6-20
6-34 Air is decelerated in an adiabatic diffuser. The velocity at the exit is to be determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 Air is an ideal gas with
constant specific heats. 3 Potential energy changes are negligible. 4 There are no work interactions. 5 The
diffuser is adiabatic.
Properties The specific heat of air at the average temperature of (20+90)/2=55°C =328 K is cp = 1.007
kJ/kg˜K (Table A-2b).
Analysis There is only one inlet and one exit, and thus m 1 m 2 m . We take diffuser as the system,
which is a control volume since mass crosses the boundary. 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 Ê0 (steady)
Rate of change in internal, kinetic,
potential, etc. energies
E in
m (h1 V12 / 2)
h1 V12 / 2
0
100 kPa
20qC
500 m/s
E out
m (h2 + V 22 /2)
AIR
200 kPa
90qC
h2 + V 22 /2
Solving for exit velocity,
V2
>V
2
1
2(h1 h2 )
@
0.5
>V
2
1
2c p (T1 T2 )
@
0.5
ª
§ 1000 m 2 /s 2
«(500 m/s) 2 2(1.007 kJ/kg ˜ K)(20 90)K¨¨
«¬
© 1 kJ/kg
330.2 m/s
·º
¸»
¸»
¹¼
0.5
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