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Isentropic Efficiency - Thermodynamics - Solved Quiz, Exercises of Thermodynamics

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

2013/2014

Uploaded on 02/01/2014

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Solution Isentropic Efficiency
A pump in a steam power plant has a feed rate of 100 kg/s of water at a temperature and
pressure of T1 = 50oC and P1 = 20 kPa. The pump has negligible heat transfer, an exit
pressure, P2 = 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 P3 = 8 MPa and T3 = 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
in kinetic and potential energy the work is given by the following equation:
outinuhhmW
.
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(T1 = 50oC, P1 = 20 kPa) vf(50oC) = 0.001012 m3/kg by eyeball
interpolation.
1
The inlet enthalpy is similarly approximated as the enthalpy of the saturated liquid
at the given temperature.
kg
kJ
CC
CC
kg
kJ
kg
kJ
kg
kJ
hoo
oo
in
71.211
01.4550
01.4506.60
78.19190.251
78.191
We find the ideal (isentropic) work as follows.
kW
mkPa
skW
kPakPa
kg
m
s
kg
PPmhhmW ssu 8.807
1
20000,20
001012.0100
3
3
21,21,
The actual pump work is then found as
kW
kW
W
W
s
su
u5.928
%87
8.807
,
kWWinput 5.928
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
becomes
inout hhmQ
. We have hout = h3 = h(P3 = 8 MPa and T3 = 500oC) = 3397.6 kJ/kg.
The value of hin = h2 must be found from the actual pump work.
kg
kJ
skW
kJ
s
kg
kW
kg
kJ
m
W
hh u
in
00.2211
1005.92871.211
2
We can now find heat transfer in the steam generator.
MW317.7Q
1
Actual interpolation gives:
kg
m
CC
CC
kg
m
kg
m
kg
m
voo
oo
in
3
33
30010123.0
01.4506.50
01.4506.60
001010.0001017.0
001010.0
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Solution Isentropic Efficiency

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

in kinetic and potential energy the work is given by the following equation: Wu  mh inhout.

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.

      kW

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

becomes Q  m hout hin. We have hout = h 3 = h(P 3 = 8 MPa and T 3 = 500oC) = 3397.6 kJ/kg.

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

W

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

     Q  317.7MW

(^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

  1. 0010123
  2. 06 45. 01
  3. 06 45. 01

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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 .  

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