02/13/14 ME313 1

Power Cycles Most power-producing devices operate

continuously, and therefore utilize cycles. For example, our lives operate on a daily cycle:

eating, working, resting, repeat. Being engineers, if we wanted to quantify a

person's work output per energy input, we would approximate (idealize) each step in the cycle.

We may approximate the rate of work output as the total daily work done per time, and the rate of energy input as all our meals per time.

Now we can begin to model an average person's efficiency, recognizing that the actual performance will differ. The ideal cycle serves as a benchmark from which we can compare the performance of any real device.

The purpose is to keep our model as simple as possible, but not too simple so as to lose the utility of the model.

02/13/14 ME313 2

Ideal Power Cycles Common assumptions in ideal power cycles

are (1) no friction, (2) quasi-equilibrium compression/expansion, (3) no heat loss between each step in the cycle. (Heat loss during some steps will obviously occur).

The efficiency of power-producing devices is =desired

required=

W netQ in

Recall from Chapter 6 that the most efficient cycle is the Carnot Cycle

=T HT L

T H=1

T LT H

The efficiency is greatest when heat addition occurs at high temperatures, and heat rejection occurs at low temperatures.

Although the Carnot cycle is the most efficient, it is not used in general because of engineering difficulties in building the engine.

For example, heat addition & rejection occur isothermally with negligible temperature differences which is hard to accomplish.

Qcond=k AdTdx

Qconv=hA T sT f

02/13/14 ME313 3

Property Diagrams It is helpful in the analysis of power cycles to

use T-s and P-v property diagrams, because the net work for a cycle is represented by the area within the process curves for both.

The straight lines for constant value processes (isothermal, isentropic) are helpful reminders when analyzing cycles.

The overall efficiency is improved whenever the area can be increased for a given qin.

A real Carnot engine would look like this

W= v dP Q=T dS

02/13/14 ME313 4

Some Cycle Analysis Tips When analyzing a cycle, draw both the P-v and

T-s diagram appropriate for that cycle. Label the 4 (or more) points between step. Note what intensive variables you know at

each point. Recall only 2 intensive variables are required to completely specify the state (that is why 2-dimensional property diagrams are useful).

Find what thermodynamic relations are appropriate for each process step, either 1st or 2nd law relations. For example

q=u2u1=C v T 2T 1

T 2T 1 = v1v2 k1

Constant volume heat addition of an ideal gas:

Isentropic compression/expansion of an ideal gas with constant Cp and Cv: For many gas power cycles, the gas behaves

ideally and some thermodynamic properties like (u, h) are independent of pressure.

However, entropy is still pressure-dependent, even for an ideal gas.

02/13/14 ME313 5

Air Standard Assumption Reciprocating engines are very common in our

everyday life: cars, chain saws, diesel trucks. The air/fuel mixture going in and the exhaust

leaving at each revolution (or two) complicates the analysis, so a common air-standard assumption is used

This assumes (1) working fluid is air, (2) processes are internally reversible, (3) combustion is approximated by heat addition, (4) exhaust is approximated by heat rejection.

One last assumption that can be made is that heat capacities are constant (no temperature dependence), and the cold air standard uses values at 25C.

Using all these assumption greatly simplifies the analysis, but will only provide approximate values. It is useful for back-of-the-envelope type calculations, but modern engine design is more complicated because these assumptions cannot be made.

02/13/14 ME313 6

Otto Cycle The Otto cycle is named after Nikolaus Otto

who built a 4-cycle engine in 1876. The Otto cycle is an idealization of the internal

combustion or spark-ignition engine.

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02/13/14 ME313 7

Reciprocating Engines There is some common terminology used

when describing reciprocating engines. Top dead center (TDC) and bottom dead

center (BDC). Stroke is the distance between TDC and BDC. Bore is the diameter of the cylinder. Clearance volume is the volume at TDC. Displacement volume is the volume between

TDC and BDC (=stroke x bore). Compression ratio is the ratio of the maximum

to minimum volume Mean effective pressure is a fictitious pressure

that would produce an equivalent amount of work when operating over the displace volume.

