convection process
TRANSCRIPT
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3.0 SYLLABUS CONTENT
3.1` Convection Principlesheat transfer coefficient
3.2 Convection boundary layer theory
3.3 Forced convection over exterior surface (laminarflow)
3.4 Forced convection in turbulent flow (Reynolds
analogy.
3.5 Principle of dynamic similarity applied to forcedconvection
3.6 Free convection and laminar profile over vertical
plates
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Convection Principles(Nusselt Number)
Nusselt Number
Developed by Wilhelm Nusselt
(1882-1957) from Germany
In convection analysis, it iscommon practice to non-
dimensionalized the governing
equations and combine the
variables, which group together indimensionless numbersto reduce
the number of variables.
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Convection Principles(Nusselt Number)
The Nusselt number is a non-dimensionalized h,
defined as:
k
hLNu c Lc - Characteristic Length
k - Thermal conductivity of fluid
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Convection Principles(Nusselt Number)
Since:
Heat transfer by conduction occurs when the fluid
is motionless and
Heat transfer by convection occurs when the fluidinvolves some motion.
In either case, the heat flux is the rate of heat
transfer per unit time per unit surface area.
L
Tkq
Thq
cond
conv
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Convection Principles(Nusselt Number)
Taking the ratio of these two equations:
Thus Nu represents the enhancement of heat transfer through a
fluid layer as a result of convection relative to conduction
across the same fluid layer. The larger Nu, the more effective
the convection.
Nu= 1 for a fluid layer, represents pure conduction.
Nu
k
LhTh
q
q
LTkcond
conv
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Dynamic viscosity ()The shear force per unit arearequired to drag on layer of fluid with unit velocity passed
another layer a unit distance away from the fluid.
Kinematic viscosity ()The ratio of dynamic viscosityto density.
Convection Principles(Viscosity)
dydu
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3.2 Convection boundarylayer theory
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Convection Principles(Velocity Boundary Layer)
Velocity boundary development on a flat plate:
The boundary layer thickness () is normally defined
as where:
uu 99.0
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Convection Principles(Velocity Boundary Layer)
The dashed line, divides the flow
over the plate into two regions:
Boundary layer region
In which the viscous effects and
velocity changes are significant.
Inviscid flow region
In which the friction effects arenegligible and the velocity
remains constant.
uy
x
Heated Surface
Boundary
Layer
Inviscid
Flow
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Convection Principles(Velocity Boundary Layer)
Flow regions in velocity boundary of a flat plate:
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Convection Principles(Velocity Boundary Layer)
Comparison of a laminar and turbulent velocity
boundary layer profile:
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Convection Principles(Thermal Boundary Layer)
Likewise there is a thermal
boundary layer
No temperature jump condition
Because velocity of the fluid
is zero at the point of contact
with the solid surface, the
fluid and solid surface must
have the same temperature
at the point of contact.
y
x
T
Heated Surface
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Convection Principles(Thermal Boundary Layer)
Thermal boundary development on a flat plate:
The thickness of the thermal boundary layer (t) at any location
along the surface is defined as the distance from the surface at
which:
T=T-Ts=0.99(T-Ts)
Ts+0.99(T-Ts)
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Convection Principles(Prandtl Number)
Prandtl Number
Developed by Ludwig Prandtl (1875-1953) of
Germany.
The relative thickness of the velocity and
thermal boundary layers is best described by
a dimensionless Prandtl number (below):
k
C
HeatofyDiffusivitMolecular
MomentumofyDiffusivitMolecular
p
Pr
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Prandtl Number ?
The Prandtl numbers of gases are about 1, which
indicates that both momentum and heat dissipate
through the fluid at about the same rate.
Heat diffuses very quickly in liquid metals (Pr > 1) relative to momentum.
Consequently the thermal boundary layer is much thicker
for liquid metals and much thinner for oils relative to the
velocity boundary layer,
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Convection Principles(Reynolds Number)
Reynolds Number
Derived by Osbourne Reynolds (1842-1912)
of Britain
The transition from laminar to turbulent flow
depends on the surface geometry, surface
roughness, free stream velocity, surface
temperature, and type of fluid (among other
things).
However, the flow regime primarily dependsupon the ratio of inertia forces to viscous
forces in a fluid. This is a dimensionless
quantity, known as Reynolds number (Re).
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Convection Principles(Reynolds Number)
The Reynolds number is defined as:
LVLV
ForcesViscousForcesInertia Re
Vupstream velocityLcharacteristic length
=
kinematic viscosity of fluid
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Convection Principles(Reynolds Number)
A large Re (inertia forces large) Means that the viscous forces cannot contain random and
rapid fluctuations (turbulent).
A small Re (viscous forces large)
Keeps the fluid in-line (laminar).
The Reynolds number where the flow becomes turbulent is
called the critical Reynolds number (Recrit)
LVLV
ForcesViscous
ForcesInertia
Re
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Convection Principles(Reynolds Number)
For flow over a flat plate, the generally accepted
value of Recritis:
Flat Plate:
where: xcrit= Distance between the leading edgeof the plate to the transition point
from laminar to turbulent flow takes place.
5
105Re
critcrit
xu
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3.3 Forced convection over an
exterior surface(laminar and turbulent flow)
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External Flow
The convection equations for an external flow can be
derived from the conservation of mass, conservation
of energy, and the conservation of momentum
equations.
