medium, and the thermal conductivity k of the medium is constant. Now we divide the x-y plane of the region into a rectangular mesh of nodal points which are spaced ∆x and ∆y apart in the x and y directions, respectively, and consider a general interior node (m, n) whose coordinates are m∆= yn∆= . Noting that the volume element centered about the general interior node n) involves heat conduction from four sides (right, left, top, and bottom) and expressing them at previous time step i, thtransient explicit finite difference formulation for a general interior node can be expressed as

x y

tTT

cyx

yTT

xkxyx ∆∆∆

TTyk

TTxk

TTyk

inm

inm

p

inm

inm

inm

inm

inm

inm

inm

inm

∆

−×∆×∆=

∆

−×∆

−×∆+

−×∆

−×

+

−++−

,1

,

,1,,,1,1,,,1

)1(

)1(+)1()1(+)1

ρ

Taking a square mesh (∆x = ∆y = l) and dividing each term by k gives, after simplifying,

∆(

τ

inm

inmi

nmi

nmi

nmi

nmi

nmTT

TTTTT ,1

,4−

=−++++

−++− ,1,1,,1,1

where pck ρα /= is the thermal diffusivity f the material and 2/ lt∆=ατ is the dimension o less mesh Fourier number. It can also be expressed in terms of the temperatures at the neighboring nodes in the following easy-to-remember form:

τ

iiiiiii TT

TTTTT node1

nodenodebottomrighttopleft 4

−=−+++

+

Discussion We note that setting gives the steady finite difference formulation. ii TT node1

node =+


5-81

o heat

m, e

5-93 Starting with an energy balance on a volume element, the two-dimensional transient implicit finite difference equation for a general interior node in rectangular coordinates for ),,( tyx for the case of constant thermal conductivity and ngeneration is to be obtained.

T

Analysis (See Figure 5-49 in the text). We consider a rectangular region in which heat conduction is significant in the x and y directions, and consider a unit depth of ∆z = 1 in the z direction. There is no heat generation in the medium, and the thermal conductivity k of the medium is constant. Now we divide the x-y plane of the region into a rectangular mesh of nodal points which are spaced ∆x and ∆y apart in the x and y directions, respectively, and consider a general interior node (m, n) whose coordinates are and yn∆= . Noting that the volume element centered about the general interior node (n) involves heat conduction from four sides (right, left, top, and bottom) and expressing them at previous time step i, thtransient implicit finite difference formulation for a general interior node can be expressed as

xmx ∆= y

tTT

cyx

yTT

xkx

TTyk

yTT

xkx

TTyk

inm

inm

p

inm

inm

inm

inm

inm

inm

inm

inm

∆

−×∆×∆=

∆

−×∆

∆

−×∆+

∆

−×∆

∆

−×∆

+

++−

++

+++

++−

,1

,

1,

11,,

1,1

1,

11,

1,

1,1

)1(

)1(+ )1()1(+)1(

ρ

Taking a square mesh (∆x = ∆y = l) and dividing each term by k gives, after simplifying,

τnmnmnmnm −+

im

imiiii

nmTT

TTTTT−

=−++++

++++i++−

111111

,1 4

urier number. It can also be expressed in terms of the temperatures at the neighboring nodes in the following easy-to-remember form:

,1,1,,1

where pck ρα /= is the thermal diffusivity of the material and 2/ lt∆=ατ is the dimensionless mesh Fo

τnodebottomrighttopleft

iiiiiii TT

TTTTT node1

node11111 4−

=−++++

+++++

iscussion We note that setting gives the steady finite difference formulation.

ii TT node1

node =+D


5-82

5-94 Starting with an energy balance on a disk volume element, the one-dimensional transient explicit finite difference equation for a general interior node for in a cylinder whose side surface is insulated for the case of constant thermal conductivity with uniform heat generation is to be obtained.

),( tzT

Analysis We consider transient one-dimensional heat conduction in the axial z direction in an insulated cylindrical rod of constant cross-sectional area A with constant heat generation 0 and constant conductivity k with a mesh size ofg& z∆ in the zdirection. Noting that the volume element of a general interior node m involves heat conduction from two sides and the volume of the element is zA∆=element , the transient explicit finite difference formulation for an interior node can be

expressed as V

txx ∆∆∆TT

xcAxAeTT

kATT

kAi

mi

mp

im

im

im

im −

∆=∆+−− +

+−1

011 + ρ&

Canceling the surface area A and multiplying by ∆x/k, it simplifies to

)()( 122

01

im

im

im

im TTxxe

T −∆

=∆

++ ++

& 21

im tk

TT∆

−− α

Using the definition of the dimensionless mesh Fourier number

where pck ρα /= is the thermal diffusivity of the wall material.

2xt

∆

∆=ατ , the last equation reduces to

τ

ii −+12mmi

mi

mi

mTT

kxe

TTT =∆

++− +−0

11 2&

We note that setting gives the steady finite difference formulation.

ght boundary (node 2). The complete

one-dimensional, and the thermal conductivity to t tran s negligible. 3 There is no heat generation.

Analysis Using the energy balance approach with a unit area A = 1 and taking the direction of all heat transfers to be towards the node under consideration, the finite difference formulations become

Discussion ii TT m1

m =+

5-95 A composite plane wall consists of two layers A and B in perfect contact at the interface where node 1 is at the interface. The wall is insulated at the left (node 0) and subjected to radiation at the ritransient explicit finite difference formulation of this problem is to be obtained.

Assumptions 1 Heat transfer through the wall is given to be transient and

• • m-1 m m+1

InsulationDisk

Insulated

∆x

1

ε• ••0 2

A

Tsurr

RadiationB

Interface

be constant. 2 Convection hea sfer i


tTT

cxx 2TT

kii

ApA

ii

∆−∆

=∆− +

01

0,

01 ρ

Node 1 (at interface):

A

tTT

cxcxxTT

kxTT

kii

1BBApA

ii

B

ii

A ∆−

⎟⎠⎞

⎜⎝⎛ ∆

+∆

=∆−

+∆− +

11

,1210

22ρρ

Node 2 (at right boundary):

tTT

cxxTT

kTTii

BpB

ii

Bi

∆−∆

=∆−

+−+

21

2,

2142

4surr 2

])([ ρεσ


5-83

α .

5-96 A uranium plate initially at a uniform temperature is subjected to insulation on one side and convection on the other. The transient finite difference formulation of this problem is to be obtained, and the nodal temperatures after 5 min and under steady conditions are to be determined.

Assumptions 1 Heat transfer is one-dimensional since the plate is large relative to its thickness. 2 Thermal conductivity is constant. 3 Radiation heat transfer is negligible.

Properties The conductivity and diffusivity are given to be k = 28 W/m⋅°C and = /sm 105.12 26−×

Analysis The nodal spacing is given to be ∆x = 0.015 m. Then the number of nodes becomes 1/ +∆= xLM = 0.09/0.015+1 = 7. This problem involves 7 unknown nodal temperatures, and thus we need to have 7 equations. Node 0 is on insulated boundary, and thus we can treat it as an interior note by using the mirror image concept. Nodes 1, 2, and 3 are interior nodes, and thus for them we can use the general explicit finite difference relation expressed as

τ

im

im

imi

mi

mi

mTT

kxe

TT−−1 2T−

=∆

+++

+

12

1&

→ k

xeTTTT

imi

mi

mi

mi

m

2

111 )21()(

∆+−++= +−

+ &τττ

The finite difference equation for node 4 on the right surface subjected to convection is obtained by applying an energy balance on the half volume element about node 4 and taking the direction of all heat transfers to be towards the node under consideration:

tTT

cxxeTT

kTThiiii

i −∆=

∆+

−+−

+6

1665)( :n)(convectioNode ρ&

x

kxe

TTTT

kxe

TTTT

kxe

TT

kxe

TTTT

kxe

TTTT

p

iiii

iiii

iiii

iiii

iiii

∆∆

∆+−++=

∆+−++=

∆+−+=

∆+−++=

∆+−++=

∞

+

+

+

+

+

06

20

5641

5

0453

14

20

231

2

20

1201

1

20

0111

0

22 6

)21()( :(interior) 5 Node

)21()( :r)

)21() :(interior) 2 Node


)21()( :)(insulated 0 Node

τττ

τττ

ττ

τττ

τττ

&

&

&

&

&

or

TT +1(τ

kxe

TTTT iiii ∆+−++=+

2

20

3421

3

(interio 4 Node

)21()( :(interior) 3 Node τττ&

kkk ⎠⎝

where

xeTxhTTxhT iii

20

561

6)(

22221∆

+∆

++⎟⎞

⎜⎛ ∆

−−= ∞+ &

τττττ

& , and m2/s.

er ro. The coefficient of is smaller in this case, and thus the stability criteria for this problem can be

C20 C, W/m35 C, W/m28 , W/m10 m, 015.0 2360 °=°⋅=°⋅===∆ ∞Thke 6105.12 −×=αx

The upper limit of the time step ∆t is determined from the stability criteria that requires all primary coefficients to be greatthan or equal to ze iT4

expressed as

)/1(2

)/1(2

0221kxh

tkxhk ∆+

≤∆→∆+

≤→≥−−α

τττ

since 2/ xt ∆∆=ατ . Substituting the given quantities, the maximum allowable the time step becomes

1 2xxh ∆∆

s 8.8C)] W/m.28/(m) 015.0)(C. W/m35(1/s)[m 105.12(2

)m 015.0( 2

226 =°°+×

≤∆−

t

Therefore, any time step less than 8.8 s can be used to solve this problem. For convenience, let us choose the time step to be ∆t = 7.5 s. Then the mesh Fourier number becomes

h, T∞∆x

Insulated

• • • • • • • 0 1 2 3 4 5 6

e&


5-84

4167.0)m 015.0( 22 ==

∆=

xτ s) /s)(7.5m 105.12( 26×∆ −tα

Substituting this value of τ and other given quantities, the nodal temperatures after 5×60/7.5 = 40 time steps (5 min) are ined to be determ

After 5 min:

T = 229.9°C, T = 229.7°C, T = 229.0°C, T = 227.7°C, T = 225.8°C, T = 223.3°C, and T6 = 220.0 °C

the number of time steps until the nodal temperatures no longer change. Using EES, we increased time steps for 16.8 hours (60500 seconds). The temperatures seem to remain constant at about this time. Then, the nodal temperatures under steady conditions are

T0 = 2736°C, T1 = 2732°C, T2 = 2720°C, T4 = 2700°C, T5 = 2672°C, T6 = 2636°C, and T7 = 2591 °C

0 1 2 3 4 5

b) The time needed for transient operation to be established is determined by increasing (


5-85

5-97 Prob. 5-96 is reconsidered. The nodal temperatures after 5 min and under steady conditions are to be determined.

ThT Input window for 1-Dimensional Transient Problem, the problem parameters and the

Analysis e problem is solved using SS-T-CONDUCT, and the solution is given below. (a) On the SS-T-CONDUCboundary conditions are entered into the appropriate text boxes.



5-86

) The time needed for steady state conditions to be established is determined by using the Implicit method, since the number of time steps required to compute by the E cit me , with e st at do violate the stability criteria, exceed the maximu owab mber of time ste SS-T UDUOn the SS-T-COND Inp ow im T m parameters and the boundary condit re ente th ria box e t mp ethod is selected.

(bxpli thod a tim ep th es not

m all le nu ps in -CO CT. UCT

red intout winde approp

for 1-Dte text

ensional ransient Problhat the I

elicit m

, the problemions a es. Not

The nodal temperatures under steady conditions are determined to be

T1 = 2736°C, T2 = 2732°C, T3 = 2720°C, T4 = 2700°C, T5 = 2672°C, T6 = 2636°C, and T7 = 2591°C

The time needed for steady state conditions to be established is about 60500 s.


5-87

5-98 Prob. 5-96 is reconsidered. The effect of the cooling time on the temperatures of the left and right sides of the plate is to be investigated.


g_dot=1E6 [W/m^3] ity=20 [C]

h=35 [W/m^2-C]

M=L/DELTAx+1 "Number of nodes" "DELTAt=7.5 [s]" tau=(alpha*DELTAt)/DELTAx^2 "The technique is to store the temperatures in the parametric table and recover them (as old temperatures) using the variable ROW. The first row contains the initial values so Solve Table must begin at row 2. Use the

UPLICATE statement to reduce the number of equations that need to be typed. Column 1 contains the time, value of T[2], etc., and column 9 the Row."

Time=TableValue(Row-1,#Time)+DELTAt Duplicate i=1,7 T_old[i]=TableValue(Row-1,#T[i]) end

"Using the explicit finite difference approach, the six equations for the six unknown temperatures are determined

tau)*T_old[3]+tau*(g_dot*DELTAx^2)/k "Node 2" tau)*T_old[4]+tau*(g_dot*DELTAx^2)/k "Node 3"

T[5]=tau*(T_old[4]+T_old[6])+(1-2*tau)*T_old[5]+tau*(g_dot*DELTAx^2)/k "Node 4" T[6]=tau*(T_old[5]+T_old[7])+(1-2*tau)*T_old[6]+tau*(g_dot*DELTAx^2)/k "Node 5"

1-2*tau-2*tau*(h*DELTAx)/k)*T_old[7]+2*tau*T_old[6]+2*tau*(h*DELTAx)/k*T_infinity+tau*(g_dot*DELTAx^2)/k "Node 6, convection"

"GIVEN" L=0.09 [m] k=28 [W/m-C] alpha=12.5E-6 [m^2/s] T_i=100 [C]

T_infin

DELTAx=0.015 [m] "time=300 [s]" "ANALYSIS"

Dcolumn 2 the value of T[1], column 3, the

to be" T[1]=tau*(T_old[2]+T_old[2])+(1-2*tau)*T_old[1]+tau*(g_dot*DELTAx^2)/k "Node 0, insulated" T[2]=tau*(T_old[1]+T_old[3])+(1-2*tau)*T_old[2]+tau*(g_dot*DELTAx^2)/k "Node 1" T[3]=tau*(T_old[2]+T_old[4])+(1-2*T[4]=tau*(T_old[3]+T_old[5])+(1-2*

T[7]=(


5-88

] [C] [C] [C] [C] [C] [C]

Time [s]

T1 [C

T2 T3 T4 T5 T6 T7 Row

0 7.5 15

100 103.3 106.7

100 103.3 106.7

100 103.3 106.7

100 103.3 106.7

100 103.3 106.7

100 103.3 106.2

100 103.3

1 2

22.5 110 110 110 110 130 37.5 45 52.5 60

113.4 116.7 120.1 123.4 126.8

113.4 116.7 120.1 123.4

113.4 116.7 120 123.3

113.3 116.6 119.8 123.1

113.1 116.2 119.4 122.6

112.3 115.5 118.5 121.7

11311611912

67.5 130.1 126.7 130

126.6 129.9

126.3 129.5

09.8

125.7 128.9

109.3

124.8 127.9

106.7 109.8

.1

.2

.4 2.6

125.7 128.9

3 4 5 6 7 8 9 10

… … … … … … … … … … … … … … … … … … 3525 3533 3540 3548 3555 3563 3570 3578 3585 3593 3600

1276 1277 1279 1281 1283 1285 1287 1288 1290 1292 1294

1274 1276 1277 1279 1281 1283 1285 1287 1288 1290 1292

1268 1270 1272 1274 1276 1277 1279 1281 1283 1285 1286

1259 1261 1263 1265 1266 1268 1270 1272 1274 1275 1277

1246 1248 1250 1252 1254 1255 1257 1259 1261 1262 1264

1230 1232 1233 1235 1237 1239 1240 1242 1244 1246 1247

1209 1211 1213 1215 1216 1218 1220 1222 1223 1225 1227

471 472 473 474 475 476 477 478 479 480 481

0 500 1000 1500 2000 2500 3000 35000

200

400

600

800

1000

1200

1400

Time [s]

T [C

]

T1

T7


5-89

-99E A plain window glass initially at a uniform temperature is subjected to convection on both sides. The transient finite difference formulation of this problem is to be obtained, and it is to be determined how long it will take for the fog on the windows to clear up (i.e., for the inner surface temperature of the window glass to reach 54°F).

ssumptions 1 Heat transfer is one-dimensional since the window is large relative to its thickness. 2 Thermal conductivity is

Properties The conductivity and diffusivity are given to be k = 0.48 Btu/h.ft⋅°F and

lysis The nodal spacing is given to be ∆x = 0.125 in. Then the number of nodes become

5

Aconstant. 3 Radiation heat transfer is negligible.

