CHAPTER 6
HEAT TRANSFER IN CHANNEL FLOW
6.1 Introduction
(1) Laminar vs. turbulent flow
Flow through tubes, transition Reynolds number Re D t is
Re Dt
uD
2300
(6.1)
(2) Entrance vs. fully developed region
Classification based on velocity and temperature profiles:
(i) Entrance region
(ii) Fully developed region
(3) Surface boundary conditions
Two common boundary conditions::
(i) Uniform surface temperature
(ii) Uniform surface heat flux
(4) Objective.
Objective depends on surface thermal boundary condition:
(i) Uniform surface temperature. Determine axial variation of:
(1) Mean fluid temperature
(2) Heat transfer coefficient
(3) Surface heat flux
(ii) Uniform surface flux. Determine axial variation of:
(1) Mean fluid temperature
(2) Heat transfer coefficient
(3) Surface temperature
6.2 Hydrodynamic and Thermal Regions: General Features
Fluid enters with uniform velocity and temperature.
(1) Entrance region. Extends from the inlet to the section where the boundary layer
thickness reaches the center of channel.
(2) Fully developed region. This zone follows the entrance region.
2
6.2.1 Flow Field
(1) Entrance Region (Developing Flow, 0
x
Lh ).
Name: hydrodynamic entrance region.
Length: L h (hydrodynamic entrance length).
Streamlines are not parallel.
Core velocity u c increases with distance
Pressure decreases with distance ( dp / dx 0 ).
D/ 2.
r
uc
Vi
x
u
u
Lh
(2) Fully Developed Flow Region. x
fully developed
Lh
Fig. 6.1
Streamlines are parallel ( vr 0).
u / x 0 for two-dimensional incompressible fluid.
6.2.2 Temperature Field
(1) Entrance Region (Developing Temperature, 0
x
Lt )
Name: Thermal entrance region.
Length: Lt (thermal entrance length).
Core temperature Tc is uniform, Tc Ti .
D/2
t
(2) Fully Developed Temperature Region. x
r
Tc
Ts
Ts
T
Vi
Ti
Lt
x
Ts
t
Temperature varies radially and axially,
T / x 0.
fully developed
Lt
Fig. 6.2
6.3 Hydrodynamic and Thermal Entrance Lengths
6.3.1 Scale Analysis
(1) Hydrodynamic Entrance Length Lh .
Starting with external flow result (4.16)
1
x
Applying (4.16) to a tube at x
(4.16)
Re x
Lh :
D and Re Lh
Substituting (b) into (4.16) and rearranging
Re D
Lh
D
(b)
3
1/ 2
Lh / D
Re D
(6.2)
~1
(2) Thermal Entrance Length Lt .
Starting with external flow result (4.24)
t
Applying (4.24) at x
~ L Re L1 / 2 Pr
1/ 2
(4.24)
Lt :
t
D and Re Lt
Re D
Lt
D
(b)
Substituting (b) into (4.24) and rearranging
Lt / D
Re D Pr
1/ 2
(6.3)
~1
(6.2) and (6.3) give
Lt
~ Pr
Lh
(6.4)
6.3.2 Analytic and Numerical Solutions: Laminar Flow
(1) Hydrodynamic Entrance Length L h .
Results for L h :
Lh
De
Table 6.1
Entrance length coefficients C h and C t [1]
(6.5)
C h Re D e
Ct
Table 6.1 gives C h
Compare with scaling:
Lh / D
Re D
geometry
uniform
surface flux
uniform
surface
temperature
0.056
0.043
0.033
0.09
0.066
0.041
0.085
0.057
0.049
0.075
0.042
0.054
0.011
0.012
0.008
1/ 2
(6.2)
~1
a
a
b
Rewrite (6.5)
a/b =1
a
1/ 2
Lh / De
Re De
Ch
Ch
1/ 2
(a)
Example: Rectangular channel,
aspect ratio 2, Table 6.1 gives
C h 0.085. Substituting this value
into (a), gives
a/b = 2
b
a
b
a/b = 4
4
1/ 2
Lh / De
Re D e
0.085 1 / 2
(b)
0.29
Scaling replaces 0.29 by unity.
