7-26
7-57 Steam enters a nozzle at a specified state and leaves at a specified pressure. The process is to be sketched on the T-s
diagram and the maximum outlet velocity is to be determined.
Analysis (b) The inlet state properties are
6000 kPa
T
P1  6000 kPa  h1  2784.6 kJ/kg
(Table A - 5)

x1  1
 s1  5.8902 kJ/kg  K
1200 kPa
1
For the maximum velocity at the exit, the entropy will be constant
during the process. The exit state enthalpy is (Table A-6)
2
s 2  s f 5.8902  2.2159
x2 

 0.8533
P2  1200 kPa

s fg
4.3058

s 2  s1  5.8902 kJ/kg  K  h2  h f  xh fg  798.33  0.8533  1985.4  2492.5 kJ/kg
s
We take the nozzle as the system, which is a control volume since mass crosses the boundary. Noting that one fluid stream
enters and leaves the nozzle, the energy balance for this steady-flow system can be expressed in the rate form as
E  E
inout


Rate of net energy transfer
by heat, work, and mass
E system 0 (steady)



0
Rate of change in internal, kinetic,
potential, etc. energies
E in  E out


V2 
V2 
m  h1  1   m  h2  2 


2 
2 


 V 2  V12
h1  h2   2

2

(since W  Q  pe  0)




Solving for the exit velocity and substituting,
 V 2  V12
h1  h2   2

2






V2  V12  2(h1  h2 )

0.5

 1000 m 2 /s 2
 (0 m/s) 2  2(2784.6  2492.5) kJ/kg
 1 kJ/kg






0.5
 764.3 m/s
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