/sft 102.4 26−×=α .

s 1/ +∆= xLMAna = 0.375/0.125+1 = 4. This problem involves 4 unknown nodal temperatures, and thus we need to have 4 equations. Nodes 2

rior nodes, and t f hem we can use the general explicit finite difference relation expressed as and 3 are inte hus or t

τ

im

im

imiii TTxe

TTT−

=∆

++−+12

2&

→ iiii TTTT )21()(1 ττ −++=+ mmm k+− 11 mmm 11 +−

since there is no heat generation. The fi tion for nodes 1 and 4 on the surfaces subjected to convection balance on the half volume n i ion of all hea f tow ode r tion

m

nite difference equais obtained by aut the

pplying an energy ode, and tak element abo ng the direct

t trans ers to be ards the n unde considera : hoTo

∆x

Fog

• • • 1 2 3 4

Window glass

hiTi

tc

xkTTh oo ∆

=∆

+− 4 2)( :n)(convectio 4 Node ρ

TTxTT

TTT

TT

Tk

xhT

tT

kT

iiiii

ii

ii

iii

i

i

−∆−

−+

∆+⎟⎟

⎠

⎞⎜⎜⎝

⎛=

∆−

=−

+

+

41

443

34

23

1

11

)21()(ior)Nod

)(ior)d

221

ectiod

ττ

ττ

τ

ρ

T i =+13

2

:

Ti +T i =+1 :

T i +22τk

xhiτT i+11 or

Tcx i

p∆ 1

x∆

∆TT ii − 12Thi ( :n)

i2

1

i +1 )

+

−+ 21()

−− 2τ

(inter 3 e

(inter 2 eNo

2(conv 1 eNo

ooiioi Tk

xhTT

kxh

T∆

++⎟⎟⎠

⎞⎜⎜⎝

⎛ ∆−−=+ ττττ 22221 34

14 or

where ∆x = 0.125/12 ft , k = 0.48 Btu/h.ft⋅°F, hi = 1.2 Btu/h.ft2⋅°F, Ti =35+2*(t/60)°F (t in seconds), ho = 2.6 Btu/h.ft2⋅°F, and To =35°F. The upper limit of the time step ∆t is determined from the stability criteria that requires all primary coefficients to be greater than or equal to zero. The coefficient of is smaller in this case, and thus the stability criteria for this problem can be expressed as

iT4

)/1(2

)/1(2

1 02212

kxhxt

kxhkxh

∆+∆

≤∆→∆+

≤→≥∆

−−α

τττ

since . Substituting the given quantities, the maximum allowable time step becomes 2/ xt ∆∆=ατ

s 2.12F)]Btu/h.ft. 48.0/(m) 12/125.0)(F.Btu/h.ft 6.2(1/s)[ft 102.4(2

)ft 12/125.0(226

2=

°°+×≤∆

−t

Therefore, any time step less than 12.2 s can be used to solve this problem. For convenience, let us choose the time step to be ∆t = 10 s. Then the mesh Fourier number becomes

3871.0)ft 12/125.0(

s) /s)(10ft 102.4(2

26

2=

×=

∆

∆=

−

xtατ

τ and other given quantities, the time needed for the inner surface temperature of the window glass Substituting this value ofto reach 54°F to avoid fogging is determined to be never. This is because steady conditions are reached in about 156 min, and the inner surface temperature at that time is determined to be 48.0°F. Therefore, the window will be fogged at all times.


5-90

ough the wall is steady and one-dimensional. 2 Thermal properties are constant. 3 Heat

n.

5-100 A stainless steel plane wall experiencing a uniform heat generation is subjected to constant temperatures on both sidesurfaces. Using SS-T-Conduct (or other) software, the nodal temperatures are to be determined, and compared with analytical solution.

Assumptions 1 Heat transfer thrtransfer by radiation is negligible. 4 The heat generation in the body is uniform.

Properties The thermal conductivity is given as 15.1 W/m·K.

Analysis (a) On the SS-T-Conduct Input window for 1-Dimensional Steady State Problem, the problem parameters and the boundary conditions are entered into the appropriate text boxes. Note that with a uniform nodal spacing of 10 cm, there are 11 nodes in the x directio



5-91

(b) From Chapter 2, the temperature variation in a plane wall with uniform heat generation is given as

⎟⎠

⎞⎜⎝

⎛ ++⎟

⎠

⎞⎜⎝

⎛ −+⎟

⎟⎠

⎞⎜⎜⎝

⎛−=

221

2)( 1212

2

22 TTLxTT

Lx

kLe

xT gen&

where 3 W/m10000= , K W/m1.15 ⋅=k , m 5.0=L , C 100gene& 1 °=T C 202 °=T ,


T(x),°C T(x),°C

x, m Analytical Numerical x, m Analytical Numerical

-0.5 100.00 100.00 0.1 131.47 131.47

-0.4 121.80 121.80 0.2 113.54 113.54

-0.3 136.98 136.98 0.3 88.98 88.98

-0.2 145.54 145.54 0.4 57.80 57.80

-0.1 147.47 147.47 0.5 20.00 20.00

0 142.78 142.78



5-92

x, m

T, °

160

AnalyticalNumerical140

100

120

C

0

20

40

60

80

-0.5 -0.4 - -0.2 -0.1 0. 0.1 0.2 0.3 0.4 0.50.3 0

iscussion The results computed by the SS-T-Conduct software match with the analytical solution from Chapter 2. The mperature variation plot shows that the temperature profile within the wall, for the case with asymmetrical boundary

onditions (T1 > T2), is not symmetric and the maximum temperature occurs to the left of the centerline.

Dtec


5-93

5-101 A hot 10-cm thick brass plate with uniform heat generation and both side surfaces are being cooled by liquid. Using o duct (o ther) software with implicit method, the surface temperature and the temperature at the center as a

function of time as it varies from 0 to 10 minutes are to be plotted. The nodal temperatures when steady conditions are are to be determined.

Assump n ensi are constant. 3 Convection heat transfer coefficient is uniform .

Properties −6 m2/s

SS-T-C n r o

achieved

tions 1 Transient heat conductio is one-dim onal. 2 Thermal properties. 4 Heat transfer by radiation is negligible. 5 Heat generation is uniform

The properties of the brass plate are given as k = 110 W/m·K and α = 33.9 × 10

Analysis On the SS-T-Conduct Input window for 1-Dimensional Transient Problem, the problem parameters and the boundary conditions are entered into the appropriate text boxes. Note that with a uniform nodal spacing of 2.5 cm, there are 5nodes in the x direction.


5-94

y clicking on the Calculate Temperature button, the computed results are as follows. B

The temperatures at the surface and the center versus time are plotted as follows.

t, sec

0 100 200 300 400 500 600

T, °

C

0

100

200

300

600

500

400

Tsurface Tcenter

Discussion Since both sides of the plate are exposed to the same liquid temperature and convection heat transfer coefficient, it is possible to solve half of the plane wall by treating the centerline of the plane wall as symmetry line and get the same results.

The nodal temperatures when steady conditions are achieved are

T(0) = T(0.1 m) = 90°C, T(0.025 m) = T(0.075 m) = 115.6°C, T(0.05 m) = 124.1°C,


5-95

ons 1 Heat transfer is one-dimensional since the exposed surface of the wall large relative to its thickness. 2

⋅°C, , and

5-102 The passive solar heating of a house through a Trombe wall is studied. The temperature distribution in the wall in 12 h intervals and the amount of heat transfer during the first and second days are to be determined.

AssumptiThermal conductivity is constant. 3 The heat transfer coefficients are constant.

/sm 1044.0 26−×=α 76.0=κProperties The wall properties are given to be k = 0.70 W/m . The hourly variation ux incident on a vertical surface is given to be of monthly average ambient temperature and solar heat fl

Time of day Ambient Temperature, °C

Solar insolation W/m2

7am-10am 10am-1pm

4am-7am

0 4 6

-4

375 750

0

1pm-4pm 4pm-7pm 7pm-10pm 10pm-1am 1am-4am

1 -2 -3 -4

580 95 0 0 0

Analysis The nodal spacing is given to be ∆x = 0.05 m, Then the number of nodes becomes = 0.30/0.05+1 = blem involves 7 unknown nodal temperatures, and thus we need to have 7 equations. Nodes 1, 2, 3, 4, and 5 are des, and t us for them we can use the general explicit finite difference relation expressed as

Trombe ∆x

• • • • • • • 0 1 2 3 4 5 6

hinTin

houtTout

wall

Glazing

Sun’s rays

Heat loss Heat

in

gahoutTout

hinTin

1/ +∆= xLM7. This prointerior no h

τ

im

im

imi

mi

mi

mTT

kxe

TTT i −

=∆

++−+

+−

12

11 2&

→ mi

mi

mi

m TTTT )21()( 111 ττ −++= +−+

e difference equation for boundary nodes 0 and 6 are obtained by applying an energy balance on the half volume elements a de under consideration:

Node 0:

The finitnd taking the direction of all heat transfers to be towards the no

tTT

cxAxTT

kATTAhii

p

iiii

∆−∆

=∆−

−+

01

0010inin 2

+)( ρ

or iiii Tk

xhTT

kxh

T inin

10in1

0 22+221∆

+⎟⎟⎠

⎞⎜⎜⎝

⎛ ∆−−=+ ττττ

iiii

iiii

TTTTm

TTTTm

2311

2

1201

1

)21()( :)2( 2 Node

)21()( :)1( 1 Node

ττ

ττ

−++==

−++==+

+

iiii

iiii

iiii

TTTTm

TTTTm

TTTTm

5641

5

4531

4

3421

3

)21()( :)5( 5 Node

)21()( :)4( 4 Node

)21()( :)3( 3 Node

ττ

ττ

ττ

−++==

−++==

−++==

+

+

+


5-96

Node 6 t

TTcxA

xTT

kAqATTAhii

p

iiiii

∆−∆

=∆−

+−+

61

665solar6outout 2

+)( ρκ &

or k

xqT

kxh

TTk

xhT

iiiii ∆∆

+⎟⎟⎞

⎜⎜⎛ ∆

−−=+ou

out56

out16 22+221

&κττττ +

⎠⎝solar

t 2τ

where

L = 0.30 m, k = 0.70 W/m.°C, /sm 1044.0 T26−×=α , out and solarq& are as given in the table,

=κ 0.76 , hout = 3.4 W/m2.°C, Tin = 20°C, hin = 9.1 W/m2.°C, and ∆x = 0.05 m.

Next we need to determine the upper limit of the time step ∆t from the stability criteria since we are using the explicit method. This requires the identification of the smallest primary coefficient in the system. We know that the boundary nodes

cient in this case is the coefficient of in the formulation of node 0 since hin >

are more restrictive than the interior nodes, and thus we examine the formulations of the boundary nodes 0 and 6 only. The smallest and thus the most restrictive primary coeffi iT0

hout, and thus

kxh

kxh ∆

−−<∆

−− outin 221221 ττττ

Therefore, the stability criteria for this problem can be expressed as

)/1(2

)/1(2

1 0221in

2

in

in

kxhxt

kxhkxh

∆+∆

≤∆→∆+

≤→≥∆

−−α

τττ

since . Substituting the given quantities, the maximum allowable the time step becomes 2/ xt ∆∆=ατ

s 1722C)] W/m.70.0/(m) 05.0)(C. W/m1.9(1/s)[m 1044.0(2

)m 05.0(226

2=

°°+×≤∆

−t

Therefore, any time step less than 1722 s can be used to solve this problem. For convenience, let us choose the time step to be ∆t = 900 s = 15 min. Then the mesh Fourier number becomes

1584.0)m 05.0(

s) /s)(900m 1044.0(2

26

2=

×=

∆

∆=

−

xtατ

Initially (at 7 am or t = 0), the temperature of the wall is said to vary linearly between 20°C at node 0 and 0°C at node 6. Noting that there are 6 nodal spacing of equal length, the temperature change between two neighboring nodes is (20 - 0)°C/6 = 3.33°C. Therefore, the initial nodal temperatures are

Substituting the given and calculated quantities, the nodal temperatures after 6, 12, 18, 24, 30, 36, 42, and 48 h are calculated and presented in the following table and chart.

C0 C,33.3 C,66.6 C,10 C,33.13 C,66.16 C,20 06

05

04

03

02

01

00 °=°=°=°=°=°=°= TTTTTTT


5-97

Time Time Nodal temperatures, °C step, i T0 T1 T2 T3 T4 T5 T6

0 h (7am) 0 20.0 16.7 13.3 10.0 6.66 3.33 0.0 6 h (1 pm) 24 17.5 16.1 15.9 18.1 24.8 38.8 61.5 12 h (7 pm) 48 21.4 22.9 25.8 30.2 34.6 37.2 35.8 18 h (1 am) 72 22.9 24.6 26.0 26.6 26.0 23.5 19.1 24 h (7 am) 96 21.6 22.5 22.7 22.1 20.4 17.7 13.9 30 h (1 pm) 120 21.0 21.8 23.4 26.8 34.1 47.6 68.9 36 h (7 pm) 144 24.1 27.0 31.3 36.4 41.1 43.2 40.9 42 h (1 am) 168 24.7 27.6 29.9 31.1 30.5 27.8 22.6 48 h (7 am) 192 23.0 24.6 25.5 25.2 23.7 20.7 16.3

t transfeThe rate of heaewton’s law o

r from the Trombe wall to the interior of the house during each time step is determined from f cooling using the average temperature at the inner surface of the wall (node 0) as

iiii ∆−+=∆−=∆= − ]2/)[()( in1

00inin0in wallTrumbe wallbe&

ount of heat transfer during the first time step (i = 1) or during the first 15 min period is 22

in0 −=°−+×°=∆− tT

loss. e

=

∆ii

t11

]

ified time period. In this case I = 48 for 12 h, 96 for 24 h, etc. wing the approach described above using a computer, the amount of heat transfer between the Trombe wall and the

kWh after 12 h

r

eat loss at night. Therefore, it may be worthwhile to cover the outer surface at night to minimize the heat losses.

Also the house loses 3421 kWh through the Trombe wall the 1st daytime as a result of the low start-up temperature, but delivers about 13,500 kWh of heat to the house the second day. It can be shown that the Trombe wall will deliver even more heat to the house during the 3rd day since it will start the day at a higher average temperature.

N

Qium tTTTAhtTTAhtQTr

Therefore, the am

[( 010in

1 wallTrumbe += TTAhQ kWh 8.96h) 25.0](C702/)703.68)[(m 7C)(2.8. W/m1.9(]2/)

The negative sign indicates that heat is transferred to the Trombe wall from the air in the house which represents a heateach timThen the total heat transfer during a specified time period is determined by adding the heat transfer amounts for

step as

∑∑ − −+== iii TTTAhQQ in1

00in wallTrumbe wallTrumbe 2/)[(II

=

where I is the total number of time intervals in the specFollointerior of the house is determined to be

QTrombe wall = - 3421

0 10 20 30 40 50

80

QTrombe wall = 1753 kWh after 24 h

QTrombe wall = 5393 kWh after 36 h

QTrombe wall = 15,230 kWh after 48 h

Discussion Note that the interiotemperature of the Trombe wall drops in early morning hours, but then rises as the solar energy absorbed by the exterior surface diffuses through the wall. The exterior surface temperature of the Trombe wall rises from 0 to 61.5°C in just 6 h because of the solar energy absorbed, but then drops to 13.9°C by next morning as a result of h

0

10

20

30

40

50

60

70

Time [hour]

e

T0T0T1T1T2T2T3T3T4T4T5T5T6T6]

[CTe

mpe

ratu

r


5-98

-103 Heat conduction through a long L-shaped solid bar with specified boundary conditions is considered. The temperature at t cor node f the b after and 3 to be determ with t ansient explicit finite difference me .

Assumptions tra rou bod en trans tw nsi Thermal conductivity is constant. 3 H era nif

Properties Th duct d d ty a n tobe 5 W/ d 1× .