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External Flow Equations(Conservation of Mass)
Conservation of Mass
dxx
u
u
dx
dy
dydy
dv
v
u
v
AreaUnit
y
AreaUnit
x
dxvm
dyum
1
1
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External Flow Equations(Conservation of Mass)
Rate of mass
flow into
control volume
Rate of mass
flow out of
control volume
=
dydxyvdxvdydx
xudyudxvdyu
dxdyy
vvdydx
x
uudxvdyu
0
y
v
x
u~ 2-D Continuity Equation
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External Flow Equations(Conservation of Momentum)
Conservation of Momentum
ma = Net Force
dxx
PP
P
dyy
dx
dy
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External Flow Equations(Conservation of Momentum)
In the x-direction:
In the y-direction:
volume
unitper forceBody
x
forcesshearandviscousofeffectNet
forcepressureNet
du
gy
u
x
u
x
P
y
uv
x
uu
2
2
2
2
volumeunitperforceBody
y
forcesshearandviscousofeffectNet
force
pressureNetdv
gy
v
x
v
y
P
y
vv
x
vu
2
2
2
2
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External Flow Equations(Conservation of Energy)
Conservation of Energy
dx
dy
Eheat out, y Emass out, y
Emass in, yEheat in, y
Emass in, x
Eheat in, x
Emass out, x
Eheat out, x
0 outin
EE
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External Flow Equations(Conservation of Energy)
General 2-D energy equation
For 2-D inviscid flow:
222
2
2
2
2
2
x
v
y
u
y
v
x
u
y
T
x
Tk
y
Tv
x
TuCp
2
2
2
2
yT
xTk
yTv
xTuCp
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Convection over a Flat Plate
Consider laminar flow over a flat plat. When viscousdissipation is negligible, the convection equations
reduce for steady, incompressible laminar flow (with
constant properties) over a flat plate.
x
y
T
, u
u(x,0)= 0
v(x,0)= 0
T(x,0)= Ts
dy
dx
Boundary layer
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Convection over a Flat Plate
Consider elemental control volume for force balance
in the laminar boundary layer.
Continuity:
Momentum:
Energy:
0
y
v
x
u
2
2
y
u
y
uv
x
uu
2
2
y
T
y
Tv
x
Tu
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Convection over a Flat Plate
Boundary conditions:
At x= 0: u(0,y)= u
, T(0,y)= T
At y= 0: u(x,0)= 0, v(x,0)= 0, T(x,0)= Ts
At y=
: u(x,)= u
, T(x,)= T
Define a dimensionless similarity variable:
x
uy
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Convection over a Flat Plate
Recall, that the stream function is defined as:
Dependent variable:
xv
yu
;
yu
u
xu
f
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Convection over a Flat Plate
Therefore:
fd
df
x
u
fxu
u
dx
df
u
xu
xxv
d
dfu
x
u
d
df
u
xu
yyu
2
1
2
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Convection over a Flat Plate
So:
3
32
2
2
2
2
2
2
2
fd
x
u
y
u
d
fd
x
uu
y
u
d
fd
x
u
x
u
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Convection over a Flat Plate(Momentum Equation)
Substituting these into the momentum equation and simplifying
gives:
A 3rdorder non-linear differential equation. Therefore the system
of partial differential equations is transformed into a single
ordinary differential equation by use of a similarity variable.
022
2
3
3
d
fdf
d
fdEQN 6-49
text
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Convection over a Flat Plate(Momentum Equation)
Using the definitions for f and , the boundary equations in
terms of the similarity variables can be found.
1
0
00
0
d
df
d
df
f However, the transformed equation
with its similarity variable cannot besolved analytically.
Therefore, an alternative solution is
necessary.
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Convection over a Flat Plate(Momentum Equation)
The non-dimensional velocity profile can be obtained by
plotting u/u
vs. . The results agree experimentally.
A value of: corresponds to:
Recall that the definition of a velocity boundary layer is when:
992.0u
u
d
df
0.5
99.0
u
u
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Convection over a Flat Plate(Momentum Equation)
So substituting these values into the definition for , gives the
boundary layer thickness for a flat plate.
d
d
xuwhere
x
x
u
x
u
x
uy
Re:Re
0.50.5
5
d y;0.5
For laminar
flat plate:
EQN 6-51
text
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Energy Equation
Knowing the velocity profile, we can now solve the energy
equation.
Introduce dimensionless temperature:
Note: both Tsand T
are constant.