MEP=wnet

vmaxvmin

02/13/14 ME313 8

Otto Thermal Efficiency The thermal efficiency turns out to be a fairly

simple function of the compression ratio.

th=wnetq in

=1qoutqin

Using air standard assumption (i.e. ideal gas), the heat addition & rejection steps are simply functions of temperature

q in=u3u2=cv T 3T 2qout=u4u1=c vT 4T 1

Putting these 1st Law relations into the definition of efficiency

th=1T 4T 1T 3T 2

=1T 1T 4 /T 11T 2T 3 /T 21

The rearrangement in the last term allows us to use a 2nd Law relation for isentropic process

T 1T 2

= v2v1 k1

= v3v 4 k1

=T 4T 3

The middle terms are equal because the compression ratio is constant.

th=1T 1T 2T 4 /T 11T 3/T 21

=1 1rk1

From above (or seen on left), these ratios are equal (volume ratios are equal), making the 2nd term equal to 1.

compression ratio

02/13/14 ME313 9

Compression Ratio & Efficiency Obviously the efficiency increases with

increasing compression ratio. There are other factors that limit how high the

compression ratio can be. Auto-ignition (knocking) occurs as high compression ratio when temperatures are equally high.

Lead was added to prevent knocking until its environmental hazard became apparent.

Higher octane levels also prevent knocking at higher compression ratio.

The efficiency also decreases with decreasing k-values, and larger k-values are associated with larger molecules.

02/13/14 ME313 10

Example 9-2Consider a Otto cycle with r=8, and initial conditions of 100 kPa, 17C, and qin=800 kJ/kg.Use a variable heat capacity.

P

v

T

s1

4

1

2

3

42

3

Starting from point 1, we can move around the cycle using the initial information and thermodynamic relationships. Isentropic compression from 12

T 2T 1 = v1v2 k1

Constant heat capacity

v1rv2r = v1v2 =r=8Variable heat capacity

vrTPr= Texp s/R

Recall from chapter 7

Using the variable heat capacity approachv1r=676.1 from Table A-17 at 290 KTherefore v2r=676.18=84.5 which corresponds with T=652K

The constant heat capacity relation gives T = 666K

The volume is a 2nd intensive variable (at point 2)

v2=v18=

RT18 P1

=0.287290

8100=0.104 m3 /kg v1=0.832m

3 /kg

We know two intensive values here.

02/13/14 ME313 11

Example 9-2 (cont)Now consider step 2-3, the heat addition step due to combustion of the fuel mixture. This is a constant volume process, so no boundary work and the 1st Law for the closed system reduces to

qin=u3u2= c vT 3T 2

The last equality assumes constant heat capacity. Using the variable approach, find u2 on table A-17.

u2=475.11 kJ/kg by interpolation

Therefore, at state 3u3=q inu2=800475.11=1275.11 kJ/kg

Since u directly corresponds to T for an ideal gas, use Table A-17 again to get a 2nd intensive variable

T 3=1575KWe can get all the properties now. The highest pressure is, using the ideal gas law

P3=RT 3v3

=RT 3v2

=0.2871575

0.104=4346kPa

Continuing to move around the cycle (step 3-4), the 2nd Law isentropic relation is used once again

v4rv3r =8 v 4r=86.108=48.86 T 4=796 K

02/13/14 ME313 12

Example 9-2 (cont)The net work produced in this cycle is

wnet=qinqout=qinu1u4

The u values are found on Table A-17 using Twnet=qinqout=800588.74206.91=418.2 kJ/kg

The thermal efficiency is

th=wnetqin

=418.2800

=0.52=52 %

Recall that the cold air standard gives the thermal efficiency as simply a function of the compression ratio.

th=11

rk1=1 1

80.4=0.565=56.5%

It was the variable specific heat that precluded us from using this more simple relationship to calculate the efficiency.

02/13/14 ME313 13

Diesel Cycle The diesel cycle is the idealization for

compression ignition cycles. The engine was invented by Rudolf Diesel, a

contemporary of Otto's. Only air is compressed, often to such a high a

degree that is above the auto-ignition temp. Knocking does not occur without any fuel.