An is The nodal spacing is given to be ∆x 0.0Th icit f ffe ua e d ed bas the e ala the nt c res

5he top thod

ner ( #3) o ody 2, 5, 0 min is ined he tr

1 Heateat gen

nsfer thtion is u

gh the orm.

y is giv to be ient and o-dime onal. 2

e con ivity an iffusivi re give k = 1 m⋅°C an /sm 0 26−2.3=α

alys =∆x=l= 15 m. e expl inite di rence eq tions ar etermin on the is of nergy b nce for transie ase exp sed as

tT i

m−+1TcE

im

p∑ elementiele

si All

Vρ&

Th itie &, , do not change with tim hus not us pe t i for them. Also, the ene lan ss be ied the ons ma ivit d the dimensionless

me rier r τ x

Qi&∆

=+ mentdes

e quant s Th , Rq& and

• • •

14

32

6 7 8 • • • •

, T∞

0°C

1

4 5 •

h

InsulatedLq&

e∞ e, and t we do need to e the su rscriprgy ba ce expre ions can simplif using definiti of ther l diffus y kα = pcρ an/

2/ l lt∆=α where sh Fou numbe y =∆=∆ . e th de un des pt node 5 that is an inte de ore ll h ely rgy es n th di eq s. Using energy balances, the finite difference equations for e the s ar ed ws

Node 1:

We not at all no s are bo dary no excerior no . Theref , we wi ave to r on ene balanc to obtai e finite fference uation

ach of 8 node e obtain as follo :

t∆T ii −+

11T

lTll

=−

+−∞1

2

422

ode 2:

c4

ρ l 2li1 e+ 0&

Tl i4k

2lTT ii − 12kT +1 )Th+ (

2lqL&

i

tTT

cllel

TTkl

lTTlk

lTTlkTThl

ii

pNiiii

i

∆−

=+−

+−

+−

+−∞2

12

2

0252321

2 2222)( ρ&

Node 3:

ii +2

tTT

cllel

TTlkl

TTlkTThliiiiii

i

∆−

=+−

+−

+−+

∞3

13

22

03632

3 4422)( ρ&

⎟⎟⎠

⎞⎜⎜⎝

⎛++++⎟

⎠⎞

⎜⎝⎛ −−= ∞

+

kle

TkhlTTT

khlT iiii

222441

20

6431

3&

τττ ) (It can be rearranged as

tTT

cllel

TTkl

lTlk

lTTlklq

iiiiiii

L ∆−

=+−

+−

+−

++

41

422

045441

22140

22ρ&& Node 4:

Node 5 (interior): ( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛+++++−=+

kle

TTTTT iiiii2

06425

15 14041

&ττ

tTT

cllel

TTlkl

Tkl

lTT

kll

TTlkTThliiiiiiiii

i

∆−

=+−

+−

+−

+−

+−+

∞6

16

22

06766563

6 43

43

2140

2)( ρ& Node 6:

tTT

cllel

Tkl

lTTlk

lTTlkTThl

iiiiiiii

∆−

=+−

+−

+−

+−+

∞7

17

22

077876

7 22140

22)( ρ& Node 7:

tTT

cllel

Tlkl

TTlkTTlhiiiii

i

∆−

=+−

+−

+−+

∞8

18

22

0887

8 44140

22)(

2ρ& Node 8:

where

l = 0.015 m, k =15 W/m⋅°C, h = 80 W/m2⋅°C, and T∞ =25°C. , W/m8000 , W/m102 2370 =×= Lqe &&


5-99

The upper limit of the time step ∆t is determined from the stability criteria that requires the coefficient of imT in the 1+i

mT expression (the primary coefficient) be greater than or equal to zero for all nodes. The smallest primary coefficient in the 8 equations above is the coefficient of iT3 in the 1

3+iT expression since it is exposed to most convection per unit volume (this

can be verified), and thus the stability criteria for this problem can be expressed as

1 4 4 0 14 1 4 1

2− − ≥ → ≤

+→ ≤

+τ τ τ

αhlk hl k

t lhl k

( / ) ( /

∆

2

)

since =τ . Substituting the given quantities, the maximum allowable value of the time step is determined to be / lt∆α

s 3.16C)] W/m.15/(m) 015.0)(C. W/m80(1/s)[m 102.3(4

)m 015.0(226

2=

°°+×≤∆

−t

Therefore, any time step less than 16.3 s can be used to solve this problem. For convenience, we choose the time step to be ∆t = 15 s. Then the mesh Fourier number becomes

2133.0)m 015.0(

10.3(2

6

2=

×==

−

lτ (for ∆t = 15 s)

Using the specified initial condition as the solution at time t = 0 (for i = 0), sweeping through the 9 equations above will givthe solution at intervals of 15 s

s) /s)(15m 2 2∆tα

e . Using a computer, the solution at the upper corner node (node 3) is determined to be 441,

9°C at 2, 5, and 30 min, respectively. It can be shown that the steady state solution at node 3 is 531°C. 520, and 52


5-100

5-104 Prob. 5-103 is reconsidered. The temperature at the top corner as a function of heating time is to be plotted.

T_i=140 [C] k=

T_infinity=25 [C] h=80 [W/m^2-C] q_dot_L=8000 [W/m^2] DELTAx=0.015 [m] DELTAy=0.015 [m] time=120 [s] "ANALYSIS" l=DELTAx DELTAt=15 [s] tau=(alpha*DELTAt)/l^2 RhoC=k/alpha "RhoC=rho*C" "The technique is to store the temperatures in the parametric table and recover them (as old temperatures) using the variable ROW. The first row contains the initial values so Solve Table must begin at row 2. Use the DUPLICATE statement to reduce the number of equations that need to be typed. Column 1 contains the time, column 2 the value of T[1], column 3, the value of T[2], etc., and column 10 the Row." Time=TableValue('Table 1',Row-1,#Time)+DELTAt Duplicate i=1,8 T_old[i]=TableValue('Table 1',Row-1,#T[i]) end "Using the explicit finite difference approach, the eight equations for the eight unknown temperatures are determined to be" q_dot_L*l/2+h*l/2*(T_infinity-T_old[1])+k*l/2*(T_old[2]-T_old[1])/l+k*l/2*(T_old[4]-T_old[1])/l+e_dot*l^2/4=RhoC*l^2/4*(T[1]-T_old[1])/DELTAt "Node 1" h*l*(T_infinity-T_old[2])+k*l/2*(T_old[1]-T_old[2])/l+k*l/2*(T_old[3]-T_old[2])/l+k*l*(T_old[5]-T_old[2])/l+e_dot*l^2/2=RhoC*l^2/2*(T[2]-T_old[2])/DELTAt "Node 2" h*l*(T_infinity-T_old[3])+k*l/2*(T_old[2]-T_old[3])/l+k*l/2*(T_old[6]-T_old[3])/l+e_dot*l^2/4=RhoC*l^2/4*(T[3]-T_old[3])/DELTAt "Node 3" q_dot_L*l+k*l/2*(T_old[1]-T_old[4])/l+k*l/2*(T_bottom-T_old[4])/l+k*l*(T_old[5]-T_old[4])/l+e_dot*l^2/2=RhoC*l^2/2*(T[4]-T_old[4])/DELTAt "Node 4" T[5]=(1-4*tau)*T_old[5]+tau*(T_old[2]+T_old[4]+T_old[6]+T_bottom+e_dot*l^2/k) "Node 5" h*l*(T_infinity-T_old[6])+k*l/2*(T_old[3]-T_old[6])/l+k*l*(T_old[5]-T_old[6])/l+k*l*(T_bottom-T_old[6])/l+k*l/2*(T_old[7]-T_old[6])/l+e_dot*3/4*l^2=RhoC*3/4*l^2*(T[6]-T_old[6])/DELTAt "Node 6" h*l*(T_infinity-T_old[7])+k*l/2*(T_old[6]-T_old[7])/l+k*l/2*(T_old[8]-T_old[7])/l+k*l*(T_bottom-T_old[7])/l+e_dot*l^2/2=RhoC*l^2/2*(T[7]-T_old[7])/DELTAt "Node 7" h*l/2*(T_infinity-T_old[8])+k*l/2*(T_old[7]-T_old[8])/l+k*l/2*(T_bottom-T_old[8])/l+e_dot*l^2/4=RhoC*l^2/4*(T[8]-T_old[8])/DELTAt "Node 8"


"GIVEN"

15 [W/m-C] alpha=3.2E-6 [m^2/s] e_dot=2E7 [W/m^3] T_bottom=140 [C]


5-101

Time [s]

T1 [C]

T2 [C]

T3 [C]

T4 [C]

T5 [C]

T6 [C]

T7 [C]

T8 [C]

Row

0 140 140 140 140 140 140 140 140 1 15 203.5 200.1 196.1 207.4 204 201.4 200.1 200.1 2 30 265 259.7 252.4 258.2 253.7 243.7 232.7 232.5 3 45 319 312.7 300.3 299.9 293.5 275.7 252.4 250.1 4 60 365.5 357.4 340.3 334.6 326.4 300.7 265.2 260.4 5 75 404.6 394.9 373.2 363.6 353.5 320.6 274.1 267 6 90 437.4 426.1 400.3 387.8 375.9 336.7 280.8 271.6 7 105 464.7 451.9 422.5 407.9 394.5 349.9 286 275 8 120 487.4 473.3 440.9 424.5 409.8 360.7 290.1 277.5 9 135 506.2 491 456.1 438.4 422.5 369.6 293.4 279.6 10 … … … … … … … … … … … … … … … … … … … … 1650 596.3 575.7 528.5 504.6 483.1 411.9 308.8 288.9 111 1665 596.3 575.7 528.5 504.6 483.1 411.9 308.8 288.9 112 1680 596.3 575.7 528.5 504.6 483.1 411.9 308.8 288.9 113 1695 596.3 575.7 528.5 504.6 483.1 411.9 308.8 288.9 114 1710 596.3 575.7 528.5 504.6 483.1 411.9 308.8 288.9 115 1725 596.3 575.7 528.5 504.6 483.1 411.9 308.8 288.9 116 1740 596.3 575.7 528.5 504.6 483.1 411.9 308.8 288.9 117 1755 596.3 575.7 528.5 504.6 483.1 411.9 308.8 288.9 118 1770 596.3 575.7 528.5 504.6 483.1 411.9 308.8 288.9 119 1785 596.3 575.7 528.5 504.6 483.1 411.9 308.8 288.9 120

0 200 400 600 800 1000 1200 1400 1600 1800100

150

200

250

300

350

400

450

500

550

Time [s]

T 3 [C

]


5-102

0

dimensional. 2 Heat is generated uniformly in

. explicit finite

5-105 A long solid bar is subjected to transient two-dimensional heat transfer. The centerline temperature of the bar after 2min and after steady conditions are established are to be determined. Assumptions 1 Heat transfer through the body is given to be transient and two-the body. 3 The heat transfer coefficient also includes the radiation effects. Properties The conductivity and diffusivity are given to be k = 28 W/m⋅°C and =αAnalysis The nodal spacing is given to be ∆x=∆x=l=0.1 m. The

/sm 1012 26−×

difference equations are determined on the basis of the energy balance for thetransient case expressed as

• • •

4 5 6

• • •

• • •h, T∞

1 2 3

7 8 9

h, T∞

h, T∞

e& tTT

cEQ mmp

i

∆−

=+∑ elementielement

sides All

Vρ&&

The quantities 0 and , , eTh &∞

ii+1

do not change with time, and thus we do not need to use the superscript i for them. The general explicit finite difference form of an

h, T∞

interior node for transient two-dimensional heat conduction is expressed as

kle

TTTTTT i2i

nodeinode

ibottom

iright

itop

ileft

1node )41()(

&τττ +−++++=+

There is symmetry about the vertical, horizontal, and diagonal lines passing through the center. Therefore, and , and are the only 3 unknown noda tures, and thus we need

ations to determine them uniquely. Also, we can replace the symmetry lines by insulation and utilize the mirror-image concept when writing the finite difference equations for the interior nodes. The finite difference equations for

applying an energy balance on the volume elements and taking the direction of all heat be towa s the node under consideration:

9731 TTTT === 8642 TTTT === 521 and ,, TTT l temperaonly 3 equ

boundary nodes are obtained bytransfers to rd

tTT

cllel

TTlkl

TTlkTThliiiiii

i

∆−

=+−

+−

+−+

∞1

11

22

01412

1 4422)( ρ& Node 1:

tTT

cllel

TTlkl

TTlkTTlhiiiiii

i

∆−

=+−

+−

+−+

∞2

12

22

02521

2 4422)(

2ρ& Node 2:

Node 5 (interior): ( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛++−=+

kle

TTT iii2

025

15 441

&ττ

350 W/m108×=where e& , l = 0.1 m, and k = 28 W/m⋅°C, h = 45 W/m2⋅°C, and T∞ = 30°C.

he upper limit of the time step ∆t is determined from the stability criteria that requires the coefficient of in the

expression (the primary coefficient) be greater than or equal to zero for all nodes. The smallest primary coefficient in

tions abov is the coefficient of in the expression since it is exposed to most convection per unit volume (this can be verified), and thus the stability criteria for this problem can be expressed as

T imT

1+imT

iT11

1+iTthe 3 equa e

)/1(4

)/1(4

1 04412

khllt

khlkhl

+≤∆→

+≤→≥−−

ατττ

since ∆=ατ um allowable value of the time step is determined to be 2/ lt . Substituting the given quantities, the maxim

s 179C)] W/m.28/(m) 1.0)(C. W/m45(1/s)[m 1012(4 226 °+×

≤∆−

t)m 1.0( 2

=°

Therefore, any time step less than 179 s can be used to solve this problem. For convenience, we choose the time step to be ∆t = 60 s. Then the mesh Fourier number becomes

072.0)m 1.0(

s) /s)(60m 1012(2

26

2=

×=

∆=

−

ltατ (for ∆t = 60 s)

Using the specified initial condition as the solution at time t = 0 (for i = 0), sweeping through the 3 equations above will give the solution at intervals of 1 min. Using a computer, the solution at the center node (node 5) is determined to be 227.5°C, 312.0°C, 387.6°C, 455.1°C, 515.5°C, 617.7°C, 699.3°C, and 764.5°C at 10, 15, 20, 25, 30, 40, 50, and 60 min, respectively. Continuing in this manner, it is observed that steady conditions are reached in the medium after about 6 hours for which the temperature at the center node is 1023°C.


5-103

di

Transient Problem, the problem parameters and the boundary conditions are entered into the appropriate text boxes.

5-106 Prob. 5-105 is reconsidered. The centerline temperature of the bar after 20 min and after steady con tions areestablished are to be determined.

Analysis The problem is solved using SS-T-CONDUCT, and the solution is given below. (a) On the SS-T-CONDUCT Input window for 2-Dimensional



5-104

(b) Steady conditions are reached in the medium after about 6 hours for which the temperature at the center node is 1023°C.


5-105

the explicit method.

neration only at

ty are given to be k = 0.84 W/m⋅°C

1 cm. The exnergy balance f

5-107 The formation of fog on the glass surfaces of a car is to be prevented by attaching electric resistance heaters to theinner surfaces. The temperature distribution throughout the glass 15 min after the strip heaters are turned on and also whensteady conditions are reached are to be determined using

Assumptions 1 Heat transfer through the glass is given to be transient and two-dimensional. 2 Thermal conductivity is constant. 3 There is heat gethe inner surface, which will be treated as prescribed heat flux.

Properties The conductivity and diffusivi26−

• • •

and /sm 1039.0 ×=α .

