02/25/14 ME313 1

Vapor Cycles Previously we analyzed cycles where the

working fluid was out in ideal gas land far from saturation.

This allowed us to use the compact form of property tables, i.e. Pv = RT

It also allowed us to use fairly simple isentropic relations, one for constant Cp, and not.

Now we'll look at cycles that occur within and close to the saturation curve, which means

q=u2u1C v T 2T 1 but use tables

Also, isentropic processes will involve find s1, equating to s2, then going back to P or T.

02/25/14 ME313 2

Rankine Cycle The Rankine cycle is the ideal continuous flow

vapor power cycle, analogous to Brayton cycle.

boiler

condenserT

s1

2

4

3

1 4

3

2

The ideal cycle has

isentropic pump of saturated liquid. Isentropic expansion to a high quality. Constant pressure boiler and condenser.

P

1 4

32

v

The net work from the cycle is both the area within the T-s diagram and the P-v diagram.

02/25/14 ME313 3

Rankine Efficiency Like all previous cycles, we can determine the

efficiency with th=

wnetq in

=1qoutqin

boiler qin=h3h2condenser qout=h4h1

turbine wout,turbine=h3h4pump w in,pump=h2h1=v P2P1

Note that v1 v2 = v.

The net work can be found either by the 1st Law, or by wnet = wturbine-wpump.

Solving problems that involve the Rankine cycle is best approached like before: start at a point with 2 intensive variables, and work your way around the cycle.

Use constant-value processes when appropriate, like P or s.

02/25/14 ME313 4

ExampleLet's solve example 10-1 for a simple Rankine cycle using EES. That will alleviate looking up values in tables and solving for quality.Be sure the correct unit system is set

givenintheproblemP[3]=3000;T[3]=350;P[4]=75;"constantpressure"P[1]=P[4];P[2]=P[3];"condenseroutlet"x[1]=0;"12perfectpump"s[1]=entropy(water,P=P[1],x=x[1]);s[2]=s[1];"34perfectturbine"s[3]=entropy(water,P=P[3],T=T[3]);s[4]=s[3];"Havethemall,nowenthalpies"h[1]=enthalpy(water,P=P[1],x=x[1]);h[2]=enthalpy(water,P=P[2],s=s[2]);h[3]=enthalpy(water,P=P[3],s=s[3]);h[4]=enthalpy(water,P=P[4],s=s[4]);"1stLaw"Q_in=h[3]h[1];Q_out=h[4]h[1];W_net=Q_inQ_out;eff=1Q_out/Q_in;

qin = 2727qout = 2018wnet = 709.8eff = 0.2602

When solving by hand, you would look these values up in the tables.Note that finding values at a given entropy requires interpolation, and it may be easier to first calculate the quality for mixtures (e.g. x4)

02/25/14 ME313 5

Deviations From Perfection Deviations from the ideal cycle occur due to

pressure drop in the boiler & condenser, and friction (not isentropic) in the pump & turbine.

T

1 4

3

2Turbine friction

Boiler pressure drop

Condenser pressure drop

Pump friction

The inefficiencies of the pump and turbine can be quantified in terms of their isentropic efficiency

P=w swa=

h2sh1h2ah1

T=waws=

h3h4ah3h4s

02/25/14 ME313 6

ExampleLet's add an 80% efficient turbine

givenintheproblemP[3]=3000;T[3]=350;P[4]=75;"constantpressure"P[1]=P[4];P[2]=P[3];"condenseroutlet"x[1]=0;"12perfectpump"s[1]=entropy(water,P=P[1],x=x[1]);s[2]=s[1];"34perfectturbine"s[3]=entropy(water,P=P[3],T=T[3]);

