8

1. Internal Flow 1
Chapter 8
Convection: Internal Flow (8.1-8.6)

2. Internal Flow 2
Introduction
In Chapter 7 we obtained a non-dimensional form for the heat transfer
coefficient, applicable for problems involving external flow:
 Calculation of fluid properties was done at surface temperature,
bulk temperature of the fluid, or film temperature
• In this chapter we will obtain convection coefficients for geometries
involving internal flow, such as flow in tubes
 Recall Newton’s law of cooling:
 For flow inside a tube we cannot define
 Must know how temperature evolves inside the pipe and find alternative
expressions for calculating heat flux due to convection.
)("
∞−= TThq ss
∞T

3. Internal Flow 3
Flow Conditions for Internal Flow
• Onset of turbulent flow at 2300Re ≈
µ
ρ
=
Dum
D
• Hydrodynamic entry length:
– Laminar flow
– Turbulent flow
Dhfd Dx Re05.0/, ≈
10/, >Dx hfd
0/ =dxdu

4. Internal Flow 4
Mean Velocity
• Velocity inside a tube varies over the cross section. For every differential
area dAc:
∫∫ ρ==∴
A
c
A
dAxrumdm ),(cdAxrumd ),(ρ=
• Overall rate of mass transfer through a tube with cross section Ac:
cm Aum ρ= and
c
m
A
m
u
ρ
=
 where um is the mean (average)
velocity
(8.1)
Combining with (8.1):
∫
∫ =
ρ
ρ
=
o
c
r
oc
A
c
m rdrxru
rA
dAxru
u
02
),(
2
),(
 Can determine average temperature at any axial location (along
the x-direction), from knowledge of the velocity profile
(8.2)

5. Internal Flow 5
Velocity Profile in a pipe
• For laminar flow of an incompressible, constant property fluid in the
fully developed region of a circular tube (pipe):














−





µ
−=
2
2
1
4
1
)(
o
o
r
r
r
dx
dp
ru
dx
dpr
u o
m
µ
−=
8
2
(8.3a)
(8.4)














−=
2
12
)(
om r
r
u
ru
(8.3b)

6. Internal Flow 6
Thermal Considerations: Mean Temperature
• We can write Newton’s law of cooling
inside a tube, by considering a mean
temperature, instead of ∞T
)("
mSx TThq −=
• The internal energy per unit mass for a
differential area is: ),()(),( xrTcudAxrTcmdEd c υυ ρ== 
• Overall rate of energy transfer :
υ
=
cm
E
Tm


where Tm is the mean (average)
velocity
Combining
with (8.6): ∫
∫ =
ρ
=
υ
υ o
c
r
om
A
c
m rdrxruT
rucm
dAxrTuc
T
02
),(
2
),(

• Integrating over the entire cross section: c
AA
dAxrTucEdE ∫∫ υρ== ),(
mTcmE υ=  and
(8.5)
(8.6)
(8.7)

7. Internal Flow 7
Example 8.1
For flow of a liquid metal through a circular tube, the velocity and
temperature profiles at a particular axial location may be approximated
as being uniform and parabolic respectively. That is, u(r)=C1 and
T(r)-Ts=C2[1-(r/ro)2
], where C1 and C2 are constants and Ts the temperature
at the surface of the tube. What is the value of the Nusselt number, NuD
at this location?
])/(1[)( 2
2 os rrCTrT −=−
1)( Cru = ro
Ans. NuD=8

8. Internal Flow 8
Fully Developed Conditions
• For internal flows, the temperature, T(r), as well as the mean
temperature, Tm always vary in the x-direction, ie.
0,0
)(
≠≠
dx
dT
dx
rdT m
?Can we claim that ?0
)(
=
dx
rdT

9. Internal Flow 9
Fully Developed Conditions
• A fully developed thermally region is possible, if one of two possible
surface conditions exist :
– Uniform temperature (Ts=constant)
– Uniform heat flux (qx”=const)
• Thermal Entry Length :
10)/(
PrRe05.0)/(
,
,
=
≈
turbtfd
Dlamtfd
Dx
Dx
• Although T(r) changes with x, the relative shape of the temperature profile
remains the same: Flow is thermally fully developed.
0
)()(
),()(
,
=





−
−
∂
∂
tfdms
s
xTxT
xrTxT
x

10. Internal Flow 10
Fully Developed Conditions
• It can be proven that for fully developed conditions, the local
convection coefficient is a constant, independent of x:
)(xfh ≠

11. Internal Flow 11
Mean temperature variation along a tube
We are still left with the problem of knowing how the mean
temperature Tm(x), varies as a function of distance, so that we can use
it in Newton’s law of cooling to estimate convection heat transfer.
 Consider an energy balance on a differential control volume inside the
tube:
– Main contributions are due to internal energy changes [= ],
convection heat transfer and flow work [=pυ], needed to move fluid.
mTcm υ
)( υ+= υ pTcdmdq mconv 
The rate of convection heat transfer to
the fluid is equal to the rate at which the
fluid thermal energy increases, plus the
net rate at which is work is done in
moving the fluid through the control
volume
P=surface perimeter
(8.8a)

