Download Unit 9: November 4

College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 370
Thermodynamics
Fall 2010 Number: 14319 Instructor: Larry Caretto
Solution to Group Exercise for Unit Nine – Ideal Gas Entropy
An adiabatic gas turbine has an inlet pressure and temperature of 1.2 MPa and 1300 K. It
has an outlet pressure of 100 kPa. The properties of the exhaust gas can be assumed to
be those of air. Use the air tables to answer the following questions.
1. What is the maximum work for the given inlet state and outlet pressure?
Assume that air behaves as an ideal gas. The maximum work occurs in a reversible process; in
an adiabatic reversible process, the entropy is constant so we can use the isentropic relations for
an ideal gas to find the outlet state for the maximum work.
The end points, T1, P1, T2, and P2, for an ideal gas isentropic process are related by the equation
P2/P1 = Pr(T2)/Pr(T1) where Pr(T) is found from the air tables. From Table A-17 on page 936 of
the course text we find that Pr(T1 = 1300 K) = 330.9 so that Pr(T2) = Pr(T1) P2/P1 = 330.9(100 kPa)
/ (1200 kPa) = 27.575. We need to find the value of T for which Pr(T) = 27.575; we find this by
interpolating in the air tables between Pr(690 K) = 27.29 and Pr(700 K) = 28.80.
T  690 K 
700 K  690 K
27.257  27.29  691.9 K
28.80  27.29
We are given that the system has no heat transfer and we see that it has one inlet and one outlet.
If we assume that this is a steady system with negligible kinetic and potential energy changes we
can write the first law for the work per unit mass as w = hin – hout. We can use the air tables to
find these enthalpies: hin =h(Tin = 1300 K) = 1395.97 kJ/kg and hout.is found by interpolation.
hout
713.27 kJ 702.52 kJ

702.52 kJ
kg
kg
691.9 K  690 K   704.56 kJ


kg
700 K  690 K
kg
We could have also found this value directly from the Pr value without computing the temperature.
hout
713.27 kJ 702.52 kJ

702.52 kJ
kg
kg
27.257  27.29  704.56 kJ


kg
28.80  27.29
kg
Either approach gives hout = 704.56 kJ/kg. With the values of hin and hout we can find the
maximum work from our first law equation.
w = hin – hout. = 1395.97 kJ/kg – 704.56 kJ/kg = 691.4 kJ/kg
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Exercise nine solutions
ME 370, L. S. Caretto, Fall 2010
Page 2
2. What is the outlet temperature and the entropy change if the work is only 80% of the
maximum work?
If the actual work is only 80% of the maximum work, we will have a work output of 80%(691.4
kJ/kg) = 553.14 kJ/kg. Since our first law still applies, we have w = hin – hout, which we can solve
for the outlet enthalpy: hout = hin – w = 1395.97 kJ/kg – 553.14 kJ/kg = 842.83 kJ/kg. We have to
find the temperature in the air tables for which the enthalpy is 842.83 kJ/kg. We interpolate
between h(800 K) = 821.95 kJ/kg and h(820 K) = 843.98 kJ/kg to find the final temperature.
Tout  800 K 
820 K  800 K
 842.83 kJ  821.95 kJ  = 819.4 K .
kg
kg 
kJ
kJ

843.98
 821.95
kg
kg
To find the entropy change we need to interpolate for the outlet value of so, using the same
factors that we used to find Tout.
2.74504 kJ 2.71787 kJ

2.71787 kJ
kg  K
kg  K
o
s Tout  

843
.
98
kJ
821
.95 kJ
kg  K

kg
kg
s2  s1  s o (Tout )  s o (Tin )  R ln
 842.83 kJ 821.95 kJ  2.74362 kJ

 

kg
kg
kg  K


Pout 2.74362 kJ 3.27345 kJ 0.2870 kJ 100 kPa 0.18334 kJ



ln

Pin
kg  K
kg  K
kg  K
1200 kPa
kg  K
When the work is less than the maximum work in an adiabatic process, the entropy change is
positive.
A closed system containing air with an initial pressure and temperature of 100 kPa and
280 K undergoes an adiabatic compression where its volume decreases by a factor of 10.
Use the air tables to answer the following questions.
3. What is the maximum work for the given initial state and final specific volume?
Assume that air behaves as an ideal gas. The maximum work occurs in a reversible process; in
an adiabatic reversible process, the entropy is constant so we can use the isentropic relations for
an ideal gas to find the outlet state for the maximum work.
The end points, T1, v1, T2, and v2, for an ideal gas isentropic process are related by the equation
v2/v1 = vr(T2)/vr(T1) where vr(T) is found from the air tables. From Table A-17 on page 936 we find
that vr(T1 = 280 K) = 738.0 so that vr(T2) = vr(T1) v2/v1 = 738.0(0.1) = 73.80. We need to find the
value of T for which vr(T) = 73.80; we find this by interpolating in the air tables between vr(680 K)
= 75.56 and vr(690 K) = 72.56.
T  680 K 
690 K  680 K
73.80  75.50  685.8 K
72.56  75.50
We see that the system has no heat transfer, so the first law for the work per unit mass in this
closed system becomes w = u1 – u2. We can use the air tables to find these internal energies: u1
=u(T1 = 280 K) = 199.75 kJ/kg and u2.= u(T2.= 685.8 K) = 501.15 kJ/kg. (The value of u2 is found
by interpolation; we could have found this directly from vr without finding the temperature.) With
these internal energy values we can find the maximum work from our first law equation.
Exercise nine solutions
ME 370, L. S. Caretto, Fall 2010
Page 3
w = u1 – u2. = 199.75 kJ/kg – 501.15 kJ/kg = –301.40 kJ/kg
4. What would the final temperature be for the same initial conditions and the same ratio
of final to initial volume (and an adiabatic compression) if the actual work input is
120% of the work1 input you found in part 3? What is the entropy change for this
process?
Here we have w = 120%(–301.40 kJ/kg) = –361.68 kJ/kg. Our final internal energy is still given
by the first law: u2 = u1 – w = 199.75 kJ/kg – (–361.68 kJ/kg) = 561.43 kJ/kg. We have to find the
temperature in the air tables for which the internal energy is 561.43 kJ/kg. We interpolate
between u(760 K) = 560.01 kJ/kg and u(780 K) = 576.12 kJ/kg to find the final temperature.
T  760 K 
780 K  760 K
 561.43 kJ  560.01 kJ  = 761.8 K .
kg
kg 
kJ
kJ