"Havethemall,nowenthalpies"h[1]=enthalpy(water,P=P[1],x=x[1]);h[2]=enthalpy(water,P=P[2],s=s[2]);h[3]=enthalpy(water,P=P[3],s=s[3]);h_4s=enthalpy(water,P=P[4],s=s[3]);0.85=(h[3]h[4])/(h[3]h_4s);"1stLaw"Q_in=h[3]h[1];Q_out=h[4]h[1];W_net=Q_inQ_out;eff=1Q_out/Q_in;

qin = 2730qout = 2125wnet = 605.9eff = 0.2219

That was easy! Note qin is the same qout is larger wnet and eff are smaller

02/25/14 ME313 7

Review Using Tables Find h at P=200 kPa and x=0.25

Go to saturation table as a function of pressure; A-5h=hfx hghf

h=504.710.252705.3504.71=1055

h=hfx hfgh=504.710.252201.6=1055

The value hfg is just the difference of hg and hf

Find h at P=200 kPa and s=8.6 kJ/kgKCheck the saturation table at 200 kPa. The value of sg=7.127, so the state is a superheated vapor.Now go to the super heated tables for P=200 kPa.

T v u h sSat. 0.88578 2529.1 2706.3 7.1270150 0.95986 2577.1 2769.1 7.2810...500 3487.7 8.5153600 3704.8 8.7793

Value we want is in between here

Label these rows

AB

Do a linear interpolationhhAhBhA

=ssAsBsA

h3487.73704.83487.7

= 8.68.51538.77938.5153

h=3557 kJ/kg

02/25/14 ME313 8

Increasing Rankine Efficiency There are three primary was to increase

efficiency

Lower condenser pressure Many operate under vacuum Small increase in qin Problems include liquid in the

turbine and vacuum leaks. Superheat steam

Wnet increases but qin does also. Since average T is higher, efficiency is greater.

Material limitations on high T. Increase boiler pressure @ same T

Efficiency increases since average T at qin increases.

Problems include liquid in the turbine, but reheating can fix this.

02/25/14 ME313 9

ExamplePlot the increase in efficiency as a function of boiler pressure for the earlier example for the range of 3 MPa to 15 MPa. Also plot the quality at the turbine outlet.Using the Rankine EES file, add x[4] and comment out P[3].

x4=quality(water,P=P[4],s=s[4])Create a parametric table. Select Tables New Parametric TableSet the number of runs to 7, and put P[3], eff, and x[4] into the table. Enter values for P[3] from 3000 to 15000, every 2000. Select Calculate Solve Table.

Now plot the data. Plots New Plot Window X-Y Plot. Put P[3] on the x-axis and both eff and x[4] on the y-axis.

efficiency increases,though

low qualityBad news forturbine blades!

rankine.EES

What is the maximum P[3] value?Psat at T[3]. Any higher and it would be a liquid and not a vapor.

02/25/14 ME313 10

Reheating One way to avoid liquid at the turbine outlet

and have a high maximum pressure is to reheat with a 2-stage (or more) turbine.

boiler

condenser

T

s

1

2

4

3

1 6

3

2

5

4

P

14

32

v6

5

5

6

Each stage of reheating brings a smaller increase in efficiency. Normally only 1 or 2 reheats are done.

What is the optimum pressure for reheating? It is not

That relation was developed for ideal gas, see section 7-11 of the text.

q in=qfirstq reheat=h3h2h5h4

wturbine=h3h4h5h6

P4P3 P6

02/25/14 ME313 11

ExampleWhat is the optimum reheat pressure if Pmax is 10 MPa?

givenintheproblemP[3]=10000T[3]=350P[6]=75reheatT[5]=T[3]constantpressurestepsP[1]=P[6]P[2]=P[3]P[4]=P[5]constantentropystepss[1]=s[2]s[3]=s[4]s[5]=s[6]condenseroutletissat.liquidx[1]=0FirstLawRelationsqin=h[3]h[2]+h[5]h[4]qout=h[6]h[1]wnet=qinqouteff=1qout/qinenthalpiesh[1]=enthalpy(water,x=x[1],P=P[1])h[2]=enthalpy(water,s=s[2],P=P[2])h[3]=enthalpy(water,s=s[3],P=P[3])h[4]=enthalpy(water,s=s[4],P=P[4])h[5]=enthalpy(water,s=s[5],P=P[5])h[6]=enthalpy(water,s=s[6],P=P[6])entropiess[1]=entropy(water,x=x[1],P=P[1])s[3]=entropy(water,P=P[3],T=T[3])s[5]=entropy(water,P=P[5],T=T[5])

T5 is T3 if we assume the reheating is complete

Note: P[4] has not been specified, so this problem cannot be solved yet.

rankine_reheat.EES

02/25/14 ME313 12

Example (cont) Make a parametric table with P[4] and eff,

and set pressure from 1000 to 10000 or so. Plot the data also.

