W. George Woodruff School of Mechanical Engineering
Quiz #3
Summer 2011
ME3322
NAME ____________Key_______________
_______________________________________________________________________________________
1) Fill-in the blanks in the following statements regarding the Clausius inequality:
Q
a. The cyclic integral
is always negative. (1)
T b
b. If the boundary temperature, Tb, is constant over the entire cycle then the Clausius inequality simply states
that Q0. (1)
2) The temperature vs. entropy (T-s) diagram of an internally reversible power cycle is shown below.
a. Find the net heat input during the cycle. (3)
s3
s3
Q= Tds Tds =Area of Triangle 123
s2
2
700
s1
Q=4002/2=400 KJ. Thus 400 KJ of Heat is added.
b. Find the net work developed during the cycle. (2)
U=Q-W=0W=Q= 400 KJ
300
1
1
3
S (KJ/oK)
3
3) Consider a closed thermal system going from the initial state 1 to final state 2.
a. Suppose that the process (12) is adiabatic (no heat transfer). What can you say about the entropy change
S2-S1? Why? (2)
By the 2nd law of thermodynamics for closed systems, S2-S1=(Q/T)b+=0
b. Is it possible to have S2<S1 if heat transfer were allowed? If no explain why and if yes explain how. (2)
Yes. If heat is removed, the entropy could go down.
c. Suppose that the process (12) is isothermal expansion of an ideal gas. What can you say about the
entropy change S2-S1? Why? (2)
S2-S1=-m ln(p2/p1)=-mln(v1/v2)>0
4) Consider a single channel open (control volume) thermal system operating at steady-state:
h2,s2
h1,s1
system
The specific enthalpies and entropies of the inlet and outlet given are by h1=2500 KJ/Kg, h2=1500 KJ/Kg, s1=2
KJ/Kg oK and s2=1 KJ/ Kg oK. The mass flowrate of the fluid through the control volume is dm/dt=10 Kg/sec.
a. Is it possible for this system to be adiabatic (perfectly insulated)? Why? (2)
No because it implies entropy destruction. To show this more precisely, consider the entropy flow
Q
equation: j + m( si se ) + cv = 0 . If Qj=0, then m(2 1) + cv = m + cv = 0 , which is
j Tj
b
impossible since m > 0 and cv 0 .
ME3322
Quiz #3
Summer 2011
NAME ____________________________
_______________________________________________________________________________________
b. Find the maximum amount of power (dWCV/dt) that this system can develop if it has a uniform
boundary temperature of Tb=500 oK. (5)
Wcv Qcv
Qcv Wcv
+ (hi he ) = 0
=
+ 1000
m
m
(1)
Energy Balance:
m
Entropy Balance:
Q
Qcv
+ ( si se ) + cv = 0 cv = 500(1 + cv ) (2)
m
mTb
Wcv
= 500(1 cv ) . Since cv 0 , the maximum work output is
m
W
achieved when cv = 0 . Thus cv = 500 KJ/Kg and dWcv/dt=5000 KJ/sec.
m
max
Combining eqs. (1) and (2) gives
ME3322
Quiz #3
Summer 2011
NAME ____________________________
_______________________________________________________________________________________
Useful Formulas
Ideal Gas Entropy Change: s2-s1=so(T2)-so(T1)-Rln(p2/p1).
2
Heat Transfer to an Internally Reversible Process: Qint = Tds
rev
1
Q
Closed System Entropy Balance: S =
+
T b
1
Steady-state Control Volume Energy and Entropy Balance (Single Channel, K.E.=P.E.=0):
Qcv Wcv
+ (hi he ) = 0
m
Q
T j + m(si se ) + cv = 0
j j
b
2