Analysis The nodal spacing is given to be ∆x = 0.2 cm and ∆y =finite difference equations are determined on the basis of the etransient case expressed as

plicit or the

tTT

EQii

i −+

+

∑1

igen,&& c mm

p ∆= elementelement

sides All

Vρ

metry. Note that we do not have a square mesh in this case, and thus we will have to rely on energy balances to obtain the

e nce equations. Using energy balances, the finite difference equations for

We consider only 9 nodes because of sym

finite diff reeach of the 9 nodes are obtained as follows:

Node 1: t

TTyxxTTyk

yTTxkTTyh

iiiii

ii∆

=∆−∆

+∆−∆

+−∆ 1214

1 22)(

2ρc p ∆

−∆ 11

22

Node 2:

ii+1

tTTyxc

yTT

xkxTTyk

xTTyk

ii

p

iiiiii

∆−∆

∆=∆−

∆+∆−∆

+∆−∆ +

21

2252321

222ρ

Node 3: t

TTyxcxTTyk

yTTxkTTyh

ii

p

iiiii

oo ∆−∆∆

=∆−∆

+∆−∆

+−∆ +

31

332363 2222

)(2

ρ

Heater 25 W/m

Thermal symmetry line

0.2 cm

1 cm

Outer surface

1 2 3

Glass

Inner surface

• • •

• • •

• • •

• • •

4 5 6

7 8 9


tTTxyc

xTT

ykyTTxk

yTTxkTTyh

ii

p

iiiiiii

ii ∆−∆

∆=∆−

∆+∆−∆

+∆−∆

+−∆+

41

44547414 222

)( ρ Node 4:

Node 5: t

yxcy

xky

xkx

ykx

yk p ∆∆∆=

∆∆+

∆∆+

∆∆+

∆∆ 55258565 ρ

TTTTTTTTTT iiiiiiiiii −−−−− +5

14

Node 6: t

TTxycxTT

ykyTTxk

yTTxkTTyh

ii

p

iiiiiii

io ∆−∆

∆=∆−

∆+∆−∆

+∆−∆

+−∆+

61

66569636 222

)( ρ

Node 7: t

TTyxcxTTyk

yTTxkTTyh p

iii ∆

−∆∆=

∆−∆

+∆−∆

+−∆

+ 7778747 2222

)(2

W5.12 ρ iiiiii +1

tTTyxc

TTxk

TTykTTyk

iiiiiii ∆∆=

−∆+

−∆+

−∆ +18858987 ρ

yxx

i

p ∆−

∆∆∆8

222 Node 8:

tTTyxc

xk

ykTTh poo =

∆+

∆+− 9 22

)(2

ρNode 9: TTyTTxy iiiii

i

∆−∆∆−∆−∆∆ +

91

99896

22

where

k = 0.84 W/m.°C, Ti = To = -3°C hi = 6 W/m2.°C,

ho = 20 W/m2.°C, ∆x = 0.002 m, and ∆y = 0.01 m.

i

/sm 1039.0/ 26−×== ck ρα ,


5-106

e stability criteria that requires the coefficient of in the xpression (the primary coefficient) be greater than or equal to zero for all nodes. The smallest primary coefficient in the 9

imT 1+i

mT The upper limit of the time step ∆t is determined from theequations above is the coefficient of iT6 in the 1

6+iT expression since it is exposed to most convection per unit volume (this

can be verified). The equation for node 6 can be rearranged as

⎟⎟⎠

⎞⎜⎜⎝

⎛

∆+

∆

++

∆∆+

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛

∆+

∆+

∆∆−=+

25

293

06221

6 21121x

Ty

TTT

xkh

tTxyxk

htT

iiioioi αα

Therefore, the stability criteria for this problem can be expressed as

⎟⎟⎠

⎞⎜⎜⎝

⎛

∆+

∆+

∆

⎟⎠

⎜⎝ ∆∆∆

22

22 112xyxk

hxyxk oα

Substituting the given quantities, the maximum allowable value of the time step is determined to be

or,

≤∆→≥⎟⎞

⎜⎛

++∆−1t 01121

ht oα

s 7.4

m) 01.0()m 002.0(m) 002.0)(C W/m84.0()/m 1039.0(2

2226

⎟⎟⎠

⎜⎜⎝

++°⋅

×× s

Therefore, any time step less than 4.8 s can be used to solve this problem. For convenience, we choose the time

11C W/m20

1t2

=⎞⎛ °⋅

≤∆

step to be ∆t 4 s. Then the temperature distribution throughout the glass 15 min after the strip heaters are turned on and when steady

ined to be (from the EES solutions in CD)

T = -0.73°C, T = -0.84°C, T = -0.98°C, T = 0.68°C, T = -0.03°C,

T6 = -2.0 C, T7 = 36.0 C, T8 = 32.0 C, T9 = 29.2 C

Steady-state: T1 = -0.69°C, T2 = -0.79°C, T3 = -0.93°C, T4 = 0.71°C, T5 °C,

T = -2.0°C, T = 36.1°C, T = 32.0°C, T = 29.2°C

=conditions are reached are determ

15 min: 1 2 3 4 5

° ° ° °

= -0.005

6 7 8 9

Discussion Steady operating conditions are reached in about 10 min.


5-107

5-108 The formation of fog on the glass surfaces of a car is to be prevented by attaching electric resistance heaters to the inner surfaces. The temperature distribution throughout the glass 15 min after the strip heaters are turned on and also

step of ∆t = 1 min. Assumptions 1 Heat transfer through the glass is given to be transient and two-

nsional. 2 Thermal conductivity is constant. 3 There is heat generation only at the inner surface, which will be treated as prescribed heat flux.

alanp sed as

when steady conditions are reached are to be determined using the implicit method with a time

dime

Properties The conductivity and diffusivity are given to be k = 0.84 W/m⋅°C and /sm 1039.0 26−×=α . Analysis The nodal spacing is given to be ∆x = 0.2 cm and ∆y = 1 cm. The implicit finite difference equations are determined on the basis of the energy b ce for the transient case ex res

• • • 1 2 3

tTT

cEQi

mi

mp

i

∆−

=++

+∑1

element1+i

elementgen,sides All

1 Vρ&&

We consider only 9 nodes because of symmetry. Note that we do not have a uare mesh in this case, and thus we will have to rely on energy balances to

btain the finite difference equations. Using energy balances, the finite sqodifference equations for each of the 9 nodes are obtained as follows:

Node 1: t

TTyxcxTTyk

yTTxkTTyh

ii

p

iiiii

ii ∆−∆∆

=∆−∆

+∆−∆

+−∆ +++++

+ 11

11

11

21

11

411 2222

)(2

ρ

Node 2: t

TTyxcyTT

xkxTTyk

xTTyk

ii

p

iiiiii

∆−∆

∆=∆−

∆+∆−∆

+∆−∆ +++++++

21

21

21

51

21

31

21

1

222ρ

Node 3: t

TTyxcxTTyk

yTTxkTTyh

iiiiii −−− +++++ 11111

pi

oo ∆∆∆

=∆

∆+

∆∆

+−∆ + 3332361

3 2222)(

2ρ

N4: t

TTxycx

yky

xkyTTxkTTyh p

iii ∆

∆∆=

∆∆+

∆∆

+∆

∆+−∆ + 444547411

4 222)( ρ

TTTT iiiiiiii −−−− +++++++ 1111111

Heater


1 cm

Outer

Glass

Inner • surfacesurface

25 W/m 0.2 cm

• •

• • •

• • •

• • •

4 5 6

7 8 9


tTT

yxcyTT

xkyTT

xkxTT

ykxTT

ykii

p

iiiiiiii

∆−

∆∆=∆−

∆+∆−

∆+∆−

∆+∆−

∆+++++++++

51

51

51

21

51

81

51

61

51

4 ρ Node 5:

N6: txyy pio ∆∆∆∆6 222

Node 7:

TTxycTT

ykTTxk

TTxkTTyhiiiiiiii

i −∆∆=

−∆+

−∆+

−∆+−∆

++++++++ 6

16

16

15

16

19

16

131 )( ρ

tTTyxTTyTTxkTTyh

iiiiiii

ii−∆∆−∆−∆

+−∆

++++++

+1111

71

417 )( W5.12 c

xk

y p ∆=

∆+

∆7778

22222ρ

ode 8:N tyxx ∆∆∆∆ 222

Node 9:

TTyxcTT

xkTTyk

TTykii

p

iiiiii −∆∆=

−∆+

−∆+

−∆ +++++++8

18

18

15

18

19

18

17 ρ

tTTyxc

xTTyk

yTTxkTTyh

ii

p

iiiii

oo ∆−∆∆

=∆−∆

+∆−∆

+−∆ +++++

+ 91

91

91

81

91

619 2222

)(2

ρ

26− 2 2where k = 0.84 W/m.°C, ×== pck ρα , Ti = To = -3°C hi = 6 W/m .°C, ho = 20 W/m .°C, ∆x = 0.002 m,

and ∆y = 0.01 m. Taking time step to be ∆t = 1 min, the temperature distribution throughout the glass 15 min after the strip heaters are turned on and when steady conditions are reached are determined to be (from the EES solutions in the CD) 15 min: T1 = -0.73°C, T2 = -0.84°C, T3 = -0.98°C, T4 = 0.68°C, T5 = -0.03°C,

T6 = -2.0°C, T7 = 36.0°C, T8 = 32.0°C, T9 = 29.2°C Steady-state: T1 = -0.69°C, T2 = -0.79°C, T3 = -0.93°C, T4 = 0.71°C, T5 = -0.005°C,

T6 = -2.0°C, T7 = 36.1°C, T8 = 32.0°C, T9 = 29.2°C Discussion Steady operating conditions are reached in about 10 min.

/sm 1039.0/


5-108

use initially at a uniform temperature is subjected to convection and radiation on both sides. The mperatures of the inner and outer surfaces of the roof at 6 am in the morning as well as the average rate of heat transfer rough the roof during that night are to be determined.

roperties, heat transfer coefficients, and the indoor and outdoor ant.

.4 W/m.° . The emissivity of roof is 0.9.

0.03 m. Then the

eior

5-109 The roof of a hoteth

Assumptions 1 Heat transfer is one-dimensional. 2 Thermal ptemperatures are constant. 3 Radiation heat transfer is signific

Properties The conductivity and diffusivity are given to be k = 1both surfaces of the concrete

C and α /sm 1069.0 26−×=

Analysis The nodal spacing is given to be ∆x = number of nodes becomes 1/ +∆= xLM = 0.15/0.03+1 = 6. This problem involves 6 unknown nodal temp ratures, and thus we need to have 6 equations. Nodes 2, 3, 4, and 5 are internodes, and thus for them we can use the general explicit finitedifference relation expressed as

τ

im

im

imi

mi

mi

mTT

kxe

TTT−

=∆

++−+

+−

12

11 2&

→ k

xeTTTT mi

mi

mi

mi

m 111 )21()(

∆+−++= +−

+ &τττ

i 2

The finite difference equations for nodes 1 and 6 subjected to convection and radiation are obtained from an energy balance by taking the direction

towards the node under consideration:

Convection

εho, To

Concrete roof

Tsky

Radiation

• • • • • •

6 5 4 3 2 1

hi, Ti

εRadiation

Convection

of all heat transfers to be

[ ]

[ ]t

TTcxTT

xTT

kTTh

TTTT iiii −++=+14534

)21()( :(interior) 5 Node ττ

TTTT

TTTT

Tt

TTcxTT

TTkTTh

ii

pi

sky

iii

o

iiii

iiii

ii

pi

wall

iii

ii

∆−∆

=+−+∆−

+−

−++=

−++=

∆−∆

=+−+−

+−

+

+

+

+

61

646

46560

5645

1342

13

2312

11

141

4121

2)273()( :n)(convectio 6 Node



)21()( :(interior) 2 Node2

)273()( :n)onvectio 1 Node

ρεσ

ττ

ττ

ρεσ

where k = 1.4 W/m.°C, Ti = 20°C, Twall = 293 K, To = 6°C, Tsky =260 K, hi = 5 W/m2.°C, ho

= 12 W/m2.°C, ∆x = 0.03 m, and ∆t = 5 min. Also, the mesh Fourier number is

(c

TTTx

iiii −++=∆

+1 ττ

/sm 1069.0/ 26−×== pck ρα ,

230.0)m 03.0(

s) /s)(300m 1069.0(2

26

2=

×=

∆

∆=

−

xtατ

τSubstituting this value of and other given quantities, the inner and outer surface temperatures of the roof after 12×(60/5) = 144 time steps (12 h) are determined to be T1 = 10.3°C and T6 = -0.97°C.

(b) The average temperature of the inner surface of the roof can be taken to be

C15.142

3.10182

AM 6 @ 1PM 6 @ 1,1 °=

+=

+=

TTT avg

Then the average rate of heat loss through the roof that night becomes

[ ]

W33,640=+−⋅××+

°×°⋅=

+−+−=

]K) 27315.14(K) 293)[(K W/m10)(5.67m 320.9(18

C14.15) -)(20m 32C)(18 W/m5(

)273()(

44428-2

22

41

4,1

iwallsaveisiavg TTATTAhQ εσ&


5-109

exact results obtained analytically because the results een a numerical solution and the exact solution (the

alled the truncation or formulation error) which is l method, and the round-off error which is caused by nificant digits and continuously rounding (or

formulation error) is due to replacing the derivatives by ion between two adjacent nodes by a straight line

ween the two solutions at each time step is called the local discretization error. The total discretization error at any step is called the global or accumulated discretization error. The local and global discretization errors are identical for the first time step.

5-112C Yes, the global (accumulated) discretization error be less than the local error during a step. The global discretization error usually increases with increasing number of steps, but the opposite may occur when the solution function changes direction frequently, giving rise to local discretization errors of opposite signs which tend to cancel each other.

The Tay

Special Topic: Controlling the Numerical Error

5-110C The results obtained using a numerical method differ from theobtained by a numerical method are approximate. The difference betwerror) is primarily due to two sources: The discretization error (also ccaused by the approximations used in the formulation of the numericathe computers' representing a number by using a limited number of sigchopping) off the digits it cannot retain.

5-111C The discretization error (also called the truncation ordifferences in each step, or replacing the actual temperature distributsegment. The difference bet

5-113C lor series expansion of the temperature at a specified nodal point m about time ti is

L+∂

∆+∂

∆+=∆+2

2 ),(1),(),(),(

txTt

txTttxTttxT imim

∂∂ 22 ttimim

The finite difference formulation of the time derivative at the same nodal point is expressed as

t

TTtxTtxTtxT im

imimimim −

=−∆+

≅∂ +1),(),(),( or

tt

t ∆∆∂ ttxTttxTttxT ≅∆+ (),( im

imim ∂∂

∆+),(),

which resembles the Taylor series expansion terminated after the first two terms.


5-110

5-114C The Taylor series expansion of the temperature at a specified nodal point m about time ti is

L+∂

∆+∂

∆+=∆+2

2 ),(1),(),(),(

txTt

txTttxTttxT imim

imim ∂∂ 22 tt

l point is expressed as The finite difference formulation of the time derivative at the same noda

t

TTt

txTttxTt

txT mmimimim

∆−

=∆

−∆+≅

∂∂ ),(),(),( or

ii+1

ttxTttxTttxT m ,(

which resembles the Taylor series expa

imimi ∂

∂+≅∆+

),(),()

nsion terminated after the first two terms. Therefore, the 3rd and following terms in ed in the finite difference approximation. For a sufficiently small time

es, and their contributions become smaller and smaller. The first term neglected in the Taylor series expansion is proportional to , and thus the local discretization error is also

proportional to .

proportional to the step size to ∆t itself since, at the worst case, the accumulated discretization error after I time steps during a time period is which is proportional to ∆t.

5-115C The round-off error is caused by retaining a limited number of digits during calculations. It depends on the number of calculations, the method of rounding off, the type of the computer, and even the sequence of calculations. Calculations that involve the alternate addition of small and large numbers are most susceptible to round-off error.

5-116C As the step size is decreased, the discretization error decreases but the round-off error increases.

5-117C The round-off error can be reduced by avoiding extremely small mess sizes (smaller than necessary to keep the discretization error in check) and sequencing the terms in the program such that the addition of small and large numbers is avoided.

5-118C A practical way of checking if the round-off error has been significant in calculations is to repeat the calculations using double precision holding the mesh size and the size of the time step constant. If the changes are not significant, we conclude that the round-off error is not a problem.

5-119C A practical way of checking if the discretization error has been significant in calculations is to start the calculations with a reasonable mesh size ∆x (and time step size ∆t for transient problems), based on experience, and then to repeat the calculations using a mesh size of ∆x/2. If the results obtained by halving the mesh size do not differ significantly from the results obtained with the full mesh size, we conclude that the discretization error is at an acceptable level.

∆

the Taylor series expansion represent the error involvstep, these terms decay rapidly as the order of derivative increas

2)( t∆2)( t∆

The global discretization error is

0t ttttttI ∆=∆∆=∆ 02

02 )/(


5-111

at

e element of size

Review Problems

5-120 Starting with an energy balance on a volume element, the steady three-dimensional finite difference equation for a n+1 general interior node in rectangular coordinates for T(x, y, z) for the case of constant thermal conductivity and uniform hegeneration is to be obtained.

