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Mar. 25, 2015
Queen’s University, Mechanical and Materials Engineering
ME 346 HEAT TRANSFER
MID TERM EXAM #2 – CLOSED BOOK -State and justify your assumptions
_______________________________________________________________________________________________________________________________________________________________
1) You are required to estimate the heat loss from a vertical smoke stack (2 m in diameter and 50 m high)
when the wind is blowing at 3 m/s, the surrounding air temperature is 20oC, and the smoke stack has a
uniform surface temperature of 80oC. For these conditions and neglecting radiation determine the following:
a) the Reynolds Number of the airflow around the smoke stack;
b) the average Nusselt number for convective heat loss from the stack;
c) the total heat loss from the outer surface of the smoke stack.
The properties of air are: at 20 C:
at 80 C:
=0.0257 W/m·K, Pr=0.713,
=0.0299 W/m·K, Pr=0.708,
=15.11e-6 m2/s, and
=20.94e-6 m2/s.
(15 Marks)
State your assumptions and show all steps for your calculations.
2) A long square steel bar (k  13.4 W/m  K , c p  468J/kg  K ,   8238kg/m3 ) with cross-sectional
dimensions 10 cm x 10 cm and is heated to a uniform temperature of 1100K. Air at 300K is suddenly blown, at
26 m/s, in a perpendicular direction over the surface of the bar, cooling it. If, during this cooling process, the
centerline temperature of the bar drops to 500K after a time period of 30 minutes, and neglecting radiation
effects, estimate the following:
a) the average convective heat transfer coefficient on the outer surface of the rectangular bar during the
cooling process; and
b) using the appropriate analytical correlation, estimate the average surface heat transfer per unit length
of the rectangular bar. For this calculation, assume the properties of air at the film temperature are:
k  0.0262 W/m  K ,  8 105 m2 / s,   106 m2 / s.
(20 Marks)
State your assumptions. A Heisler chart for an infinite plane wall is provided below.
T ( x, y, t )  T
 P1 (0, t )  P2 (0, t )
Ti  T
______________________________
Total out of 35 Marks.
1) You are required to estimate the heat loss from a vertical smoke stack (2 m in diameter and 50 m high)
when the wind is blowing at 3 m/s, the surrounding air temperature is 20oC, and the smoke stack has a
uniform surface temperature of 80oC. For these conditions and neglecting radiation determine the following:
a) the Reynolds Number of the airflow around the smoke stack;
b) the average Nusselt number for convective heat loss from the stack;
c) the total heat loss from the outer surface of the smoke stack.
The properties of air are: at 20 C:
at 80 C:
=0.0257 W/m·K, Pr=0.713,
=0.0299 W/m·K, Pr=0.708,
=15.11e-6 m2/s, and
=20.94e-6 m2/s.
State your assumptions and show all steps for your calculations.
(15 Marks)
Solution:
Calculate the properties for air at the Tf.
T T
T f   S  50C
2
Pr  PrS
Pr f  
 0.7105
2
k k
k f   S  0.0278
2
 
 f   S  18.03 106
2
VD
A) Re D 

Either use the film kinematic viscosity or the surrounding kinematic viscosity.
3 m / s *2 m
3 m / s *2 m
 3.33 105
Re D 
 3.97 105 or Re D 
6
2
18.03 105
15.1110 m / s
(5 Marks)
B) Two options depending on which equation from Table 7.9 you choose
1. If you chose NuD  C Rem Pr1/3 you need to use the properties at Tf and Table 7.2.
So ReD  3.33 105 , Pr  0.7105 and C = 0.027, m=0.805
NuD  0.027*(3.33 105 )0.805 *0.71051/3  671.92
1/4
 Pr 
2. If you chose NuD  C Re Pr 
 you need to use properties at Tinf and use Table 7.4
 Prs 
We were expecting you to use the eq’n in option 1 because you need n for this one, in this case
n=0.37 for Pr  10 . So ReD  3.97 105 , Pr  0.713 and C=0.076, m=0.7.
m
n
1/4
 0.713 
NuD  0.076*3.97 100.7 *0.7130.37 * 

