A diffuser is used to reduce the velocity of steam
from 525 ft/s to 64 ft/s. The inlet steam is saturated vapor at 285^{o}F and the effluent pressure
is 60 psia. Calculate
the temperature of the
effluent. 











Read : 
If we can determine
the enthalpy of the effluent, we can use the Steam Tables and the known value of
the pressure to
determine the temperature. 

We can simplify the 1st Law if we assume the process is adiabatic with no shaft
work. Changes in potential energy are negligible. 

We can evaluate DE_{kin} because we know both the inlet
and outlet velocities. 

We can evaluate H_{1} from the Saturated Temperature
Table of the Steam
Tables. 

This leaves only one unknown in the 1st Law, H_{2}. Once we evaluate H_{2}, by solving the 1st Law, we can use H_{2} and P_{2} and the Steam Tables to determine T_{2}. 

Diagram: 


Given: 
x_{1} 
1 
kg vap/kg 



P_{2} 
60 
psia 

T_{1} 
285 
^{o}F 

v_{2} 
64 
ft/s 

v_{1} 
525 
ft/s 






Find: 
T_{2} 
??? 
^{o}F 


Assumptions: 
1  
The fluid passes quickly through the diffuser so that heat exchange with the surroundings is negligible. 


2  
Assume changes in potential energy are negligible. Unless the diffuser is very
long and oriented vertically, this is a pretty good
assumption. 


3 
No shaft work
occurs in this process. 

Equations
/ Data / Solve: 




We could use the Steam Tables to determine T_{2} if we
knew the value of one more intensive
variable in state 2. The most likely choice
is to find H_{2}. This is the right choice because H_{2} appears in the 1st Law. 



Begin by writing the 1st Law
for an open system : 

Eqn 1 


Assume that the diffuser is adiabatic, there is no shaft
work and that changes in potential energy are negligible. This allows us to
simplify Eqn 1 to : 













Eqn 2 


We can lookup H_{1} because we know the water is a saturated vapor at 285^{o}F. 


P_{1} 
53.266 
psia 

H_{1} 
1176.3 
Btu/lb_{m} 


The specific
kinetic energy is defined as : 



Eqn 3 


Since we know both velocities, we can evaluate both the inlet
and outlet specific kinetic energies and the change as well : 











g_{c} 
32.174 
ftlb_{m} / lb_{f}s^{2} 

1 Btu = 
778.170 
ftlb_{f} 

E_{kin,1} 
4283.3 
ftlb_{f} / lb_{m} 

E_{kin,1} 
5.504 
Btu/lb_{m} 

E_{kin,2} 
63.7 
ftlb_{f} / lb_{m} 

E_{kin,2} 
0.082 
Btu/lb_{m} 

DE_{kin} 
4219.7 
ftlb_{f} / lb_{m} 

DE_{kin} 
5.423 
Btu/lb_{m} 


Now, we can solve Eqn 2 for H_{2} : 



Eqn 4 


Now, we can plug
numbers into Eqn 4 and
evaluate H_{2} : 

H_{2} 
1181.7 
Btu/lb_{m} 


Next, we need to
determine the state of the water in state
2. 









At 60 psia : 
H_{sat liq} 
262.38 
Btu/lb_{m} 

H_{sat vap} 
1178.6 
Btu/lb_{m} 


Since H_{2} > H_{sat vap}, we conclude that stream 2 is a superheated vapor. Therefore, we must use
the Superheated Steam Tables to determine the temperature of steam
at 60 psia that has a specific enthalpy of 1181.7 Btu/lb_{m}. 











From the NIST Webbook at 60 psia: 

T (^{o}F) 
H (Btu/lb_{m}) 








292.7 
1178.6 








T_{2} 
1181.7 








300 
1182.7 








350 
1209.2 













Interpolating between T_{sat} = 292.7^{o}F and 300^{o}F yields T_{2} : 

T_{2} 
298.25 
^{o}F 








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









Answers : 
T_{1} 
298 
^{o}F 