Convection over a Flat Plate(Energy Equation)
s
s
TT
TyxTyx
,,
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Convection over a Flat Plate(Energy Equation)
Substituting into the energy equation gives:
Again using the similarity variable, , so = ()
So the energy equation becomes:
2
2
yyv
xu
x
uy
2
2
2
2
1
dy
d
d
d
dy
d
d
df
d
df
x
u
dx
d
d
d
d
dfu
C i Fl Pl
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Convection over a Flat Plate(Energy Equation)
Since:
x
u
dy
d
x
u
uy
x
uy
dx
d
x
uy
2
1
2
1 23
C ti Fl t Pl t
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Convection over a Flat Plate(Energy Equation)
and:
yud
df
yu
u
xu
f
C ti Fl t Pl t
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Convection over a Flat Plate(Energy Equation)
Substituting these in gives:
2
2
2
2
1
dy
d
d
d
dy
d
d
df
d
df
x
u
dx
d
d
d
d
dfu
x
u
d
d
x
u
d
d
yuyux
uy
x
u
x
u
u
y
d
d
yuu
2
2
2
1
2
C ti Fl t Pl t
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Convection over a Flat Plate(Energy Equation)
x
u
d
d
x
u
yux
u
ux
u
x
u
x
u
ud
d
2
2
1
2
1
2
2
21
dd
ux
xyxu
xxu
udd
C ti Fl t Pl t
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Convection over a Flat Plate(Energy Equation)
2
2
2
d
d
yux
u
ux
u
u
x
d
d
2
2
Pr
2
d
d
yud
d
f
0Pr22
2
d
df
d
d
Pr
Prandtl number
EQN 6-58
text
C ti Fl t Pl t
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Convection over a Flat Plate(Energy Equation)
A closed form solution cannot be obtained for this boundary
layer problem, and it must be solved numerically.
If this equation is solved for numerous values of Pr, then for
Pr > 0.6, the non-dimensional temperature gradient at the
surface is found to be (reference Table 6-3, p. 378 in text):
31
Pr332.0
0
d
d
C ti Fl t Pl t
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Convection over a Flat Plate(Energy Equation)
The temperature gradient at the surface is:
Since: then:
Therefore substituting these values in gives:
0000
y
s
y
s
y y
TTy
TTy
T
x
u
y
x
uy
x
uTT
y
Ts
y
31
Pr332.0
0
C ti Fl t Pl t
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Convection over a Flat Plate(Energy Equation)
Therefore the local convection coefficient and Nusselt number
become:
TT
TTk
TT
k
TT
qh
s
x
u
s
s
yyT
s
s
x
3
1
Pr332.00
xukhx
31
Pr332.0
C ti Fl t Pl t
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The local Nusselt number is the dimensionless temperature
gradient at the surface. This is defined as:
Thus for Pr > 0.6, the local Nusselt number for laminar flow is:
Convection over a Flat Plate(Laminar Flow)
k
xhNu xx
21
31
RePr332.0 xNu
C ti Fl t Pl t
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Convection over a Flat Plate(Laminar Flow)
The local friction coefficient (CFx) can also be determined.
Since the wall shear stress is:
From Table 6-3 (pp. 378 in text) this is found to be:
0
2
2
0
d
fd
x
u
uy
u
y
wall
x
wall
u
Re
332.0 2
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Con ection o er a Flat Plate
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Convection over a Flat Plate(Laminar Flow)
Therefore the local skin friction coefficient is:
21
Re664.02
2,
x
wall
xF
u
C
2,
2
1 uC xFwall
Convection over a Flat Plate
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Convection over a Flat Plate(Laminar Flow)
The average heat transfer coefficient over the entire plate can be
obtained by integrating over its length:
dxh
L
h
L
x
0
1
L
k
LuL
k
xu
L
k
dx
x
u
L
kh
L
L
21
31
3
1
31
31
RePr664.0
Pr664.0
2Pr332.0
Pr332.0
0
0
Convection over a Flat Plate
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Convection over a Flat Plate(Laminar Flow)
So the average Nusselt number for laminar flow over
the entire plate is:
31
PrRe664.0 5.0
L
k
Lh
Nu
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Convection over a Flat Plate(Laminar Flow)
Solving numerically for temperature profile for
different Prandtl numbers, and using the definition of
the thermal boundary layer, it is determined that for
laminar flow over a flat plate:
31
31
PrPr
026.1
dd
dt
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Convection over a Flat Plate(Example 3.1)
Example 3.1a Calculate the heat transfer and the
thermal boundary layer thickness of the way along
a flat plate that is 50 m long. Liquid (Tsat = 40 C)
flows over it at 4 m/s. The plate is kept at a surface
temperature (Ts= 80 C).
50 m
x
y
40 C
4 m/s
Ts= 80C
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Convection over a Flat Plate(Example 3.1)
The first step is to calculate the mean film temperature of the
fluid flowing along the plate.
This is just the average of the surface temperature and the fluid
bulk temperature.
CCCTT
T sfilm
602
4080
2
50 mx
y
80 C
40 C
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Convection over a Flat Plate(Example 3.1)
For liquid water at 60 C from Table
50 mx
y
80 C
40 C
99.2Pr
654.0
67.4
3.983 3
CmW
sm
kg
m
kg
k
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Convection over a Flat Plate(Example 3.1)
First calculate the Reynolds number to determine whether theflow is laminar or turbulent.
Since Re < Recrit = 5x105 or 500,000 ~ Flow is laminar
Therefore:
8.527,10
67.4
5043.983Re
41
3
sm
kg
sm
m
kgmxu
m
m
mx
t
m
412.0
026.1
99.2609.0
026.1
Pr
609.08.527,10
5
Re
5
31
31
450
d
d
d
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Convection over a Flat Plate(Example 3.1)
Example 3.1b Now calculate the convective heat transfer.
First we must check to see whether the entire plate is in a
laminar boundary layer or not.