576.12
 560.01
kg
kg
To calculate the entropy change we first have to interpolate for the final value of so.
2.69013 kJ 2.66176 kJ

2.66176 kJ
kg  K
kg  K
o
s T2  

576.12 kJ 560.01 kJ
kg  K

kg
kg
s 2  s1  s o (T2 )  s o (T1 )  R ln
 561.43 kJ 560.01 kJ  2.66426 kJ

 

kg
kg
kg  K


T2 v1 2.66426 kJ 1.63279 kJ 0.2870 kJ  751.8 K  0.08716 kJ



ln 
10  
T1 v 2
kg  K
kg  K
kg  K
kg  K
 280 K

When the work is less than the maximum work in an adiabatic process, the entropy change is
positive.
Repeat the solution to questions 1-4 assuming constant heat capacities.
For constant heat capacities the final temperature in an isentropic process with given initial
temperature and initial and final pressures is T 2 = T1(P2/P2)(k-1)/k. For parts 1 and 2, we will use
the properties for air at 1000 K; this is less than the turbine inlet temperature and is the maximum
temperature for which we have data in Table A-2(b) on page 912. At this temperature, k = 1.336
and cp = 1.142 k/kg∙K. The outlet temperature for an isentropic process is T 2 = (1300 K)[(100
kPa)/(1200 kPa)](1.336 – 1)/1.336 = 695.88. The work is still equal to the enthalpy change, but for
constant heat capacity we have w = h = cpT = (1.142 k/kg∙K)(1300 K – 695.88 K). For
constant heat capacities then, the maximum work is w = 689.91 kJ/kg .
If the actual work is 80% of this value or 551.92 kJ/kg, the outlet temperature is found from the
equation w = hin – hout = cp(Tin – Tout) so that Tout = Tin – w/cp = 1300 K – (551.92 kJ/kg) / (1.142
kJ/kg∙K); T2 = 816.70 K.
1
When there is a work input, the mathematical result for work is negative. In such a case the
“maximum work” will have a smaller magnitude than any actual work input. For example, if you
find wmax = -100 kJ/kg, so |wmax| = 100 kJ/kg, then an actual work that is 120% of this would be an
input of |w| = 120 kJ/kg. In the first law we would express this work input as w = -120 kJ/kg. So it
would be less than the maximum work of -100 kJ/kg, but would represent a larger work input.
Exercise nine solutions
ME 370, L. S. Caretto, Fall 2010
Page 4
In this constant-heat-capacity case we find the entropy change as follows.
sout  sin  c p ln
Tout
P
1.142 kJ 816.70 K 0.2870 kJ 100 kPa 0.18231kJ
 R ln out 
ln

ln

Tin
Pin
kg  K
1300 K
kg  K
1200 kPa
kg  K
For the closed system problem in parts 3 and 4, we use a similar approach with constant heat
capacities. However, in this case we are given the volume ratio so that we use the following
equation to find the final temperature in an isentropic process T 2 = T1(v1/v2)(k-1). Here we’ll use
the data for air at 300 K in Table A-2(a) on page 885: k = 1.400 and cv = 0.718 k/kg∙K. The outlet
temperature for an isentropic process is T2 = (280 K)(10)(1.400 – 1) = 703.33 K. The work is still
equal to the internal change, but for constant heat capacity in a process with Q = 0, the first law
gives w = u = cvT = (0.718 k/kg∙K)(280 K – 703.33 K). For constant heat capacities then, the
maximum work is found to be w = –303.95 kJ/kg .
If the actual work input is 120% of the minimum work input, we have |w| = 1.2|wmin| = 1.2|wmax| =
1.2|–303.95 kJ/kg| = 364.74 kJ/kg. Because this is a work input, w = –|w| = –379.94 kJ/kg. Our
first law still applies; we have w = u1 – u2, which we can write as follows for constant heat
capacities: w = cv(T1 – T2). We can solve this equation for the final temperature: T2 = T1 – w/cv.
Plugging in the numbers gives T 2 = 280 K – (–364.74 kJ/kg) / (0.718 kJ/kg∙K); T2 = 788.0 K.
In this constant-heat-capacity case we find the entropy change as follows.
s2  s1  cv ln
T2
v
0.718 kJ 788.0 K 0.2870 kJ  v1 1  0.08208 kJ
 
 R ln 2 
ln

ln 
T1
v1
kg  K
380 K
kg  K
kg  K
 10 v1 