Analysis We consider a volum zyx ∆×∆×∆ c (mentered about a general interior node , n, r) in a region in which heat is generated at a constant rate of and the

0e&thermal conductivity k is variable. Assuming the direction of heat conduction to be towards the node under consideration at all surfaces, the energy balance on the volume element can be expressed as

0elementelementgen,back cond,front left cond, cond,bottom cond,right cond, topcond, =

∆∆

=++++++t

EEQQQQQQ &&&&&&&

for the steady case. Again assuming the temperatures between the adjacent nodes to vary linearly, the energy balance relation above becomes

0)()(+ )( ∆×∆ yxk m

)(+ )( ,,,1,,,,,1 −∆×∆

∆

−∆×∆+ −+ TT

zxkx

TTzyk rnmrnmrnmrnm

)(+

0,,1,,,,1,,

,,,,,,,,1

=∆×∆×∆+∆

−∆×∆

∆

−+

∆

∆

−∆×∆

∆

−

+−

−

zyxez

TTyxk

zTT

y

yTT

zxkx

Tk

rnrnmrnmrnm

rnmrmrnmrnm

&

ng each term by and simplifying gives

)( 1∆×∆ +Tzy n

Dividi k zyx ∆×∆×∆

0222 0

21,,,,1,,

2,1,,,,1,

2,,1,,,,1 =+

∆

+−+

∆

+−+

∆

+− +−+−+

ke

zTTT

yTTT

xT rnmrnmrnmrnmrnmrnmrnmrnm &

w ∆x = ∆y = ∆z = l, and the relation above simplifies to

− rnm TT

For a cubic mesh ith

062

0,,1,,1,,,1,,1,,,1,,1− mrnm =+−+++++ +−−−+ k

leTTTTTTT rnmrnmrnmrnmrnmrn

&

can al ressed in the following easy-to-remember form: It so be exp

062

0nodebackfrontbottomrighttopleft =+−+++++

kle

TTTTTTT&

e∆x∆y∆z

•

•

•

•

••

n

r+1r

m+1

m-1

e0

∆z ∆x

∆y


5-112

ance on a volume element, the three-dimensional transient explicit finite difference in rectangular coordinates for T(x, y, z, t) for the case of constant thermal conductivity k

nd no hea neration is to be obtained.

nalysis We consider a rectangular region in which heat conduction

l points ∆z apart in the x, y, and z directions, eneral interior node (m, n, r) whose

t centered about e general interior node (m, n, r)

ssed as

5-121 Starting with an energy balequation for a general interior node a t ge

A n+1 is significant in the x and y directions. There is no heat generation in the medium, and the thermal conductivity k of the medium is constant. Now we divide the x-y-z region into a mesh of nodawhich are spaced ∆x, ∆y, andrespectively, and consider a gcoordinates are x = m∆x, y = n∆y, are z = r∆z. Noting that the volume elemen thinvolves heat conduction from six sides (right, left, front, rear, top, and bottom) and expressing them at previous time step i, the transient explicit finite difference formulation for a general interior node can be expre

tTT

cyx

zTT

yxkz

TTyxk

yTT

zxk

xTT

zyky

TTzxk

xTT

zyk

inm

inm

irnm

irnm

irnm

irnm

irnm

irnm

irnm

irnm

irnm

irnm

irnm

irnm

∆

−×∆×∆=

∆

−∆×∆

∆

−∆×∆+

∆

−∆×∆

∆

−∆×∆+

∆

−∆×∆

∆

−∆×∆

+

+−−

++−

,1

,

,,1,,,,1,,,,,1,

,,,,1,,,1,,,,,1

)1(

)(+)()(+

)()(+)(

ρ

Taking a cubic mesh (∆x = ∆y = ∆z = l) and dividing each term by k gives, after simplifying,

τ

irnm

irnmi

rnmi

rnmi

rnmi

rnmi

rnmi

rnmi

rnmTT

TTTTTTT ,,1,,

,,1,,1,,,1,,1,,,1,,1 6−

=−++++++

+−−++−

where ck ρα /= is the thermal diffusivity of the material and is the dimensionless mesh Fourier number. It can also be expressed in terms of the temperatures at the neighboring nodes in the following easy-to-remember form:

2/ lt∆=ατ

τ

iiiiiiiii TT

TTTTTTT node1

nodenodebackfrontbottomrighttopleft 6

−=−+++++

+

Discussion We note that setting gives the steady finite difference formulation. ii TT node1

node =+

•

•

•

•

••

n

1

m+1

∆yr+r

m-1∆z

∆x


5-113

node (node 3) and the finite difference formulation for the rate of heat transfer at the left boundary (node

wall is given to be steady and onstant.

become

ight boundary node (all temperatures are in K):

5-122 A plane wall with variable heat generation and constant thermal conductivity is subjected to combined convection andradiation at the right (node 3) and specified temperature at the left boundary (node 0). The finite difference formulation of the right boundary0) are to be determined.

Assumptions 1 Heat transfer through the

Convection

∆xT0

• • • •0 1 2 3

Tsurr

Radiation

ε

)(xe&

one-dimensional. 2 The thermal conductivity is given to be c

Analysis Using the energy balance approach and taking the directionof all heat transfers to be towards the node under consideration, the finite difference formulations

R

0)2/()()( 332

34

34

surr =∆+∆−

+−+− ∞ xAexTT

kATThATTA &εσ

Heat transfer at left surface:

0)2/(001

surfaceleft =∆+∆

+ xAex

kAQ && −TT


5-114

eat uniformly with both side surfaces nce equations and the nodal temperatures are to be determined, and the surface uel element are to be compared with analytical solution.

ssumptions 1 Heat transfer through the nuclear fuel element is steady and one-dimensional. 2 Thermal properties are onstant. 3 at transfer b d tion is negligible.

ies The thermal conductivity is given as 57 W/m·K.

5-123 A nuclear fuel element, modeled as a plane wall, generates 3 × 107 W/m3 of hcooled by liquid. The finite differetemperatures of both sides of the fAc He y ra iaPropertAnalysis (a) The nodal spacing is given as ∆x = 8 mm. Then the number of nodes M becomes

61mm 8mm 401 =+=+

∆=

xLM

There are 6 unknown nodal temperatures, thus we need to have 6 equations to determine them uniquely. The finite difference equation for node 0 on t e lh eft surface subjected to convection is obtained by pplying an energy balance on the half volume element about that

node:

a

02

)( 001

0 =∆

+∆−

+−∞xe

xTT

kTTh & → 02

12

001 =∆+∆

+⎟⎠⎞

⎜⎝⎛ ∆+− ∞xT

kh

kxeTx

khT &

Nodes 1, 2, 3, and 4 are interior nodes, and we can use the general finite difference relation expressed as

02

211 =+

∆

+− +−

ke

xTTT mmmm &

→ 02 211 =∆++− +− x

ke

TTT mmmm

&

The finite difference equation for node 5 on the right surface subjected to convection is obtained by applying an energy balance on the half volume element about that node:

0)(2 55

54 =−+∆

+∆−

∞ TThxexTT

k & → 02

1 5

2

54 =∆+∆

+⎟⎠⎞

⎜⎝⎛ ∆+− ∞xT

khe

kxTx

khT &

Then

m = 0:

m = 1:

m = 2:

m = 3:

m = 4:

m = 5:

(b) The nodal temperatures under steady conditions are determined by solving the 6 equations above simultaneously with an equation solver. Copy the following lines and paste on a blank EES screen to solve the above equations: e_gen=3E7 h=8000 k=57 Dx=8E-3 T_inf=80 T_1-(1+h*Dx/k)*T_0+(Dx^2*e_gen)/(2*k)+(h*Dx/k)*T_inf=0 T_0-2*T_1+T_2+(e_gen/k)*Dx^2=0 T_1-2*T_2+T_3+(e_gen/k)*Dx^2=0 T_2-2*T_3+T_4+(e_gen/k)*Dx^2=0 T_3-2*T_4+T_5+(e_gen/k)*Dx^2=0 T_4-(1+h*Dx/k)*T_5+(Dx^2*e_gen)/(2*k)+(h*Dx/k)*T_inf=0

0)/()2/()()/1( 02


0)/(2 21210 =∆++− xkeTTT &

0)/(2 22321 =∆++− xkeTTT &

0)/(2 23432 =∆++− xkeTTT &

0)/(2 24543 =∆++− xkeTTT &

0)/()2/()()/1( 52



5-115


C 155 °=0T , C 222 °=1T , C 256 °=2T

C 256 °=3T , C 222 °=4T , C 155 °=5T

(c) Using the analytical solution from Chapter 2, for a plane wall temperature exposed to convection can be determined using

of thickness 2L with heat generation, the surface

C 155 °=⋅

×+°=+=

W103(C 807

gen LeTT

&∞ W/m8000 2 wallplane , hs

The analytical solution matches exactly with the results obtained temperatures, C 155 °== TT .

K)m 02.0)(/m3

(for both sides)

using numerical method for both sides of the surface

ent are exposed to the same liquid temperature and convection heat transfer coefficient, it is possible to solve half of the plane wall by treating the centerline of the plane wall as symmetry line and get

50

Discussion Since both side of the fuel elem

the same results.


5-116

at uniformly with both side surfaces cooled by mined by making use of the symmetry line

2 Thermal properties are constant. 3

given to be ∆x = 4 mm. Then the omes

5-124 A fuel element, modeled as a plane wall, generates 5 × 107 W/m3 of heliquid. The finite difference equations and the nodal temperatures are to be deterof the plane wall.

Assumptions 1 Heat transfer through the fuel element is steady and one-dimensional. Heat transfer by radiation is negligible.

Properties The thermal conductivity is given as 67 W/m·K.

Analysis (a) The nodal spacing isnumber of nodes M bec

61mm 4mm 201 =+=+

∆=

xLM

There are 6 unknown nodal temequations to determine

peratures, thus we need to have 6 them uniquely. The finite difference equation

ode 0 the symmetry line is obtained by applying an energy e element about that node (the symmetry

ry is similar

for n onbalance on the half volumbounda to the insulated boundary):

020

01 =∆

+∆− xexTT

k & → 02

2

001 =∆

+−kxeTT &

Nodes 1, 2, 3, and 4 are interior nodes, and we can use the general finite difference relation expressed as

02

211 =+

∆

+− +−

ke

xTTT mmmm &

→ 02 211 =∆++− +− x

ke

TTT mmmm

&

The finite difference equation for node 5 on the right surface subjected to convection is obtained by applying an energy balance on the half volume element about that node:

0)(2 55

54 =−+∆

+∆−

∞ TThxexTT

k & → 02

1 5

2

54 =∆+∆

+⎟⎠⎞

⎜⎝⎛ ∆+− ∞xT

khe

kxTx

khT &

Then

m = 0:

m = 1:

m = 2:

m = 3:

m = 4:

m = 5:

(b) The nodal temperatures under steady conditions are determined by solving the 6 equations above simultaneously with an equation solver. Copy the following lines and paste on a blank EES screen to solve the above equations:

e_gen=5E7

h=5000

k=67

Dx=4E-3

T_inf=90

T_1-T_0+(Dx^2*e_gen)/(2*k)=0

T_0-2*T_1+T_2+(e_gen/k)*Dx^2=0

T_1-2*T_2+T_3+(e_gen/k)*Dx^2=0

0)2/()( 2001 =∆+− kxeTT &

0)/(2 21210 =∆++− xkeTTT &

0)/(2 22321 =∆++− xkeTTT &

0)/(2 23432 =∆++− xkeTTT &

0)/(2 24543 =∆++− xkeTTT &

0)/()2/()()/1( 52



5-117

T_4-(1+h*Dx/k)*T_5+(Dx^2*e_gen)/(2*k)+(h*Dx/k)*T_inf=0

,

T_2-2*T_3+T_4+(e_gen/k)*Dx^2=0

T_3-2*T_4+T_5+(e_gen/k)*Dx^2=0


C 439 °=0T , C 433 °=1T , C 415 °=2T

C 386 °=3T , =4T C 290 ° C 344 ° =5T

Discussion Using the analytical solution from Chapter 2, for a plane wall of thickness 2L with heat generation, the surfatemperature exposed to convection can be determined using

ce

C290K W/m5000

C 90 2 wallplane , °=⋅

+°=+= ∞ h

TTs )m 02.0)( W/m105( 37gen ×Le&

The analytical solution matches exactly with the results obtained using numerical method for both sides of the surface temperatures, C2905 °= T .


5-118

nd

at transfer through the wall is given to be

t

n, the explicit

5-125 A plane wall with variable heat generation and variable thermal conductivity is subjected to uniform heat flux 0q& aconvection at the left (node 0) and radiation at the right boundary (node 2). The explicit transient finite difference formulation of the problem using the energy balance approach method is to be determined.

Assumptions 1 Hetransient, and the thermal conductivity and heat generation to be variables. 2 Heat transfer is one-dimensional since the plate is large relative to its thickness. 3 Radiation from the left surface, and convection from he right surface are negligible.

Analysis Using the energy balance approach and taking the direction of all heat transfers to be towards the node under consideratio finite difference formulations become

Left boundary node (node 0):

tp ∆∞ 000 2)2/()(

TTcxAxAeTThAAq

xTT

Aki

iiii

i −∆=∆+−++

∆− +

01

0010 ρ&&

terior no (node 1):

i

In de

tTT

cAeTT

AkTT

Akii

pi

iii

iii −

+−

+− +

11

11

121

101 (& xAx

xx ∆∆=∆

∆∆) ρ

Right boundary node (node 2):

t

TTcxAxAeTTA

x∆ 2TT

Akii

piii

surr

iii

∆−∆

=∆++−++− +

21

22

42

4212 )2/(])273()273[( ρεσ &

5-126 A plane wall with variable heat generation and variable thermal conductivity is subjected to uniform heat flux and convection at the left (node 0) and radiation at the right boundary (node 2). The implicit transient finite difference formulation of the problem using the energy balance approach method is to be determined.

Assumptions 1 Heat transfer through the wall is given to be transient, and the thermal conductivity and heat generation to be variables. 2 Heat transfer is one-dimensional since the plate is large relative to its thickness. 3 Radiation from the left surface, and convection from the right surface are negligible.

Analysis Using the energy balance approach and taking the direction of all heat transfers to be towards the node under consideration, the implicit finite difference formulations become

Left boundary node (node 0):

&q0

tTT

cxAxAeTThAAqxTT

Akii

pii

iii

∆−∆

=∆+−++∆− +

++∞

+++ 0

101

01

00

10

111

0 2)2/()( ρ&&

Interior node (node 1):

tTT

xcAxAexTT

AkxTT

Akii

pi

iii

iii

∆−

∆=∆+∆−

+∆− +

+++

+++

+ 11

111

11

121

1

11

101

1 )( ρ&

Right boundary node (node 2):

t

TTcxAxAeTTA

xTT

Akii

piii

surr

iii

∆−∆

=∆++−++∆− +

+++++

+ 21

212

412

411

21

112 2

)2/(])273()273[( ρεσ &

Convectio

∆x

1

ε

Convectio

∆x

1

ε• ••0 2

Tsurr

Radiationh, T∞ k(T)

)(xe&

0q&

Tsurr

Radiation h, T∞ k(T)

)(xe&

0q&

• ••0 2


5-119

to

at transfer at the right surface is negligible.

ature T0 at node we need two

ance approach and taking the ration, the finite

5-127 A pin fin with convection and radiation heat transfer from its tip is considered. The complete finite difference formulation for the determination of nodal temperatures is to be obtained.

Assumptions 1 Heat transfer through the pin fin is given to be steady and one-dimensional, and the thermal conductivitybe constant. 2 Convection heat transfer coefficient and emissivity are constant and uniform.

Assumptions 1 Heat transfer through the wall is given to be steady and one-dimensional, and the thermal conductivity and heat generation to be variable. 2 Convection he Convectio

h, T∞

• ••0 1 2D∆x

Tsurr

Radiationε

Analysis The nodal network consists of 3 nodes, and the base temper0 is specified. Therefore, there are two unknowns T1 and T2, andequations to determine them. Using the energy baldirection of all heat transfers to be towards the node under considedifference formulations become

Node 1 (at midpoint):

0])273()[())(( 11210 ∆+−∆+

∆−

+∆−

∞ xpTTxphxTT

kAxTT

kA εσ 41

4surr =+− TT

ode 2 (at fin tip): N

0])273()[2/( ∆+εσ))(2/( 42

4surr2

21 =+−+−+∆+∆−

∞ TTAxpTTAxphxTT

kA

where 4/2DA π= is the cross-sectional area and Dp π= is the perimeter of the fin.


