Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN: 9781259696527
Author: J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher: McGraw-Hill Education
Question
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Chapter 8, Problem 8.8P
Interpretation Introduction

Interpretation:

The thermal efficiency of the cycle and the fraction of the steam entering the turbine is extracted for the feed water heater should be determined.

Concept Introduction:

The thermal efficiency of the cycle is calculated as:

  η=|Wnet|QH

Here Wnet is the total work produced by the turbine and pump. QH is the difference in enthalpies of the turbine and boiler.

The entropy is constant for idealistic turbine. Hence an isentropic process is developed between the enter and exit pressures of the idealistic turbine. The idealistic condenser has constant enthalpy.

Expert Solution & Answer
Check Mark

Answer to Problem 8.8P

  η=0.311

And

  m=0.1868lbm

Explanation of Solution

Given information:

It is given thatcycle is a regenerative cycle in a steam power plant includes one feedwater heater, turbine, condenser and a boiler. Steam enters the turbine at 650psia and 900F and exhaust at 1psia to condenser and 50psia to feedwater heater. Steam for the feedwater heater is at 50psia and in condensing raises the temperature of the feedwater to within 11F . Turbine and pump efficiencies are both given as 0.78 .

Introduction to Chemical Engineering Thermodynamics, Chapter 8, Problem 8.8P , additional homework tip  1

Figure to describe the process in simplified form is given as below

Introduction to Chemical Engineering Thermodynamics, Chapter 8, Problem 8.8P , additional homework tip  2

From figure point 1 and point 6 has same operating conditions.

Considering idealistic turbine for which entropy is constant at inlet and outlet

at point 2 and at 650psia and 900F, the properties are written from superheated steam table in fps unit as,

  H2=1461.2Btulbm and S2=1.6671BtulbmRankine

Now for idealistic turbine, entropy is constant at inlet and outlet. Hence

  S4'=S3'=S2=1.6671BtulbmRankine

Exhaust coming to condenser is at pressure 1psia and entropy S4'=1.6671BtulbmRankine and turbine has thermal efficiency of 0.78 . Hence, we need to calculate the actual entropy of the steam coming to condenser inlet at point 4.

The steam coming to condenser is wet and saturated. Hence for saturated steam at 1psia, properties are written from the steam table for the saturated steam at 1psia as,

At 0.94924psia

  Hliq=67.999BtulbmandHvap=1105.1BtulbmSliq=0.1295BtulbmRankineandSvap=1.9825BtulbmRankine

At 1.00789psia

  Hliq=69.995BtulbmandHvap=1105.9BtulbmSliq=0.1331BtulbmRankineandSvap=1.9775BtulbmRankine

From linear interpolation, if M is the function of a single independent variable X then the value of M at X is intermediate between two given values, M1 at X1 and M2 at X2

  M=( X 2X X 2 X 1)M1+(X X 1 X 2 X 1)M2Hliq=(1.00789psia1psia1.00789psia-0.94924psia)×67.999Btulbm+(1psia0.94924psia1.00789psia-0.94924psia)×69.995BtulbmHliq=69.73Btulbm

Similarly

  M=( X 2X X 2 X 1)M1+(X X 1 X 2 X 1)M2HVap=(1.00789psia1psia1.00789psia-0.94924psia)×1105.1Btulbm+(1psia0.94924psia1.00789psia-0.94924psia)×1105.9BtulbmHvap=1105.79Btulbm

And

  M=( X 2X X 2 X 1)M1+(X X 1 X 2 X 1)M2SVap=(1.00789psia1psia1.00789psia-0.94924psia)×1.9825BtulbmRankine+(1psia0.94924psia1.00789psia-0.94924psia)×1.9775BtulbmRankineSvap=1.9781BtulbmRankine

  M=( X 2X X 2 X 1)M1+(X X 1 X 2 X 1)M2Sliq=(1.00789psia1psia1.00789psia-0.94924psia)×0.1295BtulbmRankine+(1psia0.94924psia1.00789psia-0.94924psia)×0.1331BtulbmRankineSliq=0.1326BtulbmRankine

And

We also know that

  S4'=Sliq+x4'(SvapSliq)x4'=S4'SliqSvapSliqx4'=1.6671Btu lb mRankine0.1326Btu lb mRankine1.9781Btu lb mRankine0.1326Btu lb mRankinex4'=0.831

Therefore,

  H4'=Hliq+x4'(HvapHliq)H4'=69.73Btulbm+0.831×(1105.79Btu lb m69.73Btu lb m)H4'=931.195Btulbm

Hence the actual enthalpy of the steam exhaust from the turbine to the condenser is calculated by the thermodynamics efficiency of the work produced from the turbine which is

  η=WactualWidealη=H4H2H4'H20.78=H41461.2Btu lb m931.195Btu lb m1461.2Btu lb mH4=1047.796Btulbm

Hence

  H4=Hliq+x4(HvapHliq)x4=H4HliqHvapHliqx4=1047.796Btu lb m69.73Btu lb m1105.79Btu lb m69.73Btu lb mx4=0.944

and

  S4=Sliq+x4(SvapSliq)S4=0.1326BtulbmRankine+0.944×(1.9781Btu lb mRankine0.1326Btu lb mRankine)S4=1.8748BtulbmRankine

Exhaust coming to feedwater heater is at pressure 50psia and entropy S3'=1.6671BtulbmRankine and turbine has thermal efficiency of 0.78 . Hence, we need to calculate the actual entropy of the steam coming to feedwater heater inlet at point 3.