The optimum pressure is around 6000 kPa. EES will do the optimization for you as well.

Select Calculate Min/Max. We want to maximize the efficiency eff with P[4] as the independent variable.

You'll need to set the bounds on P[4]. Use the full range of 75 to 10000 (or 5000 to 7000).

The optimum intermediate pressure is 5782 kPa.

02/25/14 ME313 13

Example P-v and T-s diagrams

Make a T-s diagram from Plots Property Plot. Remove all lines except for P=10000.Now add Plots Overlay Plot, and use the s and T variables.

We drew the 10 MPa isobar because EES just connects the dots and doesn't know the process followed between steps.

02/25/14 ME313 14

TPX Excel Add-In The TPX Excel add-in package can be used as

an alternative to EES if you're more comfortable using Excel.

Install using Tools Add-ins and then browse to the file TPX.XLA.

Let's solve example 10-4 using Excel. The functional form can be viewed using the

drop down menu, choosing Pure Substances.=H('water','PX',500,0.9)=P('water','HS',$A2,C2)

Enthalpy at P=500, x=0.9Pressure using cell references

You may want to change the default units under Tools Units...

Temporary URL for this software and a tutorial ishttp://www.tecnun.es/Asignaturas/Termo/SOFTWARE/TPX/index.html

02/25/14 ME313 15

Excel Example (cont) With the spreadsheet, it is easy to implement

the State Postulate, i.e. two intensive variables completely specify the system.

Move around the cycle, using constant property relations. For example,

At point 6, (P,x) h, s At point 5, T[5]=T[3] due to a perfect reheater. At point 5, s[5]=s[6]. Now fill in P, h. At point 4, P[4]=P[5] due to cont. P reheater.

Nothing more now, so go to point 1. At point 1, P[1]=P[6], now fill in s, h. At point 2, s[2]=s[1], now fill in h. At point 3, P[3]=P[2], now fill in s, h. At point 4, s[4]=s[3], now fill in h.

02/25/14 ME313 16

Regenerative Rankine Cycle Recall that the boiler inlet is at point 2, which is

relatively cold. There is a greater loss of exergy (increase in entropy) by this large T heat transfer.

One method to increase efficiency is to extract some steam from mid-way through the turbine into a feedwater heater to pre-heat the boiler inlet.

An open feedwater heater is a mixing chamber where the steam and condensate are mixed before being further pumped to the maximum pressure.

boiler

condenser

T

s

1

2

7

5

1

6

5

2Feedwaterheater

3

4

7

P

1

6

54

v

2,3

3

4

6

7

y1-y

02/25/14 ME313 17

First Law Rankine Regeneration There are 6 components of this cycle, and the

first law for each isboiler qin=h5h4

Because the stream is split within the turbine, we will define a mass fraction y that gets removed at the mid-point in the turbine.

condenser qout=1 y h7h1 For the turbine, all mass goes from 5-6 (High

Pressure), and (1y) of it goes on further from 6-7 (Low Pressure)

turbine wout=h5h61 y h6h7

There are two pumps, and one only handles the (1-y) fraction of the water

pump I w in,I=1 y h2h1=1 y v P2P1

pump II w in,II=h4h3=v P4P3 The specific volume in the pumping equations

is constant because liquid water is nearly incompressible; v1v2v3v4=v.

