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: Lh (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
Lh
(2) Fully Developed Flow Region. x  Lh


u

u
fully developed
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 .
t  D / 2
(2) Fully Developed Temperature Region. x  Lt

r
Tc
Ts
Ts
T
Vi
Ti
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)

x


1
Applying (4.16) to a tube at x  Lh :
  D and Re Lh  Re D

(4.16)
Re x
Substituting (b) into (4.16) and rearranging
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 ~ L Re L1 / 2 Pr 1 / 2

(4.24)
Applying (4.24) at x  Lt :
 t  D and Re Lt  Re D
Lt
D
(b)
Substituting (b) into (4.24) and rearranging
1/ 2
 Lt / D 


 Re D Pr 

(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 Lh .

Results for Lh :
Lh
 C h Re D e
De


Table 6.1
Entrance length coefficients C h and C t [1]
(6.5)
Ct
Table 6.1 gives C h
Compare with scaling:
geometry
Ch
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
 Lh / D 


 Re D 
~1
(6.2)
a
a
b
Rewrite (6.5)
 L h / De

 Re D
e

a
1/ 2




a/b =1
 C h 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
 Lh / De

 Re
De

1/ 2




 0.0851 / 2  0.29
(b)
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
 Ct PrRe D
De


(6.6)
Table 6.1 gives C t for both cases.
Compare with scaling. Rewrite (6.6)
1/ 2
 Lt / De 


 PrRe D 
 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  0.049. Substituting
this value into (c), gives
 Lt / De

 PrRe D
e





1/ 2
 0.049 1 / 2  0.22
(d)
Scaling replaces 0.22 be unity.

Turbulent flow: L  Lh  Lt
L
 10
D
(6.7)
6.4 Channels with Uniform Surface Heat Flux q s

Inlet mean temperature: Tmi  Tm (0) .

Determine:
(1) Total heat transfer rate q s .
(2) Mean temperature variation Tm (x).
(3) Surface temperature variation Ts (x).

Total heat transfer rate
L
Tmi
0
x
qs
Fig. 6.3
Tm (x)
5
q s  qs As  qs P x
(6.8)
As = surface area
P = perimeter

Mean temperature Tm (x) . Conservation of energy between inlet and section x:
q s  q s P x  mcp [Tm ( x)  Tmi ]
or
Tm ( x)  Tmi 

q s P
x
mc p
(6.9)
Surface temperature Ts (x) . Newton’s law of cooling gives
q s  h( x)Ts ( x)  Tm ( x)
or
Ts ( x)  Tm ( x) 
q s
h( x )
Using (6.9)
 Px
1 
Ts ( x)  Tm i  q s 


c
h( x ) 
 m p

(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) .

Determine:
Ts
(1) Mean temperature variation Tm (x).
(2) Total heat transfer rate q s between x  0 and
location x.
(3) Surface heat flux variation q s (x).

Mean temperature variation Tm (x).
0
x
Tm (x)
m
dqs
(a)
dx
Tm 
Tm
Conservation of energy to element
dq s  m c p dTm
Tmi
dx
Fig.6.4
dTm
dx
dx
6
Newton's law:
dq s  h( x)Ts  Tm ( x)Pdx
(b)
dTm
P

h( x)dx
Ts  Tm ( x) m c p
(c)
Combine (a) and (b)
Integrating (c)
 T ( x)  Ts 
P
ln  m
 c
m p
 Tm i  Ts 

x
h( x)dx
(6.11)
0
Definite h
1
h
x

x
(6.12)
h( x)dx
0
(6.12) into (6.11), solve Tm (x)
Tm ( x)  Ts  (Tm i  Ts ) exp[
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:
q s  m c p [Tm ( x)  Tmi ]

(6.14)
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 Ts , mean temperature Tm .
Fourier’s law and Newton’s law:
T ( ro , x)
k
r
h
Tm  Ts
Scaling (6.16)
r
q s
ro
0
Tm
(6.16)
Fig. 6.5
Ts
7
Tm  Ts 
k
h~
t
Tm  Ts
or
h~
k
(6.17)
t
The Nusselt number
Nu D 
hD
k
Use (6.17)
Nu D ~

D
(6.18)
t
Fully developed region:  t (x) ~ D, equation (6.18) gives
Nu D ~ 1 (fully developed)

(6.19)
Entrance region: Need to scale  t (  t (x) < D).

