# Example Problem with Complete Solution

5C-9 : Outlet Temperature From a Steam Diffuser 5 pts
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 285oF 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 DEkin because we know both the inlet and outlet velocities.
We can evaluate H1 from the Saturated Temperature Table of the Steam Tables.
This leaves only one unknown in the 1st Law, H2.  Once we evaluate H2, by solving the 1st Law, we can use H2 and P2 and the Steam Tables to determine T2.
Diagram: Given: x1 1 kg vap/kg P2 60 psia
T1 285 oF v2 64 ft/s
v1 525 ft/s
Find: T2 ??? oF
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 T2 if we knew the value of one more intensive variable in state 2.  The most likely choice is to find H2.  This is the right choice because H2 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 H1 because we know the water is a saturated vapor at 285oF.
P1 53.266 psia H1 1176.3 Btu/lbm
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 :
gc 32.174 ft-lbm / lbf-s2 1 Btu = 778.170 ft-lbf
Ekin,1 4283.3 ft-lbf / lbm Ekin,1 5.504 Btu/lbm
Ekin,2 63.7 ft-lbf / lbm Ekin,2 0.082 Btu/lbm
DEkin -4219.7 ft-lbf / lbm DEkin -5.423 Btu/lbm
Now, we can solve Eqn 2 for H2 : Eqn 4
Now, we can plug numbers into Eqn 4 and evaluate H2 : H2 1181.7 Btu/lbm
Next, we need to determine the state of the water in state 2.
At 60 psia : Hsat liq 262.38 Btu/lbm Hsat vap 1178.6 Btu/lbm
Since H2 > Hsat 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/lbm.
From the NIST Webbook at 60 psia: T (oF) H (Btu/lbm)
292.7 1178.6
T2 1181.7
300 1182.7
350 1209.2
Interpolating between Tsat = 292.7oF and 300oF yields T2 : T2 298.25 oF
Verify: None of the assumptions made in this problem solution can be verified. 