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. |
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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. |
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We can simplify the 1st Law if we assume the process is adiabatic with no shaft
work. Changes in potential energy are negligible. |
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We can evaluate DEkin because we know both the inlet
and outlet velocities. |
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We can evaluate H1 from the Saturated Temperature
Table of the Steam
Tables. |
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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. |
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Diagram: |
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Given: |
x1 |
1 |
kg vap/kg |
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P2 |
60 |
psia |
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T1 |
285 |
oF |
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v2 |
64 |
ft/s |
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v1 |
525 |
ft/s |
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Find: |
T2 |
??? |
oF |
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Assumptions: |
1 - |
The fluid passes quickly through the diffuser so that heat exchange with the surroundings is negligible. |
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2 - |
Assume changes in potential energy are negligible. Unless the diffuser is very
long and oriented vertically, this is a pretty good
assumption. |
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3- |
No shaft work
occurs in this process. |
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Equations
/ Data / Solve: |
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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. |
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Begin by writing the 1st Law
for an open system : |
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Eqn 1 |
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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 : |
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Eqn 2 |
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We can lookup H1 because we know the water is a saturated vapor at 285oF. |
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P1 |
53.266 |
psia |
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H1 |
1176.3 |
Btu/lbm |
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The specific
kinetic energy is defined as : |
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Eqn 3 |
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Since we know both velocities, we can evaluate both the inlet
and outlet specific kinetic energies and the change as well : |
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gc |
32.174 |
ft-lbm / lbf-s2 |
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1 Btu = |
778.170 |
ft-lbf |
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Ekin,1 |
4283.3 |
ft-lbf / lbm |
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Ekin,1 |
5.504 |
Btu/lbm |
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Ekin,2 |
63.7 |
ft-lbf / lbm |
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Ekin,2 |
0.082 |
Btu/lbm |
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DEkin |
-4219.7 |
ft-lbf / lbm |
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DEkin |
-5.423 |
Btu/lbm |
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Now, we can solve Eqn 2 for H2 : |
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Eqn 4 |
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Now, we can plug
numbers into Eqn 4 and
evaluate H2 : |
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H2 |
1181.7 |
Btu/lbm |
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Next, we need to
determine the state of the water in state
2. |
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At 60 psia : |
Hsat liq |
262.38 |
Btu/lbm |
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Hsat vap |
1178.6 |
Btu/lbm |
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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. |
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From the NIST Webbook at 60 psia: |
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T (oF) |
H (Btu/lbm) |
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292.7 |
1178.6 |
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T2 |
1181.7 |
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300 |
1182.7 |
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350 |
1209.2 |
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Interpolating between Tsat = 292.7oF and 300oF yields T2 : |
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T2 |
298.25 |
oF |
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Verify: |
None of the assumptions
made in this problem solution can be verified. |
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Answers : |
T1 |
298 |
oF |
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