(2) Thermal Entrance Length Lt .
Lt depends on surface boundary conditions: Two cases: (i) Uniform surface
temperature. (ii) Uniform surface flux.
Solution
Lt
De
(6.6)
C t PrRe D
Table 6.1 gives C t for both cases.
Compare with scaling. Rewrite (6.6)
Lt / De
PrRe D
1/ 2
Ct
1/ 2
(c)
Scaling gives
1/ 2
Lt / D
Re D Pr
(6.3)
~1
Example: Rectangular channel, aspect ratio 2, Table 6.1 gives Ct
this value into (c), gives
0.049. Substituting
1/ 2
Lt / De
PrRe De
0.049 1 / 2
(d)
0.22
Scaling replaces 0.22 be unity.
Turbulent flow: L
Lh
Lt
L
D
(6.7)
10
6.4 Channels with Uniform Surface Heat Flux q s
Inlet mean temperature: Tmi
Tm (0) .
L
Determine:
(1) Total heat transfer rate q s .
(2) Mean temperature variation Tm (x).
(3) Surface temperature variation Ts (x ).
Total heat transfer rate
Tmi
0
x
qs
Fig. 6.3
Tm (x)
5
qs
q s As
(6.8)
qs P x
As = surface area
P = perimeter
Mean temperature Tm (x) . Conservation of energy between inlet and section x:
qs
qs P x
mcp [Tm ( x) Tmi ]
or
Tm ( x)
qs P
x
mc p
Tmi
(6.9)
Surface temperature Ts (x) . Newton’s law of cooling gives
qs
h( x) Ts ( x) Tm ( x)
or
Ts ( x)
qs
h( x )
Tm ( x)
Using (6.9)
Ts ( x)
Tm i
qs
Px
mc p
1
h( x )
(6.10)
NOTE:
Determining Ts (x) requires knowing h(x).
To determine h(x): Must know if:
Flow is Laminar or turbulent.
Entrance or fully developed region
6.5 Channels with Uniform Surface Temperature
Inlet mean temperature: Tmi
Tm (0) .
Ts
Determine:
(1) Mean temperature variation Tm (x).
(2) Total heat transfer rate q s between x
location x.
(3) Surface heat flux variation q s (x).
Tmi
0
dx
dqs
Tm
Tm
Conservation of energy to element
m c p dTm
m
0 and
Mean temperature variation Tm (x).
dqs
Tm (x)
x
dx
(a)
Fig. 6.4
dTm
dx
dx
6
Newton's law:
dqs
h( x) Ts
Tm ( x) Pdx
(b)
P
h( x)dx
m cp
(c)
Combine (a) and (b)
dTm
Ts Tm ( x)
Integrating (c)
x
T ( x) Ts
ln m
Tmi Ts
P
m cp
Definite h
(6.11)
h( x)dx
0
x
h
1
x
h( x)dx
(6.12)
0
(6.12) into (6.11), solve Tm (x)
Tm ( x)
(Tmi Ts ) exp[
Ts
Ph
x]
mcp
(6.13)
NOTE:
Determining Tm (x) requires knowing h(x).
To determine h(x): Must know if:
Flow is laminar or turbulent.
Entrance or fully developed region
Heat transfer rate. Conservation of energy:
qs
(6.14)
m c p [Tm ( x) Tmi ]
Surface heat flux.: Newton’s law:
q s ( x)
h( x)[Ts
Tm ( x)]
(6.15)
6.6 Determination of Heat Transfer Coefficient h(x) and Nusselt Number Nu D
6.6.1 Scale Analysis
Estimate h(x) and Nu D .
Tube: radius ro , surface temperature T s , mean temperature Tm .