Since Re < Recrit = 5x105 or 500,000 ~ Flow is laminar over the
entire plate
3.111,42
67.4
5043.983Re
3
sm
kg
sm
m
kgmLu
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Convection over a Flat Plate(Example 3.1)
Therefore we can use the following equation to find h:
CmW
sm
kg
m
kg
sm
Cm
W
m
x
uk
x
ukh
2
33
1
31
31
619.0
5067.4
3.9834654.099.2332.0
Pr332.0Pr332.0
Convection over a Flat Plate
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Convection over a Flat Plate(Example 3.1)
Using this h, we can now find the convection heat transfer:
2
2
8.244080619.0
)(
m
W
Cm
W
s
CC
TThq
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Convection over a Flat Plate(Turbulent and Mixed Flows)
Turbulent
Completely Turbulent Flow
Mixed Laminar/Turbulent Flow
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Convection over a Flat Plate
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Convection over a Flat Plate(Turbulent and Mixed Flows)
Note: if it had been found that the boundary layer was notcompletely laminar another equation for h could have been
used instead.
For turbulent flow (all over the plate):
For a mixed combination of laminar and turbulent flow over the
plate:
7510Re105
60Pr6.0
3
1
PrRe037.0 8.0 LNu
31Pr871Re037.0 8.0 LNu 75 10Re10560Pr6.0
L
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Example 3.2 Oil flows over a 40-m long heated plate at freestream conditions of 5 m/s and 25C. If the plate is held at 45C.
a) Determine the velocity and thermal boundary layer
thicknesses at the middle of the plate.
b) Calculate the distance where the laminar change to
turbulence flow
c) Calculate the total convection heat transfer for a 1-m
width.
Convection over a Flat Plate(Example 3.2)
40 m
u
= 5 m/s
T= 25C
Ts= 45C
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Convection over a Flat Plate(Example 3.2)
First calculate the film temperature (Tf)
From Tables for oil at 35C, the fluid properties are:
CCCTT
T sfilm
352
4525
2
3
2
255,1
2864.0
105.3
711,3Pr
4
m
kg
Cm
W
sm
k
Convection over a Flat Plate
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Convection over a Flat Plate(Example 3.2)
a) At the middle of the plate:
Since the critical Reynolds number is 5x105, then:
The flow at the mid-point of the plate is laminar.
mm
x 202
40
54
int
1086.2105.3
205Re
2
sm
sm
pomid
mxu
critpomid ReRe
int
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Convection over a Flat Plate(Example 3.2)
The hydrodynamic (or velocity) boundary layer is:
The thermal boundary layer is:
cmorm
mxx 7.18187.0
1086.2
205
Re
5
520
d
mmormm
t
8.110118.0711,3026.1
187.0
Pr026.1
31
31
dd
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Convection over a Flat Plate(Example 3.2)
b) At the end of the plate:
Since Re > Recrit the flow is turbulent at the end
The critical distance (transition point from laminar to
turbulent is:
54
10714.5105.3
405Re
2
sm
sm
end
mLu
m
ux
sm
sm
crit
crit
35
5
105.3105
Re
245
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Convection over a Flat Plate(Example 3.2)
c) Using the mixed Nu equation for a flat plate:
7.600,9 711,38711071.5037.0
Pr871Re037.0
31
31
8.05
8.0
LNu
Cm
WCmW
m
L
kNu
h
27.6840
2864.07.600,9
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Convection over a Flat Plate(Example 3.2)
The total heat transfer per is:
W
CCmm
TTAhQ
Cm
W
ss
960,54
25451407.68 2
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Example 7-1 7-2, 7-3 pp 404-407
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Forced Convection(on Cylinders and Spheres)
Flows across cylinders andspheres, in general, involve
flow separation which is
difficult to handle analytically.
Thus these must be studied
empirically or experimentally
Several correlations have
been developed for the heattransfer coefficient (h).
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Forced Convection(on Cylinders and Spheres)
Churchill and Bernstein developed this empiricalequation for flow over a cylinder (Eqn. 7-35 in text):
Whitaker developed this empirical equation for flow
over a sphere (Eqn. 7-36 in text):
54
85
413
2
31
21
000,282
Re
11
PrRe62.0
3.0Pr4.0
k
hD
Nucyl
4
1
32
21 4.0
PrRe06.0Re4.02
s
sphk
hDNu
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Forced Convection(over Circular and Non-Circular Cylinders)
Additionally the following empirical correlations have been madeby Zukauskas and Jakob for the average Nusselt number for flow
over circular and non-circular cylinders (Table 7-1 in text):
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Forced Convection(over Circular and Non-Circular Cylinders)
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77
Forced Convection(Example 3.3)
Example 3.3 A long 10-cm diameter hexagonal steam pipewhose external surface temperature is 110C passes through
some open area that is not protected against the wind.
Determine the rate of heat loss when the air is at 1 atm
pressure and 10C and the wind is blowing across a 1-m length
of pipe at a velocity of 8 m/s.