Fuel is then sprayed into the cylinder which is slowly burned during the first part of the power stroke.

Only step 2-3 is different between Otto (v = constant) and diesel (P = constant)

02/13/14 ME313 14

Diesel Thermal Efficiency The thermal efficiency is slightly different than

that for the Otto cycle. We start the same th=

wnetq in

=1qoutqin

The 1st Law for the heat addition step must account for the boundary work done.

q inW b,out=u3u2 For cold air standard conditions (P2=const.=P3)

q in=P2v3v2u3u2=h3h2=c p T 3T 2

Putting this into the efficiency definition

th=1qoutq in

=1cv T 4T 1c pT 3T 2

=1T 1T 4/T 11

k T 2T 3 /T 21

An additional definition often used when discussing diesel efficiency is the cutoff ratio

rc=v3v2

Using the same isentropic relations from earlierT 1T 2

= v2v1 k1

= 1rk1

T 4T 3= v3v4

k1

More algebraic manipulation leads to

th=11

rk1 [ rck1

k rc1 ] [ v3v4 k1

v2v1 k11

T 3/T 21] and Pv2

T2=

Pv3T 3

Assumes constant heat capacity

Compression ratio

02/13/14 ME313 15

Diesel Thermal Efficiency It can be shown that because k>1, the term in

square brackets is always greater than 1. This means that the Otto cycle has a greater

efficiency than Diesel for the same value of r. Because Diesel engines can operate at higher

compression, they are typically more efficient. Cut-off ratios are small relative to compression

ratios, around 2. The analysis of a Diesel cycle is performed

similar to the Otto cycle. The cut-off ratio may be specified instead of qin, for example.

02/13/14 ME313 16

Dual Cycle The Otto and Diesel cycles are extreme

examples of compression ignition. A dual cycle represents the transition between the two.

In dual cycle, there is both a constant pressure and constant volume step, the extents of which are defined by pressure and cut-off ratios.

The dual cycle is analogous to how a polytropic process is the transition between isentropic and isothermal.

P

v

T

s1

4

1

2

4

52

3

5

3

qin=cv T 3T 2c pT 4T 3

=1qoutqin

=1cv T 5T 1

cv T 3T 2c pT 4T 3

qout=cv T 5T 1

=1 1rk1

r p rck1

k r prc1r p1

Heat in Heat out

Define rp=P3/P2, and a more algebraic manipulation

02/13/14 ME313 17

EES Dual Cycle Example

"to create plots with connected lines"T[6]=T[1];s[6]=s[1];P[6]=P[1];v[6]=v[1];"1st Laws - use average k"W_12=intEnergy(air,T=T[2])-intEnergy(air,T=T[1]);Q_23=intEnergy(air,T=T[3])-intEnergy(air,T=T[2]);Q_34=enthalpy(air,T=T[4])-enthalpy(air,T=T[3]);W_34=P[3]*(v[4]-v[3])*convert(psi*ft^3,Btu);-W_45=intEnergy(air,T=T[5])-intEnergy(air,T=T[4]);"efficiency"eff=W_net/Q_net;W_net=W_34+W_45-W_12;Q_net=Q_23+Q_34;

"given"v[1]/v[2]=15;P[1]=14.2;T[1]=75;"1-2"s[2]=s[1];s[1]=entropy(air,T=T[1],P=P[1]);v[1]=volume(air,T=T[1],P=P[1]);"solve""remaining state 2 variables"T[2]=temperature(air,v=v[2],s=s[2]);P[2]=pressure(air,T=T[2],v=v[2]);"solve""2 -3"P[3]/P[2]=1.1; "given"v[3]=v[2]; "constant volume""solve""get remaining state 3 variable"s[3]=entropy(air,P=P[3],v=v[3]);T[3]=temperature(air,P=P[3],v=v[3]);"solve""3 - 4"v[4]/v[3]=1.4; "given"P[4]=P[3]; "constant pressure""solve""get remaining state 4 variables"s[4]=entropy(air,P=P[4],v=v[4]);T[4]=temperature(air,P=P[4],v=v[4]);"solve""4 - 5"s[5]=s[4]; "constant entropy"v[5]=v[1]; "constant volume""solve""get remaining state 5 variables"P[5]=pressure(air,s=s[5],v=v[5]);T[5]=temperature(air,s=s[5],v=v[5]);"solve"

