Chapter Five Applications of Thermodynamic · PDF fileChapter Five Applications of...

Chapter Five Applications of Thermodynamic Cycle Updated on 2014/12/01

Power cycles: Power generation by converting heat to work. Uses: Power generation, Propulsion. Cycles: Carnot cycle, Otto cycle, Diesel cycle, Brayton cycle, Stirling cycle, Rankine cycle. Real applications: Power plant, Automobile engine, Jet engine, Waste heat recovery Refrigeration cycles: Low temperature generation by transfer heat from low temperature reservoir to high temperature reservoir. Uses: Low temperature generation, Cryogenics Cycles: Carnot cycle, Brayton cycle, Stirling cycle, Rankine cycle, Linde cycle, Real applications: Air conditioning, Refrigeration, Cryogenics. Gas cycles: Transport of fluid. Real applications: Compressor, blower, vacuum pump

Advanced Thermodynamics, ME Dept NCHU 頁 1

(5.1). Carnot Cycle (5.1.1). Carnot Power Cycle ‘Reflections on the Motive Power of Fire and on Machines Fitted to Develop that Power’ by Sadi Carnot, 1824.

Fig. 5.1.1 Carnot cycle

1 2→ reversible isothermal compression, LT T= ， 2

1

lnL Lvq RTv

− =

2 3→ reversible adiabatic compression， 3

2

ln ln 0Hv

L

T vc RT v

+ =

3 4→ reversible isothermal expansion, HT T= ， 4

3

lnH Hvq RTv

=

4 1→ reversible adiabatic expansion， 1

4

ln ln 0Lv

H

T vc RT v

+ =

(5.1.1.1). Thermal efficiency

Carnot cycle is an ideal cycle. It has the highest efficiency that no other cycle can beat it. However, it has never been realized in real world. Not because the ideal isentropic process can never be reached, but because there are some intrinsic features that prohibit the realization of this cycle.

3 4

2 1

v vv v

=


1 4

2 3

v vv v

=

H H

L L

Q TQ T

=

1 1L L

H H

Q TQ T

η = − = −

The thermal efficiency is not related to the stroke. It is determined by the temperature of reservoirs only.

Carnot cycle

0

0.2

0.4

0.6

0.8

1

0 2 4 6 8 10

TH/TL

effi

cien

cy

efficiency

Fig. 5.1.2: Efficiency of Carnot cycle

------------------------------------------------------------------------------------------------------ Example：Find the thermal efficiency of a Carnot cycle with TH = 2000 K and TL = 300 K. ------------------------------------------------------------------------------------------------------ Assignment 5.1：A Carnot cycle starts with P1=1 bar, T1=300 K, V1=0.1 m3. The high temperature is 1000 K. Calculate the properties of each state and the work output if the working fluid is air.

P (bar) T (K) V (L) 1 1 300 100 2 3 1000 4

------------------------------------------------------------------------------------------------------


(5.1.1.2) Work output The work output is determined by the stroke of the isothermal process.

4

3

ln (1 )LH H

H

v TW Q mRTv T

η= = −

However, the stroke of the isothermal process can be expressed as

4 4 1 4

3 1 3 1

v v v vv v v v

γ= = ， 4

1

1ln ln ln1

v L L

H H

v c T Tv R T k T= =

−

Where 1

3

vv

γ = is the compression ratio of the engine.

4 4

3 1

1ln ln ln ln ln1

L

H

v v Tv v k T

γ γ= + = +−

1 1ln ln (1 ) ln ln ( 1)1 1

L L L HH L

H H H L

T T T TW mRT mRTk T T k T T

γ γ

= + − = + − − −

It is noted that power output is affected by temperature ratio as well as compression

ratio. However, since compression ratio is in the logarithm form, its effect is small

compared with temperature.

Specific work: the work output per unit mass of working fluid.

1ln ln ( 1)1

L HL

H L

W T Tw RTm k T T

γ

= = + − −

Non dimensional work:

1ln ln ( 1)1

L H

L H L

W T TwmRT k T T

γ

′ = = + − −

Mean effective pressure (MEP): MEP is an index to show how much work can be done in a specific volume of displacement. The higher the value of MEP, the more work that the engine can deliver.

11

1 3 11

1 1 1(1 )

W W W Pmep w PV V m RTV

γ γγ γ

γ

′= = = =− − −−

( 1 LT T= )


1

mep wP

′≈ ( 1γ >> )

------------------------------------------------------------------------------------------------------ Example：Find the work output and the mean effective pressure of a Carnot cycle with TH = 2000 K, TL = 300 K, P1 = 100 kPa,V1 = 0.1 m3, and compression ratio of 200. ------------------------------------------------------------------------------------------------------

(5.1.1.3). Maximum work

1( 1) ln ln1L

Ww X XmRT k

γ ′ = = − − −

H

L

TXT

=

For a given value of compression ratio (γ ), the specific work is determined by the

temperature ratio( X ) only.

1 1 1ln ln ( 1) 01 1

dw X XdX k k X

γ′= − − − =

− −

1 1ln ln 11

Xk X

γ = + − −

This is the temperature ratio that will produce the highest power output provided that

LT and γ are given.

------------------------------------------------------------------------------------------------------ Example：Find the value of TH that will produce maximum output work of a Carnot cycle with TL = 300 K, and compression ratio of 200. ------------------------------------------------------------------------------------------------------ (5.1.1.4). Maximum pressure

Carnot cycle is a pressure limiting cycle. The maximum pressure increases

sharply with power for a given set of reservoirs.

1 12 1 2 1 4

max 3 2 1 1 13 2 3 2 3

k kk kk k

H H

L L

v v v v T v TP P P P P Pv v v v T v T

− − = = ⋅ = ⋅ ⋅ = ⋅ ⋅ = ⋅ ⋅


14

3

H

WmRTv e

vη= ，

1 1

3 1H

kW k

mRT H

L

TP P eT

η−

= ⋅ ⋅

P3 is the maximum pressure in the cycle. If Pmax is the limiting factor in the design

of Carnot cycle, the work output is as follows.

14 max

3 0

kk

L

H

v P Tv P T

− = ⋅

Where 1 0P P= is the reference pressure.

max 1

0

max

0

(1 )ln[ ( ) ]

( 1)[ln ln ]1

kL L k

HH H

L

T P TW mRTT P T

P kmRT X XP k

−= −

= ⋅ − −−

max

0

( 1)[ln ln ]1

P kw X XP k

′ = − −−

max

0

1ln (1 ln ) 01

dw p k XdX p k X′= − − + =

−

max

0

1 11 ln lnk PXX k P

−− + = ，the value of X can be obtained by solving this equation.

max 1

0 3

H H

L L

P v T TP v T T

γ= ⋅ =

max

0

L

H

P TP T

γ = ⋅

------------------------------------------------------------------------------------------------------ Example：For a Carnot cycle with Pmax = 100 bar, P1 = 1 bar, and TL = 300 K, find the thermal efficiency and the specific work output.

Ans：x=2.172，TH = 651.5 K ，η=0.54， 4

3

6.622vv= ，w=5527 kJ/kmole

V1 = 1 liter，n = 4.009×10-5 kmole，W = nw = 0.2216 kJ = 221.6 N-m Tq × 2π = W，Tq = 35.3 Nm in torque Mep = W/V = 221.6 kPa ------------------------------------------------------------------------------------------------------


Example：For a Carnot cycle with Pmax = 300 bar, P1 = 1 bar, V1=0.1m3, and TL = 300 K, find the efficiency, the amount of heat addition, the work output, the mean effective pressure, and the properties of each state.

1 1

1

PVmRT

= = 0.1161 kg

max

0

1 11 ln lnk pxx k p

−− + = = 1.6297

x=2.71，TH = 813 K ，η=0.633

14 max

3 0

kk

L

H

v p Tv p T

− = ⋅

=9.156, max

0

L

H

p Tp T

γ = ⋅ = 110.7

1 2 3 4

T (K) 300 300 813 813 P (bar) 1 9.156 300 32.77 V (m3) 0.1 10.92×10-3 9.03×10-4 8.27×10-3 η

HQ (kJ) W (kJ) IMEP (kPa) 0.633 59.99 37.97 383.16

4

3

lnH HvQ mRTv

= = 59.99 kJ

HW Q η= = 37.97 kJ

1 3 1

111

W WmepV V V

γ

= =− −

= 383.16 kPa

------------------------------------------------------------------------------------------------------ Assignment 5.2：For a Carnot cycle with Pmax = 1000 bar, P1 = 1 bar, and TL = 300 K, find the thermal efficiency and the specific work output. ------------------------------------------------------------------------------------------------------

With constrain of maximum pressure, the thermal efficiency of Carnot cycle is no longer the highest.

The mean effective pressure is low compared with other cycle. As a result, the engine displacement should be very large for Carnot cycle in order to obtain the same output power. (5.1.1.5). Limited heat transfer rate


Isothermal process can only be implemented in two ways. The first way is that the system undergoes an infinitely slow process such that the system and its thermal reservoir may be at the same temperature. The second way is that the heat transfer coefficient between the system and its thermal reservoir is infinitely large such that heat transfer may proceed at regular speed.

If the engine rotates at a regular speed, and the heat transfer coefficient between

the system and its thermal reservoir is NOT infinitely large, there should exist temperature deviation between reservoir and system, and the cycle is no longer isentropic.

In order to enhance heat transfer rate, temperature of reservoir should be higher than that of system in heating period, and during the cooling period, temperature of reservoir should be lower that that of system. The resulting system is no more a reversible system, and the thermal efficiency would be getting lower.

1( )H RH Hq hA T Tω

= − ⋅

1( )L L RLq hA T Tω

= − ⋅

( )1 1( )

L L RL

H RH H

Q T TQ T T

η −= − = −

−

The thermal efficiency of the cycle can be shown above in which RHT is the high thermal reservoir temperature, RLT is the low thermal reservoir temperature, HT is the high temperature of the cycle, and LT is the low temperature of the cycle. It is noted that there is a temperature difference between thermal reservoir and system such that heat transfer may occur. However, the efficiency of Carnot cycle depends on the temperature ratio only.