Fourier’s law and Newton’s law:
h
Scaling (6.16)
T (ro , x)
k
r
Tm Ts
r
qs
ro
0
Tm
(6.16)
Fig. 6.5
Ts
7
Tm
k
Ts
t
h~
Tm
Ts
or
k
h~
(6.17)
t
The Nusselt number
hD
k
Nu D
Use (6.17)
D
Nu D ~
(6.18)
t
Fully developed region:
t (x ) ~
D, equation (6.18) gives
Nu D ~ 1 (fully developed)
Entrance region: Need to scale
t
(
t (x ) <
(6.19)
D).
For external flow
t
~ x Pr
1/ 2
Re x 1/ 2
(4.24)
(4.24) into (6.18)
Nu D ~
D 1 / 2 1/2
Pr Re x
x
(c)
Expressing Re x in terms of Re D
Re x
ux
ν
uD x
ν D
Re D
x
D
(d)
Substitute (d) into (c)
D
Nu D ~
x
1/2
Pr 1 / 2Re1/2
D
(6.20a)
Rewrite
Nu D
PrRe D
x/D
1/ 2
~1
Scaling estimates (6.19) and (6.20) will be compared with exact solutions.
(6.20b)
8
6.6.2 Basic Considerations for the Analytical Determination of Heat Flux, Heat
Transfer Coefficient and Nusselt Number
r
qs
Need to determine velocity and temperature distribution.
Assume: fully developed velocity
Neglect axial conduction
Section outline:
ro
0
Tm
Definitions
Governing equations for determining:
Ts
Fig. 6.5
(i) Surface heat flux
(ii) Heat transfer coefficient
(iii) Nusselt number
(1) Fourier’s law and Newton’s law.
Surface heat flux. Fourier’s law gives surface heat flux q s
qs
k
T x, ro
r
(a)
Define dimensionless variables
T Ts
,
Ti Ts
x/D
,
Re D Pr
r
,
ro
R
vx
v
k
Ts
ro
Ti
vx
,
u
vr
vr
,
u
Re D
uD
ν
(6.21)
Substitute into (a)
qs ( )
0( ,1)
R
(6.22)
Heat transfer coefficient. Define h
h
q" s
Tm Ts
(6.23)
Combine (6.22) and (6.23)
h( )
where
m
k (Ts Ti )
ro (Tm Ts )
( ,1)
R
k
ro
1
m( )
( ,1)
R
(6.24)
is defined as
m
Tm Ts
Ti Ts
(6.25)
Nusselt number. Define:
Nu ( )
h( ) D
k
h( )2ro
k
(6.26)
( ,1)
R
(6.27)
(6.24) into (6.26)
Nu ( )
2
m( )
9
(2) The Energy Equation. Review assumptions on energy equation (2.24).
T
r
c p vr
T
z
vz
k
2
1
T
r
r r
r
T
(2.24)
z2
Replace z by x, use dimensionless variables:
vx
2 Re D Pr vr
4
R
R R
R
R
1
( Re D Pr) 2
2
2
(6.28)
where
Pe
Re D Pr , Peclet number
(6.29)
Pe
(6.30)
Neglect conduction for
PrRe D
100
Thus, under such conditions, (6.28) becomes
vx
2 Re D Pr vr
4
R
R R
R
R
(6.31)
(3) Mean (Bulk) Temperature Tm . Define:
ro
mc p Tm
c p v x T 2 rdr
(a)
v x 2 rdr
(b)
0
Mass flow rate m is given by
ro
m
0
(b) into (a), assume constant properties
ro
v x Trdr
Tm
0
(6.32a)
ro
v x rdr
0
Dimensionless form:
1
v x R dR
m
Tm Ts
Ti Ts
0
1
v x R dR
0
6.7 Heat Transfer Coefficient in the Fully Developed Temperature Region
6.7.1 Definition of Fully Developed Temperature Profile
(6.32b)
10
Far away from the entrance ( x / d
fully developed.
