5C1 : 
CrossSectional Area Requirement for an Adiabatic Nozzle 
6 pts 
Steam enters a nozzle operating at steadystate with P_{1} = 40 bar, T_{1} = 400^{o}C and a velocity of 10 m/s. The steam flows through the nozzle with negligible heat transfer and no significant change in potential energy. At the exit, P_{2} = 15 bar and the velocity is 665 m/s. The mass flow rate is 2 kg/s. Determine the exit area of the nozzle in m^{2}. 








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 sate 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} 
4000 
kPa 

Find: 



T_{1} 
400 
^{o}C 

A_{2} 
??? 
m^{2} 

v_{1} 
10 
m/s 





P_{2} 
1500 
kPa 





v_{2} 
665 
m/s 





m_{dot} 
2 
kg/s 












Diagrams : 



















































































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 look up 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} 
3214.5 
kJ/kg 





H_{2} 
2993.4 
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 1.5 MPa : 












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





250 
2923.9 
0.15201 





T_{2} 
2993.4 
V_{2} 

T_{2} 
280 
^{o}C 

300 
3038.2 
0.16971 

V_{2} 
0.16278 
m^{3}/kg 









Now, plug V_{2} into Eqn 5 : 

A_{2} 
4.896E04 
m^{2} 








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








Answers : 
A_{2} 
4.90E04 
m^{2} 



