ME 315
Exam 3
8:00 -9:00 PM
Thursday, April 16, 2009
 This is a closed-book, closed-notes examination. There is a formula sheet at the
back.
 You must turn off all communications devices before starting this exam, and
leave them off for the entire exam.
 Please write legibly and show all work for your own benefit. Please show your
final answers in the boxes provided.
 State all assumptions.
 Please arrange all your sheets in the correct order. Make sure they are all
included.
Name: ____________________________________
Last
First
CIRCLE YOUR DIVISION
Div. 1 (9:30 am)
Prof. Murthy
Problem
1
(30 Points)
2
(30 Points)
3
(40 Points)
Total
(100 Points)
Div. 2 (12:30 pm)
Prof. Choi
Score
1. (30 points) Consider a heat exchanger with a heat exchange area of 30 m2 and operating under
the conditions listed in the table below.
Heat Capacity C (kW/K)
Inlet temperature oC
Outlet temperature oC
(i)
Hot Fluid
5
60
--
Cold Fluid
2
45
57.5
Determine the outlet temperature of the hot stream in oC.
Tho =
(ii)
Is the heat exchanger operating in
a. Parallel flow
b. Counterflow
c. Can’t tell
Circle only one answer and explain your choice.
Explanation:
(iii)
Calculate the overall heat transfer coefficient U (W/m2K).
U=
(iv)
What would the effectiveness be if the heat exchanger were infinitely long?
=
2

Solution
(i)
q  m c p T  m h c p ,h (Th ,i  Th ,o )  m c c p ,c (Tc ,o  Tc ,i )
 Ch (Th ,i  Th ,o )  Cc (Tc ,o  Tc ,i )
 q  ( 2 kW / K ) (57.5 C  45 C )  ( 5 kW / K ) (60 C  Th ,o )  25 kW
Th,o  55 C
(ii)
Tc,o  57.5C

Counterflow
(iii)
q  U ATlm  Ch (Th,i  Th,o )  Cc (Tc,o  Tc,i )  25 kW


Th,o  55C  Tc,o  Th,o
Tlm 
T1  T2
ln (T1 T2 )
T1  Th,i  Tc,o  60C  57.5C  2.5C
T2  Th,o  Tc,i  55C  45C  10C
Tlm 
T1  T2
2.5C  10C

 5.41011C
ln (T1 T2 ) ln (2.5C 10C)
q  U ATlm  U (30 m 2 )(5.41011C)  25 kW

U  0.154033
kW
W

154.033
m2  K
m2  K
3
(iv)

q
qmax
qmax  Cmin (Th,i  Th,i )  (2 kW /K)(60C  45C)  30 kW
As L  , Tc,o  Th,i  60C
q  Cc (Tc,o  Tc,i )  (2 kW /K)(60C  45C)  30 kW

 
q
qmax

30 kW
1
30 kW

4
2. (30 points) You are measuring the combustion gas temperature at the turbine inlet in a gas
turbine engine using a thermocouple junction. The thermocouple has a diameter of 1 mm, and is
initially at Ti = 25 C. You may neglect radiation in the following analysis.
Hot combustion gas
(evaluated at the
free stream temperature)
k = 0.04 W/mK
ρ = 0.6 kg/m3
cp = 1000 J/kgK
ν = 50 x 10-6 m2/s
U = 20 m/s
Thermocouple
k = 300 W/mK
ρ = 9000 kg/m3
cp = 400 J/kgK
μs = 25 x 10-6 kg/ms (gas viscosity evaluated at the thermocouple surface temperature, Ts)
(i) Determine the average convective heat transfer coefficient, ℎ.
ℎ=
W/m2K
(ii) The temperature measured at the thermocouple centerline after 3 sec is 900 C. Determine
the combustion gas temperature, 𝑇∞ .
𝑇∞ =
K
5
Solution
(i)
NuD  2  (0.4 ReD1/ 2  0.06 ReD 2 / 3 ) Pr 0.4 (
ReD 
Pr 
 1/ 4
)
s
 U Lc  U D  U (20 m /s) (0.001m)



 400



50106 m 2 /s
cp

(0.6 kg/m 3 ) (1000 J /kg  K)

