8-65
8-77 Steam is accelerated in an adiabatic nozzle. The exit velocity of the steam, the isentropic efficiency, and the exergy
destroyed within the nozzle are to be determined.
Assumptions 1 The nozzle operates steadily. 2 The changes in potential energies are negligible.
Properties The properties of steam at the inlet and the exit of the nozzle are (Tables A-4 through A-6)
P1  7 MPa  h1  3411.4 kJ/kg

T1  500C  s1  6.8000 kJ/kg  K
7 MPa
500C
70 m/s
P2  5 MPa  h2  3317.2 kJ/kg

T2  450C  s 2  6.8210 kJ/kg  K
STEAM
5 MPa
450C
P2 s  5 MPa 
 h2 s  3302.0 kJ/kg

s 2 s  s1
Analysis (a) We take the nozzle to be 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
inout

E system 0 (steady)



Rate of net energy transfer
by heat, work, and mass
0
Rate of change in internal, kinetic,
potential, etc. energies
E in  E out
m (h1  V12 / 2)  m (h2 + V22 /2) (since W  Q  pe  0)
0  h2  h1 
V 22  V12
2
Then the exit velocity becomes
 1000 m 2 /s 2
V 2  2(h1  h2 )  V12  2(3411.4  3317.2) kJ/kg
 1 kJ/kg

  (70 m/s) 2  439.6 m/s


(b) The exit velocity for the isentropic case is determined from
 1000 m 2 /s 2
V 2 s  2(h1  h2 s )  V12  2(3411.4  3302.0) kJ/kg
 1 kJ/kg

  (70 m/s) 2  472.9 m/s


Thus,
N 
V 22 / 2
V 22s

/2
(439.6 m/s) 2 / 2
(472.9 m/s) 2 / 2
 86.4%
(c) 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 the actual nozzle.
It gives
S in  S out



Rate of net entropy transfer
by heat and mass
S gen

Rate of entropy
generation
 S system 0  0

Rate of change
of entropy
m s1  m s 2  S gen  0  S gen  m s 2  s1  or s gen  s 2  s1
Substituting, the exergy destruction in the nozzle on a unit mass basis is determined to be
x destroyed  T0 s gen  T0 ( s 2  s1 )  (298 K)(6.8210  6.8000)kJ/kg  K  6.28 kJ/kg
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