1 L

H

TT

η = −

As a result, the cycle temperature and the thermal reservoir temperature can be related as the following. ( )( )

L RL L

RH H H

T T TT T T

−=

−

H L H RL L RH H LT T T T T T T T− = −


2L RH

HL RL

T TTT T

=−

, /2 / 1

H L RL

RH L RL

T T TT T T

=−

2 21 1 1 1

2

L L L RL RL L

L RHH RH RH RH

L RL

T T T T T TT TT T T TT T

η −= − = − = − = + −

−

21 ( )2

RL L L RHH RH

RH RH L RL

hA T T T TW q TT T T T

ηω

= = + − − −

/1 2 (1 )/ 2 / 1

RL RL L L RL

RH RH RH RL L RL

W T T T T ThAT T T T T Tω

= + − − −

The value of L

RL

TT

should be confined to the region that efficiency is nonnegative.

1 2 0RL RL L

RH RH RL

T T TT T T

η = + − ≥

11 ( 1)2

L RH

RL RL

T TT T

≤ ≤ +

It is noted that for a given value of /RH RLT T , the output work varies as /L RLT T increases from 1.0 to its maximum allowable value, and peaks a certain value of

/RH RLT T . The efficiency decreases all the way as /L RLT T increases, and the value of HT decreases too. If we raise the value of /RH RLT T , the allowable value of /L RLT T extends to a greater range, and the work output increases too.


Performance of Carnot cycle (TRH/TRL=3.0)

0

0.2

0.4

0.6

0.8

1

1 1.2 1.4 1.6 1.8 2

TL/TRL

perf

orm

ance work

TH/TRH

efficiency

Fig. 5.1.3 Performance of Carnot cycle with limited heat transfer rate

21 ( )2

RL L L RHH RH

RH RH L RL

hA T T T TW q TT T T T

ηω

= = + − − −

2

2 2 2( ) 1 ( )2 2 (2 )

L RH RL L RH L RHRH

L RH L RL RH RH L RL L RL

dW hA T T T T T T TTdT T T T T T T T T Tω

= − − − + − − − − −

2

2 1 22 1 (2 2 )2 (2 )

L RL LL RH RL RH L RH

L RL L RL RH RH

hA T T T T T T T T TT T T T T Tω

= − + − + − − − − −

( )2

2 12 22 (2 )

LRL RH RL RL L RL

L RL L RL

hA T T T T T T TT T T Tω

= − + + + − − −

[ ]2

1 8 8 2 4 2 2(2 ) L L L RL RL RL L L RL L RL RH RL RL L RL

L RL

hA T T T T T T T T T T T T T T T TT Tω

= − + − + − + + −−

[ ]2

1 4 4 0(2 ) L L L RL RL RL RL RH

L RL

hA T T T T T T T TT Tω

= − + − + =−

4 4 ( ) 0L L L RL RL RH RLT T T T T T T− − − =

( ) ( )1 14 16 16 ( )8 2L RL RL RL RL RH RL RL RL RHT T T T T T T T T T= ± + − = ±

1 12

L RH

RL RL

T TT T

= ±


Since 1L

RL

TT

≥ , the only choice is 1 12

L RH

RL RL

T TT T

= +

1 1 12 1 ( 1)

2 221 1

RH RH

RL RLH L RL

RH L RL RHRHRH

RLRL

T TT TT T T

T T T TTTTT

+ +

= = = = +−

+ −

( )1 11 1111

RH

RLL RL RL

H RH RHRL

RH

TTT T T

T T TTT

χ

χ χχ

+ + = = = =

++

21 1 1 1RL L RL RL RH RL

RH RH RH RH RL RH

T T T T T TT T T T T T

η

= + − = + − + = −

2

12RH RL

HRH

hA T TW qT

ηω

= = −

Work of Carnot cycle

0

0.050.1

0.150.2

0.25

1 2 3 4

TL/TRL

wor

k

TRH/TRL=7.0

TRH/TRL=5.0

TRH/TRL=3.0

Fig. 5.1.4: Work of Carnot cycle with limited heat transfer rate ------------------------------------------------------------------------------------------------------ Example: A Carnot engine operates between two thermal reservoirs at 1000K and 300 K. Find the efficiency if heat transfer rate is considered.

RHT = 1000 K, RLT = 300 K, HT = 750 K, LT = 350 K, η = 0.452 ------------------------------------------------------------------------------------------------------ Assignment 5.3: A Carnot engine operates between two thermal reservoirs at 3000K


and 300 K. Find the efficiency if heat transfer rate is considered. ------------------------------------------------------------------------------------------------------

Efficiency of Carnot cycle

0

0.2

0.4

0.6

0.8

1

0 5 10

Temperature ratio

effi

cien

cy

Carnot cycle

Heat transfer

Fig. 5.1.5: Efficiency of Carnot cycle with finite temperature heat transfer

(5.1.1.6). Entropy generation For ideal Carnot cycle, the process are reversible, no entropy is generated.

However, if limited heat transfer is considered, entropy will be increased. Entropy generation of Carnot cycle:

( ) 1 ( ) 1 2H L RH H L RL H L

RH RL RH RL RH RL

q q hA T T hA T T hA T TsT T T T T Tω ω ω

− −∆ = − + = − ⋅ + ⋅ = + −

21 14 4

22 2

RL RH RL RH

RH RL RH RL

hA T T hA T TsT T T Tω ω

∆ = + − = −

------------------------------------------------------------------------------------------------------ Example: A Carnot engine operates between two thermal reservoirs at 1000K and 300 K. Find the entropy generated if heat transfer rate is considered. ------------------------------------------------------------------------------------------------------ (5.1.2). Carnot Refrigerator The Carnot refrigerator is composed of four components, including evaporator,


condenser, compressor, and expander. It’s working principle is just the reverse of

Carnot heat engine.

Fig. 5.1.6: The working principle of Carnot Refrigerator

1 4Lq h h= − , constant pressure heat addition, evaporation of refrigerant

2 3Hq h h= − , constant pressure heat removing, condensation of refrigerant

2 1Cw h h= − , isentropic compression with compressor

3 4Ew h h= − , isentropic expansion with expander

H H

L L

q Tq T

=

net C E net H Lw w w q q q= − = = −

Coefficient of performance : COP

1

1L L L

HH L H L

L

q q TCOP Tw q q T TT

= = = =− − −

1H

L

TT

→ , COP →∞

In most applications of refrigerator, H aT T≈ . The lower the value of LT , the more

difficult to remove heat from thermal reservoir. As a result, the value of COP would

be lower.


2H

L

TT

= , 0.5 150L HT T K= ≈ , 1COP =

150LT K> , 1COP > , Lw q<

150LT K< , 1COP < , Lw q>

If L LoadT T= , and H aT T= , then

H HH

a H

q qsT T

∆ = = ， L LL

Load L

q qsT T

∆ = − = −

0H Lnet H L

H L

q qs s sT T

∆ = ∆ + ∆ = − =

Carnot refrigeration cycle is an ideal cycle in which net entropy generation is zero.

------------------------------------------------------------------------------------------------------ Example: A Carnot refrigerator uses R134a as the refrigerant. It is known that TH=26.7, and TL=-40, find the amount of heat absorption and the COP. P1=0.5164 bar，T1=-40，P2=7.0 bar，T2=26.7

h2=261.85，s2=0.9080，s1=0.9080，x1=0.95，h1=211.69

h3=86.78，s3=0.3242，s4=0.3242，x4=0.3391，h4=75.58

qL=h1-h4=136.11，qH=h2-h3=175.07，wC=h2-h1=50.16，wE=h3-h4=11.20

COP=3.49

Back work ratio = wE/wC=0.223

------------------------------------------------------------------------------------------------------

In real applications, L LoadT T< , and H aT T> , then

H HH

H a

q qsT T

∆ = < ， L LL

L Load

q qsT T

∆ = − > −

0H Lnet H L

a Load

q qs s sT T

∆ = ∆ + ∆ = − >


(5.1.7). The reversible work of refrigeration

The minimum work to cool down an object from room temperature to the

desired low temperature in a reversible process is the reversible work of refrigeration.

The net entropy will not increase during this process.

Ta

QH

W

QL

Fig. 5.1.7: The reversible refrigeration process

0Hsys ev sys

a

QS S S ST

∆ = ∆ + ∆ = ∆ + =

2 1( )H a sys aQ T S T S S= − ∆ = − −

1 2LQ h h= −

2 1 2 1( )H L aW Q Q h h T S S= − = − − −

1 2

2 1 2 1( )L

a

Q h hCOPW h h T S S

−= =

− − −

------------------------------------------------------------------------------------------------------

Example: Find the reversible work to cool down 1kg of brass block from 25 to

-40.

------------------------------------------------------------------------------------------------------

Example: Find the reversible work to make 1kg of ice at -40 from water at 25


.

------------------------------------------------------------------------------------------------------

Assignment 5.4: Find the reversible work to liquefy make 1kg of hydrogen from 25

.

------------------------------------------------------------------------------------------------------


(5. 2). Otto Cycle (5.2.1). Ideal Otto cycle

1 2→ reversible adiabatic compression， 2 2

1 1

ln ln 0vT vc RT v+ = , 12

1

kTT

γ −=

2 3→ constant volume heat addition， 3 2( )H vq c T T= −

3 4→ reversible isothermal expansion， 41

3

1k

TT γ −=

4 1→ constant volume heat rejection， 4 1( )L vq c T T= −

Fig.5.2.1: Otto cycle

4

4 1 1 1

33 2 2

2

1( )1 1 1( ) 1

L v

H v

Tq c T T T T

Tq c T T TT

η−

−= − = − = −

− −

12 3

1 4

kT TT T

γ −= = , 4 3

1 2

T TT T

= , 1

11 kηγ −= −

Work of Otto cycle

3 2 1

1( ) 1H v kw q c T Tηγ −

= = − −

131

1 1

1( ) 1kk

v

w Tc T T

γγ

−−

= − −

Unlike the Carnot cycle in which the thermal efficiency is determined by the

high end temperature, the thermal efficiency of Otto cycle does not depend on the

operating temperature of the cycle. However, the output work does depend on the

high end temperature of the cycle. The higher the value of T3, the more output


work the system will have.