 (50106 m 2 /s)
 0.75

k
0.04 W /m  K
      (50106 m 2 /s)  (0.6 kg/m 3 )  3105 kg/m  s
1/ 2
 NuD  2  (0.4 ReD
 0.06 ReD
2/3
) Pr 0.4 (
 1/ 4
)
s
 2  (0.4  4001/ 2  0.06 400 2 / 3 )(0.75 0.4 )(
NuD 

3105 kg/m  s 1/ 4
)  12.502
25106 kg/m  s
hD
k
 h  NuD 
k
0.04W /m  K
 12.502 
 500.063W /m 2  K
D
0.001m

6
(ii)
Bi 
Bi 
h Lc
k solid
h Lc

k solid
Lc 
ro 0.001 m 2

b  4ac
3
6
(500.063 W / m 2  K )  (
300 W / m  K
0.001 m
)
6
 0.000278  0.1
 Lumped capacitance analysis is valid.
T  T
t
 ex p ( )
Ti  T
t
t 
 V  c p
 t 
h  As
4
V    r3
3
 V  c p
  (   r3 )  cp
h  As

4
3
h  (4    r 2 )
As  4    r 2

  r  cp
3 h

0.001
m)  (400 J / kg  K )
2
 1.20 sec
3  (500.063 W / m 2  K )
(9000 kg / m3 )  (
At t  3 sec, T  900 C

T  T 900 C  T
3 sec

 ex p (
)
Ti  T
25 C  T
1.20 sec
T  978.247 C
7
3. (40 points) Fluid enters a duct of triangular cross-section with a mass flow rate of m . The
cross-section is an equilateral triangle of side s. The fluid is heated at the walls with a heat flux
q  ax +b (W/m2) into the domain, where x is measured from the entrance to the duct.
Furthermore, a chemical reaction in the volume of the fluid causes a constant volumetric heat
generation rate of q W/m3. The inlet bulk temperature of the fluid is Tmi. You are given that the
fluid has a constant density  and a constant specific heat Cp.
dx
Flow
Differential
control
volume
x
(a) By considering the differential control volume shown in the figure, write an energy balance to
dTm
derive a symbolic expression for
as a function of a, b, s, x, m , q and the physical
dx
properties of the fluid.
dTm
=
dx
(b) Derive a symbolic expression for the variation of Tm with x in terms of a, b, s, x, m , q , the
inlet bulk temperature Tmi and the physical properties of the fluid.
Tm(x) =
(c) Is the expression you derived in part (b) for Tm(x)
(i) always valid for this problem?
(ii) only valid for this problem if the flow is fully developed but the temperature is
developing?
(iii) only valid for this problem if the flow and temperature are fully-developed?
8
Circle only one answer, and justify your choice.
Solution
(a)
q”
Tm
Tm +dx
q
dx
s
s
3s
2
s
Perimeter = 3s

Area =

9
3 s2
4
E in  E out  E gen  E store
E in  m c p Tm  q " (3s )  dx  m c p Tm  (a x  b)  (3s )  dx
dT
E out  m c p Tm  dx  m c p (Tm  m dx )
dx
3 s2
E gen  q V  q A  dx  q
 dx
4
E store  0
d Tm
3 s2
 E in  E out  E gen  m c p Tm  (a x  b)  (3s )  dx  m c p (Tm 
dx )  q
 dx  0
dx
4
(3s )  (a x  b)  dx  m c p
m c p
d Tm
3 s 2 q
dx 
 dx  0
dx
4
d Tm
3 s 2 q
 (3s )  (a x  b) 
dx
4
d Tm (3s)  (a x  b)
3 s 2 q



dx
m c p
4 m c p
10
(b)
d Tm (3s)  (a x  b)
3 s 2 q


dx
m c p
4 m c p
x  (3s )  ( a x  b)
3 s 2 q
d
T



Tm,i m 0  m c p
4 m c p

Tm
Tm  Tm ,i

 dx

 (3s) a x 2
3 s 2 q 
 
(
 b x) 
x
2
4 m c p 
 m c p
Tm  Tm,i
x

0
3s a x 2
3 s 2 q
(
 b x) 
x
m c p
2
4 m c p
3s a x 2
3 s 2 q

(
 b x) 
x
m c p
2
4 m c p
(c) It is ALWAYS VALID for the problem. The expression was derived using a simple energy
balance. Therefore, no assumptions about fully developed flow or heat transfer are necessary. In
fact, if q” varies with x, no thermally fully developed flow is possible.
11