V= 8 m/s
T
= 10C Ts=110C
10 cm
1 m
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78
Forced Convection(Example 3.3)
The properties of air at the average film temperatureof:
can be found from Table A-15 as:
CCCTT
T sfilm
60
2
10110
2
sm
Cm
W
k25
10896.1
7202.0Pr;02808.0
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Forced Convection(Example 3.3)
The Reynolds number is:
The Nusselt number can be determined from Table 7-1
in the text book:
45
10219.410896.1
10.08Re
2
sm
sm mDV
5.122
7202.010219.4153.0
PrRe153.03
1
31
638.04
638.0
Nu
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Forced Convection(Example 3.3)
Therefore:
The surface area of the hexagon is:
CmWCm
W
m
NuD
kh
24.345.12210.0
02808.0
2
346.0
60sin
110.03
60sin26
m
mm
LD
As
D/260
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81
Forced Convection(Example 3.3)
Therefore, the heat transfer is:
W
CCm
TTAhQ
Cm
W
ss
7.191,1
10110346.04.34 2
2
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Example 7-5 pp 414
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83
3.4 Principle of dynamic similarity
and dimensional analysis(applied to forced convection)
Non-dimensionalized
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84
convection equations
The continuity , momentum, and energy equations for steady,incompressible, laminar flow of a fluid with constant properties
can be non-dimensionalized by dividing all the dependent and
independent variables, as follows:
Note: the asterisks denote non-dimensional variables.
s
s
TT
TTT
V
PP
V
vv
V
uu
Lyy
Lxx
*
2
*
**
**
;
;
;;
Free stream velocity
Free stream temperature
Surface temperature
Non-dimensionalized
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convection equations
Introducing these variables the equations become:
Cont inui ty :
Momentum:
Energy:
0*
*
*
*
y
v
x
u
*
*
2*
*2
*
*
*
*
*
*
Re
1
dx
dP
y
u
y
uv
x
uu
L
2*
2
*
*
*
*
*
*
PrRe
1
y
T
y
Tv
x
Tu
L
Non-dimensionalized
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convection equations
For a plate, the boundary conditions are:
1,1,
00,00,
1,000,1,0
****
****
******
xTxu
xTxu
yTxvyu
x*
y*
Ts
u
, T
Similarity
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87
Similarity
Where:
For a given geometry, the solutions of problems with
the same Re and Nu are similar, thus Re and Nu are
called similarity parameters.
Two physical phenomena are similar if they have the
same dimensionless forms of the governing
differential equations and boundary conditions.
PrRe LV
L
Similarity
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88
Similarity
A major advantage in non-dimensionalizing is thesignificant reduction in the number of similarity
parameters.
Original equations have 6 parameters: (L, V
, T,Ts, , and n)
The non-dimensionalized equations have only 2
parameters (ReLand Pr).
Similarity
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89
Similarity
For a given geometry, problems that have the samevalues of similarity parameters (ReLand Pr) have
identical solutions.
Fig 6-28 (text)
Similarity
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90
Similarity
Example: Determining the convection heat transfercoefficient (h) for flow over a given surface willrequire numerical solutions or experiments withseveral sets of:
Velocities (V) Surface lengths (L)
Wall temperatures (Ts)
Free stream temperatures (T
).
The same information can be determined with farfewer experiments or investigations by grouping thedata into the dimensionless:
Reynolds number (Re)
Prantdl number (Pr)
Similarity
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91
Similarity
Fig 6-29 (text)
Similarity
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92
Similarity
Another advantage is that data from a large group ofexperiments can be conveniently reported in the
terms of the similarity parameters.
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93
3.5 Reynolds Analogy
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(Drag Force)
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95
(Reynolds Analogy)
In forced convection analysis, we are primarilyinterested in the determination of quantities of:
The coefficient of friction (CF) (to calculate the
shear stress at the wall)
Nusselt number (Nu) ( to calculate the heat
transfer rates).
Therefore, it is desirable to have a relation between
CFand Nu, so that we can calculate one when theother is available.
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96
(Reynolds Analogy)
Since:
The shear stress at the surface becomes:
Lyxfu Re,, **1*
Lyy
s xfL
V
y
u
L
V
y
uRe,*2
0
*
*
0 *
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97
(Reynolds Analogy)
L
L
L
y
V
yy
uL
V
Vsxf
xf
xf
y
uC
Re,
Re,
Re
2
Re
2
*
3
*
2
0*
*
*
2
0*
2
, 2
*
*
2
Substituting this into its definition gives the local
friction coefficient:
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98
(Reynolds Analogy)
Similarly, solving the energy equation for thedimensionless temperature (T*) for a given geometry
gives:
Using this definition, the convection heat transfer
coefficient (h) becomes:
Pr,Re,, **
1
*
L
yxgT
TT
kh
s
yyT
0
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99
(Reynolds Analogy)
Since:
Then:
localfor
s
s
x
y
L
yy
TT
TTT
**
**
*
*
yT
xTT
xyTTTT
yT sss
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100
(Reynolds Analogy)
Therefore:
0*
*
*
0*
*
*
y
ys
s
y
T
L
k
y
T
TTx
TTkh
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101
Substituting this into the local Nusselt numberequation gives:
We previously determined that:
Therefore:
0*
*
*
0*
*
*
y
h
y
xy
T
y
T
x
k
k
x
k
xhNu
(Reynolds Analogy)
Pr,Re,, **
1
*
LyxgT
Pr,Re,*2
0*
*
*
L
y
x xg
y
TNu
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102
(Reynolds Analogy)
Note: the Nusselt number is equivalent to thedimensionless temperature gradient at the surface, and
this is why it is sometimes called the dimensionless heat
transfer coefficient (h).