11 31

1 3 11

1( ) 11(1 )

kvk

W c T TmepV V TV

γγ

γ

−−

= = − − − −

Otto cycle

0

0.2

0.4

0.6

0.8

1

0 10 20 30 40 50

compression ratio

ther

mal

eff

icie

ncy

efficiency

Fig. 5.2.2: Efficiency of Otto cycle for air

------------------------------------------------------------------------------------------------------ Example：For an Otto cycle with P1 = 1 bar, T1 = 300 K, V1 = 0.1 m3, qH=1200 kJ/kg, γ=10, find the properties at each state, the mean effective pressure , and the thermal efficiency. (1) Assume that the working fluid is air, and air is an ideal gas with constant heat

capacity. (2) The heat capacity of air is temperature dependent. ------------------------------------------------------------------------------------------------------ Assignment 5.5: If the compression ratio has been raised toγ=15, find the properties at each state, and the thermal efficiency. ------------------------------------------------------------------------------------------------------

(5. 2.2). The effect of compression ratio γ

It can be shown that the higher the compression ratio, the higher the thermal efficiency.

If the effect of heat capacity dependence on temperature is considered, the thermal efficiency still increases as the compression ratio is raised. However, the simple relationship above does not hold any longer.


If dissociations of CO2 and H2O at high temperature are considered, the dependence of thermal efficiency on compression ratio is NOT so clear at high compression ratios.

Furthermore, if heat transfer between burned gas and cylinder wall is considered, it is not necessary that high compression ratio would result in high thermal efficiency.

The thermal efficiency is related to the compression ratio only. No matter how high or how low the temperature is, the efficiency remains the same. However, the output work is proportional to the input heat transfer.

The compression ratio of Otto engine has been raised from 3 to 10 in the past 100 years. The limiting factor in chemical related, not mechanical related. The anti knock property of fuel as well the concern on the exhaust emission of nitrogen oxides is the dominant factor to determine the compression ratio of current internal combustion engines. The compression ratio of internal combustion engine has been raised gradually since 19th century due to the improvement of fuel anti-knock quality. However, this tendency has been reversed at the late 20th century because the concern of air pollution has overwhelmed the search for high efficiency.

Fig. 5.2.3: Evolution of compression ratio


(5.2.4). The effect of ratio of specific heat k

For a fixed value of compression ratio, the thermal efficiency gets higher if the value of k increases.

1

11 kηγ −= −

In the limiting case that as k →∞, then η→1. In the other limiting case that k →1, then η→0. For monatomic gases such as helium and neon, the value of k is 1.667, which is the highest value among all gases. For multi atomic gases such CO2 and H2O, the value of k is much lower. Work in an isentropic process kPV = Ω= constant.

12 2 1 1

1 1 ( )1 1

krev kW PdV dV V PV PV

V k k−Ω

= = = Ω = −− −∫ ∫

For a compression process, 1

2

VV

γ = , 11 1 (1 )1

kC

PVWk

γ −= −−

, 1

1 1

1 (1 )1

kCWPV k

γ −= −−

For an expansion process, 2

1

VV

γ = , 1 11

1(1 )1E k

PVWk γ −= −−

, 11 1

1 1(1 )1

Ek

WPV k γ −= −

−

1p v

v v v

c c R Rkc c c

+= = = +

1vRc

k=

−

Since the gas constant R is a fixed value, the higher the value of cv is, the lower the value of k will be.

Specific heat capacity at constant volume cv is a measure of the capability of gases to absorb heat. If the molecules contain several energy modes, temperature rise caused by heat absorption would be lower because energy is distributed among different modes. The pressure rise associated with the temperature variation is thus diminished and the work produced by volume expansion is reduced.

------------------------------------------------------------------------------------------------------ Example: For an Otto cycle with P1 = 1 bar, T1 = 300 K, V1 = 0.1 m3, qH=1200 kJ/kg, γ=10, find the properties at each state if (a). the gas is air, (b). the gas is helium, (c).


the gas is carbon dioxide, and (d). the gas is ethane. He air CO2 C2H6 k 1.667 1.400 1.289 1.186 M (kg/kmole) 4.003 28.97 44.01 30.070 R (kJ/kg-K) 2.0771 0.287 0.1889 0.2765 cv (kJ/kg-K) 3.116 0.717 0.653 1.490 T1 300 300 300 300 P1 100 100 100 100 m T2 P2 QH (kJ) T3 P3 T4 P4 QL (kJ) W (kJ) η ------------------------------------------------------------------------------------------------------ Assignment 5.6: For an Otto cycle with P1 = 1 bar, T1 = 300 K, V1 = 0.1 m3, qH=1200 kJ/kg, γ=10, find the properties at each state if 25% of air is replaced with burned gases as the following reaction.

4 2 2 2 2 22( 3.76 ) 2 7.52CH O N CO H O N+ + → + +

------------------------------------------------------------------------------------------------------

(5.2.5). The amount of heat addition Hq

The amount of heat addition has no relation with the cycle efficiency. However, the maximum temperature as well as the maximum pressure of the cycle is determined by the amount of heat addition.

3 2( )H vq c T T= − 1

3 2 1kH H

v v

q qT T Tc c

γ −= + = +

13

1 1

k H

v

T qT c T

γ −= +


33 2 1 1

2 1

(1 )k Hk

v

T qP P PT c T

γγ −= = +

3

1 1

k H

v

P qP c T

γ γ= +

The amount of heat addition in real engine is determined by the fuel

characteristic as well as the equivalence ratio.

2 2 2 2 21 3.76( )( 3.76 ) ( )

4 2 4n mm m mC H n O N nCO H O n N

φ φ+ + + → + + +

( )4.76 28.971 4/12.01 1.008

a

f

mnmA Fm n mφ

+ ⋅= =

+

1 /f f f

Hf a

m Q Qq

m m A F= =

+ +

Where fQ is the heating value of fuel (kJ/kg) and /A F is the air fuel ratio of mixture. The amount of energy for the mixture of air and fuel contained in a fixed volume depends on the heating value of fuel as well as the fuel type. If the fuel is of gaseous form like propane or methane, the volume is filled with the gaseous mixture. However, if the fuel is of liquid form like gasoline of methanol, the volume is filled with air only because liquid fuel would occupy negligible fraction of volume only. ------------------------------------------------------------------------------------------------------ Example: Find the amount of heat addition and the associated work output for an Otto engine running with propane mixed with stoichiometric amount of air. The compression ratio is 10. The displacement volume is 1L. The initial condition is 300 K and 100 kP. ------------------------------------------------------------------------------------------------------ Assignment 5.7: Find the amount of heat addition and the associated work output for an Otto engine running with the fuels of methane, methanol, gasoline, and hydrogen respectively mixed with stoichiometric amount of air. The compression ratio is 10. The displacement volume is 1L. The initial condition is 300 K and 100 kP. It is noted that gasoline and methanol are in the form of fine liquid droplets in the air fuel mixture while methane and hydrogen are in the gaseous state in the mixture.


Fuel MJ/kg A/F Methane 50.009 17.24 Propane 46.357 15.67

Methanol 19.930 6.46 Hydrogen 120.971 34.47 gasoline 44.400 14.67

------------------------------------------------------------------------------------------------------

(5.2.6). Otto cycle with temperature limit

13

1 1

k H

v

T qT c T

γ −= +

max max

1 1 1

k H

v

P q TP c T T

γ γ γ= + =

In the case of P1 = 1 bar and T1 = 300 K, if Tmax=3000 K, the maximum pressure would be around 100 bars for normal compression ratio of Otto cycle.

It is noted that 3000 K is very high for normal combustion process, however, 100 bars is not a severe running condition for a real engine. As a result, it is easier to exceed the maximum temperature in an Otto engine.

T3 is the maximum temperature in the cycle. In order to have more power

output, the higher value of T3 the better. However, there is a physical limit. If

Tmax is the limiting factor in the design of Otto cycle, the work output is as follows.

Otto cycle work

-5

0

5

10

0 20 40

compression ratio

wor

k

Tmax=1000 K

Tmax=3000 K

Tmax=5000 K

Fig. 5.2.4 Work of Otto cycle


http://en.wikipedia.org/wiki/Megajoule

http://en.wikipedia.org/wiki/Methane

http://en.wikipedia.org/wiki/Propane

http://en.wikipedia.org/wiki/Methanol

http://en.wikipedia.org/wiki/Hydrogen

http://en.wikipedia.org/wiki/Sulfur

1 1max max max1 1

1 1 1 1

1 1( ) 1 1k kk k

v

w T T Tc T T T T

γ γγ γ

− −− −

= − − = − − +

The maximum work occurs at

2 max

1 1

1( 1) ( 1) 0kk

v

d w Tk kd c T T

γγ γ

− = − − + − =

12( 1)

max

1

kTT

γ−

=

,

12

1 max

1

k TT

γ − =

1 12( 2) 2( 2)

max max max max max

1 1 1 1 1

1

k kk k

v

w T T T Tc T T T T T

− −−

− − = − − +

1 1 22 2

max max max max

1 1 1 1

1 1T T T TT T T T

= − − + = −

------------------------------------------------------------------------------------------------------ Example：For an Otto cycle with P1 = 1 bar, T1 = 300 K, V1 = 0.1 m3, Tmax=3000 K, find the optimum compression ratio that would render maximum output work, the efficiency, the amount of heat addition, the work output, the mean effective pressure, and the properties of each state.

1 1

1

PVmRT

= = 0.1161 kg

12( 1)

max

1

kTT

γ−

=

= 17.782

η=0.684 1 2 3 4

T (K) 300 948.7 3000 948.7 P (bar) 1 56.23 177.82 3.16 V (m3) 0.1 5.62×10-3 5.62×10-4 0.1 η

HQ (kJ) W (kJ) IMEP (kPa) 0.684 178.88 116.88 1238.45

3 2( )H vQ mc T T= − = 178.88 kJ

HW Q η= = 116.88 kJ


1 3 1

111

W WmepV V V

γ

= =− −

= 1238.45 kPa

------------------------------------------------------------------------------------------------------ Assignment 5.8: Compare the efficiency and the work output of a Carnot engine and an Otto engine with the same condition of P1 = 1 bar, T1 = 300 K, and V1 = 0.1 m3 if the maximum temperature must not exceed 3000 K, and the maximum pressure should be under 300 bars. ------------------------------------------------------------------------------------------------------ (5.2.7). Pressure is the limiting factor P3 is the maximum pressure in the cycle. If Pmax is the limiting factor in the design

of Otto cycle, the work output is as follows.