5-120

rm

and y

ates are and . Noting that the volume element centered about the general interior node (m, xpressing them at previous time step i, the

n be expressed as

5-128 Starting with an energy balance on a volume element, the two-dimensional transient explicit finite difference equation for a general interior node in rectangular coordinates for T(x, y, t) for the case of constant thermal conductivity k and unifoheat generation 0e& is to be obtained.

Analysis (See Figure 5-24 in the text). We consider a rectangular region in which heat conduction is significant in the x directions, and consider a unit depth of ∆z = 1 in the z direction. There is uniform heat generation in the medium, and thethermal conductivity k of the medium is constant. Now we divide the x-y plane of the region into a rectangular mesh of nodal points which are spaced ∆x and ∆y apart in the x and y directions, respectively, and consider a general interior node (m, n) whose coordin xmx ∆= yny ∆=n) involves heat conduction from four sides (right, left, top, and bottom) and etransient explicit finite difference formulation for a general interior node ca

tTT

cyxyxe

TTyk

yTT

xkTT

yk

inm

inm

p

inm

inm

inm

inm

inm

∆

−×∆×∆=×∆×∆+

∆

−×∆+

∆

−×∆

−×∆

+

++−

,1

,0

,1,1,,,1

)1()1(

)1()1(+)1(

ρ&

yTT

xkxx

inm

inm

inm

∆

−×∆

∆− ,1,, )1(+

ng a sq re mesh (∆x = ∆y = l) and dividing each term by k gives, after simplifying, Taki ua

τ

inm

inmi

nmi

nmi

nmi

nmi

nmTT

kle

TTTTT ,1

,2

0,1,1,,1,1 4

−=+−+++

+

−++−&

where pck ρα /= is the thermal diffusivity of the material and 2/ lt∆=ατ is the dimensionless mesh Fourier number. It ca en also be xpressed in terms of the temperatures at the neighboring nodes in the following easy-to-remember form:

τ

iiiii T

TTT node1

node −+++

ii Tkle

TT2

0nodebottomrighttopleft 4 =+−+

&

Discussion gives the steady finite difference formulation. ii TT node1

node =+ We note that setting


5-121

s subjected to convection with a convection al conductivity with uniform heat generation

heat conduction in on in a cylindrical rod of constant cross-sectional

stant conductivity k

5-129 Starting with an energy balance on a disk volume element, the one-dimensional transient implicit finite difference equation for a general interior node for ),( tzT in a cylinder whose side surface icoefficient of h and an ambient temperature of T∞ for the case of constant thermis to be obtained.

Analysis We consider transient one-dimensionalthe axial z directi

• • • m-1 m m+1

Convectionh, T∞Disk area A with constant heat generation 0e& and con

with a mesh size of z∆ in the z direction. Notingelement of a general interior node

that the volume m involves heat conduction from

o sides, convecti om its lateral surface, and the volume of the lement is , the transient implicit finite difference rmulation for an interior

twe

on frzA∆=elementV

fo node can be expressed as

tTTzcAzAe

zTTkA

zTTkATTzhp

im

im

p

im

im

im

imi

m ∆−

∆=∆+∆−

∆−

+−∆+++

+++

−+∞

1

0

111

1111 +)( ρ&

where is the cross-sectional area. Multiplying both sides of equation by ∆z/(kA),

4/2DA π=

)()(+)()( 122

0111

111

12

im

im

pim

im

im

im

im TT

tkcz

kzeTTTTTT

kAzhp

−∆

∆=

∆+−−+−

∆ ++++

++−

+∞

ρ&

pck ρα /= and the dimensionless mesh Fourier number2zt

∆

∆=ατUsing the definitions of thermal diffusivity the equation

reduces to

τ)()2()(

12011

111

12 i

mi

mim

im

im

im

TTk

zeTTTTTkA

zhp −=

∆+−++−

∆ +++

++−

+∞

&

Discussion We note that setting gives the steady finite difference formulation. ii TT m1

m =+


5-122

perties, heat

lity uniform

roperties The conductivity and diffusivity are given to be k = 0.61 /m.°C and . The volumetric absorption

oefficients of water are as given in the problem.

nalysis The nodal spacing is given to be ∆x = 0.25 m. Then the ber of nodes become = 1/0.25+1 = 4. This

roblem involves 5 unknown nodal temperatures, and thus we eed to have 5 equations. Nodes 2, 3, and 4 are interior nodes, and us for them we can use the general explicit finite difference lation expressed as

5-130 A large pond is initially at a uniform temperature. Solar energy is incident on the pond surface at for 4 h The temperature distribution in the pond under the most favorable conditions is to be determined.

Assumptions 1 Heat transfer is one-dimensional since the pond is large relative to its depth. 2 Thermal protransfer coefficients, and the indoor temperatures are constant. 3 Radiation heat transfer is significant. 4 There are no convection currents in the water. 5 The given time step ∆t = 15 min is less than the critical time step so that the stabicriteria is satisfied. 6 All heat losses from the pond are negligible. 7 Heat generation due to absorption of radiation isin each layer.

qs, W/m2

Solar radiation

x

Top layer

45°

Solar pond

Black

• • • • •

0 1 2 3 4

Upper mid layerLower mid layer

Bottom

P/sm 1015.0 26−×=αW

c

As 1/ +∆= xLMnum

pnthre

τ

im

im

imi

mi

mi

mTT

kxe

TTT−

=∆

++−+

+−

12

11 2&

→ k

xeTTTT

imi

mi

mi

mi

m

2

111 )21()(

∆+−++= +−

+ &τττ

Node 0 can also be treated as an interior node by using the mirror image concept. The finite difference equation for node 4 subjec ea btain n energy balance by taking the direction of all heat transfers to be towards the node:

ted to h t flux is o ed from a

tTT

cxkxexTT

kq &

kxeTTT

kxeTTT

kxeTTT

kxeTTT

ii

p

ii

iii

iiii

iii

iiii

∆−∆

=∆+∆−

+

∆+−++=

∆+−++=

∆+−++=

∆+−++=

+

+

+

41

424

43b

23342

22231

12

211201

20011

1

2/)(:n)nvectio6 Node

/)()21()( :rior)3 Node

/)()21()( :erior)2 Node

/)()21()( :n)ulatio0 Node

/)()21()( :n)atio0 Node

ρτ

τττ

τττ

τττ

τττ

&

&

&

&

&

= 0.61 W/m.°C, , ∆x = 0.25 m, and ∆t = 15 min = 900 s. Also, the mesh Fourier

T i+13

T

T

T i+1

0

(co

(inte

(int

(ins

(ins ul

/sm 1015.0/ 26−×== pck ραwhere k number is

002160.0)m 25.0(

s) /s)(900m 1015.0(2

26

2=

×=

∆

∆=

−

xtατ

The values of heat generation rates at the nodal points are determined as follows:

32

00 W/m946

m) )(0.25m (1 W500473.0

Volume=

×==

Ee

&&

32

11 W/m534

m) )(0.25m (1 W500]2/)061.0473.0[(

Volume=

×+==

Ee

&&

32

44 W/m48

m) )(0.25m (1 W500024.0

Volume=

×==

Ee

&&

Also, the heat flux at the bottom surface is . Substituting these values, the nodal temperatures in the pond after 4×(60/15) = 16 time steps (4 h) are determined to be

T0 = 18.3°C, T1 = 16.9°C, T2 = 15.4°C, T3 = 15.3°C, and T4 = 20.2°C

22 W/m5.189 W/m500379.0 =×=bq&


5-123

temperature of 15°C throughout. Solar energy is incident on the pond rface at 4 ° at an average rate of 500 W/m2 for a period of 4 h The temperature distribution in the pond under the most vorable c nditions is to be determined.

t transfer is one-dimensional since the pond is large relative to its depth. 2 Thermal properties, heat transfer coefficients, and the indoor temperatures are constant. 3 Radiation heat transfer is significant. 4 There are no onvection currents in the water. 5 The given time step t = 15 min is less than the critical time step so that the stability

criteria is satisfied. 6 All heat losses from the pond are negligible. 7 Heat generation due to absorption of radiation is uniform

are given to be k = 0.61 W/m.°C nd . The volumetric absorption coefficients of water

. Analysis The nodal spacing is given to be ∆x = 0.25 m. Then the number of odes become = 1/0.25+1 = 4. This problem involves 5

peratures, and thus we need to have 5 equations. Nodes 2, 3, and 4 are interior nodes, and thus for them we can use the general explicit

xpressed as

5-131 A large 1-m deep pond is initially at a uniformsu 5fa oAssumptions 1 Hea

c ∆

in each layer. Properties The conductivity and diffusivity

qs, W/m2

Solar radiation

x

Top layer

45°

Solar pond

Black

• • • • •

0 1 2 3 4

Upper mid layerLower mid layer

Bottom

/sm 1015.0 26−×=αaare as given in the problem

nunknown nodal tem

s 1/ +∆= xLM

finite difference relation e

τ

im

im

imi

mi

mi m

TTk

xeTTT

−=

∆++−

+

+−

12

11 2&

→ k

xeTTTT

imi

mi

mi

mi

m

2

111 )21()(

∆+−++= +−

+ &τττ

Node 0 can also be treated as an interior node by using the mirror image concept. The finite difference equation for node 4 subjected to heat flux is obtained from an energy balance by taking the direction of all heat transfers to be towards the node:

tTT

cxkxexTT

kq

kxeTTTT

kxeTTTT

kxeTTTT

kxeTTTT

ii

p

ii

b

iiii

iiii

iiii

iiii

∆−∆

=∆+∆−

+

∆+−++=

∆+−++=

∆+−++=

∆+−++=

+

+

+

+

+

41

424

43

23342

13

22231

12

21120

11

20011

10

2/)( :n)(convectio 6 Node

/)()21()( :(interior) 3 Node

/)()21()( :(interior) 2 Node

/)()21()( :n)(insulatio 0 Node

/)()21()( :n)(insulatio 0 Node

ρτ

τττ

τττ

τττ

τττ

&&

&

&

&

&

where k = 0.61 W/m.°C, , ∆x = 0.25 m, and ∆t = 15 min = 900 s. Also, the mesh Fourier number is

/sm 1015.0/ 26−×== pck ρα

002160.0)m 25.0(

s) /s)(900m 1015.0(2

26

2=

×=

∆

∆=

−

xtατ

The absorption of solar radiation is given to be

where is the solar flux incident on the surface of the pond in W/m2, and x is the distance form the free surface of the pond, in m. Then the values of heat generation rates at the nodal points are determined to be

Node 0 (x = 0):

Node 1(x=0.25):

Node 2 (x=0.50):

Node3 (x=0.75):

Node 4 (x = 1.00):

Also, the heat flux at the bottom surface is . Substituting these values, the nodal temperatures in the pond after 4×(60/15) = 16 time steps (4 h) are determined to be

T0 = 16.5°C, T1 = 15.6°C, T2 = 15.3°C, T3 = 15.3°C, and T4 = 20.2°C

)278.2339.6704.6415.3859.0()( 432 xxxxqxe s +−+−= &&

sq&

34320 W/m5.429)0278.20339.60704.60415.3859.0(500 =×+×−×+×−=e&

34321 W/m1.167)25.0278.225.0339.625.0704.625.0415.3859.0(500 =×+×−×+×−=e&

34322 W/m8.88)5.0278.25.0339.65.0704.65.0415.3859.0(500 =×+×−×+×−=e&

34323 W/m6.57)75.0278.275.0339.675.0704.675.0415.3859.0(500 =×+×−×+×−=e&

34324 W/m5.43)1278.21339.61704.61415.3859.0(500 =×+×−×+×−=e&

22 W/m5.189 W/m500379.0 =×=bq&


5-124

d

.

Node 1:

5-132 A two-dimensional bar shown in the figure is considered. The simplest form of the matrix equation is to be written anthe grid notes with energy balance equations are to be identified on the figure.

Assumptions 1 Heat transfer through the body is given to be steady and two-dimensional. 2 Thermal conductivity is constant. 3 There is no heat generation

Insulation

Analysis From symmetry, we have only three unknown temperatures at nodes 1, 2, and 3. The finite difference formulations are

AABBABA

BB

BBA

A

TkTkTkkTkTTkTTkTk

LL

TTkL

LTTT

k

−=++++=−+−+

=−

+−

+

321

13121

1312

)()0)()()

022

Node 2:

TB

kB

kA

1

3

2

TB

TB

TA TA

AAA

kTkTT

−−+− 21

(2()(

A kLL

TTkL

LT −

+− 121

22

BBAAABABA

BABBBBBA

AA

TkkTkTkkTkTTkT

LTT

k

)3(2)(40()

21

22221212

2

+−−=+−+

−

Node 3:

AA

BA

BB

BBBA

kkTTkTTkTTTTkTk

LL

TTkL

LTT

kLL

TTkL

LTT

kLL

TTkL

)()()()(2)()(2

02222

2222121

+=−+−+−+−+−−

=−

+−

+−

+−

+−

+

B

BB

TTTTTTTTT

340)(2

31

3331

−=−+=−+−+−

BB

BBB L

kLL

kL

k 022

3331 =++LTTTTLTT −−−

The matrix equation is

Discussion Note that the results do not depend on L (size of the system). If you don’t use the symmetry and get a 4×4 linear system, two of the equations must be equivalent.

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−+−−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−+−+

++−

B

BBAAA

AA

BABA

BBABA

TTkkTk

Tk

TTT

kkkkkkkkk

3)3(2

4010)(4

)(2

3

2

1


5-125

e steps for stability condition and the temperature field at certain times are to be determined.

nsional. 2 All surafces of the bar except the bottom is constant. 4 There is no heat generation.

the unknown temperatures in the grid using the explicit method are

Node T (10 s) T (20 s)

5-133 A two-dimensional long steel bar shown in the figure is considered. The finite difference equations for the unknown temperatures in the grid using the explicit method is to be written and dimensionless parameters are to be identified. Also, thrange of time

Assumptions 1 Heat transfer through the body is transient and two-dimesurface are maintained at a constant temperature. 3 Thermal conductivity

Analysis (a) the finite difference equations for

Ts = 10°C

1 10 10 2 443.3 234.4 3 10 10 4 10 10 5 315 168.6 6 10 10 7 10 10

Node 5:

30FoFo)Fo41(

)(4

251

5

51

5

2

54267

51

5254525657

×++−=

−∆∆

=−+++

∆−

∆=∆∆−

+∆∆−

+∆∆−

+∆∆−

+

+

+

iii

iiiiiii

ii

p

iiiiiiii

TTT

TTtkxcTTTTT

tTT

xcxxTT

kxxTT

kxxTT

kxxTT

k

ρ

ρ

(1)

where 2

Foxctk

p∆

∆=ρ

Node 2:

20FoFo2)Fo41(

)(4

222

521

2

21

2

2

2351

21

22

232521

×++−=

−∆

∆−=−++

∆−∆

=∆

∆−

+∆∆−

+∆

∆−

+

+

+

iii

iipiiii

ii

p

iiiiii

TTT

TTtk

xcTTTT

tTTxcx

xTT

kxxTT

kxxTT

k

ρ

ρ

(2)

1 8 3

6 5 4

7

5 cm 5

2

5 cm


5-126

(b) For both steps, stability condition is

s 13.44==∆

≤∆

≤∆

∆⎯→⎯≤⎯→⎯≥−

)40(4)025.0)(430)(8000(

4

44Fo0Fo41

22

2

kxc

t

xc

p

p

ρ

ρ

(c) For ∆t = 10 s,

11 tk

186.0 )025.0)(430)(8000(

)10)(40(Fo2

==∆

∆=

xctk

pρ

Then, Eq. (1) and (2) become

58.5186.0256.0

125

15 ++=+

+

iii

iii TTT

2

522 ++= TTT

Substituting at ∆t = 10 s,

°=++= 58.5)700(186.0)700(256.01

5T

°=++ 58.5)443(186.0)315(256T

72.3372.0256.0

C315

C443.3°=++= 72.3)700(372.0)700(256.012T

Substituting at ∆t = 20 s,

= .02

5 C168.6

C234.4°=++= 72.3)315(372.0)3.443(256.012T


5-127

hot surface is to be cooled by aluminum pin fins. The nodal tem tthe explicit finite difference method. Also to be determined is the time it takes for steady conditions to be reached.