Problem 9.64E An ideal dual cycle has a compression ratio of 15 and a cutoff ratio of 1.4. The pressure ratio during constant volume heat addition process is 1.1. The state of the air at thebeginning of the compression is P1 =14.2 psia and T1 =75F. Calculate the cycles netspecific work, specific heat addition, and thermal efficiency. Use constant specific heatsat room temperature.

W 12=CV (T 2T 1)Q23=C V (T 3T 2)

Q34W 34=CV (T 4T 3)W 34=P3(v4v3)Q34=C P (T 4T 3)

combined with this

or this

W 45=CV (T 5T 4)

All W's and Q's are positive as written in the equations above

02/13/14 ME313 18

Sterling and Ericsson Cycles Although the ideal Otto and Diesel cycles are

internally reversible, they are externally irreversible because the non-isothermal heat transfer steps require finite temperature differences. (exergy is destroyed)

The ideal Sterling and Ericsson cycles are externally reversible by creating isothermal heat transfer steps using a regenerator.

Carnot Stirling Ericsson

02/13/14 ME313 19

Regenerator A Sterling regenerator may be a porous plug

with high thermal mass (mcp) that temporarily stores energy that is transferred between the cold and hot steps.

An Ericsson regenerator (P=const) is a counter flow heat exchanger with little pressure drop.

A perfect heat exchanger has negligible dT between the fluids as any stage, approaching complete reversibility.

Ericsson engine

Sterling engineT q

34 1

2

regenerator

TH

Tl

flow direction

volume increases (expansion) during heat addition to keep T=constant

volume decreases (compression) during heat removal to keep T=constant

02/13/14 ME313 20

Sterling & Ericsson Efficiency The isothermal heat transfer steps mean the

efficiency of both can approach the Carnot.

=1qoutqin

=1T LT H

Real Ericsson engines must deal with a finite pressure drop and temperature difference in the heat exchanger.

Real Sterling engines must deal with non-isothermal regenerators.

Although not widely used, these external combustion engines have several attributes, notably that the energy source can come from a variety of sources (solar, waste heat).

02/13/14 ME313 21

Stirling ExampleA Stirling cycle operates between 600 and 10 psia. The volume ratio is 10 and minimum temperature is 100F. How much work is produced. Use air standard assumptions.

P

v

T

s

1 2

34

1

2

3

4

Given: P1=600 psiaP3=10 psia

v3v 4=

v2v1=10

T 3=T 4=100F=560R

The work produced is determined by the 1st Law

q inqout=wnet

The heat addition and rejection steps are neither constant pressure nor constant volume and cannot be determined by simple Cv & Cp relations.Recall the isothermal work relation from Chapter 4 for an ideal gas

q in=w1-2=R T 1 lnv2v1

P dV

qout=w3-4=R T 3 lnv4v3=0.06855560R ln 1

10=88.4 Btu/lbm

We still need T1. Two ideal gas relations get us there (3 4 1)

We can get qout now

P4=P3v3v4=10 psia 10=100 psia

T 1=T 4P1P4=560R 600 psia

100 psia=3360R

Constant temperature

Constant volume

q in=0.068553360 ln10=530 Btu/lbm

wnet=53088=442 Btu/lbmFinally

qin

qout

q

wout

win

02/13/14 ME313 22

Brayton Cycle The Brayton cycle is the steady flow counter

part to reciprocating cycles. It can be internally reversible, but is externally irreversible because of non-isothermal heat transfer steps.

An open-system burner system brings in fresh air, compresses it, further heats it in a combustor, turns the turbine, and spits out.

To calculate the efficiency, we'll model the cycle as a closed system with a heat exchanger to remove qout.