1 1 13 max max max1 1

1 1 1 1 1

1 1 1 1 1( ) 1 ( ) 1 1k k kk k k

v

w T P P Pc T T P P P

γ γ γγ γ γ γ γ

− − −− −

= − − = − − = − − +

The maximum work occurs at 2max max

2 11 1 1

1 1 ( 1) 0kk

v

d w P Pk kd c T P P

γγ γ γ

−+

= − + − − =

max max1

1 1

1 ( 1) kk

P Pk kP P

γγ − = − +

max1

1

( 1) ( 1) kk

P k kP

γγ − − = −

max

11

11 1

k

k

Pkk Pγ

γ −

=− −

1 1 0kkγ − − >

11 2.319kkγ −< =

------------------------------------------------------------------------------------------------------ Example: The maximum pressure of Otto engine is 300 bars, find the optimum compression ratio that would render maximum output work. ------------------------------------------------------------------------------------------------------


(5.2.8). Entropy generation of Otto cycle:

3 2 4 1 4 2

3 1 3 1 1 3

2H Lv v

q q T T T T T Ts c cT T T T T T

− −∆ = − + = − + = + −

3 maxT T= , 12 1

kT Tγ −= , 4 max 1

1kT T

γ −=

2

1 1max 1 max 11 1

1 max 1 max

1 12k kk k

v

s T T T Tc T T T T

γ γγ γ

− −− −

∆= − − = −

12( 1)

max

1

kTT

γ−

=

21144

max 1

1 maxv

s T Tc T T

∆ = −

The reason that entropy would be generated in the Otto cycle is that during the constant volume heat addition process, the system temperature must be lower than the thermal reservoir temperature such that heat transfer may occur. In addition, during the heat removing process, the system temperature must be higher than the thermal reservoir temperature for the same reason. Entropy would be generated if a finite temperature difference heat transfer occurs. ------------------------------------------------------------------------------------------------------ Example: Find the entropy generated of an Otto engine running at Tmax= 3000 K and T1 = 300 K. ------------------------------------------------------------------------------------------------------

(5.2.9). Exergy analysis of Otto cycle

1 2→ reversible adiabatic compression,

1 2, 2 1 0 2 1( )revw u u T s s→ = − + + −

1 2, 2 1aw u u→ = − +

1 2 0 2 1( ) 0I T s s→ = − =


2 3→ constant volume heat addition

02 3, 3 2 0 3 2( ) (1 )rev H

H

Tw u u T s s qT→ = − + + − + −

2 3, 0aw → =

02 3 3 2 0 3 2( ) (1 )H

H

TI u u T s s qT→ = − + + − + −

3 2 3 2( )H vq u u c T T= − + = − +

3 33 2

2 2

ln( ) ln( )vT vs s c RT v

− = +

3 22 3 0

2 3

[ln( ) (1 )]vT TI c TT T→ = − −

3 4→ reversible adiabatic expansion

3 4, 4 3 0 4 3( )revw u u T s s→ = − + + −

3 4, 4 3aw u u→ = − +

3 4 0 4 3( ) 0I T s s→ = − =

4 1→ constant volume heat rejection

4 1, 1 4 0 1 4( )revw u u T s s→ = − + + −

4 1, 0aw → =

1 44 1 1 4 0 1 4 0

4 1

( ) [ln( ) (1 )]vT TI u u T s s c TT T→ = − + + − = − −

Cycle Performance

0(1 )rev HH

Tw qT

= −

4 3 2 1aw u u u u= − + − +

1

11 1Lk

H

qq

ηγ −= − = −


20 0(1 )

a a H H H Hnd

rev H rev HH

H

w w q q T TTw q w T T TqT

η η η η= = = = =− ∆−

------------------------------------------------------------------------------------------------------ Example: For an Otto engine running at P1 = 1 bar, T1 = T0 = 298 K, γ=10, Pmax =100 bars, find the thermal efficiency and the second law efficiency of the cycle. T2 = 748.5 K，P2 = 25.119 bar，T3 = T2 ×P3 /P2 = 2980 K，T4 = 1186.4 K

qH=1601.101 kJ/kg

qL=637.427 kJ/kg

w=963.674 kJ/kg

0(1 )rev HH

Tw qT

= − =1440.991 kJ/kg

1

11 kηγ −= − = 0.6019

20

a Hnd

rev H

w Tw T T

η η= = =−

0.6688

3 22 3 0

2 3

[ln( ) (1 )]vT TI c TT T→ = − − =135.298 kJ/kg

1 44 1 0

4 1

[ln( ) (1 )]vT TI c TT T→ = − − =341.994 kJ/kg

The major irreversibility occurs at the heat rejection process.

------------------------------------------------------------------------------------------------------

(5.2.10). Miller cycle

The expansion stroke is longer than the compression stroke. The expansion

ratio is higher than the compression ratio.

Valve timing control:

Intake valve closes at late timing, and exhaust valve opens at the end of

expansion stroke.

1→2: Constant pressure process


2→3: Isentropic compression

3→4: Constant volume combustion

4→5: Isentropic expansion

5→1: Constant volume heat rejection

4

3

5

2 1

Geometric compression ratio: 3

1g

VV

γ =

Effective compression ratio: 3

2e

VV

γ =

Piston control:

The compression process and the expansion process are taken place in different

cylinder. Air is sucked into the compression cylinder as the piston moves down,

and then is compressed. At the same time, the piston in the expansion cylinder

moves up to expel the burned gas out of the cylinder. The expansion piston is a few

degrees ahead of the compression piston. When the expansion piston moves to the

TDC, all the burned gas has been pushed out. At the same time, the compression

piston is close to TDC, and the inlet air has been compressed to a high pressure.

The valve connecting between these two cylinders then opens to let the compressed

air flow from the compression side to the expansion side. The expansion piston

moves downwards as the compression piston moves upwards such that the air


transfer process is undertaken at constant volume process. When the compression

piston reaches TDC, all the air has been transferred, and the connecting valve closes.

Combustion occurs in the expansion side and the burned gas expands to the end of

stroke. Since compression process and expansion process take place in different

cylinder, the stroke may be different for each process.

------------------------------------------------------------------------------------------------------

Example: A piston control type Miller cycle is combined with Otto cycle. The

compression ratio is 10. The maximum temperature is 2500K.

(1). Fill up the following table.

(2). Determine the mean effective pressure and the thermal efficiency.

(3). Compare the efficiency of this cycle with a standard Otto cycle, and make your

own comments.


P (bar) T (K) V (L)

1 1 300 1

2

3 2500

4 1

------------------------------------------------------------------------------------------------------

(5.2.11). Real internal combustion engine Comparison of ideal engine and real engine

ideal engine real engine

Inlet process Constant pressure air inlet Pressure drop through valve

Compression Polytropic compression Heat transfer and leakage

during compression

Combustion Instantaneous combustion Finite rate of combustion

Incomplete combustion

Expansion Polytropic expansion Heat transfer and leakage

during expansion

Outlet process Constant pressure air outlet Exhaust blow down

Pressure drop through valve

Incomplete gas exchange


Fig. 5.2.5: Ideal and real Otto engine

Irreversibility in the cycle of an internal combustion engine

Mixing: mixing of fuel and air, mixing of residual and fresh mixture

Throttling: free expansion without constrain

Flow restriction

Friction

Heat loss

Leakage

Chemical reaction

Blow down

Deviations from ideal cycle:


Fig. 5.2.6: Deviations from ideal cycle


(5.3). Diesel Cycle (5.3.1). Ideal Diesel cycle


1 1

ln ln 0vT vc RT v+ = , 12

1

kTT

γ −=

2 3→ constant pressure heat addition， 3 2( )H pq c T T= − ,

23 3

2 2 2 2

1

H

p Hc

p

qTcv T q

v T T c Tγ

+= = = = +

3 4→ reversible isothermal expansion， 14 3

3 4

( )kT vT v

−=

4 1→ constant volume heat rejection， 4 1( )L vq c T T= −

Fig. 5.3.1 Diesel cycle

1

1 11( 1)

kc

kck

γηγ γ−

−= −

−, 3

2c

vv

γ γ= ≤ : cutoff ratio

The thermal efficiency is related to the stroke and the heat addition as well.

Diesel efficiency

0

0.2

0.4

0.6

0.8

0 10 20 30 40 50

cut off ratio

effi

cien

cy

eff

Fig. 5.3.2 Efficiency of Diesel cycle ( 50γ = )


Cutoff ratio is an index of how much heat is added, or in practice much fuel is burned, into the engine during the isothermal expansion process. 1 cγ γ≤ ≤ The minimum value of cγ is one, which indicates that no fuel is burned. The maximum value of cγ is γ , which indicates that fuel is burned all the way as piston moves down.

1 >1( 1)

kc

ckγγ−−

, 1 1

1 1 11 1( 1)

kc

k kck

γηγ γ γ− −

−= − < −

−,

It is noted that the efficiency of Diesel cycle is lower than that of Otto cycle provided that the compression ratio is the same. Why is that the fuel economy of Diesel car is better than that of gasoline car?

If 1cγ → , 1

11 kηγ −→ − , the thermal efficiency of Diesel cycle is approaching that of

Otto cycle as heat addition is very little.

13 2 11 1

1 1 1 1( ) 1 ( 1) 1( 1) ( 1)

k kkc c

H p p ck kc c

w q c T T c Tk kγ γη γ γ

γ γ γ γ−

− −

− −= = − − = − − − −

1

1

1( 1)k

k cc

p

wc T k

γγ γ− −= − −

1 1

1

0k kc

c p

d wd c T

γ γγ

− −

= − =

The maximum work occurs at cγ γ= Note: If there is no constraint on the operation of Diesel cycle, the way to obtain the highest efficiency is to add heat as little as possible, and the resulting power is minimum. On the other side, the way to obtain the highest power is to add heat as much as possible, and the resulting efficiency is minimum.