turbine wout=w t,HP+ w t,LP

02/25/14 ME313 18

EES Example"given"P[1]=10;x[1]=0;P[3]=15000;T[3]=600;"1-2"h[1]=enthalpy(water,P=P[1],x=x[1]);s[1]=entropy(water,P=P[1],x=x[1]);s[2]=s[1];P[2]=P[3];h[2]=enthalpy(water,s=s[2],P=P[2]);"3-4"s[3]=entropy(water,P=P[3],T=T[3]);h[3]=enthalpy(water,P=P[3],T=T[3]);s[4]=s[3];P[4]=P[1];h[4]=enthalpy(water,P=P[4],s=s[4]);W_net=(h[3]-h[4])-(h[2]-h[1]);Q_in=h[3]-h[2];eff=W_net/Q_in; eff = 0.4303

02/25/14 ME313 19

Example Regenerative RankineP[5]=15000 given in the problemT[5]=600P[7]=10P[6]=1200 open FWH pressureconstant pressure stepsP[1]=P[7]P[4]=P[5]P[2]=P[3]P[3]=P[6]constant entropy stepss[1]=s[2] ; s[3]=s[4]s[5]=s[6] ; s[6]=s[7]condenser outlet is sat. liquidx[1]=0x[3]=0first law relationsqin=h[5]-h[4]qout=(1-y)*(h[7]-h[1])y*h[6]+(1-y)*h[2]=h[3] FWH wnet=qin-qouteff=wnet/qinenthalpies & entropiesh[1]=enthalpy(water,s=s[1],P=P[1])h[2]=enthalpy(water,s=s[2],P=P[2])h[3]=enthalpy(water,s=s[3],P=P[3])h[4]=enthalpy(water,s=s[4],P=P[4])h[5]=enthalpy(water,s=s[5],P=P[5])h[6]=enthalpy(water,s=s[6],P=P[6])h[7]=enthalpy(water,s=s[7],P=P[7])s[1]=entropy(water,x=x[1],P=P[1])s[3]=entropy(water,x=x[3],P=P[3])s[5]=entropy(water,P=P[5],T=T[5])

y = 0.2273eff = 0.4632

The P,T pair is ok here because it is superheated

The individual components of turbine and pump work can be solved for here if desired.

02/25/14 ME313 20

Optimum Feedwater Pressure Let's investigate how the feedwater pressure

effects the efficiency.

Make a parametric table for P[6] and eff.

There appears to be a maximum near 1200 kPa. Use the Min/Max function to get the exact feedwater pressure that gives max. efficiency.

Use Calculate Min/Max. Select Maximum. Set the bounds on pressure from 500 to 1500 (too large range gives convergence problems.)

The optimum feedwater pressure is 1339 kPa. In practice, the pressure is difficult to control

with this precision, but a close value is used and then the fraction y is adjusted with a valve.

Add y to the parametric table and solve.

02/25/14 ME313 21

Reheat and RegenerationLet's add a reheater to this regenerative system with P[8]=1339. We also need to specify the reheater pressure P[6], which is obviously greater than P[8], otherwise the 2nd stage of the turbine would produce no work.

boiler

condenser1

2

9

5

Feedwaterheater

T

s1

6

5

23

4 8

3

4

67

8

7

9

P

1 9

54

v

2,36 7

8

02/25/14 ME313 22

EES Reheat & RegenerationP[5]=15000T[5]=600P[9]=10P[6]=3000P[8]=1339complete reheatT[5]=T[7]constant pressure stepsP[1]=P[9] ; P[4]=P[5]P[2]=P[3] ; P[3]=P[8] ; P[6]=P[7]constant entropy stepss[1]=s[2] ; s[3]=s[4]s[5]=s[6] ; s[7]=s[8] ; s[8]=s[9]condenser outlet is sat. liquidx[1]=0 ; x[3]=0first law relationsqin=h[5]-h[4]+h[7]-h[6]qout=(1-y)*(h[9]-h[1])y*h[8]+(1-y)*h[2]=h[3] wnet=qin-qouteff=wnet/qinenthalpiesDUPLICATE j=1,9 h[j]=enthalpy(water,P=P[j],s=s[j])ENDentropiess[1]=entropy(water,x=x[1],P=P[1])s[3]=entropy(water,x=x[3],P=P[3])s[5]=entropy(water,P=P[5],T=T[5])s[7]=entropy(water,P=P[7],T=T[7])

y = 0.1965eff = 0.4757

02/25/14 ME313 23

Closed Feedwater Heater A closed feedwater heater is a heat exchanger

where the fluid can be at different pressures.