Fig 6.30 (text)
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103
(Reynolds Analogy)
The average friction and heat transfer coefficientsare determined by integrating the local CF,xand Nux
over the surface of the given body with respect to x*
(from 0 to 0.1), which removes the dependence on x*
and thus gives:
These relations allow experimenters to study aproblem with a minimum amount of experiments and
report their results in terms of just Re and Pr.
Pr,ReRe 34 LLF gNuandfC
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104
(Reynolds Analogy)
The experimental data for heat transfer is oftenrepresented (with reasonable accuracy) by a simple
power law relation of the form:
Where m and n are constant exponents (normally between 0
and 1), and the value of C depends on geometry.
nmLCNu PrRe
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105
(Reynolds Analogy)
Summary, so far (Fig 6-31 in text book):
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106
(Reynolds Analogy)
Now if we simplify the momentum and energyequations by assuming:
Pr = 1 (which is approximately true for gases)
(true when u = u= V= constant)0*
*
x
P
For Pr = 1, the
thermal andvelocity boundary
layers coincide
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107
(Reynolds Analogy)
The equations then become:
Momentum:
Energy:
Note: These two equations are exactly in the same
form for u* and T*.
2*
*2
*
*
*
*
*
*
2*
*2
*
*
*
*
*
*
Re
1
Re
1
y
T
y
Tv
x
Tu
y
u
y
uv
x
uu
L
L
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108
Since the boundary conditions are also identical:
Recall:
Then:
(Reynolds Analogy)
1,1,00,00,
1,01,0
****
****
****
xTxu
xTxu
yTyu
0*
*
*
0*
*
*
yy y
Tyu Equation *
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109
(Reynolds Analogy)
Since as previously derived:
Rearranging these equations gives:
*
*
*
*
Re
2
y
T
L
khand
y
uC
F
x
y
xF
y
NukLh
yTandC
yu
0*
*
*,
0*
*
*
2Re
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110
(Reynolds Analogy)
Therefore substituting these values into Equation*gives:
0*
*
*
0*
*
*
yy y
T
y
u
2
2
Re
,
,
xF
x
xF
x
CSt
or
CNu
Reynolds Analogyfor
Pr = 1
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111
(Stanton Number)
Reynolds Analogy can also beexpressed in terms of the Stanton
number (St).
This was derived by Sir ThomasEdward Stanton (1865-1931) from
England
PrRe
Nu
VC
hSt
P
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112
(Reynolds Analogy)
Reynolds Analogy is important because it allows us
to determine the heat transfer coefficient (h) for fluids
where Pr = 1, from knowledge of the friction
coefficient (which is easier to measure).
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113
(Chilton-Colburn Analogy)
However, the Reynolds number is of limited usebecause of the restrictions:
Pr = 1
Therefore it is desirable to have an analogy that is
applicable over a wide range of Pr.
This is done by adding a Prandtl number correction.
0*
*
x
P
Forced Convection(Chilton Colburn Analogy)
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114
(Chilton-Colburn Analogy)
Recall as previously derived:
Taking their ratio and rearranging give the relation
known as the Chilton-Colburn analogy or the modifiedReynold's analogy:
21
31
21
RePr332.0Re664.0, xxxxF NuandC
HLx
xFjNu
C
1,RePr
2
31
32
Pr2
,
VC
hCj
p
xxF
H
For 0.6 < Pr < 60
Colburn j-factor
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115
(Chilton-Colburn Analogy)
The Chilton-Colburn Analogy is derived using: Laminar flow
Over a flat plate ( )
However, experimental studies however show that it is also
approximately applicable to turbulent flow over a surface in
the presence of pressure gradients.
For laminar flow it is not applicable unless it is a flat plate,
therefore it cannot be applied to laminar flow in a pipe.
Also the analogy above can be used for local or average
quantities.
0
x
P
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116
(Example 3.4)
Example 3.4 Laminar flow profileover a vertical plate. A 2 x 3 m plate
is suspended in a room and subject
to air flow parallel to its surfaces
along its 3 m side. The total dragforce acting on the plate is 0.86 N.
Determine the average heat transfer
coefficient (h) for the plate:
The properties of air at 1 atm (Table A-15 in
text book) at Tfilm= 20C:
3 m
2 m
Air FlowT
= 15C
V
= 7 m/s
7309.0Pr
007.1;204.13
kg
kJC
m
kgp
Ts=25C
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117
(Example 3.4)
Set L= 3 m ~ Characteristic length
Since both sides of the plate are exposed to the air (and
considering the thickness negligibly small) the total surface area
is:
212322
2
mmm
LwAs
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118
For all flat plates:
Drag = Frict ion Force
Therefore:
(Example 3.4)
00243.0712204.1
86.022222
3
sm
m
kgs
Fm
N
VA
DC
221
VACDF sFfriction
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119
(Example 3.4)
Then from the modified Reynolds analogy (Chilton-Colburn) the average heat transfer coefficient (h) can
be calculated:
Cm
W
CkgJ
sm
m
kg
pF CVCh
2
32
3
32
7.127309.0
10077204.1
2
00243.0
Pr2
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120
3.6 Convection in an
internal flow
Internal Flow
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121
Internal flow relates to flow through fixed conduitssuch as pipes or ducts.