For external flow
 t ~ x Pr 1/ 2 Re x1/ 2

(4.24) into (6.18)
Nu D ~

(4.24)
D 1 / 2 1/2
Pr Re x
x
(c)
uD x
x
 Re D
ν D
D
(d)
Expressing Re x in terms of Re D
Re x 
ux
ν

Substitute (d) into (c)
D
Nu D ~  
x
1/2
Pr 1 / 2Re 1/2
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




Need to determine velocity and temperature distribution.
Assume: fully developed velocity
Neglect axial conduction
Section outline:


q s
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
q s  k
T x, ro 
r
(a)
Define dimensionless variables

T  Ts
,
Ti  Ts

x/D
,
Re D Pr
R
r
,
ro
Substitute into (a)
q s ( ) 

v x  v
vx
,
u
v r 
vr
,
u
k
Ts  Ti  0( ,1)
ro
R
Re D 
uD
ν
(6.21)
(6.22)
Heat transfer coefficient. Define h
q"s
Tm  Ts
(6.23)
k (Ts  Ti )  ( ,1)
k
1  ( ,1)

ro (Tm  Ts ) R
ro  m ( ) R
(6.24)
h  
Combine (6.22) and (6.23)
h( ) 
where  m is defined as
m 

Tm  Ts
Ti  Ts
(6.25)
Nusselt number. Define:
Nu( ) 
h( ) D h( )2ro

k
k
(6.26)
 2  ( ,1)
 m ( ) R
(6.27)
(6.24) into (6.26)
Nu ( ) 
9
(2) The Energy Equation. Review assumptions on energy equation (2.24).


 c p  vr
 1   T   2T 
T
T 
 vz
  k
r


r
z 
 r r  r  z 2 
(2.24)
Replace z by x, use dimensionless variables:
v x

 4    
1
 2
 2 Re D Pr vr

R




R R R  R  ( Re D Pr) 2  2
(6.28)
where

Pe  Re D Pr , Peclet number
(6.29)
Pe  PrRe D  100
(6.30)
Neglect conduction for
Thus, under such conditions, (6.28) becomes
v x

 4    
 2 Re D Pr v r

R


R R R  R 
(6.31)
(3) Mean (Bulk) Temperature Tm . Define:

ro

ro
mc p Tm 
 c p v x T 2rdr
(a)
0
Mass flow rate m is given by
m
 v x 2rdr
(b)
0
(b) into (a), assume constant properties
Tm 


ro
v x Trdr
0
(6.32a)
ro
v x rdr
0
Dimensionless form:
1
m 
Tm  Ts

Ti  Ts


v x R dR
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  0.05Re D Pr ), temperature profile becomes
fully developed.

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
x
(6.35)

  T ( x)  T ( r , x) 
  s
0
x x  Ts ( x)  Tm ( x) 
(6.36a)
(6.34) gives
(6.33) and (6.35) give
Expand and use the definition of  in (6.33)
dTs T
dT 
 dT

  (r )  s  m   0
dx x
dx 
 dx
(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 

d (ro )
hD
 D
k
dr
Scaling estimate based on limiting case of entrance region:
Nu D ~ 1 (fully developed)

(6.38)
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)
Nu D ~ 1 (fully developed)
(6.40)
6.7.3 Fully Developed Region for Tubes at Uniform Surface flux

r
Determine:
qs
T
(i) Surface temperature Ts (x).
(ii) Heat transfer coefficient.
u
0
D
x
qs
Fig. 6.6

Newton’s law
q s  hTs ( x)  Tm ( x)
(a)
Since q s and h are constant it follows that
Ts ( x)  Tm ( x)  constant
(b)
Differentiate
dTs dTm
.

dx
dx
(c)
(c) into (6.36b)
T dTs

x
dx
(d)
(c) and (d)
T dTs dTm
(for constant q s )


x
dx
dx

Unknowns: T ( r , x ), Tm (x) and Ts (x)

Conservation of energy:
dT


q s Pdx  mc pTm  mc p Tm  m dx 
dx


or
m
(6.41)
Tm 
Tm
dx
qs
Fig. 6.7
dTm
dx
dx
12
dTm q s P

= constant
dx
mc p
(6.42)
T dTs dTm q s P
=
= constant


x
dx
dx mc p
(6.43)
Substitute (6.42) into (6.41)
Integrate(6.43)
Tm ( x) 
q s P
x  C1
mc p
(e)
Use inlet condition
Tm (0)  Tmi
Solution (e) becomes
Tm ( x)  Tmi 
q s P
x
mc p
(f)
(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)
 c p vx
T k   T 

r

x r r  r 
(6.45)
Fully developed flow axial velocity
 r2 
v x  2u 1  2 
 ro 
(6.46)
(6.43) and (6.46) into (6.45)

 c p 2u 1 

r 2  q s P k   T 

r


ro2  m c p r r  r 
(g)
However, m   ro2  u and P  2 ro , equation (g) becomes
4q s  r 2  k   T 
r

1   
ro  ro2  r r  r 
Boundary conditions are:
Integrate (6.47)
T (0, x)
0
r
T (ro , x)
k
 q s
r
(6.47)
(6.48a)
(6.48b)
13
r 2
4
r4 
T
q s   2   kr
 f x 
ro
r
 2 4ro 
(h)
Boundary condition (6.48a) gives f ( x)  0. Equation (h) becomes
T 4q s  r r 3 

