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Some of the topic in thermodynamics are: Property tables and ideal gases, First law for closed and open (steady and unsteady) systems, Entropy and maximum work calculations, Isentropic efficiencies, Cycle calculations (Rankine, refrigeration, air standard) with mass flow rate ratios. This quiz is about: Isentropic Efficiency, Steam Power Plant, Heat Transfer, Isentropic, First Law, Volume Flow Rate of Steam, Actual Pump Work
Typology: Exercises
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A pump in a steam power plant has a feed rate of 100 kg/s of water at a temperature and pressure of T 1 = 50oC and P 1 = 20 kPa. The pump has negligible heat transfer, an exit pressure, P 2 = 8 MPa, and an isentropic efficiency of 87%. The pump exit is then fed to a steam generator where heat is added to produce steam at P 3 = 8 MPa and T 3 = 500oC. Determine (a) the power input to the pump.
The pump is a work input device so the work, w, is the ideal work divided by the isentropic
efficiency: w = ws/s. If we assume that the pump is a steady flow system with negligible changes
This is valid since we neglect the heat transfer and have only one inlet and one outlet. For a pump where the specific volume is nearly constant, the isentropic work is given by the following equation: hout,s – hin = vin(Pout – Pin). Here we see that the inlet is a slightly subcooled liquid so we sue the approximation that, vin = v(T 1 = 50oC, P 1 = 20 kPa) vf(50oC) = 0.001012 m^3 /kg by eyeball interpolation. 1 The inlet enthalpy is similarly approximated as the enthalpy of the saturated liquid at the given temperature.
kg
kJ C C C C
kg
kJ
kg
kJ
kg
kJ h
o o in o o
We find the ideal (isentropic) work as follows.
kPa m
kW s kPa kPa kg
m
s
kg W (^) u s mh h s mP P 807. 8
3
3
, 1 2 , 1 2
The actual pump work is then found as kW
W kW W s
us u 928.^5 87 %
, 807.^8
Winput ^928.^5 kW
Determine (b) the heat input to the steam generator.
With the same assumptions for the first law as before (steady, one-inlet, one-outlet, negligible changes in kinetic and potential energies) and no useful work in the steam generator, the first law
The value of hin = h 2 must be found from the actual pump work.
kg
kJ
kW s
kJ
s
kg
kW
kg
kJ
m
h h u in
We can now find heat transfer in the steam generator.
kJ
MW s
kg
kJ
kg
kJ
s
kg Q mh h 1000
3 2
(^1) Actual interpolation gives:
kg
m C C C C
kg
m
kg
m
kg
m v
o o in o o
3
3 3
3
Determine (c) the volume flow rate of steam leaving the steam generator.
The mass flow rate and the volume flow rate are related by the specific volume. Here we want the
volume flow rate at the outlet state where v = v(P 3 = 8 MPa and T 3 = 500oC) = 0.04175 m^3 /kg. We
thus have
kg
m
s
kg V mv
3
3 3
s
4175 m V
3 .