0.05Re D Pr ), temperature profile becomes
To define fully developed temperature, introduce the dimensionless temperature
Ts ( x ) T ( r , x )
Ts ( x) Tm ( x)
(6.33)
Fully developed temperature is defined as a profile in which
is independent of x:
(r )
(6.34)
0
(6.35)
(6.34) gives
x
(6.33) and (6.35) give
Ts ( x) T (r , x)
x Ts ( x) Tm ( x)
x
Expand and use the definition of
(6.36a)
0
in (6.33)
dTs
dx
T
x
(r )
dTs
dx
dTm
dx
0
(6.36b)
6.7.2 Heat Transfer Coefficient and Nusselt Number
Examine h and Nu in the fully developed region.
Fourier’s and Newton’s law:
T (ro , x)
r
Tm Ts
k
h
(6.16)
Use (6.33) to eliminate T (ro , x) / r . (6.16) gives
h
k
d (ro )
= constant
dr
(6.37)
IMPORTANT CONCLUSOIN:
THE HEAT TRANSFER COEFFICIENT IN THE FULLY DEVELOPED REGION
IS CONSTANT INDEPENDET OF LOCATION.
Nusselt number
Nu D
hD
k
D
d (ro )
dr
(6.38)
Scaling estimate based on limiting case of entrance region:
Nu D ~ 1 (fully developed)
Scale estimate based on fully developed region:
(6.19)
11
Scale T (ro , x) / r as
T (ro , x) Ts Tm
~
r
D
Substitute into (6.16)
h~
k
D
(6.39)
Substitute (6.39) into (6.38)
(fully developed)
Nu D ~ 1
(6.40)
6.7.3 Fully Developed Region for Tubes at Uniform Surface flux
r
Determine:
qs
T
(i) Surface temperature Ts (x).
(ii) Heat transfer coefficient.
u
0
D
x
qs
Fig. 6.6
Newton’s law
qs
h Ts ( x) Tm ( x)
(a)
Since q s and h are constant it follows that
Ts ( x) Tm ( x) constant
(b)
Differentiate
dTs
dx
dTm
.
dx
(c)
dTs
dx
(d)
(c) into (6.36b)
T
x
(c) and (d)
T
x
dTs
dx
dTm
(for constant q s )
dx
(6.41)
Unknowns: T (r, x), Tm (x) and Ts (x)
Conservation of energy:
q s Pdx mc pTm
or
mc p Tm
m
dTm
dx
dx
Tm
Tm
dx
qs
Fig. 6.7
dTm
dx
dx
12
dTm
dx
qs P
= constant
mc p
(6.42)
dTs
dx
dTm q s P
=
= constant
dx mc p
(6.43)
Substitute (6.42) into (6.41)
T
x
Integrate(6.43)
qs P
x C1
mc p
Tm ( x)
(e)
Use inlet condition
Tm (0)
Tmi
(f)
Solution (e) becomes
Tm ( x)
qs P
x
mc p
Tmi
(6.44)
Need to determine T (r, x) and Ts (x). This requires solving the differential form of
the energy equation.