The steam coming to feedwater heater is superheated. Hence for superheated steam at 50psia, properties are written from the steam table for the superheated steam at 50psia as,

  50psia and S3'=1.6671BtulbmRankine lies in between 50psia and S3'=1.6586BtulbmRankine with enthalpy 1174.1Btulbm and 50psia and S3'=1.6720BtulbmRankine with enthalpy 1184.1Btulbm

From linear interpolation, if M is the function of a single independent variable X then the value of M at X is intermediate between two given values, M1 at X1 and M2 at X2

  M=( X 2X X 2 X 1)M1+(X X 1 X 2 X 1)M2H3'=(1.6720 Btu lbm Rankine1.6671 Btu lbm Rankine1.6720 Btu lbm Rankine-1.6586 Btu lbm Rankine)×1174.1Btulbm+(1.6671 Btu lbm Rankine1.6586 Btu lbm Rankine1.6720 Btu lbm Rankine-1.6586 Btu lbm Rankine)×1184.1BtulbmH3'=1180.44Btulbm

Hence the actual enthalpy of the steam exhaust from the turbine to the feedwater heater is calculated by the thermodynamics efficiency of the work produced from the turbine which is

  η=WactualWidealη=H3H2H3'H20.78=H31461.2Btu lb m1180.44Btu lb m1461.2Btu lb mH3=1242.21Btulbm

And entropy at H3=1242.21Btulbm and 50psia from superheated steam table in fps unit after the interpolation is

  S3=1.7431BtulbmRankine

Now,

The idealistic condenser has constant enthalpy and hence the enthalpy of saturated liquid at inlet point 4 of condenser is same as enthalpy at the outlet of condenser at point 5.

  H5=Hliq=69.73Btulbm

And at 1psia, the volume of saturated liquid at outlet of condenser point 5 is written from the saturated steam table in fps units after the interpolation as

  V5=0.0161ft3lbm

Hence work required for circulation of the steam from the condenser to the feedwater heater is calculated fromusing the pump efficiency is

  η=WidealWactualWactual=Wpump=WidealηWpump=V5( P 6 P 5)ηWpump=0.0161 ft 3 lb m(650 lbf in2 1 lbf in2 )× 12 2 in 21 ft 20.78Wpump=1920.11lbf.ftlbm×1Btu778.17lbf.f=2.47Btulbm

And the work-done by pump is also calculated as

  Wpump=H6H5=2.47BtulbmH6=2.47Btulbm+69.73BtulbmH6=72.197Btulbm

Now it is given that steam for the feedwater heater is at 50psia and in condensing raises the temperature of the feedwater to within 11F .

Hence at point 7 the steam is saturated liquid from feedwater heater going towards the condenser. Hence for 50psia and saturated steam, for saturated steam table, properties are written as,

  Hliq=H7=250.21BtulbmandSliq=S7=0.4112BtulbmRankineT7=281.01F

Hence the temperature at the point 1 according to the given condition and in condensing raises the temperature of the feedwater to within 11F is

  T1=281.01F-11FT1=270.01F

And at T1=270.01F, for saturated liquid from feedwater heater going towards the boiler, the properties are written from saturated steam table after interpolation as,

  Hliq=238.96BtulbmandSliq=0.3960BtulbmRankinePsat=41.87psia,Vliq=0.1717ft3lbm

The volume expansivity for boiler is calculated as

  β=1V(VT)=10.1717 ft 3 lb m×(0.01738 ft3 lbm 0.017193 ft3 lbm 292 F- 272 F)β=5.445×1051Rankine

Hence the enthalpy and entropy across the boiler is

  ΔH=V(1βT)ΔPH1Hliq=Vliq(1βT1)(P1Psat)H1238.96Btulbm=0.1717ft3lbm×(1(5.445× 10 5 1 Rankine×( 270.01 F+459.67)Rankine))×(650Psia41.87psia)×122in21ft2×1Btu778.17lbf.fH1=257.51Btulbm

And

  ΔS=βVΔPS1Sliq=βVliq(P1Psat)S10.3960BtulbmRankine=0.1717ft3lbm×5.445×1051Rankine×(650Psia41.87psia)×122in21ft2×1Btu778.17lbf.fS1=0.397BtulbmRankine

Let mass of steam exhaust from the turbine going towards feedwater heater is m hence mass of steam exhaust from turbine going towards condenser is (1m)

Basis 1lbmof steam exhaust of turbine

Applying energy balance across the feedwater heater. Hence fraction of the steam entering the turbine is extracted for the feedwater heater is

  mH3+1lbmH6=(1m)H7+mH1mH3+1lbmH6=mH7+1lbmH1m(H3H7)=1lbm(H1H6)m=H1H6H3H7m=257.51Btu lb m72.197Btu lb m1242.21Btu lb m250.21Btu lb mm=0.1868lbm

And

The thermal efficiency of the cycle is calculated as:

  η=|Wnet|QH

Here Wnet is the total work produced by the turbine and pump. QH is the difference in enthalpies of the turbine and boiler.

  Wturbine-1st=η(H3'H2)=218.99Btulbm×1lbm=218.99BtuWturbine-2nd=(1m)(H4H3)=(1lbm0.1868lbm)×(1047.796Btu lb m1242.21Btu lb m)Wturbine-2nd=158.098BtuWpump=2.47Btulbm×1lbm=2.47Btuη=|W turbine-1st+W turbine-2nd+W pump|( H 2 H 1)η=|( 218.99Btu)+( 158.098Btu)+2.47Btu|(1461.2 Btu lbm 257.51 Btu lbm )η=0.311

Conclusion

  η=0.311

And

  m=0.1868lbm

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