8-43
8-57 Air is accelerated in a nozzle while losing some heat to the surroundings. The exit temperature of air and the exergy
destroyed during the process are to be determined.
Assumptions 1 Air is an ideal gas with variable specific heats. 2 The nozzle operates steadily.
Properties The gas constant of air is R = 0.287 kJ/kg.K (Table A-1). The properties of air at the nozzle inlet are (Table A17)
T1  338 K 
 h1  338.40 kJ/kg
s1o  1.8219 kJ/kg  K
Analysis (a) We take the nozzle as the system, which is a control volume. The energy balance for this steady-flow system
can be expressed in the rate form as
E  E out
in



Rate of net energy transfer
by heat, work, and mass
Esystem0 (steady)

0
3 kJ/kg
Rate of change in internal, kinetic,
potential, etc. energies
E in  E out
m (h1  V12 / 2)  m ( h2 + V22 /2) + Q out
35 m/s
AIR
240 m/s
or
0  q out + h2  h1 
V 22  V12
2
Therefore,
h2  h1  q out 
V22  V12
(240 m/s) 2  (35 m/s) 2  1 kJ/kg

 338.40  3 

  307.21 kJ/kg
2
2
2
2
 1000 m / s 
T2  307.0 K = 34.0C and s 2o  1.7251 kJ/kg  K
At this h2 value we read, from Table A-17,
(b) The exergy destroyed during a process can be determined from an exergy balance or directly from its definition
X destroyed  T0S gen where the entropy generation Sgen is determined from an entropy balance on an extended system that
includes the device and its immediate surroundings so that the boundary temperature of the extended system is Tsurr at all
times. It gives
S  S out
in


Rate of net entropy transfer
by heat and mass
ms1  m s 2 
S gen

Rate of entropy
generation
 S system 0  0

Rate of change
of entropy
Q out
 S gen  0
Tb,surr
Q
S gen  m s 2  s1   out
Tsurr
where
s air  s 2o  s1o  R ln
P2
95 kPa
 (1.7251  1.8219)kJ/kg  K  (0.287 kJ/kg  K) ln
 0.1169 kJ/kg  K
P1
200 kPa
Substituting, the entropy generation and exergy destruction per unit mass of air are determined to be
x destroyed  T0 s gen  Tsurr s gen

q
 T0  s 2  s1  surr
T
surr


3 kJ/kg 

  (290 K) 0.1169 kJ/kg  K +
  36.9 kJ/kg

290 K 


PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
8-44
Alternative solution The exergy destroyed during a process can be determined from an exergy balance applied on the
extended system that includes the device and its immediate surroundings so that the boundary temperature of the extended
system is environment temperature T0 (or Tsurr) at all times. Noting that exergy transfer with heat is zero when the
temperature at the point of transfer is the environment temperature, the exergy balance for this steady-flow system can be
expressed as
X  X out
in

Rate of net exergy transfer
by heat, work,and mass
 X destroyed  X system0 (steady)  0  X destroyed  X in  X out  m 1  m  2  m (1  2 )


 
Rate of exergy
destruction
Rate of change
of exergy
 m [(h1  h2 )  T0 (s1  s2 )  ke  pe0 ]  m [T0 (s2  s1)  (h2  h1  ke)]
 m [T0 (s2  s1)  qout ] since, from energy balance,  qout  h2  h1  ke

Q 
 T0  m (s2  s1)  out   T0Sgen
T0 

Therefore, the two approaches for the determination of exergy destruction are identical.
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.