1 1 11

1 21

1 1( 1) ( 1)1 11(1 ) (1 )

k kp k kc c

c c

c Tw k Pmepv v k k kv

γ γγ γ γ γ

γ γ

− − − −= = − − = − − − − − −

------------------------------------------------------------------------------------------------------ Example：For a Diesel cycle with P1 = 1 bar, T1 = 300 K, V1 = 0.1 m3, qH=1200 kJ/kg, γ=10, find the properties at each state, the thermal efficiency, the work output, and the mean effective pressure. (1). Assume that the working fluid is air, and air is an ideal gas with constant heat capacity. (2). The heat capacity of air is temperature dependent. ------------------------------------------------------------------------------------------------------ Assignment 5.9: If the compression ratio has been raised toγ=15, find the properties at each state, the thermal efficiency. ------------------------------------------------------------------------------------------------------ (5.3.2). Temperature limiting cycle If there is any constraint on the operation of Diesel cycle, the most possible one is temperature limitation. Its maximum temperature is subjected to a upper bound such that the engine would not be overheating. T3 is the maximum temperature in the cycle. If Tmax is the limiting factor in the

design of Diesel cycle, the work output is as follows.

max max1

2 1c k

T TT T

γγ −= =

1 max

1

kc

TT

γ γ − = , 1 cγ γ≤ ≤

For example, if Tmax= 3000 K and T1 = 300 K, the value of cγ varies in the range

1~5.18, and the corresponding values of γ are 316.2~5.18.

1 1max max max max1 1 ( 1)

1 1 1 1 1

1 1 1 1( 1) 1k k

k kk k k k

p

w T T T Tc T T k T T k T k

γ γγ γ γ

− −− − −

= − − − = − − +


2 max( 1) 1

1 1

1( 1) ( 1) 0k

kk k

p

d w Tk kd c T T

γγ γ

−− +

= − − + − =

2max( 1) 1

1

1k

kk k

TT

γγ

−− +

=

, ( 1)( 1)max

1

kk kT

Tγ − +

=

( 1)( 1)max

1

kk kT

Tγ

− + =

11 1

max 1 max

1 max 1

kk k

cT T TT T T

γ+ +

= =

1max1

( 1) 1 ( 1)1max max max max

1 1 1 1 1

11 1( 1)

kk

k kk k k

p

TTw T T T k T

c T T T k T k k T

+

+ + +

− + = − − = + −

1 1

1 111

1 1 1max max

1 1 1 1

1 11 1 1 1 1 11 1 1

11

k kk k

max maxkc

k kkc k k k

max max

T TT T

k k kT T T TT T T T

γηγ γ

+ +

−+ + +

− − − = − = − = −

− − −

( 1) ( 1)max max max max

max 1 1 11 1 1 1

1 1 1 1W1

k kk k

pT k T k T k Tmc T PVT k k T k T k k T

+ +

+ + = + − = + − −

------------------------------------------------------------------------------------------------------ Example：For a Diesel cycle with P1 = 1 bar, T1 = 300 K, V1 = 0.1 m3, Tmax=3000 K, find the optimum compression ratio that would render maximum output work, the efficiency, the amount of heat addition, the work output, the mean effective pressure, and the properties of each state.

1 1

1

PVmRT

= = 0.1161 kg

( 1)( 1)max

1

kk kT

Tγ

− + =

= 28.73

11

max

1

k

cTT

γ+

=

= 2.61


1

1 11( 1)

kc

kck

γηγ γ−

−= −

−= 0.672

1 2 3 4 T (K) 300 1149 3000 1149 P (bar) 1 110.07 110.07 3.83 V (m3) 0.1 3.48×10-3 9.08×10-3 0.1 η

HQ (kJ) W (kJ) IMEP (kPa) 0.672 215.87 145.06 1502.94

3 2( )H pQ mc T T= − = 215.87 kJ

HW Q η= = 145.06 kJ

1 3 1

111

W WmepV V V

γ

= =− −

= 1502.94 kPa

------------------------------------------------------------------------------------------------------ Assignment 5.10: In a turbocharged Diesel engine without intercooler, the inlet air is boosted to 200 kPa. Assume that the compressor has an efficiency of 90%，and the maximum allowable temperature is 3000K, find the maximum power output. ------------------------------------------------------------------------------------------------------ (5.3.3). Heat release rate analysis

12 2 1 1 11 1

3 2( ) ( 1) ( 1) ( 1)1

kp p k

H p c c c

c PV c P V k PVq c T TmR mR k m

γ γγ γ γ γ

−−= − = − = − = −

−

The amount of heat addition is related to the compression ratio as well as the cut off ratio.

11 1 ( 1)

1k

H H ckQ mq PV

kγ γ−= = −

−

i i e edUQ m h m h Wdt

+ = + +

0i em m= = dVW Pdt

=

b um m m= +


b b u uU mu m u m u= = + b b u u

b b u udU du dm du dmm u m udt dt dt dt dt

= + + +

b udm dmdt dt

= −

( )b b ub b b u u u

dU dT dm dTm c u u m cdt dt dt dt

= + − +

Assume the mixture of burned and unburned gas is uniformly distributed inside the cylinder.

b uT T T= =

( ) ( ) ( )b bb b u u b u v b u

dU dT dm dT dmm c m c u u mc u udt dt dt dt dt

= + + − = + −

( ) bv b u

dU dU dT dm dVQ W mc u u Pdt dt dt dt dt

= + = = + − +

( ) bb u v

dm dT dVu u Q mc Pdt dt dt

− = − −

PVTmR

=

dT P dV V dPdt mR dt mR dt

= +

( ) pb vb u v

cdm P dV V dP dV dV c dPu u Q mc P Q P Vdt mR dt mR dt dt R dt R dt

− = − + − = − −

1pc k

R k=

−

11

vcR k=

−

( )1 1

bb u

dm k dV k dPu u Q P Vdt k dt k dt

− = − −− −

This is the apparent heat release rate of internal combustion engine. It is determined by the heat transfer rate and the variations of pressure. For ideal Diesel cycle, we have

0Q =

2 1kP P Pγ= =

2( )1

bHRR b u

dm k dVQ u u Pdt k dt

= − − =−


0 (1 cos )2DVV V θ= + +

0

0

DV VV

γ +=

sin sin2 2D DdV V d V

dt dtθθ ω θ= − = −

2 sin1 2

DHRR

k PVQk

ω θ= −−

2 3 2( )1H HRR

kQ Q dt P V Vk

= = −−∫

(5.3.4). Entropy generation of Diesel cycle

3 2 4 1

3 1 3 1

( ) ( )pH L vc T Tq q c T TsT T T T

− −∆ = − + = − +

3 maxT T= , 12 1

kT Tγ −= , 4 max 1

1kc

T Tγ −=

( 1)2 4 1 max max

13 1 max 1 1 1

max

1

1 1 1 1 11 1

kk

kp k

s T T T T Tc T k T k T T k T kT

T

+

−+

∆= − + + − = − + + −

1 21 1

max max

1 1

1 11k kT T

T k T k

−+ +

= − + + −

------------------------------------------------------------------------------------------------------ Example: Find the entropy generated of a Diesel engine running at Tmax= 3000 K and T1 = 300 K. ------------------------------------------------------------------------------------------------------

(5.3.5). Exergy analysis of Diesel cycle

1 2→ reversible adiabatic compression,

1 2, 2 1 0 2 1( )revw u u T s s→ = − + + −


1 2, 2 1aw u u→ = − +

1 2 0 2 1( ) 0I T s s→ = − =

2 3→ constant pressure heat addition

02 3, 3 2 0 3 2( ) (1 )rev H

H

Tw u u T s s qT→ = − + + − + −

2 3, 3 2a Hw u u q→ = − + +

2 3 0 3 2 0 3 23

( ) ( )H H

H

q qI T s s T s sT T→ = − − = − −

3 2 3 2( )H pq h h c T T= − + = − +

33 2

2

ln( )pTs s cT

− =

3 22 3 0

2 3

[ln( ) (1 )]pT TI c TT T→ = − −

3 4→ reversible adiabatic expansion

3 4, 4 3 0 4 3( )revw u u T s s→ = − + + −

3 4, 4 3aw u u→ = − +

3 4 0 4 3( ) 0I T s s→ = − =

4 1→ constant volume heat rejection

4 1, 1 4 0 1 4( )revw u u T s s→ = − + + −

4 1, 0aw → =

1 44 1 1 4 0 1 4 0

4 1

( ) [ln( ) (1 )]vT TI u u T s s c TT T→ = − + + − = − −

Cycle Performance

0(1 )rev HH

Tw qT

= −


1 4a Hw u u q= − +

1 42

0 0(1 )a H H H

ndrev H

HH

w u u q T TTw T T TqT

η η η− += = = =

− ∆−

------------------------------------------------------------------------------------------------------ Example: For a Diesel engine running at P1 = 1 bar, T1 = 298 K, γ=20, Tmax =2500K, find the thermal efficiency and the second law efficiency of the cycle. T0 = 298 K, P0 = 100 kPa. ------------------------------------------------------------------------------------------------------ (5.3.6). Intake and exhaust process 1 2→ constant pressure intake process, 1 2 0T T T= = , 1 2 0P P P= = , 2 3→ reversible adiabatic compression 3 4→ constant pressure heat addition 4 5→ reversible isothermal expansion

5 6→ blow down process,

1

6 6

5 5

kkT P

T P

−

=

6 7→ constant pressure push out process, 6 7T T= , 6 7 0P P P= = The work required to suck air into cylinder: 0 2 1( )iW P V V= − The work required to push air out of cylinder: 0 7 6( )oW P V V= −

0i oW W+ = No net work is required for the gas exchange process theoretically. Practically, the cylinder pressure is lower than the atmospheric pressure during the intake stroke, and is higher than the atmospheric pressure during the exhaust stroke. As a result, the net work is negative, indicating that work is required to complete the gas exchange process. This amount of work is called the pumping work.

7 6 2 1( ) ( ) ( ) 0i o e e i eW W P V V P V V P P V+ = − + − = − ∆ < The exergy loss during the blow down process:


PV mRT=

1

5 5

km Pm P

=

1

55

1 kP dPdm mk P P

=

1 1

5 5 5

k kkT P m

T P m

−−

= =

1 1

55 5 5 5

5 5 5

1 11

kk k

p p pP P dP dP VdH c Tdm c T m c T m dPP k P P k P k

−

= = = = −

The enthalpy contained in the exhaust flowing out of the cylinder during the blow down process is as the following.