 0.708 
 557.75
C) Two options depending on what you did in part B
1. If you used method 1 above: use kf for Tf.
h
NuD * k f
D
=
671.92*0.0278W / mK
 9.34W / m2 K
2m
q  hA(Ts  T )
A   DL   *2*50  314m
q  9.34W / m2 K *314m *(80  20)  176049W
q  176 kW
2. If you used method 2 above: use kf for Tinf.
h
NuD * k f
D
=
557.75*0.0257 W / mK
 7.17 W / m2 K
2m
q  hA(Ts  T )
A   DL   *2*50  314m
q  7.17W / m2 K *314m *(80  20)  135097W
q  135 kW
Solution to MECH 346 Midterm 2 Question 2
A long square steel bar (k  13.4 W/m  K , c p  468J/kg  K ,   8238kg/m3 ) with cross-sectional
dimensions 10 cm x 10 cm is heated to a uniform temperature of 1100K. Air at 300K is suddenly blown,
at 26 m/s, in a perpendicular direction over the surface of the par, cooling it. If, during this cooling
process, the centerline temperature drops to 500K after a time period of 30 minutes, and neglecting
radiation effects, estimate the following.
a) the average convective heat transfer coefficient on the outer surface of the rectangular bar
during the cooling process;
Hint:
T ( x, y, t )  T
 P1 (0, t )  P2 (0, t )
Ti  T
With the knowledge that this is an “Infinite Rectangular Bar” scenario from the supplementary notes, and
knowing this is a symmetrical situation, we obtain the following equation:
 P(0, t )
2
 T  T   500  300 
 o  

 Ti  T  1100  300 
P(0, t )   * 
500  300
 0.5
1100  300
We can also calculate the Fourier number:
Fo 
 t
L2
where  

3.48 106  30  60
 0.05
2
 2.51  2.5 ,
k
13.4
m2

 3.48 106
  c p 8238  468
s
From the Heisler chart, one might guess an inverse Biot number of roughly 3. From that we can find h:
hL
k
Bi  k (1/ 3) 13.4
W
h

 89.33  90 2
L
.05
m K
Bi 
Common errors are: incorrect implementation of characteristic length as described on the Heisler chart,
incorrect use or lack of use of a square root to calculate P, using alpha or conduction coefficient for air
from part b even though you’re given all of the components to calculate alpha anyway.
ALTERNATE SOLUTION FOR PART A
Knowing the Fourier number and the temperature difference as calculated above, one might deem the
following equation to be appropriate:
o* 
To  T
 C1 exp  12 Fo
Ti  T


Then we’d have to perform an iterative solution. Ideally, one guesses a value for C1 and calculates ζ1. If
we’re clever, we’ll see that for varying values of C1, we usually obtain a number slightly greater than 0.5,
which should lead us in the general direction of the final solution for the Biot number, which is roughly
0.3. One might also perform a linear interpolation at that point to get a more precise answer of 0.33. The
question would continue on to the same answer from there.
b) using the appropriate analytical correlation, estimate the average surface heat transfer per
unit length of the rectangular bar. For this calculation assume the properties of air at the
film temperature are:
 k  2.624 10
2
W/m  K ,  8 105 m2 / s ,   106 m2 / s

We know the velocity and the properties of the air. From this we can calculate the Reynolds number:
Re 
VL


26  0.1
 165816  1.66 105
8 105
This is within the acceptable range of the parameters given in Table 7.3. Using the numbers for a fluid
passing over a cylinder, and using the appropriate equation from Table 7.9 i.e. THE EQUATION FOR A
NON-CIRCULAR CYLINDER (AS STATED IN THE DESCRIPTION OF TABLE 7.3), we obtain the
following:

Nu  C  Rem  Pr1/3  0.102  1.66 105

0.675
 0.7071/3  303
Finally using the standard equation for Nusselt numbers we can find the convective heat transfer
coefficient.
Nu 
hL
Nu  k 303.1 2.624 102
W
, h

 79.5  80 2
k
L
.1
mK
We now calculate the total heat loss for the system.
q ''  hAT  80  (4  0.1)  (1100  300)  25.6 kW per unit length of the par, which one can simple
equate to 1 metre. Marks were given very generously for part b considering the ambiguity around the
temperature values needed to calculate q.
Common errors are: incorrect use of length for the Reynolds number and Nusselt number, incorrect
equation for Nusselt number, using values of viscosity from question 1 even though the film temperature
properties are given in this question.
ALTERNATE SOLUTION FOR PART B
One could again attempt to use the one-term approximation here.
o* 
To  T
 C1 exp  12 Fo
Ti  T


Instead of using the equation however, if one simply assumes the Biot number from part a, they quickly
find the value for zeta and the constant. Plugging those values in, one achieves a number of roughly
P  0.5 , which is the same value obtained in part a.
Then, one could find the heat lost from the bar during cooling at 500K in the following way
Qo   cV (Ti  T )  8238  468  (0.1 0.11)  (500  300)  7,710,768W
This can be thought of as the energy of the surface at T=500K relative to the air temperature.
For the surface loss over t=30 minutes:
 sin( 1 ) * 
 sin(0.5218) 
Q  Qo 1 
o   7710768 1 
 .5   4, 027,971W
0.5218
1




It should be noted that there are a lot of assumptions involved using this alternate method, such as Biot
number going to infinity. It is recommended you use the other method for solving this problem.