Internal Flow(Non-Circular Tubes)
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122
(Non-Circular Tubes)
For flow through non-circular tubes Re and Nu,
are based on the hydraulic
diameter Dh.
Where p is the perimeter, Vmis
the mean velocity, and Acis the
cross-sectional area.
hm
ch
DV
pAD
Re
4
Internal Flow(Mean Velocity)
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(Mean Velocity)
Because the velocity varies over the cross-section itis necessary to work with a mean velocity (Vm) when
dealing with internal flows.
c
m
mc
A
mV
VAm
Internal Flow(Circular Tubes)
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124
(Circular Tubes)
In a circular tube:
Re < 2,300 laminar flow2,300 < Re < 10,000 transitional flow
Re > 10,000 turbulent flow
DV
DDp
AD
m
D
c
h
Re
444
2
Internal Flow(Entrance Region)
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(Entrance Region)
Internal Flow Equations(Example 3 5)
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(Example 3.5)
Example 3.5 - Temperature rise of oil in a bearing
(a) Find the temperature and velocity distributions
(b) Find the maximum temperature in the oil
(c) Find the maximum heat flux in the oil
u(y)L= 2 mm
V= 12 m/sUpper plate moving
Oil
k= 0.145 W/(mK)
= 0.8 kg/(ms)
Lower plate stationary
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127
(Example 3.5)
Assumptions: Steady operating conditions
Oil is incompressible with constant properties
Body forces such as gravity are negligible
The plates are large, so no variation in the z-direction
u(y)L= 2 mm
V= 12 m/sUpper plate moving
Oilk= 0.145 W/(mK)
= 0.8 kg/(ms)
Lower plate stationary
Internal Flow Equations(Example 3 5)
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128
(Example 3.5)
(a) Find the temperature and velocity distributions Solution:
Flow only in the x-direction v = 0
Continuity Equation:
The x-component of velocity does not change. Since also, the
flow is maintained by the upper plate and not the pressure gradient.
)(
0
0
yuu
x
u
y
v
x
u
0
0
x
P
Internal Flow Equations(Example 3 5)
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129
(Example 3.5)
x-momentum equation:
This is a 2ndorder differential equation. So integrating twicegives:
0 0 000
xgx
P
y
u
x
u
y
vv
x
uu
2
2
2
2
02
2
y
u
21 CyCu
Internal Flow Equations(Example 3.5)
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130
(Example 3.5)
The boundary conditions are:
u(0)= 0
u(L)= V= 12 m/s
Using these boundary conditions to solve for the constants C1and C2gives:
0
00
2
21
C
CC
L
V
C
LCV
1
1 0
Internal Flow Equations(Example 3.5)
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131
(Example 3.5)
Therefore the equation becomes:
Frictional heating due to viscous dissipation in this case is
significant because of the high viscosity of oil and large plate
velocity. The plates are isothermal and there is no change in
flow direction, so the temperature changes with y only T= T(y).
VL
yu
Internal Flow Equations(Example 3.5)
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0 0
0 0 0
(Example 3.5)
So the energy equation for this system is:
2
2
2
0
y
u
y
Tk
0
222
2
2
2
2
2x
v
y
u
y
v
x
u
y
T
x
Tk
y
Tv
x
TuC
p
Internal Flow Equations(Example 3.5)
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133
(Example 3.5)
2
2
2
LV
yTk
Since:
Therefore the equation becomes:
L
V
y
u
VLyu
Internal Flow Equations(Example 3.5)
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134
(Example 3.5)
Now integrating the equation twice:
Applying boundary conditions:
T(0) = T0
T(L) = T0
43
2
2CyCV
L
y
kT
40:0 CTy
2
3
03
2
0
2
2:
VkL
C
TLCVL
L
kTLy
Internal Flow Equations(Example 3.5)
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135
( a p e 3 5)
Substituting these constants in to the equation gives:
2
22
0
0
22
2
2
2
22
L
y
L
y
k
VT
TVkL
yV
L
y
kT
Internal Flow Equations(Example 3.5)
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136
( p )
(b) Find the maximum temperature in the oilThe temperature gradient is found by differentiating T(y) with
respect to y.
Now to find the maximum temperature, maximize T by setting
the above equation equal to 0.
021
2
2
L
y
kL
V
y
T
mmL
y
L
y
001.02
002.0
2
21
Internal Flow Equations(Example 3.5)
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137
( p )
This means that the maximum temperature will occur at the mid-plane (y= 1 mm), which is not surprising since both planes are
maintained at the same temperature.
The maximum temperature at y= 1 mm is:
C
WC
k
VT
LLk
VTT
smN
CmW
sm
m
sN
LL
1191
1
145.08
128.020
8
2
2
2
0
2
2
22
2
0max
2
Internal Flow Equations(Example 3.5)
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138
( p )
(c) Find the maximum heat flux in the oilThe heat flux at the plates is determined from the definition of a
heat flux.
2
22
2
0
0
800,28
1
1
002.02
128.0
2
212
2
m
W
W
mL
V
L
y
kL
V
kdy
dT
kq
s
mN
sm
m
sN
y
Internal Flow Equations(Example 3.5)
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139
( p )
As a check, we can also calculate the heat flux at y= L (shouldbe equal but opposite sign).
