Steam at 3.5 MPa and 350^{o}C flows steadily into an adiabatic nozzle at a mass flow rate of 3.15 kg/s. The steam
leaves the nozzle at 500 kPa with a velocity of 571 m/s. 
If the inlet velocity
of the steam is 5 m/s, determine the exit diameter of the nozzle in cm. 


Read : 
The key here is that
we know both the mass flow rate and velocity of the effluent stream. If we can determine the specific volume of the effluent, we can determine the crosssectional area for flow at
the effluent, A_{2}. We are given the value
of one intensive variable for the effluent, P_{2}, but we need to know another in order to completely determine
the state of the effluent. Once know the state of the effluent, we can use the Steam Tables to determine the specific volume and then the crosssectional area. We must apply the steadystate form of 1st Law for open systems to this process. If we assume that heat transfer and changes in potential energy are negligible and that no shaft
work occurs, we can solve for the specific enthalpy of the effluent and thereby fix the state of the system. This allows us to
complete the problem. 










Given: 
P_{1} 
3500 
kPa 



P_{2} 
500 
kPa 

T_{1} 
350 
^{o}C 



v_{2} 
571 
m/s 

v_{1} 
5 
m/s 



m_{dot} 
3.15 
kg/s 










Find: 
D_{2} 
??? 
cm 
















Diagram: 













































































































































Assumptions: 
1  
The nozzle operates at steadystate. 


2  
Heat
transfer is negligible. 


3  
No shaft work crosses the system boundary. 


4  
The change in the potential energy of the fluid from the inlet to the outlet is negligible. 










Equations
/ Data / Solve: 


















Let's begin by writing
the steadystate form of the 1st Law for open systems. 






Eqn 1 

Based on the
assumptions listed above, we can simplify Eqn 1 as follows : 
















Eqn 2 











The only unknown in Eqn 2 is H_{2} because we can lookup H_{1} and the velocities are both
given. 











So, let's look up H_{1} and solve Eqn 2 for H_{2} : 












Eqn 3 



H_{1} 
3104.8 
kJ/kg 







H_{2} 
2941.8 
kJ/kg 











We could use H_{2} and P_{2} to determine T_{2} using the Steam Tables, but we are more
interested in V_{2} because : 












Eqn 4 

or : 


Eqn 5 











Once we know the specific volume at state 2, we can use Eqn 5 to determine the crosssectional area of the effluent pipe. 











Interpolating on the Steam Tables at 500 kPa : 















T (^{o}C) 
H (kJ/kg) 
V (m^{3}/kg) 







200 
2855.8 
0.42503 







T_{2} 
2941.8 
V_{2} 



T_{2} 
240.9 
^{o}C 

250 
2961.0 
0.47443 



V_{2} 
0.46541 
m^{3}/kg 











Now, plug V_{2} into Eqn 5
: 



A_{2} 
2.568E03 
m^{2} 











Determine D_{2} from A_{2} : 




Eqn 6 











Solving for D_{2} yields: 




Eqn 7 











Plug values into Eqn 7: 




D_{2} 
0.05718 
m 








5.718 
cm 










Verify: 
None of the
assumptions made in this problem solution can be verified. 










Answers : 
D_{2} 
5.72 
cm^{2} 