Set vr
0 in energy equation (2.24)
T
x
c p vx
k
T
r
r r
r
(6.45)
Fully developed flow axial velocity
vx
r2
2u 1
ro2
(6.46)
(6.43) and (6.46) into (6.45)
c p 2u 1
However, m
ro2 u and P
r2
2
ro
qs P
m c p
k
T
r
r r
r
(g)
2 ro , equation (g) becomes
4q s
r2
1 2
ro
ro
k
T
r
r r
r
(6.47)
Boundary conditions are:
T (0, x)
0
r
T (ro , x)
k
qs
r
Integrate (6.47)
(6.48a)
(6.48b)
13
4
r2
qs
ro
2
r4
T
r
kr
2
4ro
Boundary condition (6.48a) gives f (x)
(h)
f x
0. Equation (h) becomes
4q s
kro
T
r
r3
r
2
4ro2
Integrate again
r2
4
4q s
kro
T (r , x)
r4
(6.49)
g ( x)
16ro2
The integration “constant” is g (x) . Use Tm (x) to determine g (x). Substitute (6.46) and
(6.49) into (6.32a)
7 ro q s
(6.50)
Tm ( x)
g ( x)
24 k
Equate (6.44) and (6.50) gives g (x)
g ( x)
Tmi
7 ro q s
24 k
Pq s
x
mc p
r4
16ro
7 ro q s
24 k
11 ro q s
24 k
Pq s
x
mc p
(6.51)
(6.51) into (6.49)
T (r , x) Tmi
Set r
4q s
kro
r2
4
2
Pq s
x
mc p
(6.52)
ro in (6.52) to obtain Ts (x)
Ts ( x)
Tmi
(6.53)
(6.44), (6.52) and (6.53) into (6.33) gives (r )
24 1 2
(r ) 1
r
11 ro2
r4
24 Pq s
7
x
x
11 mc p
11
2
4ro
(6.54)
Differentiate (6.54) and substitute into (6.38) gives
Nu D
48
11
4.364
(6.55)
NOTE:
(6.55) applies to laminar fully developed velocity and temperature in tubes with
uniform surface heat flux.
The Nusselt number is independent of Reynolds and Prandtl numbers.
Scaling gives Nusselt as
Nu D ~ 1
(6.40)
14
This compares favorable with (6.55).
6.7.4 Fully Developed Region for Tubes at Uniform Surface Temperature
Determine: Nusselt number
Solve the energy equation for the fully developed region
Neglect axial conduction and dissipation.
Energy equation: set vr 0 in (2.24)
c p vx
T
x
k
T
r
r r
r
(6.45)
Boundary conditions
T (0, x)
0
r
T (ro , x) Ts
(6.56a)
(6.56b)
Axial velocity for fully developed flow is
vx
2u 1
r2
(6.46)
ro2
Use (6.36a) to Eliminate T / x in (6.45)
Ts ( x) T (r , x)
x Ts ( x) Tm ( x)
x
For uniform Ts ( x)
0
(6.36a)
Ts , above gives
T
x
Ts T (r , x) dTm
Ts Tm ( x) dx
(6.57)
(6.46) and (6.57) into (6.45)
2 cpu 1
r 2 Ts T (r , x) dTm
ro2 Ts Tm ( x) dx
k
T
r
r r
r
(6.58)
Solution: (6.58) was solved using an infinite power series. Solution gives the Nusselt
number as
Nu D
3.657
(6.59)
6.7.5 Nusselt Number for Laminar Fully Developed Velocity and Temperature in
Channels of Various Cross-Sections
Table 6.2 lists Nusselt numbers for channels of various cross-sections.
Two cases: (1) uniform surface heat flux and (2) uniform surface temperature.
Nusselt number of Non-circular channels is based on the equivalent diameter.
Scaling estimate:
Table 6.2
Nu D ~ 1 (fully developed)
(6.40)
Nusselt number for laminar fully developed
conditions in channels of various cross-sections [3]
Nusselt number Nu D
Table 6.2: Nusselt number ranges from 2.46
to 8.235.
a
b
Channel geometry
Uniform
surface
flux
Uniform
surface
temperature
4.364
3.657
1
3.608
2.976
2
4.123
3.391
4
5.331
4.439
8
6.49
5.597
8.235
7.541
3.102
2.46
6.8 Thermal Entrance Region: Laminar Flow
through Tubes
a
6.8.1 Uniform Surface Temperature: Graetz
Solution
b
a
Laminar flow.
b
Fully developed inlet velocity.
b
a
Neglect axial conduction (Pe > 100).
a
Uniform surface temperature Ts .
b
Fully developed flow:
vr
0
15
(3.1)
Axial velocity
vz
1 dp 2 2
(r ro )
4 dz
(3.12)
r
Ts
(3.12) expressed in dimensionless form
vx
vx
u
2
2(1 R )
T
Ti
(6.61)
u
0
x
t
(3.1) and (6.61) into energy equation (6.31)
1
1 R2
2
1
R
R R
R
Fig. 6.8
(6.62)
Boundary conditions
( ,0)
R
0
(6.63a)
( ,1)
0
(6.63b)
(0, R) 1
(6.63c)
Analytic and numerical solutions to this problem have been obtained.
