Problem session. Examples on forced convection heat transfer
Example 1
Water flows in a pipe that has a diameter D = 0.1 m that is maintained at a
uniform wall temperature of 100oC. The mass flow rate is 10 kg/s. The inlet
water temperature is 20 oC.
1. Estimate the required tube length to bring water temperature to 40 oC at
the exit of the pipe.
2. What will be the outlet temperature if the tube length is 20 m?
Solution
Objective
Determine pipe length for a given outlet temperature .Then
estimate the outlet temperature for a specified length.
This problem is divided to two parts. The first one is a design
problem that required the calculation of tube length for prescribes
physical conditions.
The second part represents a rating problem that asks for the
outlet temperature for given pipe dimensions. Both cases are for
uniform surface temperature boundary condition
Schematic
Water flow
Assumption/Conditions
Steady state, Uniform area and properties
Properties
Water at 30 oC,  = 998 kg/m3,  = 0.00101 kg/ms, cP=4186.8 kJ/kg K, k
= 0.603 W/m oC
Analysis
First we have to check whether the flow is laminar or turbulent. This is done
through calculating the Reynolds number
m
4m
4 *10
Ub 


 1.2758 m / s
2
A D
998 * 3.14159 * 0.12
U b D 998 * 1.2758 * 0.1
Re 

 126083.96

0.00101
Since Re > 10000, Flow is fully turbulent
Then Prandtl number may be calculated as follows
c
0.00101 * 4186.8
Pr  P 
 7.01
k
0.603
Now, we may use the correlation for calculating Nusselt number for turbulent
flow, assuming fully developed conditions;
Nu=0.023 Re0.8 Pr0.333 = 529.9 = hD/k
Then, h = 3195.26 W/m2 oC
This part represents a design problem, thus, the log mean temperature
difference method can be used:
LMTD =
(To  T1 )  (To  T2 )
80  60

 69.5o C
(T  T1 )
ln( 80 / 60)
ln o
(To  T2 )
Therefore, the rate of heat transfer is calculated as:

q = h As LMTD = m cP (T2  T1 ) =10*4186.8*(40-20) = 837360 W
Therefore, As = 3.77 m2 =  DL
L = 12 m
b. This part now represents a rating problem where the effectiveness method is
recommended:

m =10 kg/s
L = 20 m
h = 3195.26 W/m2 oC
As =  D L = 6.28 m2
(using the same correlation given above)
NTU =
h As

= 0.48
m cP
T2  T1
Then T2 = 50.5 oC
To  T1
The heat transfer can be evaluated as follows:
  1  exp(  NTU ) = 0.38 =
 cP (T2  T1 )  10  4186.8 (50.5  20)  1271484 W
qm
Example 2:
Air at 27 oC enters a 10-cm diameter tube of 10-m length with surface
temperature at 100 oC. The mass flow rate is 0.06 kg/s.
1.
2.
3.
4.
5.
6.
Calculate the Bulk stream velocity Ub.
Calculate the Reynolds number Re. Is the flow laminar or turbulent?
Show that Prandtl number Pr = 0.7071.
Calculate the friction factor f and heat transfer coefficient h.
Calculate the outlet temperature and the rate of heat transfer.
Calculate the power required to overcome friction losses.
Solution
Objective
This is a rating problem that asks for the outlet temperature and
heat transfer rate for given pipe dimensions. Uniform surface
temperature boundary condition is considered
Schematic
Air flow
Assumption/Conditions
Steady state, Uniform area and properties
Properties
Air properties at 27 oC are:  = 1.18 kg/m3, cP =1 kJ/kg oC,  = 1.57x10-5 m2/s,
k = 0.0262 W/m oC.
Analysis
a. Bulk stream velocity:
m
4m
4 * 0.06
Ub 


 6.474 m / s
2
A D
1.18 * 3.14159 * 0.12
b. Reynolds number:
U D 6.474 * 0.1
Re  b 
 41236.3

1.57  10 5
c. Prandtl number
c
c P 1.18  1.57  10 5  1000
Pr  P 

 0.707
k
k
0.0262
d. friction factor and heat transfer coefficient
2
f  1.58 ln Re  3.28  0.00548
Nu=0.023 Re0.8 Pr0.333 = 100.88 = hD/k
Then, h = 13.21 W/m2 oC
e. outlet temperature and rate of heat transfer
Use the effectiveness method
As  DL    0.1  10  3.14157 m 2
  1  exp(  NTU )
where Ntu 
h As

m cP
T T
T  27
  1  exp(  NTU )  0.5 = 2 1  2
To  T1 100  27
T2 = 63.5 oC

13.21  3.14157
 0.692
0.06  1000
 cP (T2  T1 )  0.06 1000 (63.5  27)  2190 W
qm
f. Pumping power, neglecting entrance and exit losses;
U b 4 Lf
1.18  6.474  4 10  0.00548
 8.372 kPa
2 D
2  0.1
m P 0.06  8.372
W 