Closed system Brayton cycles are sometime used in hostile environments like nuclear power plants where the working fluid is somewhat dangerous.

02/13/14 ME313 23

Brayton Efficiency The efficiency is calculated similar to before.

=1qoutqin

=1c p (T 4T 1)c p(T 3T 2)

=1T 1(T 4/T 11)T 2(T 3 /T 21)

A compact form is arrived at by using the pressure ratio

r pP2P1=

P3P4

The isentropic compressor and turbine relations are used also.

T 2T 1

=P2P1 k1 /k T 3

T 4= P3P4

k1 /k

=1 1r p(k1)/k

Therefore, if k = constant, the temperature ratios are also equal.

The Brayton efficiency reduces

Assumes constant heat capacity

02/13/14 ME313 24

Increasing Efficiency Higher pressure ratios yield higher efficiency,

and the trade-off is the expense of a very-high pressure compressor.

Note also that high pressure ratio means a corresponding higher temperature ratio.

Typical pressure ratios are rp ~ 10-15.

A higher temperature at the combustor outlet also increase efficiency, and is limited by material properties of the turbine blades.

There is an ideal pressure ratio that maximizes the work for a given temperature ratio.

T 2T 1= P2P1

k1/k

r p=(T maxT min )k /[2(k1)]

02/13/14 ME313 25

=1 1r pk1/ k

=1 1200.4/1.4

=0.575

Example

P

v1

2 3

4

s

Find the work produced on a per-mass basis of air, and find the thermal efficiency of the cycle.

T1 = 27C = 300 KT3 = 727C = 1000 K

P1 = 100 kPaP2 = 2 MPa

The Brayton cycle is a simple HEHE

wnet=qinqout=295.5125.6=170 kJ/kg

=w netq in

=1q outq in

We can determine q directly assuming air acts as an ideal gas.q=mC p T

The isentropic compression & expansion steps have

T

1

24

3

qin

qout

T 2=T 1 P2P1 k1 /k

=300K 2000kPa100kPa 0.4 /1.4

=706 K

T 4=T 3 P4P3 k1 /k

=1000K 100kPa2000kPa 0.4/1.4

=425K

The heat addition & rejection steps have

q in=1.005kJ/kgK1000706K=295.5kJ/kgqout=1.005kJ/kgK425300K=125.6kJ/kg

The work produced, and the thermal efficiency are

=1125.6295.5

=0.575=57.5 %

alternatively

carnot=13001000

=0.7

02/13/14 ME313 26

Brayton Isentropic Efficiency The ideal Brayton cycle has isentropic

compression & expansion steps. Inefficiencies in these steps will reduce the net work.

Compression (1-2) increases entropy. Heat addition (2-3) has a pressure drop. Turbine expansion (3-4) increases entropy. Heat rejection (4-1) has a pressure drop.

Compare apples to apples (same P1 & P2).

For ideal gases, h CpT, so T-S diagram is similar to h-S from Ch. 7 isentropic efficiency.

isobar

More compressor work

Less turbine work out

02/13/14 ME313 27

Example (cont)Repeat the previous example with 90% efficient compressor.

T

1

2s4

3

qin

qout

T 2s=300 K2000 kPa100 kPa 0.4 /1.4

=706K

The isentropic temperature is still

From the isentropic efficiency of a compressor

=h2sh1h2ah1

=C p T 2sT 1C pT 2aT 1

=0.9 T 2a=751K

Because lines 2-3 and 4-1 are isobars, the rest of the cycle is the same. Remember, the pressure drop remains the same when comparing the isentropic to actual process.

=1 1r pk1/ k

wnet=qinqout=250125.6=124.4 kJ/kg

The new heat addition step is

q in=1.005 kJ/kgK1000751K=250kJ/kgqout=1.005kJ/kgK425300K=125.6kJ/kg

The work produced, and the thermal efficiency are

=1125.6250

=0.498=50%

We cannot use this relation to determine the efficiency because it was developed using isentropic relations for compression & expansion.

Obviously it would (incorrectly) give the same result because the compression ratio is the same.