Review analytic solution leading to:
16
(i) Mean temperature,
m(
)
Gn
8
m
2
2
exp( 2
(6.66)
)
n
n
n 0
(ii) Local Nusselt number, Nu ( )
G n exp( 2
2
n
)
n 0
Nu
(66.7)
Gn
2
2
n 0
exp( 2
2
n
)
n
(iii) Average Nusselt number, Nu ( )
h ( )D
k
(f)
and G n for 0
n 10. Table 6.4 gives Nu ( ) and
Nu ( )
RESULTS
Table 6.3 lists values of
n
Nu ( ) at selected values of the axial distance .
Fig. 6.9 gives the variation of Nu ( ) and Nu ( ) along a tube.
Table 6.4
Table 6.3
Uniform surface temperature [4]
n
0
1
2
3
4
5
6
7
8
9
10
n
2.70436
6.67903
10.67338
14.67108
18.66987
22.66914
26.66866
30.66832
34.66807
38.66788
42.66773
Gn
0.74877
0.54383
0.46286
0.41542
0.38292
0.35869
0.33962
0.32406
0.31101
0.29984
0.29012
Local and average Nusselt
number for tube at uniform
surface temperature [5]
=
x/D
Re D Pr
0
0.0005
0.002
0.005
0.02
0.04
0.05
0.1
Nu ( )
Nu ( )
12.8
8.03
6.00
4.17
3.77
3.71
3.66
3.66
19.29
12.09
8.92
5.81
4.86
4.64
4.15
3.66
17
Average Nu
Nusselt number
Local Nu
Fig. 6.9
x/D
ReD Pr
Local and average Nusselt number for
tube at uniform surface temepratu re [4]
NOTE:
(1) The average Nusselt number is greater than the local Nusselt number.
(2) Asymptotic value of Nusselt number of 3.657 is reached at
0.05 . Thus
(6.69)
Nu( ) 3.657
(3) Evaluate fluid properties at the mean temperatures Tm , defined as
Tm
Tmi
Tmo
6.8.2 Uniform Surface Heat Flux
Repeat Graetz entrance problem
replacing the uniform surface
temperature with uniform heat flux.
(6.70)
2
r
qs
T
Ti
u
0
D
x
t
qs
Inlet velocity is fully developed.
Fig. 6.10
Energy equation is
1
1 R2
2
1
R
R R
R
(6.62)
18
Boundary conditions
( ,0)
)
R
(6.71a)
0
q s ro
k (Ti Ts )
( ,1)
R
(6.71b)
(6.71c)
(0, R) 1
Solution.
Local Nusselt number:
1
Nu ( )
hx
k
11
48
1
2
An exp( 2
2
Table 6.5
(6.72]
)
n
n 1
n
The average Nusselt number is given by
Nu ( )
hx
k
11
48
1
2
The eigenvalues
in Table 6.5
Limiting case:
An
n 1
2
n
1 exp( 2
2
2
n
1
)
2
n
(6.73]
and the constant An are listed
(fully developed)
Nu ( )
Uniform surface flux [4]
48
4.364
11
1
2
3
4
5
6
7
8
9
10
2
n
25.6796
83.8618
174.1667
296.5363
450.9472
637.3874
855.8495
1106.3290
1388.8226
1703.3279
(6.74)
Average Nu
Nusselt number
Local Nu
x/D
ReD Pr
Fig. 6.11 Local and average Nusselt number for
tube at uniform surface heat flux [4]
An
0.198722
0.069257
0.036521
0.023014
0.016030
0.011906
0.009249
0.007427
0.006117
0.005141