5-22
5-37 CO2 gas is accelerated in a nozzle to 450 m/s. The inlet velocity and the exit temperature are to be determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 CO2 is an ideal gas with variable specific
heats. 3 Potential energy changes are negligible. 4 The device is adiabatic and thus heat transfer is negligible. 5 There are
no work interactions.
Properties The gas constant and molar mass of CO2 are 0.1889 kPa.m3/kg.K and 44 kg/kmol (Table A-1). The enthalpy of
CO2 at 500C is h1  30,797 kJ/kmol (Table A-20).
1  m
2  m
 . Using the ideal gas relation, the specific volume is
Analysis (a) There is only one inlet and one exit, and thus m
determined to be
v1 


RT1
0.1889 kPa  m 3 /kg  K 773 K 

 0.146 m 3 /kg
P1
1000 kPa
1
CO2
2
Thus,
m 
1
v1
A1V1 
 V1 


m v1 6000/3600 kg/s  0.146 m3 /kg

 60.8 m/s
A1
40  10  4 m 2
(b) We take nozzle 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 out
in


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
  pe  0)
m (h1  V12 / 2)  m (h2 + V22 /2) (since Q  W
0  h2  h1 
V 22  V12
2
Substituting,
h2  h1 
V 22  V12
M
2
 30,797 kJ/kmol 
450 m/s2  60.8 m/s2 
2
1 kJ/kg
 1000 m 2 /s 2


44 kg/kmol


 26,423 kJ/kmol
Then the exit temperature of CO2 from Table A-20 is obtained to be
T2 = 685.8 K
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