2a

02/13/14 ME313 28

Intercooling and Reheating Recall that intercooling during compression

reduces the work requirement between P1 & P2.

P

v

2

1

T

1

2qout

Saved work

s

There is an economic trade-off for the number of intercooling steps.

The optimum intermediate pressure between intercooling steps is

P=P1 P2 P i=nP1ni P2i

Also, more work can also be extracted by using a re-heater, which sends the exhaust from a high-pressure turbine back to the combustion chamber.

T 3 4

qout

combustor

02/13/14 ME313 29

Two-stage Intercooling & Reheat

Additionally, the regenerator is a way of recycling some of the energy. A high number of steps/stages approaches the Ericsson cycle, whose efficiency is the Carnot.

regenerator

02/13/14 ME313 30

ExampleConsider the two-stage intercooling and reheat engine using an ideal gas (e.g. air). What is the minimum number of variables that must be specified to solve for the maximum net work?Assume: isentropic compression and expansion.

Ideal pressure ratios.Negligible pressure drop in pipes.Perfect regenerator.

1

2

3

4

5

6

7

8

9

10

T

s

The minimum and maximum operating pressures are definitely required.

PH=P 4 PL=P11 2

The minimum and maximum temperatures seem like obvious specifications as well.

1

3

4 5 6

7 8

910

P

v

2

T H=T 6 T L=T 13 4The first law relates some temperatures with the net work output. Qin=QoutW net

The isentropic compression and expansion relations are

T 4T 3 = P4P3 k1/ k

T 2T 1 = P2P1 k1/ k

The ideal regenerator means inlet/outlet to the heat exchanger are equal T 5=T 9=T 7 and T 4=T 10 =T 2

T 7T 6 = P7P6 k1/ k

T 9T 8 = P9P8 k1/ k

Work is minimized (maximized) in the compressor (turbine)P2=P1P4 and P7=P6P9

Qin=mC PT 6T 4mC P T 8T 7Qout=mC PT9T 1mC PT 2T 3

02/13/14 ME313 31

Brayton Cycle Using EES

specifyconditionsP[1]=100P[4]=800T[1]=300T[6]=1300cp=1.005airk=1.4air

FirstLawQin=Qout+WnetQin=cp*(T6T4)+cp(T8T7)Qout=cp*(T9T1)+cp(T2T3)

isentropicrelationsT4/T3=(P4/P3)^((k1)/k)T2/T1=(P2/P1)^((k1)/k)T7/T6=(P7/P6)^((k1)/k)T9/T8=(P9/P8)^((k1)/k)

Idealintermediatepres.P[2]=(P[1]*P[4])^0.5P[7]=(P[6]*P[9])^0.5P[4]=P[5]P[5]=P[6]P[9]=P[10]P[10]=P[1]

idealregeneratorT1=T3T4=T2T2=T10T6=T8T7=T9

Let's solve example 9-8 using EESBe sure to set the unit system properly

Qin=1236Qout=773.5Wnet=463

The solution is

The book examples uses variable specific heat and therefore givesQin=1334

Wnet=477

02/13/14 ME313 32

Brayton Cycle Using EESNow modify the EES program to account for variable specific heat capacities. There are two different areas that need to be modified. The heat addition and rejection steps must use enthalpy The isentropic relations need to be modified.

EES does not have a relative volume function, but we can simply specify that the entropy values are equal.FirstLawQin=enthalpy(air,T=T6)enthalpy(air,T=T4)+enthalpy(air,T=T8)enthalpy(air,T=T7)Qout=enthalpy(air,T=T9)enthalpy(air,T=T1)+enthalpy(air,T=T2)enthalpy(air,T=T3)

IsentropicRelationsentropy(air,T=T3,P=P3)=entropy(air,T=T4,P=P4)entropy(air,T=T1,P=P1)=entropy(air,T=T2,P=P2)entropy(air,T=T7,P=P7)=entropy(air,T=T6,P=P6)entropy(air,T=T9,P=P9)=entropy(air,T=T8,P=P8)

Qin=1334Wnet=477

Now we get the same answer as the example in the textbook

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