second semester ay 2019-2020 · 2020-01-22 · thickness 3. constant wall area and uniform...

Second Semester AY 2019-2020

Department of Chemical EngineeringUniversity of the Philippines Diliman 1

ChemE 131Thermal Systems

2

Metals: added energy increases the kinetic energy of free electrons

Atomic/

molecular

motion

excited by

the added

energy

3

What is heat conduction?

Heat conduction flux

Thermal conductivity

1D heat conduction

Transfer of energy of motion

between adjacent molecules

4




1D heat conduction

OUTSIDE INSIDE

0T 0T

1T

→t

=heating rate q

5




1D heat conduction

0T

1T

0 Lx

T

thickness L 0 L= − =

surface area A=

Aheat flux

q=

q dE

A dx −

P

From thermodynamics:

dE C dT=

P

P

C dTq k

A C dx

= −

dE

dx= −

thermal diffusivity =

P

k

C= q dT

kA dx= −

Fourier's Law

6




1D heat conduction

Quantity SI unit English unit

heat flow rate, q J/s or W Btu/h, Btu/min

heat flux, q/A W/m2 Btu/ft2h

temperature, T ºC, K ºF, ºR

thermal conductivity, k W/mK Btu/ftºF

direction/dimension mm, m in, ft

q dTk

A dx= −

7




1D heat conduction

xq

yqzq

q T T Tk

A x y z

= − + +

qk T

A= −

If conduction occurs

within the wall in

all directions:

8




1D heat conduction

9




1D heat conduction

10




1D heat conduction

Find the steady-

state temperature

profile across the

given plane wall as

a result of heat

conduction.

11




1D heat conduction

12




1D heat conduction

Assumption Implication

steady state conditions q = constant

constant transfer area q/A may be constant

isotropic material with constant properties

material properties do not vary significantly within the

temperature range

wall thickness is significantly smaller than the width and height

one-dimensional (1D) heat conduction along the wall

thickness

heat convection from the bulk surroundings occurs at a significantly higher rate than the heat conduction

Twall = Tsurroundings

k = constant

13




1D heat conduction

Fourier's law of conduction

q dTk

A dx= −

s ,2

s ,1

TL

0 T

q dx k dTA

→ = −

s ,2 s ,1T

L q kA

T−

= −

→

x q k

T

dA

d→ = −

Calculate the heat loss per m2 ofsurface area through a 25.4-mm thickinsulating wall made of fiber insulatingboard (k = 0.048 W/m·K), where theinside temperature is 352.7 K and theoutside temperature is 297.1 K.

14




1D heat conduction

15




1D heat conduction

OUTx 0.0254 m=

q dTk

A dx= −

INSIDE OUTSIDE

INT 352.7 K= OUTT 297.1 K=

INx 0 m=

OUT OUT

IN IN

x T

x T

qdx k dT

A

= −

OUT IN

OUT IN

T Tqk

A x x

−= − −

Wk 0.048

m K=

2

q W105.1

A m=

Calculate the heat loss per m2 of surface areafor an insulating wall composed of a 25.4-mmthick fiber insulating board (k = 0.048 W/mK),where the inside wall temperature is 403.5 Kand the outside temperature is 276.5 K.If the insulation is used to cover a 50-ft2 wall,how thick must the fiber board be to avoid anincrease in the heat loss flux?

16




1D heat conduction

17

inner wall

Ti = 352.7 K

fiber

insula

ting b

oard

ambient

surroundings

To = 276.5 K

heat transfer




1D heat conduction

Assumptions:

1. steady state heat conduction

2. heat transfer is only along the direction of wall thickness

3. constant wall area and uniform thickness

4. fiber insulating board wall is isotropic

5. fiber insulating board has constant properties (not temperature dependent)

6. no convective heat transfer resistance near the surface of the outer wall

7. perfect contact between inner wall and the insulating wall

18




1D heat conduction

k0.048

W/mK

Ti

403.5 K

To

276.5 K

xi = 0 xo = 0.0254 m

2basis: 1 m wall area

q dTk

A dx= − constant=

Tk

x

= −

o i

o i

T Tk

x x

−= −

−

19




1D heat conduction

( )( )( )

2o i

o i

276.5 403.5 KT T Wq kA 0.048 1 m

x x m K 0.0254 0 m

−− = − = −

− −

2q 240 W (for 1 m wall surface area)=

( )

2

2

2

for a 50-ft wall:

276.5 403.5 K240 W W 0.048

m K x0.3048 m50 ft

1 ft

− = −

x 0.1178 m (or 117.8) mm wall thickness =

A furnace wall is insulated by a material withthermal conductivity expressed as

During operation, the temperature inside thefurnace reaches 1800F, releasing heat at 3500Btu/hr·ft2. Determine the insulation thicknesssufficient to bring the temperature outside thewall to 300F.

20




1D heat conduction

( )F

Btuk 0.08 1 0.003T

hr ft F

= +




1D heat conduction

x

INSIDE OUTSIDE

INT 1800 F= OUTT 300 F=

( )Tk 0.08 1 0.003= +

Assumptions

steady state conditions

constant transfer area

isotropic material

material properties do not vary significantly within the temperature range

wall thickness is significantly smaller than

the width and height

heat convection from the bulk surroundings occurs at

a significantly higher rate

2

Btuq 3500

hr ft=




1D heat conduction

q dTk

A dx= − ( )

OUT OUT

IN IN

x T

x T

q dx k T dT

A

→ = −

( ) ( )OUT OUT

IN IN

x T

x T 3500 dx 0.08 1 0.003T dT→ = − +

( ) ( )3002

1800 3500 x 0.08 T 0.0015T→ = − +

x 0.1423 ft (1.71 in) =

Determine the heat transfer rate across the frustum shown at the right, whose thermal conductivity is measured to be 27 Btu/hrftF.

23




1D heat conduction

24




1D heat conduction

x b 4 3 r x

2r 6 b 4

+ = → =

+

b 4 b b 8 in

6 4

+= → =

1r x

4=

25




1D heat conduction

q dTk

A dx= −

dx q k dT

A→ = −

( )( )

o

o

12 in 200 F

21

8 in 400 F4

dxq 27 dT

x

= −

( )o

o

12 ft

12 200 F

400 F8 ft

12

16 1 q 27T

x

→ − = −

26




1D heat conduction

( ) o

o

Btu27 200 400 F

hr ft Fq

16 12 12 1

12 8 ft

− −

=

− −

Btuq 2120.58

hr=

ChemE 131Thermal Systems

27

Determine the heat transfer across the 1 inch-thick slab shown at the right, whose thermal conductivity varies linearly with temperature according to the equation

28

o2 o F

Btuk 0.2 0.005T

h ft F

= +

second semester ay 2019-2020 · 2020-01-22 · thickness 3. constant wall area and uniform...

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