| Steam at 350oC and 650
kPa is mixed with subcooled water at 30oC and 650
kPa in an open
feedwater heater (FWH) as a way to produce saturated
liquid water at the same pressure. |
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| Assuming the open FWH is adiabatic, determine the mass flow rate of steam
required per kilogram of subcooled liquid water fed to the open FWH. |
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| Read : |
The feedwater heater is just a fancy mixer. When we write the MIMO form of the 1st Law at steady-state, there are three unknowns: the three mass flow rates. The states of all three streams are fixed, so we can determine the specific enthalpy of each of
them.
Mass conservation
tells us that m3 = m1 + m2. We can use this to
eliminate m3 from the 1st Law. Then we can solve the 1st Law for m1 / m2 ! |
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| Given: |
T1 |
30 |
oC |
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P3 |
650 |
kPa |
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P1 |
650 |
kPa |
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x3 |
0 |
kg vap/kg |
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T2 |
350 |
oC |
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Q |
0 |
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P2 |
650 |
kPa |
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| Find: |
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mdot1
/ mdot,2
= |
??? |
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| Diagram: |
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| Assumptions: |
1 - |
The feedwater heater operates at steady-state. |
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2 - |
Changes in potential and kinetic energies are negligible. |
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3 - |
Heat
transfer is negligible. |
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4 - |
No shaft work crosses the system boundary in this process. |
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| Equations
/ Data / Solve: |
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An open
feedwater heater is essentially a mixer in which superheated vapor is used to raise
the temperature of a subcooled liquid. We can begin our analysis with the steady-state form of the 1st Law. |
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Eqn 1 |
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The assumptions in the list above allow
us to simplify the 1st Law considerably: |
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Eqn 2 |
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Conservation
of mass on the feedwater
heater operating at steady-state tells us that : |
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Eqn 3 |
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We can solve Eqn 3 for mdot,3 and use the result
to eliminate mdot,3 from Eqn 2. The result is: |
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Eqn 4 |
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The easiest way to
determine mdot,1 / mdot,2 is to divide Eqn 4 by mdot,2. |
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Eqn 5 |
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Now, we can solve Eqn 5 for mdot,1 / mdot,2 : |
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Eqn 6 |
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Now, all we need to do
is to determine the specific enthalpy of all three streams and plug these values into Eqn
6 to complete the problem. |
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First we must
determine the phase(s)
present in each stream. |
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Tsat(650kPa) = |
161.98 |
oC |
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Therefore: |
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Stream
1 is a subcooled liquid because T1 < Tsat |
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Stream
2 is a superheated
vapor because T2 > Tsat |
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T3 = Tsat because it is a saturated liquid. |
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Data from the Steam Tables of the NIST Webbook (using the default
reference state) : |
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H1 |
126.32 |
kJ/kg |
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H2 |
3165.1 |
kJ/kg |
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H3 |
684.1 |
kJ/kg |
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Now, plug these values
into Eqn 6 to obtain : |
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mdot1
/ mdot,2
= |
4.448 |
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| Verify: |
None of the assumptions
made in this problem solution can be verified. |
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| Answers : |
mdot1 / mdot,2 = |
4.45 |
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