2
22
2
800,28
1
1
002.02
128.0
2
21
2
2
m
W
W
mL
V
L
L
kL
Vk
dy
dTkq
smN
sm
m
sN
Ly
L
Correct !
Internal Flow Equations(Example 3.5)
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140
( p )
Discussion of example
A temperature rise of 99C confirms that viscous dissipation is
very significant
T=119CL= 2 mm
V= 12 m/sUpper plate moving
Lower plate stationaryT=20C
T=20C
Internal Flow Equations(Example 3.5)
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141
( p )
Discussion of example
Heat flux is equivalent to the mechanical energy rate ofdissipation. Therefore, mechanical energy is being convertedinto thermal energy to overcome friction in oil. This accounts forthe temperature flux.
L= 2 mm
V= 12 m/s
2
8.28
m
kWq
28.28m
kWq
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3.7 Free (natural) convection
Free Convection
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Hot air rises due to the buoyancyeffect.
This causes fluid motion
(possibly in a circulating pattern)
that causes natural or freeconvection
Warm air
Cold
can Cold air
Heat
Transfer
Free Convection(Volume Expansion Coefficient)
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In heat transfer, the primary variable is thetemperature, so it is desirable to express the net
buoyancy force in terms of a temperature difference.
This requires knowledge of a property that represents the
variation of the density of a fluid with temperature at constantpressure.
This is called the volume expansion coefficient () which is
defined as:
PP TT
11
Free Convection(Volume Expansion Coefficient)
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In natural convection studies, the condition of the fluidsufficiently far from the hot or cold surface is indicated by the
subscript
to indicate that the presence of the surface is not
felt.
In such cases, can be expressed approximately by replacing
the differential equations by differences, such as:
TTT
11
TT
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For an ideal gas:
Thus for an ideal gas the discharge coefficientbecomes:
TR
P
TTT
RTP
RTP
P
111
Free Convection(Grashof Number)
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The velocity and temperature for naturalconvection over a vertical plate are
shown in the figure.
As in forced convection, the
boundary layer thickness increases
in the flow direction
Unlike forced convection, the fluid
velocity (u) is 0 at the outer edge of
the boundary layer as well as the
surface of the plate.
This is expected since the fluid
beyond the boundary layer is
motionless.
Free Convection(Grashof Number)
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Recall that the x-momentum equations is:
Now the momentum equation outside the boundary layer can be
obtained from this relation as a special case by setting u = 0,
giving:
gx
P
y
u
y
uv
x
uu
2
2
gx
P
Free Convection(Grashof Number)
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Since:
Then the momentum equation becomes:
g
x
P
x
PxPxPP
TTgy
u
y
vv
x
uu
TTgy
u
y
v
vx
u
u
gy
u
y
vv
x
uu
2
2
2
2
2
2
EQN 9-13
in text
Free Convection(Grashof Number)
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If we now non-dimensionalize this x-momentumequation, we get:
2*
*2
2
*
2
3
*
*
*
*
*
*
Re
1
Re y
uTLTTg
y
u
vx
u
uLL
cs
Grashof Number
Free Convection(Grashof Number)
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The Grashof number is derived byFranz Grashof (1826-1893) from
Germany.
2
3
csL
LTTgGr
Free Convection(Grashof Number)
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Gr is a measure of the relativemagnitudes of the buoyancy force
and the opposing viscous force
acting on the fluid
Free Convection(Raleigh Number)
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Lord Raleigh (1842-1919) fromEngland derived the Raleigh Number
PrGrRa
Pr
2
3
csL
LTTgRa
Free Convection(Example 3.6)
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Example 3.6 A 6-m long section of 8-cm diameterhorizontal hot water pipe passes through a large
room. The pipe surface temperature is 70 C.
Determine the heat loss from the pipe by natural
convection.
D= 8 cm
L= 6 m
Ts= 70 CT
= 20 C
Free Convection(Example 3.6)
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Assume: Steady operating conditions
Air is an ideal gas
The local atmospheric pressure is 1 atm
From Table A-15, the properties of air are:
7241.0Pr;10749.1;02699.0sec
5 2 m
CmWk
C
CCTTT sfilm
45
2
2070
2
Free Convection(Example 3.6)
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The volumetric expansion coefficient () is:
The characteristic length is the outer diameter of the
pipe:
KCTf 318
1
27345
11
mDLc 08.0
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Therefore the Raleigh Number is:
6
25
3
3181
2
3
10869.1
10749.1
7241.008.029334381.9
Pr
2
2
sm
Ks
m
sD
mKK
DTTgRa
Free Convection(Example 3.6)
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Table 9-1 in the text book gives average Nusseltnumbers for natural convection over surfaces.
For a horizontal cylinder:
Free Convection(Example 3.6)
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Thus Nu is:
4.17
7241.0
559.01
10869.1387.060.0
Pr
559.01
387.060.0
2
6
2
278
169
61
278
169
61
DD
RaNu
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Then:
The surface area of the cylinder is:
Cm
WCmW
mNu
D
kh
2869.54.1708.0
02699.0
2508.1608.0 mmmLDAs
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Therefore the heat transfer is:
W
CCm
TTAhQ
Cm
W
ss
5.442
2070508.1869.5 2
End Of Convection SectionC C
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