The advantage of a closed FW heater is they don't require separate pumps. A trap is simply a throttle that lowers the pressure. Open FW heater (mixing chambers) are cheaper and have better heat transfer characteristics.

Pressure drop across the trap.

Mixing chamber and heat exchanger!

x=0

superheated

P7P9

02/25/14 ME313 24

Combined Cycle Combined cycles take advantage of high

temperatures in the Brayton, and use of the lower temperature exit in a Rankine.

Ideally, all heat from 8-5 goes into 1-3.

If the heat exchanger is counter flow, T8 approaches T3.

The Brayton may be open cycle, 5' 5

02/25/14 ME313 25

Cogeneration All analysis so far has the rejected heat as

waste, but it can often be used as process heat.

This is not uncommon in rural AK where a community center will use the rejected heat.

Cogeneration simply means production of two desired products from the same source.

The efficiency of a cogeneration system can be 100% is we can utilize the low temperature energy completely. This is often called the utilization factor.

U=W netQprocess

Q in However, this means that no mass is going to

the low pressure part of the turbine. By the 2nd Law, if any mass goes to the lower pressure turbine, u

02/25/14 ME313 26

Cogeneration Example 10-8

givenP[1]=7000P[5]=500P[6]=5isobarsP[9]=P[10]P[10]=P[11]P[11]=P[1]P[4]=P[5]P[5]=P[7]P[6]=P[8]entropiess[5]=s[1]s[6]=s[5]s[8]=s[9]s[7]=s[10]qualityfixesx[7]=0x[8]=0throttleh[4]=h[1]

enthalpiesh[1]=enthalpy(water,P=P[1],s=s[1])h[4]=enthalpy(water,P=P[4],s=s[4])h[5]=enthalpy(water,P=P[5],s=s[5])h[6]=enthalpy(water,P=P[6],s=s[6])h[7]=enthalpy(water,P=P[7],s=s[7])EEShastroubleconvergingwiththish[8]=enthalpy(water,P=P[8],s=s[8])Thisaccomplishesthesameandworkss[8]=entropy(water,P=P[8],x=x[8])h[9]=enthalpy(water,P=P[9],s=s[9])h[10]=enthalpy(water,P=P[10],s=s[10])h[11]=enthalpy(water,P=P[11],s=s[11])

h[1]=enthalpy(water,P=P[1],T=500)need 3h[7]=enthalpy(water,P=P[7],x=x[7])now 2h[8]=enthalpy(water,P=P[8],x=x[8])now 1mixingchamber(x+y)*h[10]+(1xy)*h[9]=h[11]back to 2x=0.1y=0.7yippee!

02/25/14 ME313 27

Combined Gas-Vapor Cycles While raising the temperature of the Brayton

cycle increased efficiency, it does not increase the work generated greatly.

Using a combined cycle allows for the low temperature side of Brayton to be the hot side of the Rankine.

The total work is increased per amount of heat input in addition to an increased efficiency.

boiler

condenser

Heatexchanger

T

1 4

3

2

1

2 3

4

5

6 7

8

5

7

6 8

Ideally, the heat exchanger is perfect, so that T3=T8 and T5=T2. In practice, there will be some temperature difference between them.

Because the heat exchanger is closed, the pressures may be very different, P5 P2.

Air outlet temperature T9 is between T2 and T3.

heat

02/25/14 ME313 28

Set-up Homework Problem 10-86 Ideal Brayton cycle with Pr, TL & TH specified.

Add inefficient turbine and compressor. Add ideal Rankine cycle. Make Rankine turbine inefficient. Calculate work output for both the air and the

steam on a per mass basis as the difference between the compressor and turbine work.

Determine mair/msteam for the heat exchanger.

T 9T 2

T 8T 3

This is NOT an ideal heat exchanger, so T2