Thermodynamics homework solution

College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 370
Thermodynamics
Fall 2010 Course Number: 14319

Unit Ten Homework Solutions, November 18, 2010

1 0 oF and leaves at a pressure of 40
Steam enters an adiabatic turbine at 800 psia and 900
psia. Determine the maximum amount of work that can be delivered by this turbine.

Assume that the turbine is a steady flow device with negligible changes in kinetic and potential
energies. For this adiabatic turbine, with one inlet and one outlet, the first law gives the work as
w = hin – hout. The maximum work in an adiabatic process occurs when the outlet entropy is the
same as the inlet entropy. Use the steam tables to find the properties.
In this case, hin = h(800 psia 900oF) = 1456.0 Btu/lbm and sin = s(800 psia 900oF) = 1.6413
Btu/lbm∙R. At the outlet pressure of 40 psia, the entropy value of 1.6413 Btu/lbm∙R is seen to be
in the mixed region, so we have to compute the quality to find the outlet enthalpy.

sout  s f ( 40 psia ) 1.6413 Btu  0.39213 kJ
lbm  R kg  K
xout    0.9725
s fg ( 40 psia ) 1.28448 kJ
kg  K

236.14 Btu  933.69 Btu  1144.2 Btu
hout  h f (40 psia )  xout h fg (40 psia )   (0.9725)  
lbm  lbm  lbm

We can now compute the maximum work.

1456.0 Btu 1144.2 Btu 311.9 Btu
w  hin  hout   w
lbm lbm lbm


2 Air is compressed steadily by a 5-kW compressor from 100 kPa and 17oC to 600 kPa and
167oC at a rate of 1.6 kg/min. During this process some heat transfer takes place between
the compressor and the surrounding medium at 17oC. Determine the rate of entropy
change of air during this process.

It appears that we have a large amount of extraneous information in this problem. Since we are
only asked to find the entropy change of the air and the process is steady, the desired entropy
change is given by the following equation if we use the air tables to find the entropy change,
accounting for the variable heat capacity.

 o  P 
Sair  m air sout  sin   m air  sair (T2 )  sair
o
(T1 )  R ln  2 
  P1 

Using the values of so from the air tables and R = 0.287 kJ/kg∙K for air gives the answer as
follows.




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, d. 2



 o  P  1.6 kg min  2.0877 kJ
Sair  m air  sair ( 440.15 K )  sair
o
( 290.15 K )  R ln  2  
  P1  min 60 s  kg  K


1.66802 kJ 0.287 kJ  600 kPa  kW  s Sair  
0.00250 kW
  ln  
kg  K kg  K  100 kPa  1 kJ K


3 Air enters a nozzle steadily at 280 kPa and 77oC with a velocity of 50 m/s and exits at 85
kPa and 320m/s. The heat losses from the nozzle to the surrounding medium at 20oC are
estimated to be 3.2 kJ/kg. Determine (a) the exit temperature and (b) the total entropy
change for this process.

Here we have a steady flow system with one inlet and one outlet. There is no useful work in the
nozzle. We neglect changes in potential energy and write the first law as follows

 V2   V2 
q   h     h  
 2  out  2 in
We use the air tables to account for the temperature variation of the heat capacity and we note
that q = -3.2 kJ/kg since this is a heat loss. With the inlet enthalpy, hin = h(350 K) = 350.49 kJ/kg
from the air tables, we can solve our first law for the outlet enthalpy as follows.
2 2
 50 m   320 m 
   
Vin2  Vout
2
3.2 kJ 350.49 kJ  s   s  kJ  s 2
hout  q  hin    
2 kg kg 2 1000 kg  m 2
This gives hout = 297.34 kJ/kg. Interpolating in the air tables to find this enthalpy, we find that it
occurs at Tout = 297.2 K .

The total entropy change, stotal is the sum of the entropy change of the air, sair, plus that of the
surroundings, ssurround. We find the entropy change of the air from the usual equation for the air
tables.

P   85 kPa 
sair  sair
o
(T2 )  sair
o
(T1 )  R ln  2   sair
o
( 297.2 K )  sair
o
(350 K )  R ln  
 P1   280 kPa 
1.6924 kJ 0.287 kJ  85 kPa  0.1775 kJ
  ln   
kg  K kg  K  280 kPa  kg  K

The entropy change of the surroundings (per kilogram of air) is the heat transfer divided by the
constant temperature of 293.15 K for the surroundings. Note that the heat transfer to the
surroundings is the negative of the heat transfer to the nozzle: qsurround = 3.2 kJ/kg.

3.2 kJ
qsurr kg 0.0109 kJ
s surr   
Tsurr 293.15 K kg  K

nge.




Adding the entropy change of the air and the surroundings gives the total entropy change.