12. Internal Flow 12
Mean temperature variation along a tube
Considering perfect gas, or incompressible liquid:
)( ,, imompconv TTcmq −= 
 qconv is related to mean temperatures at inlet and outlet.
(8.9)
mpconv dTcmdq =
By integrating:
(8.8b)
Combining equations 8.5 and 8.8b:
)(
"
ms
pp
sm
TTh
cm
P
cm
Pq
dx
dT
−==
 where P=surface perimeter =
πD for circular tube, width for flat plate
 Integration of this equation will result in an expression for the variation
of Tm as a function of x.
(8.10)

13. Internal Flow 13
Case 1: Constant Heat Flux
• Integrating equation (8.10):
)(""
LPqAqq ssconv ⋅==
x
cm
Pq
TxT
p
s
imm

"
,)( += (8.11)
where P=surface perimeter
=πD for circular tube,
=width for flat plate
constqs ="

14. Internal Flow 14
Example (Problem 8.15)
A flat-plate solar collector is used to heat atmospheric air flowing
through a rectangular channel. The bottom surface of the channel is
well insulated, while the top surface is subjected to a uniform heat flux,
which is due to the net effect of solar radiation absorption and heat
exchange between the absorber and cover plates.
For inlet conditions of mass flow rate=0.1 kg/s and Tm,i=40°C, what is
the air outlet temperature, if L=3 m, w=1 m and the heat flux is 700
W/m2
? The specific heat of air is cp=1008 J/kg.K
Ans: Tm,o=60.8°C

15. Internal Flow 15
Case 2: Constant Surface Temperature,
Ts=constant
From eq.(8.10) with Ts-Tm=∆T: Th
cm
P
dx
Td
dx
dT
p
m
∆=
∆
−=

)(
Integrating from x to any downstream location:








−=
−
−
h
cm
Px
TT
xTT
pims
ms

exp
)(
,
For the entire length of the tube:








−=
∆
∆
=
−
−
h
cm
PL
T
T
TT
TT
pi
o
ims
oms

exp
,
,
lmsconv TAhq ∆= where
)/ln( io
io
lm
TT
TT
T
∆∆
∆−∆
=∆(8.13) (8.14)
(8.12)
As is the tube surface area, As=P.L=πDL

16. Internal Flow 16
Example 8.3
Steam condensing on the outer surface of a thin-walled circular tube
of 50 mm diameter and 6-m length maintains a uniform surface
temperature of 100°C. Water flows through the tube at a mass flow
rate of 0.25 kg/s and its inlet and outlet temperatures are Tm,1=15°C
and Tm,o=57°C. What is the average convection coefficient associated
with the water flow?
D=50 mm
L=6 m
Ts=100°C
Tm,i=15°C
Tm,o=57°C
KmWhAns 2
/756. =

17. Internal Flow 17
Case 3: Uniform External Temperature
 Replace Ts by and by (the overall heat transfer coefficient,
which includes contributions due to convection at the tube inner and
outer surfaces, and due to conduction across the tube wall)
∞T h U








−=
−
−
=
∆
∆
∞
∞
p
s
im
om
i
o
cm
AU
TT
TT
T
T

exp
,
,
lms TAUq ∆=(8.15) (8.16)

18. Internal Flow 18
Summary (8.1-8.3)
• We discussed fully developed flow conditions for cases involving
internal flows, and we defined mean velocities and temperatures
• We wrote Newton’s law of cooling using the mean temperature,
instead of
• Based on an overall energy balance, we obtained an alternative
expression to calculate convection heat transfer as a function of mean
temperatures at inlet and outlet.
• We obtained relations to express the variation of Tm with length, for
cases involving constant heat flux and constant wall temperature
∞T
)("
mS TThq −=
)( ,, imompconv TTcmq −= 
x
cm
Pq
TxT
p
s
imm

"
,)( +=








−=
∆
∆
=
−
−
h
cm
PL
T
T
TT
TT
pi
o
ims
oms

exp
,
,
(8.9)

19. Internal Flow 19
Summary (8.1-8.3)
• We can combine equations (8.13-8.16) with (8.9) to obtain values of
the heat transfer coefficient (see solution of Example 8.3)
 In the rest of the chapter we will focus on obtaining values of the heat
transfer coefficient h, needed to solve the above equations
• We used these definitions, to obtain appropriate versions of Newton’s
law of cooling, for internal flows, for cases involving constant wall
temperature and constant surrounding fluid temperature
lmsconv TAhq ∆= lms TAUq ∆=
)/ln( io
io
lm
TT
TT
T
∆∆
∆−∆
=∆
(8.13-8.16)