5 5 6( )1

V P PHk−

∆ =−

The maximum work that can be recovered back is as the following.

6( )pdW c T T dm= −

6 6 5 55 5 6 6 5 6 5 5 6 6

6 5

1 1( ) ( ) ( ) ( )1 1p p

PV PVW V P P c T m m V P P c Tk k RT RT

= − − − = − + −− −

16 5 5 6 6

5 5 6 6 6 5 55 5 5

1 ( ) ( 1) 1 ( )1 1

kkT PV P PV P P k PV PV k k

k T k P P

− = − + − = − + − − −

If 6 0P P= , the final temperature in cylinder would be

1

06 5

5

kkPT T

P

−

=

1 15 5 0 0 5 5 0

0 55 5 5

( 1) 1 ( ) 1 ( )1 1

k kk kPV P P PV PW k k PV k

k P P k P

− − ∆ = − + − = + − − −

However, more work is needed to push out the remaining gas. 0 5oW PV=


As a result, the net work during the exhaust process would be less than the value shown above.

15 5 0

5

1 ( )1

kkPV PW k

k P

− ∆ = − −

This is the maximum work that can be recovered back. Usually, this part of work is recovered by means of turbocharger. High pressure blow down The piston pushes the gas out of cylinder such that the pressure inside cylinder remains the same.

5 6P P P= =

vmc dT udm hdm PdV+ = −

vmc dT RTdm PdV= −

.mRTP constV

= =

0dm dT dVm T V

+ − =

dm dTdV V Vm T

= +

5 0vdm dTmc dT RTdm P V Vm T

− + + =

0vmc dT RTdm RTdm mRdT− + + =

0pmc dT =

5 .T T const= =

5 6( )pdW c T T dm= −


1

05 6 5 5 5

5

( ) 11

kk

T pk PW c T T m PV

k P

− = − = − −

However, piston has to do extra work to push gas out. The required work is 5 5pW PV=

1 1

0 5 5 05 5 5 5

5 5

1 11 1

k kk kk P PV PW PV PV k

k P k P

− − ∆ = − − = − − −

It can be seen that the work of high pressure blow down equals to that of the natural blow down.

1 1

5 5 0 11 0 0

5 0

1 ( ) 11

k kk k

p pPV P PW k mc T T mc Tk P P

− − ∆ = − = − = − −


(5.3.7). Comparison of Otto cycle and Diesel cycle Comparison of Otto cycle and Diesel cycle：( maxT the same)

Otto cycle Diesel cycle

γ 12( 1)

max

1

kTT

−

( 1)( 1)

max

1

kk kT

T

+ −

cγ 11

1

kmaxTT

+

max

1v

wc T

2

max

1

1TT

−

1

max

1 1

1 1(1 )

kk

maxT TT T k k

+ − + +

η 12

1

max

1 TT

−

1

1

1max

1 1

111

kk

max

kk

max

TT

k T TT T

+

+

−

−

−

------------------------------------------------------------------------------------------------------ Example: (T1=300K) Tmax=3000K Tmax=4000K

Otto Diesel Otto Diesel

γ 17.78 28.73 25.48 43.71

cγ 2.61 2.94

max

1v

wc T

4.675 4.147 7.030 6.280

η 0.684 0.672 0.726 0.713


------------------------------------------------------------------------------------------------------ If Otto cycle and Diesel cycle have the same value of maximum temperature and the

same amount of heat addition, then the required compression ratio would be

different.

1,3 2 1

kH HOtto Otto

v v

q qT T Tc c

γ −= + = +

1,3 2 1

kH HDiesel Diesel

p p

q qT T Tc c

γ −= + = +

,3 ,3Diesel OttoT T=

1 11 1

k kH HDiesel Otto

p v

q qT Tc c

γ γ− −+ = +

1 1

1

1 1k k HDiesel Otto

v p

qT c c

γ γ− −

= + ⋅ −

One the other side, if Otto cycle and Diesel cycle have the same value of maximum

pressure and the same amount of heat addition, then the required compression ratio

would be also different.

23

,3 2 1 1 1 12 2 2 1

1 1

H

k k kv H HOtto Otto Otto Otto k

v v Otto

qTT c q qP P P P PT T c T c T

γ γ γγ −

+

= ⋅ = ⋅ = ⋅ + = ⋅ +

,3 2 1k

Diesel DieselP P Pγ= =

,3 ,3Otto DieselP P=

1 111

1k kHOtto Dieselk

v Otto

qP Pc T

γ γγ −

⋅ + =

1

k k HDiesel Otto Otto

v

qc T

γ γ γ= +

1

11

1k

HDiesel Otto k

v Otto

qc T

γ γγ −

= ⋅ +


------------------------------------------------------------------------------------------------------ Example: For an Otto cycle running with the following condition,

γ=10.0, Hq = 1800 kJ/kg, T1 = 300 K, P1 = 100 kPa

Find the compression ratio of a Diesel cycle if

(1). Both the maximum temperature

(2). Both the maximum pressure are the same.

Tmax = 3262.3 K，Pmax = 108.74 bar

Otto (γ=10.0) Diesel(γ=28.482) Diesel(γ=53.179)

Same maxP Same maxT

T(K) P(Bar) T(K) P(bar) T(K) P(Bar)

1 300 1.0 300 1.0 300 1.0

2 753.6 25.12 1145.4 108.74 1470.3 260.64

3 3262.3 108.74 2937.3 108.74 3262.3 260.64

4 1298.7 4.33 1121.3 3.74 915.5 3.05

η 0.6019 0.6726 0.7547

ηOtto 0.6019 0.7381 0.7960

------------------------------------------------------------------------------------------------------ Assignment 5.11: If the maximum allowable temperature is 3000K for both the Otto cycle and the Diesel cycle, find the maximum work output for each cycle. ------------------------------------------------------------------------------------------------------


(5.4). Brayton Cycle (5.4.1). The ideal Brayton cycle


1 1

ln ln 0vT Pc RT P− = ,

12

1

kkT

Tφ

−

=

2 3→ constant pressure heat addition， 3 2( )H pq c T T= − , 2

3

2 2 2

1

H

p H

p

qTcT q

T T c T

+= = +

3 4→ reversible isothermal expansion，1

4 3

3 4

( )kkT P

T P

−

=

4 1→ constant pressure heat rejection， 4 1( )L pq c T T= −

Fig. 5.4.1 Brayton cycle

4

4 1 1 1

33 2 2

2

1( )1 1 1

( ) 1

pL

H p

Tc T Tq T T

Tq c T T TT

η−

−= − = − = −

− −

12 3

1 4

kkT T

T Tφ

−

= = , 4 3

1 2

1 1T TT T− = −

1( 1) /2 1

1 11 1( / ) kk k

kP Pη

φ−−= − = −

Let 1k

kX φ−

= , 11X

η = −

The thermal efficiency is related to the pressure ratio only. No matter how high or how low the temperature is, the efficiency remains the same. However, the output work is proportional to the input heat transfer.


33 2 1

1

1 1( ) 1 ( ) 1H p pTw q c T T c T X

X T Xη = = − − = − −

3 3

1 1 1

1 1p

w T T Xc T T T X

= − − +

------------------------------------------------------------------------------------------------------ Example：For a Brayton cycle with P1 = 1 bar, T1 = 300 K, P2 = 10 bar, qH=1200 kJ/kg, find the properties at each state, the thermal efficiency, and the work output. ------------------------------------------------------------------------------------------------------

(5.4.2). Temperature limiting Brayton cycle

T3 is the maximum temperature in the cycle. If Tmax is the limiting factor in the

design of Brayton cycle, the work output is as follows.

max max

1 1 1

1 1p

w T T Xc T T T X

= − − +

1 1max

1 11 1

1 1 1 0kk

kp k

d w k T kd c T k T k

φφ φ

−−

−+

− −= − =

1 1 2( 1)1 1max

1

k k kk k kT

Tφ φ

− − −− + +

= =

2( 1)max

1

kkT

Tφ

− =

, max

1

TXT

=

1

maxmax

1

1 11 1 1 TX TT

T

η = − = − = −

1 22

max max max max max1

1 1 1 1 12max

1

1 1 1p

w T T T Tc T T T T TT

T

= − − + = −


------------------------------------------------------------------------------------------------------ Example: For a Brayton cycle with P1 = 1 bar, T1 = 300 K, P2 = 10 bar, Tmax= 1500 K, find the properties at each state, the thermal efficiency, and the work output. ------------------------------------------------------------------------------------------------------ (5.4.3). Entropy generation of Brayton cycle

3 2 4 1 4 2

3 1 3 1 1 3

2H Lp p

q q T T T T T Ts c cT T T T T T

− −∆ = − + = − + = + −

3 maxT T= , ( 1) /2 1

k kT Tφ −= , 4 max ( 1) /

1k kT T

φ −=

2

( 1) / ( 1) /max 1 max 1( 1) / ( 1) /

1 max 1 max

1 12k k k kk k k k

p

s T T T Tc T T T T

φ φφ φ

− −− −

∆= − − = −

2( 1)max

1

kkT

Tφ

− =

,

21144

max 1

1 maxp

s T Tc T T

∆ = −

------------------------------------------------------------------------------------------------------ Example: Find the entropy generation in the previous example. ------------------------------------------------------------------------------------------------------

(5.4.4). Brayton Cycle with regenerator


Fig. 5.4.2 Brayton cycle with regenerator

Fig. 5.4.3 The Ts diagram of Brayton cycle with regenerator

1 2→ 1

2

1

kkT

Tφ

−

=

2 5→ 5 2 4 6( ) ( )R p pq c T T c T T= − = − 5 3→ 3 5( )H pq c T T= −

3 4→ 1

4

3

1( )kkT

T φ

−

=

4 6→ 5 2 4 6( ) ( )R p pq c T T c T T= − = − 6 1→ 6 1( )L pq c T T= −

In the limiting case, the regenerator works ideally such that 4 5T T= , and 2 6T T= .