ransfer through the pin fi ven t . 2 The

iatio sfer is nis considered.

temperat al temperat gy balance approach and taking the direction of all heat transfers to be towards the node under consideration, the explicit transient finite difference formulations become

5-134 A peratures after 10 min are to be de ermined using

Assumptions 1 Heat t n is gi o be one-dimensional Convectio

thermal properties of the fin are constant. 3 Convection heat transfer coefficient is constant and uniform. 4 Rad n heat tran egligible. 5 Heat loss from the fin tip h, T∞

• •4

•0 2 D

∆x

• •31Analysis The nodal network of this problem consists of 5 nodes, and the base ure T0 at node 0 is specified. Therefore, there are 4 unknown nodures, and we need 4 equations to determine them. Using the ener

Node 1 (interior): t

xcAx

kAx

kATTxhp pi

∆∆=

∆+

∆+−∆ ∞

1110121 )( ρ

TTTTTT iiiii −−− +1

Node 2 (interior): t

TTxcA

xTT

kAxTT

kATTxhpii

p

iiiii

∆−

∆=∆−

+∆−

+−∆+

∞2

122123

2 )( ρ

tTT

xcAxTT

kAxTT

kATTxhpii

p

iiiii

∆−

∆=∆−

+∆−

+−∆+

∞3

133234

3 )( ρ Node 3 (interior):

Node 4 (fin tip): t

cxAx

kATTAxph pi

∆∆=

∆+−+∆ ∞

44434 )2/())(2/( ρ

where 4/2DA π=

TTTT iiii −− +1

is the cross-sectional area and Dp π= is the perimeter of the fin. Also, D = 0.008 m, k = 237 W/m.°C, 6−== pck ρα °C, T0 = Ti = 120°C, ho = 35 W/m2.°C, and ∆t = 1 s. Also, the mesh

Fourier number is

/sm 101 2× , ∆x = 0.02 m, T∞ = 15.97/

24275.0)m 02.0(

s) /s)(1m 101.97(2

26

2=

×=

∆

∆=

−

xtατ

Substituting these values, the nodal temperatures along the fin after 10×60 = 600 time steps (4 h) are determined to be

T0 = 120°C, T1 = 110.6°C, T2 = 103.9°C, T3 = 100.0°C, and T4 = 98.5°C.

Printing the temperatures after each time step and examining them, we observe that the nodal temperatures stop changing after about 3.8 min. Thus we conclude that steady conditions are reached after 3.8 min.


5-128

itions are to be

he number of ce

ves 3 s, and thus we need have 3 equations to

determine them uniqu and thus for ressed as

5-135E A plane wall in space is subjected to specified temperature on one side and radiation and heat flux on the other. The finite difference formulation of this problem is to be obtained, and the nodal temperatures under steady conddetermined.

Assumptions 1 Heat transfer through the wall is given to be steady and one-dimensional. 2 Thermal conductivity is constant. 3 There is no heat generation. 4 There is no convection in space.

Properties The properties of the wall are given to be k=1.2 Btu/h⋅ft⋅°F, ε = 0.80, and αs = 0.6.

qsT0

Analysis The nodal spacing is given to be ∆x = 0.1 ft. Then tnodes becomes 1/ +∆= xLM = 0.3/0.1+1 = 4. The left surfatemperature is given to be T0 = 520 R = 60°F. This problem involunknown nodal temperature to

ely. Nodes 1 and 2 are interior nodes, them we can use the general finite difference relation exp

)0 (since 02 021 −− TT mm112

1 ==+−→=+∆

++−

+ eTTTk

ex

Tmmm

mm &&

, for m = 1 and 2

The finite difference equation for node 3 on the right surface subjected to convection and solar heat flux is obtained by applying an energy balance on the half volume element about node 3 and taking the direction of all heat transfers to be towards the node under consideration:

0])460([ :surface)(right 3 Node 3243

4spaces =

∆−

++−+

0 :(interior) 2 Node02 :(interior) 1 Node

321

210

=2 +−=+−

xTT

kTTqs εσα &

TTTTT

where k = 1.2 Btu/h.ft.°F, ε = 0.80, αs = 0.60, Tspace = 0 R, and σ = 0.1714×10-8 Btu/h.ft2.R 4 The system s with 3 unknown temperatures constitute the finite difference formulation of the problem.

(b The nodal temperatures under steady conditions are determined by solving the 3 equations above simultaneously with an equation solver to be

T1 = 67.6°F = 527.6 R, T2 = 75.2°F = 535.2 R, and T3 = 82.8°F = 542.8 R

•

∆x

Radiation

1• ••0 2 3 Tsurr

T

2s Btu/h.ft 350=q& ,

of 3 equation

)


5-129

F lacing them on a black-a takes to defrost the steaks is to be determined.

eral

remain constant during defrosting. 3 Heat transfer through the bottom surface of the plate is negligible. 4 The thermal contact resistance between the steaks and the plate is negligible Evaporation from the steaks and thus evaporative cooling is negligible. 6 The heat storage capacity of the plate is small relative to the amount of total heat transferred to the steak, and

p

ε = 0.95, and hif = 187 kJ/kg. The he defrosting plate are k = 237 W/m.°C, , and ε = 0.90. The ρcp (volumetric

eaks and of the defrosting plate

5-136 rozen steaks are to be defrosted by p nodized circular aluminum plate. Using he explicit method, the time it t

Assumptions 1 Heat transfer in both the steaks and the defrosting plate is one-dimensional since heat transfer from the latsurfaces is negligible. 2 Thermal properties, heat transfer coefficients, and the surrounding air and surface temperatures

. 5

thus the heat transferred to the plate can be assumed to be transferred to the steak.

Properties The thermal properties of the steaks are ρ = 970 kg/m3, c = 1.55 kJ/kg.°C, k = 1.40 W/m.°C,

5

• • •

12 3 4 ••

6/sm 1093.0 26−×=α ,

thermal properties of t/sm 101.97 26−×=α

specific heat) values of the st

•

are

CkW/m 1504C)kJ/kg )(1.55kg/m 970()( 33steak °⋅=°⋅=pcρ

Analysis The nodal spacing is given to be ∆x = 0.005 m in the steaks, and ∆r = 0.0375 m in the plate. This problem involves

CkW/m 2441/m 101.97

)(26plate °⋅=

×==

−psα

6 unknown nodal t m eratures, and t o have 6 equations. Nodes 2 and 3 are interior nodes in a plain wall, and thus for t licit finite difference relation expressed as

C W/m237 3°⋅kcρ

e p hus we need them we can use the general exp

τ

im

im

imi

mi

mi

mTT

kxe

TTT−

=∆

++−+

+−

12

11 2&

→

The finite difference equations for other nodes are obtained from an energy balance by taking the direction of all heat transfers to be towards the node under consideration:

Node 1:

im

im

im

im TTTT )21()( 11

1 ττ −++= +−+

tTTxc

xTT

kTTTThii

p

iiii

∆−∆

=∆−

++−++−+

∞∞1

11

steak12

steak4

14

steak1 2)(])273() 273[()( ρσε

iiii

iiii

TTTT

TTTT

3steak42steak1

3

2steak31steak1

2



ττ

ττ

−++=

−++=+

+

Node 4:

tTT

rcxrcrTT

rk

xTT

rkTTTThrr

ii

pp

ii

iiii

∆−

+∆=∆−

+

∆−

++−++−−

+

∞∞

41

4245plate

24steak

4545plate

4324steak

44

4plate4

24

245

)]()()2/()[()2(

)(]})273() 273[()(){(

δπρπρδπ

πσεπ

Node 5:

tTT

rcrTT

rk

TTTThrrii

p

ii

ii

∆−

=∆−

+

+−++−∆+

∞∞

51

525plate

5656plate

45

4plate55

)]()()2(

]})273() 273[()({2

δπρδπ

σεπ

Node 6:

tTT

rrrcrTT

rk

TTTThrrrii

p

ii

ii

∆−

∆+=∆−

+

+−++−∆++

∞∞

61

6656plate

6556plate

46

4plate6656

)2/](2/)(2[)()2(

]})273() 273[()(){2/](2/)[(2

δπρδπ

σεπ


5-130

×= , and εplate = 0.90, T∞ = 20°C,

where C,kW/m 1504)( C,kW/m 2441)( 3steak

3plate °⋅=°⋅= pp cc ρρ ksteak = 1.40 W/m.°C, , εsteak = 0.95,

/sm 1093.0 26steak

−×=α , hif = 187 kJ/kg, kplate = 237 W/m.°C, 26plate

−α /sm 101.97

h =12 W/m2.°C, δ = 0.01 m, ∆x = 0.005 m, ∆r = 0.0375 m, and ∆t = 5 s. Also, the mesh Fourier number for the steaks is

186.0)m 005.0( 22steak ==

∆=

xτ

s) /s)(5m 1093.0( 26×∆ −tα

The vari i are ous radi r4 =0.075 m, r5 =0.1125 m, r6 =0.15 m, r45 = (0.075+0.1125)/2 m, and r = (0.1125+0.15)/2 m.

The total amount of heat transfer needed to defrost the steaks is

πV

akifsteaklatentsensible

°°=

56

kg 257.0)]m 015.0(m) (0.075)[kg/m 970( 23steak === ρm

kJ 55.2=kJ/kg) kg)(187 257.0(+C)](-18-C)[0kJ/kg. kg)(1.55 257.0(

)()( stesteak total, +∆=+= mhTmcQQ p

The amount of heat transfer to the steak during a time step i is the sum of the heat transferred to the steak directly from its p surfa irectly through the plate, and is expressed as

]})273() 273[()(){(

]})273() 273[()(){2/](2/)[(2

]})273() 273[()({

41

4steak1

44

4plate4

24

245

46

4plate6656

45

4plate5

+−++−

+−++−−

+−++−∆++

+−++−∆=

∞∞

∞∞

∞∞

∞∞

ii

ii

ii

iii

TTTThr

TTTThrr

TTTThrrr

TTTThrr

σεπ

σεπ

σεπ

σεπ

ined by finding the amount of heat transfer during each time step, and adding them up until we 5.2 ultiplying the number of time steps N by the time step ∆t = 5 s will give the defrosting time. In this case it is

∆tdefrost = N∆t = 44(5 s) = 220 s

Q

to ce, and ind

2 5steakQ

+

21+ ]})273() 273[()({

The defrosting time is determobtain 5 kJ. Mdetermined to be


5-131

defrosting

at transfer in both the steaks and the defrosting plate is one-dimensional since heat transfer from the lateral ts, and the surrounding air and surface temperatures m surface of the plate is negligible. 4 The thermal contact

an be assumed to be transferred to the steak.

roperties The thermal properties of the steaks are ρ = 970 kg 3, cp = 1.55 kJ/kg.°C, k = 1.40 W/m.°C,

ε = 0.95, and hif = 187 kJ/kg. The ermal properties of the defrosting plate are k = 401 W/m.°

e A-3). The ρcp

(volumetric specific heat) values of the steaks and of the defrosting plate are

Analysis The nodal spacing is given to be ∆x = 0.005 m in the steaks, and ∆r = 0.0375 m in the plate. This problem involves 6 unknown nodal temperatures, and thus we need to have 6 equations. Nodes 2 and 3 are interior nodes in a plain wall, and thus for them we can use the general explicit finite difference relation expressed as

5-137 Frozen steaks at -18°C are to be defrosted by placing them on a 1-cm thick black-anodized circular copper plate. Using the explicit finite difference method, the time it takes to defrost the steaks is to be determined.

Assumptions 1 Hesurfaces is negligible. 2 Thermal properties, heat transfer coefficienremain constant during defrosting. 3 Heat transfer through the bottoresistance between the steaks and the plate is negligible. 5 Evaporation from the steaks and thus evaporative cooling is negligible. 6 The heat storage capacity of the plate is small relative to the amount of total heat transferred to the steak, and thus the heat transferred to the plate c

P/m

5

• • • •

12 3 4 ••

6/sm 1093.0 26−×=α ,

thC, /sm 10117 26−×=α , 3kg/m 8933=ρ ,

Ckg/kg 385.0 °⋅=pc , and ε = 0.90 (Tabl

CkW/m 1504C)kJ/kg )(1.55kg/m 970()(

CkW/m 3439C)kJ/kg )(0.385kg/m 8933()(33

steak

33plate

°⋅=°⋅=

°⋅=°⋅=

p

p

c

c

ρ

ρ

τ

im

im

imi

mi

mi

mTT

kxe

TTT−

=∆

++−+

+−

12

11 2&

→

The finite difference equations for other nodes are obtained from an energy balance by taking the direction of all heat transfers to be towards the node under consideration:

Node 1:

im

im

im

im TTTT )21()( 11

1 ττ −++= +−+

tTTxc

xTT

kTTTThii

p

iiii

∆−∆

=∆−

++−++−+

∞∞1

11

steak12

steak4

14

steak1 2)(])273() 273[()( ρσε

iiii

iiii

TTTT

TTTT

3steak42steak1

3

2steak31steak1

2



ττ

ττ

−++=

−++=+

+

Node 4:

tTT

rcxrcrTT

rk

xTT

rkTTTThrr

ii

pp

ii

iiii

∆−

+∆=∆−

+

∆−

++−++−−

+

∞∞

41

4245plate

24steak

4545plate

4324steak

44

4plate4

24

245

)]()()2/()[()2(

)(]})273() 273[()(){(

δπρπρδπ

πσεπ

Node 5:

tTT

rcrTT

rk

TTTThrrii

p

ii

ii

∆−

=∆−

+

+−++−∆+

∞∞

51

525plate

5656plate

45

4plate55

)]()()2(

]})273() 273[()({2

δπρδπ

σεπ

Node 6:

tTT

rrrcrTT

rk

TTTThrrrii

p

ii

ii

∆−

∆+=∆−

+

+−++−∆++

∞∞

61

6656plate

6556plate

46

4plate6656

)2/](2/)(2[)()2(

]})273() 273[()(){2/](2/)[(2

δπρδπ

σεπ


5-132

here ksteak = 1.40 W/m.°C, εsteak = 0.95,

hif = 187 kJ/kg, kplate = 401 W/m.°C, , and εplate = 0.90, T∞ = 20°C,

h =12 W/m2.°C, δ = 0.01 m, ∆x = 0.005 m, ∆r = 0.0375 m, and ∆t = 5 s. Also, the mesh Fourier number for the steaks is

w C,kW/m 1504)( C,kW/m 3439)( 3steak

3plate °⋅=°⋅= pp cc ρρ

/sm 1093.0 26steak

−×=α , /sm 10117 26plate

−×=α

186.0)m 005.0(

s) /s)(5m 1093.0(2

26

2steak =×

=∆

∆=

−

xtατ

The various radii are r4 =0.075 m, r5 =0.1125 m, r6 =0.15 m, r45 = (0.075+0.1125)/2 m, and r56 = (0.1125+0.15)/2 m.

The total amount of heat transfer needed to defrost the steaks is

kg 257.0)]m 015.0(m) (0.075)[kg/m 970( 23steak === πρVm

kJ 55.2=kJ/kg) kg)(187 257.0(+C)](-18-C)[0kJ/kg. kg)(1.55 257.0(

)()( steakifsteaklatentsensiblesteak total,

°°=

+∆=+= mhTmcQQQ p

The amount of heat transfer to the steak during a time step i is the sum of the heat transferred to the steak directly from its top surface, and indirectly through the plate, and is expressed as

]})273() 273[(

]})273() 273[()(){2/](2/)[(2

]})273() 273[()({2

41

44

4

46

4plate6656

45

4plate55steak

+

+−+

+−++−∆++

+−++−∆=

∞

∞∞

∞∞

i

i

ii

iii

TT

TTTThrrr

TTTThrrQ

σ

σεπ

σεπ

we btain 55.2 kJ. Multiplying the number of time steps N by the time step ∆t = 5 s will give the defrosting time. In this case it is etermined to be

∆tdefrost = N∆t = 47(5 s) = 235 s

)(){( plate42

42

45 +−−+ ∞iTThrr επ

() 273[()({ 4steak1

21 −++−+ ∞∞

i TTTThr σεπ ]})273

The defrosting time is determined by finding the amount of heat transfer during each time step, and adding them up until od


5-133

d

etermined.

s are T1, T2, and T3. Thus, we need to have 3 equations to determine them uniquely.