 0.42 W

1.18
P 

Example 3
Oil is heated in a tube that is maintained at a uniform wall temperature (Ts =
150 oC). The tube diameter is 5 cm and it is 25 m long. Oil flow rate is 0.5 kg/s
and its inlet bulk stream temperature is 20 oC. Estimate the heat transfer rate
and oil exit temperature.
Solution
Objective
This is a rating problem that asks for the outlet temperature for
given pipe dimensions. Boundary condition is uniform surface
temperature.
Schematic
Water flow
Assumption/Conditions
Steady state, Uniform area and properties
Properties
Oil at 30 oC,  = 852 kg/m3,  = 0.032 kg/ms, cP=2131 J/kg K, k = 0.138
 c P 0.032  2131
W/m oC, Pr =

 494
k
0.138
Analysis
a. The bulk stream velocity is calculated first as follows:
m
4m
4 * 0.5
Ub 


 0.3 m / s
2
A D
852  3.14159 * 0.05 2
b.
Reynolds number
Re 
U b D 852  0.3  0.05

 400

0.032
Laminar Flow
Since the flow is laminar, we need to know whether it is developing or
fully developed. The condition for hydrodynamically fully developed flow
x
x
 0.05 Re Pr
is:  0.05 Re and for thermally fully developed flow
D
D
0.05 * Re = 20
L/D = 20/0.05 = 400
0.05 * Re * Pr = 9880
This means that the flow is hydrodynamically fully developed, but
thermally developing (thermal entry length) and hence, Nusselt number is
given by the following relation:
Re Pr
L/ D
Nu D  3.66 
 13.09
2/3
 Re Pr 
1  0.04

 L/ D 
k
h  Nu   36.13W / m 2o C
D
Now, use the effectiveness method to estimate the heat transfer rate and oil exit
temperature
As  DL    0.05  20  3.14159 m 2
36.13  3.14159
  1  exp(  NTU ) where Ntu  h As 

 0.106
0.5  2131
mc
0.0668
P
  1  exp(  NTU )  0.1=
T2  T1 T2  20

To  T1 150  20
T2 = 33 oC
 cP (T2  T1 )  0.5  2131(33  20)  13851.5 W
qm
Example 4
Air enters a tube with a flow rate of 0.02 kg/s and temperature T1 = 20 oC .The
tube diameter D = 0.05 m and it is maintained at constant heat flux (q”= 1000
W/m2) boundary conditions. Assume fully developed conditions apply for the
whole pipe length (L = 3 m).
1.
2.
3.
4.
5.
Estimate the bulk stream velocity.
Calculate the heat transfer coefficient.
Calculate the heat transfer rate.
Calculate the exit bulk stream temperature.
Estimate the surface temperature at the inlet and exit of the pipe.
Solution
Objective
Exit bulk stream temperature and surface temperature at inlet and exit of
the pipe that is maintained at UHF boundary conditions. (Dimensions are given).
Schematic
Air flow
Assumption/Conditions
Steady state, Uniform area and properties
Properties
Air at 300 K,  = 1.16 kg/m3,  = 1.846*10-5 kg/ms, cP=1007 J/kg K, k =
0.0263 W/m oC and Pr = 0.707
Analysis
a. Bulk stream velocity:
m
4m
4 * 0.05
Ub 


 8.78 m / s
2
A D
1.16 * 3.14159 * 0.05 2
b. Reynolds number:
U b D 1.16  8.78  0.05

 27589.16

1.846  10 5
Fully turbulent flow
c. friction factor and heat transfer coefficient
Re 
Nu=0.023 Re0.8 Pr0.333 = 73.14 = hD/k
Then, h = 38.47 W/m2 oC
d. rate of heat transfer
As  DL    0.05  3  0.47 m 2
q = q” * As = 1000 * 0.47 = 470 W
e. outlet temperature
 cP (T2  T1 )  0.02 1007 (T2  20)
q  470  m
T2 = 43.33 oC
f. Surface temperature at x= 0 and x = L
Apply Newton law of cooling
q"  hTs  Tb 
q"
Ts   Tb
h
1000
 20  46 o C
Therefore, Ts ,inlet 
38.47
1000
Ts ,outlet 
 43.33  69.32 o C
38.47
Example 5
Wind flows across a steam pipe (D = 10 cm) that has an outside surface
temperature of 127 oC. The wind temperature is 27 oC at a velocity of 10 m/s.
Estimate the heat loss from the pipe per unit length.
Solution
Objective
Heat transfer rate from a pipe exposed to cross flow of air.
Schematic
Air flow
steam
Air flow
.
Assumption/Conditions
Steady state, Uniform area and properties
Properties
T  T
Air at Tm  o
 350 K ,  = 2.076*10-5 m2/s, k = 0.03 W/m oC and Pr
2
= 0.697,  = 2.075*10-5 kg/ms
 = 2.286*10-5 kg/ms and ∞ = 1.846*10-5 kg/ms
Analysis
a. Reynolds number:
U D
Re   
10  0.1
 48169.56

2.076  10 5
Laminar flow
b. Heat transfer coefficient
Nu D  (0.4 Re
0 .5
D
 0.06 Re
 hD / k  139.6
Then, h = 41.88 W/m2 oC
2/3
D

) Pr  
 o
0. 4



0.25
c. rate of heat transfer per unit length
As  DL    0.1  1  0.314159 m 2
q  hAs To  T   41.88  0.314159  (127  27)  1315.63 W