2

6 1 2 1 1 1 11

43 5 3 4 3 3 3

3

1( ) 11 1 1 1 1 11( ) 1 1

pL

H p

Tc T Tq T T T T X TT XTq c T T T T T T T

T X

η−

− − −= − = − = − = − = − = −

− − − −


It is noted that the condition that efficiency with regenerator is greater than efficiency

without regenerator is that

1

3

11 1T XT X

− > −

1

3

1 T XX T>

2 3

1

TXT

<

Fig. 5.4.4 The Efficiency of Brayton cycle with regenerator

------------------------------------------------------------------------------------------------------ Example: The maximum temperature of Brayton engine is 1500 K, and the pressure ratio is 5.0. Find the improvement in cycle efficiency if a regenerator is adopted. ------------------------------------------------------------------------------------------------------


1 1 11 1

3 4 33 3

1( ) 1 1 ( ) 1k k kk k k

H p pT Tw q c T T c TT T

η φ φφ

− − − = = − − = − −

1 11 1 1 1

3 33 3 3 3

1 11 ( ) 1k kk k

p pT T T Tw c T c TT T T Tφ

φ

− − = − − + = − Χ − + Χ

133 2

3

1 0pdw Tc Td T

= − + = Χ Χ

1max

1

kkT

Tφ

−

Χ = =

2( 1)max

1

kkT

Tφ

− =

The optimum value of φ for Brayton cycle with regenerator is identical to that

without regenerator. 1

1 21 1 max 1

max max 1 max

1 1 1kkT T T T

T T T Tη φ

− = − = − = −

It is noted that the maximum work occurs at the point that the efficiency with

regenerator is the same with without regenerator.

(5.4.5). Regenerative gas turbine with inter cooling and reheat

QH1

10 9

2 4 5 6

1

3 7 8

QH2

Fig. 5.4.5 Regenerative gas turbine with inter cooling and reheat


1 3 0T T T= =

6 8 HT T T= =

2 4 0T T T X= =

7 91

HT T TX

= =

5 9T T=

1 6 7( )T pW c T T= − , 2 8 9( )T pW c T T= −

1 2 1( )C pW c T T= − , 2 4 3( )C pW c T T= −

012 (1 ) 2 ( 1)net p H pW c T c T XX

= − − −

5 4 8 71( ) ( ) 2 (1 )H p p p HQ c T T c T T c TX

= − + − = −

00

12 (1 ) 2 ( 1)112 (1 )

p H pnet

H Hp H

c T c T XW TX XQ Tc T

X

η− − −

= = = −−

------------------------------------------------------------------------------------------------------ Example: The maximum temperature of Brayton engine is 1500 K, and the pressure ratio is 10. Find the improvement in cycle efficiency if intercooler and re-heater are adopted. ------------------------------------------------------------------------------------------------------

If the compression process is divided into three stages with intercoolers installed in

between, and the expansion process is also divided into three stages with re-heater

installed in between, then the efficiency would be as the following. 1

0 31H

T XT

η = −


The limiting case is that both the compression process and the expansion process are

isothermal. In that case, the combustor is no longer needed because the regenerator

may raise the inlet air to the temperature of TH.

5 4

2 3

1

QC QH

Fig. 5.4.6 Limiting case of regenerative gas turbine with inter cooling and reheat

20

1

lnCPW RTP

=

3

4

lnT HPW RTP

=

net T CW W W= −

H TQ W=

01net

H H

W TQ T

η = = −

The efficiency of the limiting case is that of Carnot cycle.

P T

V S

Fig. 5.4.7 The cycle of limiting case


------------------------------------------------------------------------------------------------------ Assignment 5.12: The maximum temperature of Brayton engine is 1500 K, and the pressure ratio is 16. Find the improvement in cycle efficiency if three sets of intercoolers and re-heaters are adopted. ------------------------------------------------------------------------------------------------------

(5.4.6). Real gas turbine engine

For a gas turbine equipped with a regenerator and burning jet fuel, the

irreversibility of the cycle is obtained step by step analysis as the following.

Compressor efficiency, turbine efficiency, intercooler effectiveness are

considered in a real gas turbine engine.

------------------------------------------------------------------------------------------------------

Example: The maximum temperature of Brayton engine is 1500 K, and the pressure

ratio is 10. It is known that compressor efficiency is 80%, turbine efficiency is 85%,

intercooler effectiveness is 70%. Find the cycle efficiency.

------------------------------------------------------------------------------------------------------

Example: A regenerative gas turbine with inter cooler and reheating is shown in

the figure. The inlet air is at 300K and 100 kPa. Air is compressed to the pressure

of 400 kPa in the first stage and then is cooled down by the intercooler. Air is

compressed again in the second stage to 1600 kPa and then flows into the regenerator

and the combustor consecutively. The maximum temperature is 1400K. These

two compressors are driven by the high pressure turbine. The low pressure turbine

drives a generator. In the reheating chamber, air is heated to the maximum

temperature again. The compressor efficiency is 85%. The turbine efficiency is

90%. Both the regenerator and the intercooler have an effectiveness of 0.7. Assume

that air is an ideal gas with constant heat capacity.

(1). Find the work output of this system.

(2). Find the thermal efficiency.

(3). Find the thermal efficiency if all the components are perfect with efficiency 1.0.


QH1

10 9

2 4 5 6

1

3 7 8

QH2

Real case Ideal case T (K) P(kPa) T (K) P(kPa) 1 300 100 300 100 2 471.5 400 445.8 400 3 351.4 400 300 400 4 552.3 1600 445.8 1600 5 830.7 1600 800.8 1600 6 1400 1600 1400 1600 7 1027.6 469.4 1108.4 706.5 8 1400 469.4 1400 706.5 9 950.0 100 800.8 100 10 671.6 100 445.8 100

Real case

8 9( )pw c T T= − =452 kJ/kg

1 6 5( )H pq c T T= − =571.9

2 8 7( )H pq c T T= − =372.4

1 2H H Hq q q= + =944.3

H

wq

η = =0.479

Ideal case

8 9( )pw c T T= − =599.2 kJ/kg


1 6 5( )H pq c T T= − =599.2

2 8 7( )H pq c T T= − =261.6

1 2H H Hq q q= + =890.8

H

wq

η = =0.673

------------------------------------------------------------------------------------------------------ (5.4.7). Reversible work of gas turbine engine without a regenerator

1 2→ ，adiabatic compression

1 2, 2 1 0 2 1( )revw h h T s s→ = − + + −

1 2, 2 1aw h h→ = − +

1 2 0 2 1( ) 0I T s s→ = − =

2 3→ ，constant pressure heat addition

02 3, 3 2 0 3 2[ ] (1 )rev H

H

Tw h h T s s qT→ = − + + − + −

3 2Hq h h= − ， 3HT T≥

0 3 02 3, 0 3 2 0 3 2

2

( ) ln( ) ( )rev H p pH H

T T Tw T s s q T c c T TT T T→ = − − = − −

If 3HT T=

3 22 3, 0

2 3

[ln( ) (1 )]rev pT Tw c TT T→ = − −

2 3, 0aw → =

3 22 3 0

2 3

[ln( ) (1 )]pT TI c TT T→ = − −

3 4→ ，isentropic expansion

3 4, 4 3 0 4 3( )revw h h T s s→ = − + + −

3 4, 4 3aw h h→ = − +

3 4 0 4 3( ) 0I T s s→ = − =


4 1→ ，constant pressure heat rejection

4 1, 1 4 0 1 4[ ]revw h h T s s→ = − + + −

4 1, 0aw → =

1 44 1 1 4 0 1 4 0

4 1

[ ] [ln( ) 1]pT TI h h T s s c TT T→ = − + + − = + −

Cycle performance

0, 2 1 0 2 1 3 2 0 3 2( ) ( ) (1 )cycle rev H

H

Tw h h T s s h h T s s qT

= − + + − − + + − + −

4 3 0 4 3 1 4 0 1 4( ) ( )h h T s s h h T s s− + + − − + + −

10 0

, 00

(1 ) [ (Pr) ][1 ]k

H kcycle rev H p

H H

T T Tw q c TT T T

−

= − = − −

, 2 1 4 3cycle aw h h h h= − + − +

1

20 0 0

1[1 ( ) ](1 )

ka a H H H Hk

nd T Trev H rev H H

HH

w w q q T TTw q w T T T TqT

η η ηφ

−

= = = = = −− −−

0 02 1 4 3 4 1 3 2(1 ) ( ) ( )cycle H

H H

T TI q h h h h h h h hT T

= − − − + − + = − − −

1 1 1 10 0

0 0 00

1 1[ ( ) ] [ ] [ ( ) 2 ]k k k k

Hk k k kp H p H p

H H

T T Tc T T c T T c TT T T

φ φφ φ

− − − −

= − − − = − +

------------------------------------------------------------------------------------------------------ Example：Calculate the reversible work of an ideal Brayton cycle with TH=1200 K,

and Pr=10. Assume that air is an ideal gas with constant heat capacity

P1 = 100 kPa，T1 = 298 K，P2 = 1000 kPa ，T2 = 575.3 K

P3 = 1000 kPa，T3 = 1200 K，P4 = 100 kPa ，T4 = 621.5 K

wc =-278.55 kJ/kg，wT =581.10 kJ/kg，wa =302.55 kJ/kg

qH= 627.51 kJ/kg，qL= -324.96 kJ/kg


0(1 )rev HH

Tw qT

= − =471.68 kJ

rev aI W W= − =169.13 kJ/kg

3 22 3 0

2 3

[ln( ) (1 )]pT TI c TT T→ = − − =220.071 – 155.83 = 64.24 kJ/kg

1 44 1 0

4 1

[ln( ) 1]pT TI c TT T→ = + − =104.93 kJ/kg


2ndη =0.6414， Tη =0.4821

------------------------------------------------------------------------------------------------------ Example：Calculate the reversible work of an ideal Brayton cycle with TH=1200 K,

and Pr=10. The heat capacity of air is not a constant.