5-138 A square cross section with uniform heat generation is undergoing a steady two-dimensional heat transfer. The top anright surfaces are subjected to convection while the left and bottom surfaces maintain a constant temperature. The finite difference equations and the nodal temperatures are to be dAssumptions 1 Steady heat conduction is two-dimensional. 2 Thermal properties are constant. 3 The heat generation in the body is uniform. Properties The conductivity is given to be k = 25 W/m·K. Analysis (a) There is symmetry about the diagonal line passing through the center. Therefore, T1 = T4, and the unknown temperature

Node 1: 02

20022

)( 13121 ∆

−∆+

∆−

∆+∆−∆

+−∆ ∞ xyk

yTT

xkxTTykTTxh

Node 2:

11 =

∆∆+

yxeT

&

02222

)(22

1242 +

∆−∆

+∆−∆

+−⎟⎠⎞

⎜⎝⎛ ∆

+∆

∞ xTyk

yTTxkTTyxh

Node 3:

22 =

∆∆ yxeT

&

022002200

243 +−+

+− eTTTT & ,2

13 =+∆∆ kyx

nm

Or Node 1: ⎟⎟⎠

⎞⎜⎜⎝

⎛ ∆+

∆+++

∆+= ∞ k

xeT

kxhTT

kxhT

21

321 22200/24

1 &

Node 2: ⎟⎟⎠

⎞⎜⎜⎝

⎛ ∆+

∆+

∆+= ∞ k

xeT

kxhT

kxhT

222

/221 2

212

&

)/2400(25.0 2313 kxeTT ∆++= &

Then

Node 3:

Node 1: 2.4/)122.02200( 321 ++++= ∞TTTT

Node 2: 2.2/)62.02( 12 ++= ∞TTT

Node 3:

here °=k and

b) By lettin itial guesses as T1 = T2 = T3 = 200°C, the results obtained from successive iterations are listed in the llowin

dal temperature,°C

)122400(25.0 13 ++= TT

w /2node∆xe& C 12 2.0/2 =∆ kxh

( g the info g table:

NoIteration T2 T3T1

1 198.1 191.9 202.0 2 197.1 191.0 201.6 3 196.7 190.6 201.4

196.5 190.5 201.3 5 4

196.4 190.4 201.2 6 196.4 190.3 201.2 7 196.4 190.3 201.2

Hence, the converged nodal temperatures are T1 = T4 = 196.4°C, T2 = 190.3°C, T3 = 201.2°C

Discussion The finite difference equations can also be calculated using the EES. Copy the following lines and paste on a

T_3=0.25*(400+2*T_1+12) Solving by EES software, we get the same results: T1 = T4 = 196.4°C, T2 = 190.3°C, T3 = 201.2°C

blank EES screen to solve the above equations: T_1=(200+T_2+2*T_3+0.2*100+12)/4.2 T_2=(2*T_1+0.2*100+6)/2.2


5-134

ional. 2 Thermal properties are constant. 3 The heat generation in the ody is uni

is On the SS-T-Conduct Input window for 2-Dimensional Steady State Problem, the problem parameters and the oundary conditions are entered into the appropriate text boxes. Note that with a uniform nodal spacing of 1 cm, there are 3

nodes in each the x and y directions.

5-139 Prob. 5-138 is repeated. Using SS-T-Conduct (or other) software, the nodal temperatures are to be solved.

Assumptions 1 Steady heat conduction is two-dimensb form.

Properties The thermal conductivity is given to be k = 25 W/m·K.

Analysb

B puted results are as follows. y clicking on the Calculate Temperature button, the com


5-135

he converged nodal temperatures are

1 4 2 3

Discussion The temperature contour for this problem can be plotted by selecting the Graphical Output tab, as follows.

Hence, t

T = T = 196.4°C, T = 190.3°C, T = 201.2°C


5-136

-140 A hot brass plate is having its upper surface cooled by impinging jet while its lower surface is insulated. The explicit finite difference equations, the maximum allowable value of the time step, and the temperature at the center plane of the brass plate after 1 minute of cooling are to be determined.

Assumptions 1 Transient heat conduction is one-dimensional. 2 Thermal properties are constant. 3 Convection heat transfer coefficient is uniform. 4 Heat transfer by radiation is negligible. 5 There is no heat generation.

Properties The properties of the brass plate are given as ρ = 8530 kg/m3, cp = 380 J/kg·K, k = 110 W/m·K, and α = 33.9 × 10−6 m2/s.

Analysis The nodal spacing is given to be ∆x = 2.5 cm. Then the number of nodes becomes M = L/∆x +1 = 10/2.5 + 1 = 5. This problem involves 5 unknown nodal temperatures, and thus we need to have 5 equations. The finite difference equation for node 0 on the top surface subjected to convection is obtained by applying an energy balance on the half volume element about node 0 and taking the direction of all heat transfers to be towards the node under consideration:

5

tTT

cxxTT

kTThii

p

iii

∆−∆

=∆−

+−+

∞0

1001

0 2)( ρ

⎟⎠⎞

⎜⎝⎛ ∆

++⎟⎠⎞

⎜⎝⎛ ∆

−−= ∞+ T

kxhTT

kxhT iii 22221 10

10 τττ or

Node 4 is on insulated boundary, and thus we can treat it as an interior node by using the mirror image concept. Nand 3 are interior nodes, and thus for them we can use the general explicit finite difference relation expressed as

odes 1, 2,

ii+1

τmmi

mi

mi

mTT

TTT−

=+− +− 11 2

or

Thus, the explicit finite difference equations are

Node 0:

im

im

im

im TTTT )21()( 11

1 ττ −++= +−+

⎟⎠⎞

⎜⎝⎛ ∆

++⎟⎠⎞

⎜⎝⎛ ∆

−−= ∞+ T

kxhTT

kxhT iii 22221 10

10 τττ

Node 1:

Node 2:

Node 3:

Node 4:

where

∆x = 2.5 cm, k = 110 W/m·K, h = 220 W/m2·K, T∞ = 15°C, α = 33.9 × 10−6 m2/s, and h∆x/k = 0.05.

(b) The upper limit of the time step ∆t is determined from the stability criterion that requires all primary coefficients to be greater than or equal to zero. The coefficient of is smaller in this case, and thus the stability criterion for this problem can be expressed as

iiii TTTT 1201

1 )21()( ττ −++=+

iiii TTTT 2311

2 )21()( ττ −++=+

iiii TTTT 3421

3 )21()( ττ −++=+

iiii TTTT 4331

4 )21()( ττ −++=+

iT0

0221 ≥∆

−−k

xhττ → )/1(2

1kxh∆+

≤τ → )/1(2

2

kxhxt∆+

∆≤∆

α

Since . Substituting the given quantities, the maximum allowable value of the time step is determined to be

2/ xt ∆∆= ατ

s 779.8)]K W/m110/()m 025.0)(K W/m220(1)[s/m 109.33(2

)m 025.0(226

2=

⋅⋅+×≤∆

−t


5-137

e this problem. For convenience, let us choose the time step to e ∆t = 6 s. Then the mesh Fourier number becomes

Therefore, any time step less than 8.779 s can be used to solvb

32544.0)m 025.0(

)s 6)(s/m 109.33(2

26

2 =×

=∆

∆=

−

xtατ (for ∆t = 6 s)

(c) With the initial nodal temperatures of 650°C, the results obtaine iterationsd from successive are listed in the following table:

Nodal temperature,°C

Time step, i

Time,

s iT0 iT1 iT2 iT3 iT4

0 0 650 650 650 650 650

1 6 629.3 650 650 650 650

650 650 650

647.8 650 650

24 611.4 634.4 645.6 649.3 650

627.0 640.7 647.0 648.7

599.5 623.7 638.3 645.5 647.6

6.2 620.6 635.9 643.9 646.3

617.6 633.5 642.1 644.7

10 60 590.3 614.8 631.1 640.1 643.0

2 12 622.8 643.3

3 18 616.3 638.8

4

5 30 607.0 630.6 643.2 648.3 649.5

6 36 603.1

7 42

8 48 59

9 54 593.2

The temperature at the center plane of the brass plate after 1 minute of cooling is

g en as

C 631.1 °== )s 60 ,m 05.0(2 TT

(d) From Chapter 4, the approximate analytical solution is iv

10

)/cos(),(

11wall21 LxeA

TTTtxT

iλθ τλ−

∞

∞ =−−

=

where

2.0K W/m110

)m 10.0)(K W/m220( 2=

⋅⋅

==k

hLBi

2.02034.0)m 10.0(

)s 60)(/sm 109.33(2

26

2 >=×

==τ −

Ltα

and (from Table 4-2)

Hence,

4328.01 =λ 0311.11 =A

∞−

∞ +−= TLxeATTtxT i )/cos()(),( 1121 λτλ

[ ]C 630.6 °=

°+°−°= − C 15)5.0)(4328.0(cos)0311.1)(C 15C 650()s 180 ,m 05.0( )2034.0()4328.0( 2eT

Discussion The comparison between the approximate analytical and numerical solutions is within ±0.08% agreement.


5-138

lowed to cool in a room. Using SS-T-Conduct (or other) software center of the bar to 100°C is to be determined.

conduction is two-dimensional. 2 Thermal properties are constant. 3 Convection heat transfer ransfer by radiation is negligible. 5 There is no heat generation.

ate are given as k = 63.9 W/m·K and α = 18.8 × 10−6 m2/s.

Input window for 2-P m, the problem parameters

entered into the with a uniform nodal

cm, there are 5 nodes in the x direction and 3 y direction.

5-141 A long hot rectangular cross section steel bar is alwith explicit method, the duration required to cool the

Assumptions 1 Transient heatcoefficient is uniform. 4 Heat t

Properties The properties of the brass pl

Analysis On the SS-T-ConductDimensional Transient robleand the boundary conditions areappropriate text boxes. Note thatspacing of 5 nodes in the



5-139

From the results computed by the SS-T-Conduct software, the surface temperature reached 100°C at t ≈ 12000 s.

Discussion Similar result (t ≈ 12000 s) can also be obtained using the implicit method.


5-140

Fundamentals of Engineering (FE) Exam Problems

5-142 The unsteady forward-difference heat conduction for a constant area, A, pin fin with perimeter, p, exposed to air whose temperature is T B0B with a convection heat transfer coefficient of h is

*20

2*

1*

121* 21 m

ppmm

pm T

Achp

xckT

Axhp

TTxc

kT⎥⎥⎦

⎤

⎢⎢⎣

⎡−

∆−−

⎥⎥⎦

⎤

⎢⎢⎣

⎡ ∆++

∆= +−

+

ρρρ

In order for this equation to produce a stable solution, the quantity Ac

hpxc

k

pp ρρ+

∆ 22 must be

(a) Negative (b) zero (c) Positive (d) Greater than 1 (e) Less than 1

Answer (d) Greater than 1

5-143 Air at T B0B acts on top surface of the rectangular solid shown in Fig. P5-143 with a convection heat transfer coefficient of h. The correct steady-state finite-difference heat conduction equation for node 3 of this solid is

(a) T B3B = [(k/2∆)( T B2 B+ T B4 B+ T B7B) + hT B0B] / [(k/∆) + h]

(b) T B3B = [(k/2∆)( T B2 B+ T B4 B+ 2T B7B) + hT B0B] / [(2k/∆) + h]

(c) T B3B = [(k/∆)( T B2 B+ T B4 B) + hT B0B] / [(2k/∆) + h]

(d) T B3B = [(k/∆)( T B2 B+ T B4 B+ T B7B) + hT B0B] / [(k/∆) + h]

(e) T B3B = [(k/∆)( 2T B2 B+ 2T B4 B+ T B7B ) + hT B0B] / [(k/∆) + h]

Answer (b)

5-144 What is the correct unsteady forward-difference heat conduction equation of node 6 of the rectangular solid shown in Fig. P5-127 if its temperature at the previous time (∆t) is *

6T ?

[ ] [ ][ ] [ ][ ] [ ][ ] [ ][ ] [ ] *

62*

10*7

*2

*5

216

*6

2*10

*7

*2

*5

216

*6

2*10

*7

*2

*5

216

*6

2*10

*7

*2

*5

216

*6

2*10

*7

*2

*5

216

)/(41)()/(2 )(

)/(21)()/(2 )(

)/(2)()/( )(

)/(1)()/( )(

)/(41)()/( )(

TctkTTTTctkTe

TctkTTTTctkTd

TctkTTTTctkTc

TctkTTTTctkTb

TctkTTTTctkTa

ppi

ppi

ppi

ppi

ppi

∆∆−++++∆∆=

∆∆−++++∆∆=

∆∆++++∆∆=

∆∆−++++∆∆=

∆∆−++++∆∆=

+

+

+

+

+

ρρ

ρρ

ρρ

ρρ

ρρ

Answer (a)

T B0B, h

• • • • B 1 5 9 B

2 3 B

10 B

6 7 B

11 B

• • • • B

• • • • B

4 8 12 B

∆x = ∆y = ∆B

• • • • B 1 5 9 B

2 3 B

10 B

6 7 B

11 B

• • • • B

• • • • B

4 8 12 B

∆x = ∆y = ∆B


5-141

5-145 What is the correct steady-state finite-difference heat conduction equation of node 6 of the rectangular solid shown in Fig. P5-145?

(a) T B6B = (T B1B+ T B3B+ T B9 B+ T B11B) / 2

(b) T B6B = (T B5 B+ T B7 B+ T B2 B+ T B10B) / 2

(c) T B6B = (T B1 B+ T B3B+ T B9 B+ T B11B) / 4

(d) T B6B = (T B2 B+ T B5 B+ T B7 B+ T B10B) / 4

(e) T B6B = (T B1 B+ T B2 B+ T B9 B+ T B10B ) / 4

Answer (d)

5-146 The height of the cells for a finite-difference solution of the temperature in the rectangular solid shown in Fig. P5-146 is one-half the cell width to improve the accuracy of the solution. The correct steady-state finite-difference heat conduction equation for cell 6 is

(a) T B6B = 0.1(T B5 B+ T B7B) + 0.4(T B2B + T B10B)

(b) T B6B = 0.25(T B5 B+ T B7B) + 0.25(T B2B + T B10B)

(c) T B6B = 0.5(T B5 B+ T B7B) + 0.5(T B2B + T B10B)

(d) T B6B = 0.4(T B5 B+ T B7B) + 0.1(T B2B + T B10B)

(e) T B6B = 0.5(T B5 B+ T B7B) + 0.1(T B2B + T B10B)

Answer (a)

5-147 The height of the cells for a finite-difference solution of the temperature in the rectangular solid shown in Fig. P5-147 is one-half the cell width to improve the accuracy of the solution. If the left surface is exposed to air at T B0B with a heat transfer coefficient of h, the correct finite-difference heat conduction energy balance for node 5 is

(a) 2T B1 B + 2T B9B + T B6B – T B5B + h∆/k (T B0B – T B5B) = 0

(b) 2T B1B + 2T B9 B + T B6B – 2T B5B + h∆/k (T B0B – T B5B) = 0

(c) 2T B1 B + 2T B9B + T B6B – 3T B5B + h∆/k (T B0B – T B5B) = 0

(d) 2T B1B + 2T B9 B + T B6B – 4T B5B + h∆/k (T B0B – T B5B) = 0

(e) 2T B1 B + 2T B9B + T B6B – 5T B5B + h∆/k (T B0B – T B5B) = 0

Answer (e)

5-148 ….. 5-151 Design and Essay Problems

1 2 3 4 B

∆y = ∆/2 B

• • • • B

• • • • B

• • • • B

∆x = ∆B 5 6 7 8 B

9 10 11 12 B

1 2 3 4 B

∆y = ∆/2 B

• • • • B

• • • • B

• • • • B

∆x = ∆B 5 6 7

9 10 11 12 B

T B0B, h

• • • • B 1 5 9 B

2 3 B

10 B

6 7 B

11 B

• • • • B

• • • • B

4 8 12 B

∆x = ∆y = ∆B

heat 4e sm chap05

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