 

r
kro  2 4ro2 
Integrate again
T (r , x) 
r2
r4 
 
  g ( x)
2
 4 16ro 
4q s
kro
(6.49)
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 )
7 ro q s Pq s

x
24 k
mc p
(6.51)
4q s  r 2
r 4  7 ro q s Pq s

x
 

kro  4 16ro2  24 k
mc p
(6.52)
11 ro q s Pq s

x
24 k
mc p
(6.53)
g ( x)  Tmi 
(6.51) into (6.49)
T (r , x)  Tmi 
Set r  ro in (6.52) to obtain Ts (x)
Ts ( x)  Tmi 
(6.44), (6.52) and (6.53) into (6.33) gives  (r )
24 1  2 r 4  24 Pq s
7
 (r )  1 
x x
r  2  
2
11 ro 
11
4ro  11 mc p
(6.54)
Differentiate (6.54) and substitute into (6.38) gives
Nu D 
48
 4.364
11
(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


Boundary conditions
T k   T 
r


x r r   r 
T (0, x)
0
r
(6.56a)
T (ro , x)  Ts
(6.56b)
Axial velocity for fully developed flow is
 r2 
v x  2u 1  2 
 ro 

(6.45)
(6.46)
Use (6.36a) to Eliminate T / x in (6.45)

  T ( x)  T ( r , x) 
  s
0
x x  Ts ( x)  Tm ( x) 
(6.36a)
For uniform Ts ( x)  Ts , above gives
T Ts  T (r , x) dTm

x Ts  Tm ( x) dx
(6.57)
(6.46) and (6.57) into (6.45)
 r 2  T  T (r , x) dTm k   T 
r

2  c p u 1  2  s

r r   r 
 ro  Ts  Tm ( x) dx
(6.58)
Solution: (6.58) was solved using an infinite power series. Solution gives the Nusselt
number as
NuD  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]
Table 6.2: Nusselt number ranges from 2.46
to 8.235.
Nusselt number Nu D
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
(r  ro2 )
4 dz
(3.12)
r
Ts
(3.12) expressed in dimensionless form
v x
v
 x  2(1  R 2 )
u
(6.61)
T
Ti
u
0
x
(3.1) and (6.61) into energy equation (6.31)


1
 1    
1 R2

R

2
 R R  R 

(6.62)
t
Fig. 6.8
Boundary conditions
 ( ,0)
0
R
(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 ( )


 m    8
Gn
n 0
2
exp( 22n )
(6.66)
n
(ii) Local Nusselt number, Nu( )

G
Nu   
n 0

n

2
(66.7)
Gn
n 0
exp( 2 2n )
2
exp( 2 2n )
n
(iii) Average Nusselt number, Nu( )
Nu( ) 
h ( ) D
k
(f)
RESULTS

Table 6.3 lists values of  n and G n for 0  n  10. Table 6.4 gives Nu( ) and
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
n
Gn
0
1
2
3
4
5
6
7
8
9
10
2.70436
6.67903
10.67338
14.67108
18.66987
22.66914
26.66866
30.66832
34.66807
38.66788
42.66773
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
x/D
ReD Pr
Local and average Nusselt number for
tube at uniform surface temepratu re [4]

Fig. 6.9
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
Nu()  3.657
(6.69)
(3) Evaluate fluid properties at the mean temperatures Tm , defined as
Tm 
Tmi  Tmo
2
6.8.2 Uniform Surface Heat Flux

r
Inlet velocity is fully developed.

Energy equation is
qs
T
Ti
Repeat Graetz entrance problem
replacing the uniform surface
temperature with uniform heat flux.

(6.70)
u
0
D
x
t
qs
Fig. 6.10


1
 1    
1 R2

R

2
 R R  R 
(6.62)
18
Boundary conditions


(6.71a)
qsro
 ( ,1)

R
k (Ti  Ts )
(6.71b)
 (0, R)  1
(6.71c)
Solution.
Local Nusselt number:
hx  11 1
Nu( ) 
 
k  48 2


 ( ,0)
)0
R


n 1

An exp( 2  n2  )


1
Table 6.5
(6.72]
The average Nusselt number is given by
hx  11 1
Nu( ) 
 
k  48 2



n 1
1  exp( 2  n2  ) 

An

2  n2 

1
(6.73]

The eigenvalues  n2 and the constant An are listed
in Table 6.5

Limiting case:    (fully developed)
Nu() 
48
 4.364
11
Uniform surface flux [4]
n
 n2
An
1
2
3
4
5
6
7
8
9
10
25.6796
83.8618
174.1667
296.5363
450.9472
637.3874
855.8495
1106.3290
1388.8226
1703.3279
0.198722
0.069257
0.036521
0.023014
0.016030
0.011906
0.009249
0.007427
0.006117
0.005141
(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]