P1 = 100 kPa，T1 = 298 K，h1 = 298.18 kJ/kg ，s1=1.69528 kJ/kg-K

P2 = 1000 kPa，T2 = 570.5 K，h2 = 576.11 kPa ，s2=2.35531 kJ/kg-K

P3 = 1000 kPa，T3 = 1200 K，h3 = 1277.79 kJ/kg，s3=3.17888 kJ/kg-K

P4 = 100 kPa，T4 = 665.0 K，h4 = 675.81 kJ/kg，s4=2.51787 kJ/kg-K


qH= 701.68 kJ/kg，qL= -377.63 kJ/kg

0(1 )rev HH

Tw qT

= − =527.43 kJ


02 3, 0 3 2( )rev H

H

Tw T s s qT→ = − − = 245.424-174.25=71.17 kJ/kg

4 1 1 4 0 1 4[ ]I h h T s s→ = − + + − =132.50 kJ/kg


2ndη =0.6144， Tη =0.4618

------------------------------------------------------------------------------------------------------


For fixed values of TH and T0，the pressure ratio φ could be varied to

minimize the minimum value of Icycle.

00

0

1[ 2 ] 0Hp

H

d T Tc T XdX T X T

− + = ，1k

kX φ−

=

02

0

1 0H

H

T TT X T

− + =

1

0

kH kTX

Tφ

−

= = ， 1

0

( )k

H kTT

φ −=

1 10

00

1[ ( ) 2 ] 0k k

H k kcycle p

H

T TI c TT T

φφ

− −

= − + =

02

0

[1 ] 1Hnd

H H

T TT T T

η = − =−

However, at the pressure ratio obtained by the approach mentioned above, the outlet

temperature of compressor equals to that of the maximum temperature, and no heat

addition is required. The output work is thus zero. This is a trivial situation,

because no work can be obtained. It seems that the higher the pressure, the higher

the total efficiency as well as the second law efficiency. However, the higher the

pressure ratio, the lower the heat addition and the resulting work output is getting

lower too.

There is an optimized value of pressure that would maximize the output work. 1 1

0 00

1 1[1 ( ) ] [ (Pr) ] (1 )( )Pr

k kHk k

a T H p H pTw q c T T c T X

X Tη

− −

= = − − = − − ，1

(Pr)kkX−

=

0 20

1( 1 ) 0a Hp

dw Tc TdX X T

= − + =

0

HTXT

= ， 2( 1)

0

Pr ( )kkHT

T−=


20 0

0 0 0

(1 )( ) ( 1)o H H Ha p p

H

T T T Tw c T c TT T T T

= − − = −

101[1 ( ) ] [1 ]

Pr

kk

TH

TT

η−

= − = −

10 0

, 0 00 0 0

[ (Pr) ][1 ] [ ][1 ]k

H H Hkcycle rev p p

H H

T T T T Tw c T c TT T T T T

−

= − − = − −

02

0

[1 ] Hnd

H H

T TT T T

η = −−

------------------------------------------------------------------------------------------------------ Example：Calculate the best pressure ratio for an ideal Brayton cycle with TH=1200

K.

2( 1)

0

Pr ( )kkHT

T−= =11.447

P1 = 100 kPa，T1 = 298 K，P2 = 1144.7 kPa ，T2 = 598.0 K

P3 = 1144.7 kPa，T3 = 1200 K，P4 = 100 kPa ，T4 = 598.0 K


qH= 604.71 kJ/kg，qL= -301.35 kJ/kg

0(1 )rev HH

Tw qT

= − =454.54 kJ


2ndη =0.6674， Tη =0.5017

------------------------------------------------------------------------------------------------------

(4.6.8). Reversible work of real gas turbine engine

For a gas turbine equipped with a regenerator and burning jet fuel, the

irreversibility of the cycle is obtained step by step analysis as the following.


1 2→ ，adiabatic compression

1 2, 2 1 0 2 1( )revw h h T s s→ = − + + −

For a compressor with efficiency of cη , the actual work is

1 21 2, 2 1

sa

c

h hw h hη→−

= − + =

1

22 1

1

11 1

kk

c

PT TPη

− = + −

1

1 2 0 2 1 01( ) ln 1 1 ( 1)ln

kk

pc

I T s s T c kφ φη

−

→

= − = + − − −

2 3→ ，constant pressure combustion

Air + fuel → products

R PH H=

2( ) ( ) ( )f f f a a p p Hm h T m h T m h T+ =

Since TH is known, the mass of fuel is determined by the energy balance of reaction.

0 2( ) ( ) ( )R p p H f f a aS m s T m s T m s T∆ = − +

RS∆ is the increase of entropy associated with the combustion reaction.

2 3, 3 2 0 3 2 0( )rev Rw h h T s s T s→ = − + + − = ∆ ，since 2 3h h=

2 3, 0aw → =

2 3 0 RI T s→ = ∆

3 4→ ，adiabatic expansion

3 4, 4 3 0 4 3( )revw h h T s s→ = − + + −

3 4, 4 3aw h h→ = − +


3 4 0 4 3( )I T s s→ = −

4 1→ ，constant pressure gas exchange

There is no process 4 1→ in a real gas turbine engine, since the exhaust gas does

not return back to the inlet. The cycle is complete with a constant pressure gas

exchange process.

4 5→ ，constant pressure cooling

The products are cooled in the atmosphere to T0 without doing any work.

4 5, 5 4 0 5 4( )revw h h T s s→ = − + + −

4 5, 0aw → =

4 5 5 4 0 5 4( )I h h T s s→ = − + + −

Cycle performance

, 2 1 0 2 1 3 2 0 3 2 4 3 0 4 3( ) ( ) ( )cycle revw h h T s s h h T s s h h T s s= − + + − − + + − − + + −

1 5 0 5 1( )h h T s s= − + −

Where h1 and s1 are the properties of air and fuel at state 1, and h4 and s4 are the

properties of products at state 4.

, 2 1 4 3 4 1cycle aw h h h h h h= − + − + = − ，since 2 3h h=

1 42

1 5 0 5 1( )a

ndrev

w h hw h h T s s

η −= =

− + −

4 5 0 5 1( )cycleI h h T s s= − + −

------------------------------------------------------------------------------------------------------ Example：Calculate the reversible work of a gas turbine engine with TH=1200 K, and

Pr=10. The fuel is methane and methane is compressed separately to the same

pressure of air. It is assumed that all the compression and expansion processes are


of the efficiency of 0.85.

4 2 2 2 2 2 22 2 7.52( 3.76 ) 2 ( 2)CH O N CO H O O Nφ φ φ

+ + → + + − +

14.2

φ = ， 4 2 2 2 2 2 28.4 31.58 2 6.4 31.58CH O N CO H O O N+ + → + + +

P1 = 100 kPa，T1 = 298 K，h1=-64.17 kJ/kg，s1=6.982 kJ/kg-K

P2 = 1000 kPa，T2 = 611.5 K，h2=266.265 kJ/kg，s2=7.066 kJ/kg-K

P3 = 1000 kPa，T3 = 1198 K，h3=266.265 kJ/kg，s3=7.879 kJ/kg-K

P4 = 100 kPa ，T4 = 757.0 K，h4=-255.69 kJ/kg，s4=8.008 kJ/kg-K

P5 = 100 kPa ，T5 = 298 K，h5=-750.10kJ/kg，s5=7.010 kJ/kg-K


1 5 0 5 1( )revw h h T s s= − + − = 694.274 kJ

rev aI W W= − =502.752 kJ/kg

1 2 0 2 1( )I T s s→ = − =25.032 kJ/kg

2 3 0 RI T s→ = ∆ =242.274 kJ/kg

3 4 0 4 3( )I T s s→ = − =38.442 kJ/kg

4 5 5 4 0 5 4( )I h h T s s→ = − + + − =197.006 kJ/kg

The major irreversibility is accomplished by the reaction. The second one is the

exhaust energy released to the atmosphere. It is noted that the loss of exergy in

reaction can hardly be improved because that is the most convenient way to heat up

gas to the required temperature. However, one of the reason that the loss of exergy

in reaction is so huge is that dilute air is too much. If the dilute air can be reduced,

the loss of exergy can be improved.

2ndη =0.2759

The lower heating value of methane is 50144 kJ/kg, and the air fuel ratio is 72.40, so

the averaged heating value of the mixture is about 683.20 kJ/kg.


Tη =0.2803

------------------------------------------------------------------------------------------------------

Example 5.5.5：

A regenerative gas turbine with inter cooling and reheat runs with the following

condition:

TH=T6=T8=1200 K，T1=T3=T0=298 K，T5=T9，T4=T10，Pr=10

P2/P1=P4/P3，P7/P6=P9/P8，

T2=T4= 414 K，T7=T9=863.6 K

qH=2×1.0045×(1200-863.6)=675.828 kJ/kg

wT=2×1.0045×(1200-863.6)=675.828 kJ/kg

wC=2×1.0045×(414-298)=233.044 kJ/kg

wa=442.784 kJ/kg

wrev=507.997 kJ/kg 1

0 21 (Pr)k

kT

H

TT

η−

= − =0.6552

2a

ndrev

wq

η = =0.8716

Single stage without regeneration: 1

00

0

[ (Pr) ](1 )k

H krev p

H

T Tw c TT T

−

= − −

1 1

01[1 (Pr) ] [1 ( ) ]Pr

k kk k

a p p Hw c T c T− −

= − + −


1

00

[ (Pr) ]k

H kH p

Tq c TT

−

= −

111

(Pr)T k

k

η −= − ， 20

Hnd T

H

TT T

η η=−

Single stage with regeneration: 1

01[1 ( ) ](1 )Pr

kk

rev p HH

Tw c TT

−

= − −

1 1

01[1 (Pr) ] [1 ( ) ]Pr

k kk k


= − + −

11[1 ( ) ]Pr

kk

H p Hq c T−

= −

101 (Pr)

kk

TH

TT

η−

= − ， 20

Hnd T

H

TT T

η η=−

Double stage with regeneration: 1

20

0

12 [1 ( ) ]( 1)Pr

kHk

rev pTw c TT

−

= − −

1 12 2

012 [(Pr) 1] 2 [1 ( ) ]Pr

k kk k


= − − + −

1212 [1 ( ) ]

Pr

kk

H p Hq c T−

= −

10 21 (Pr)

kk

TH

TT

η−

= − ， 20

Hnd T

H

TT T

η η=−


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