Applied Thermodynamics

1

Gas Turbine Cycles

N S Senanayake

Air standard cycles• Air standard cycles refers to thermodynamic cycle

with certain assumptions so as to use the principles of thermodynamics conveniently.

• Assumptions– Air is the working fluid and behaves as a perfect gas– Mass and composition of the working fluid will not

change in the cycle– Processes are reversible– Specific heat capacity of the working fluid does not

change

Otto cycle (air standard)

1 – 2 Adiabatic compression

2 – 3 Constant volume heat addition

3– 4 Adiabatic expansion

4– 1 Constant volume heat rejection

Spark Ignition (SI) engines are based on this cycle

Otto cycle …

rVV

VVn ratioCompressio

3

4

2

1

)( , 23 TTcqaddedHeat vin

)()( , 1441 TTcTTcqrejectedHeat vvout

)()( , 1423 TTcTTcwworkNet vvnet

inputHeatworkNetEfficiencyThermal th

,

Otto cycle…

)()(1

)()}(){(

23

14

23

1423

TTTT

TTcTTTTc

v

vth

Consider process 1 – 2

CpV CVpV 1)( CTV 1

122

111

VTVT 11

2

1

1

2

rVV

TT

Consider process 3 – 4

144

133

VTVT 11

3

4

4

3

rVV

TT

2

3

1

Otto cycle…From equations 2 and 3

4

1

3

2

4

3

1

2

TT

TT

TT

TT

4

1

3

2 11TT

TT

4

14

3

23

TTT

TTT

1

4

3

14

23 r

TT

TTTT

From equation 1

123

14 11)()(1

rTT

TTth

Otto cycle efficiency vs. compression ratio

0 5 10 15 20 25 30 35 400

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

compression ratio (r)

Eff

icienc

y

( )

= 1.2

= 1.4

123

14 11)()(1

rTT

TTth

Mean effective pressure (MEP)- Otto cycle

This is the mean pressure which is developed in the cylinder. Defined as the ratio of net work done to the displacement of volume of the piston.

Mean effective pressure (MEP)- Otto cycle

ngeVolume chaNetMEP work

11

11224433

VpVpVpVpworkNet

( ) ( ) 1122443311 VpVpVpVpworkNet

11

11

11

2211

44

3344 Vp

VpVpVpVpVpworkNet

Mean effective pressure (MEP)- Otto cycle

1111

11

1

211

4

344 rp

pVprp

pVpworkNet

rVV

ppVpVp

2

1

1

22211

rVV

pp

3

4

4

3 Similarly

( ) ( ) 111

1 111

114

rVprVpworkNet

Consider process 1 – 2

( ) 14

11

11 pprVworkNet

11

Terminology : Reciprocating Engine

volume swept out by the piston when it moves from TDC to BDC is called the displacement volume.

distance from TDC to BDC is called stroke

• The piston is said to be at the top dead center (TDC) when it has moved to a position where the cylinder volume is minimum. This volume is called a clearance volume.

Spark Ignition vs Compression Ignition

Spark-ignition engines: mixture of fuel and air are ignited by a spark plug.

Compression ignition engines: Air is compressed to high enough pressure and temperature that combustion occurs spontaneously when fuel is injected.

Air-Standard Diesel Cycle The Air-Standard Diesel Cycle is the ideal cycle that approximates the compression ignition engine i.e. Compression Ignition (CI) engines are based on this cycle Process Description 1-2 Isentropic Compression 2-3 Constant Pressure Heat Addition 3-4 Isentropic Expansion 4-1 Constant Volume Heat Rejection

P

V

T

S

2 3

4

1 1

4

2

3P = P2 3

P 1

S = S1 2 S = S3 4

P 4

V =V4 1O O

T1

T4

T2

T3

y

x

V2

Diesel cycle)( , 2323 TTcqaddedHeat p

)( , 4141 TTcqrejectedHeat v

2

1 , VVrrationCompressio

2

3 , VVratiooffCut

23

4123

qqq

addedheatheatnet

th

)(

)()(

23

1423

TTcTTcTTc

p

vpth

Diesel cycle

)()(11

23

14

TTTT

th

( ) 112

11

2

1

1

2

rTTrVV

TT

Process 1 - 2

Process 2 - 3

11323

2

3

2

3 rTTTTTT

VV

1

2

3

Diesel cycleProcess 3 - 4

11

3

2

2

1

1

3

2

2

4

1

3

4

4

3 ..

r

VV

VV

VV

VV

VV

TT

4( )

1

11

1

1

34 Tr

rTr

TT

Substituting from eq. 3

Diesel cycleSubstituting for all Ts in equation 1.

)1()1(11

)()(11 11

11

1

11

rrTrTTT

th

)1()1(

11 1

rth

Diesel cycle efficiency vs. compression ratio

0 5 10 15 20 25 300.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

4

Effic

ienc

y (

)

Compression ratio (r)

=2 = 1.4

)1()1(

11 1

rth

Gas turbines

19

20

Elements of simple gas turbine Power Plant

21

The simple gas turbine power plant mainly consists of a gas turbine coupled to a rotary type air compressor and combustion chamber which is placed between the compressor and turbine in the fuel circuit.

Auxiliaries, such as cooling fan, water pumps, etc. and the generator itself, are also driven by the turbine.

Other auxiliaries are starting device, lubrication system, duct system, etc.

A modified plant may have in addition to the above, an inter-cooler, a regenerator and a re-heater

22

Flow diagram – Gas turbine power plant

Gas turbine cycle• Gas-turbines usually operate on an open

cycle

• A compressor takes in fresh ambient air (state 1), compresses it to a higher temperature and pressure (state 2).

• Fuel and the higher pressure air from compressor are sent to a combustion chamber, where fuel is burned at constant pressure. The resulting high temperature gases are sent to a turbine (state 3).

• The high temperature gases expand to the ambient pressure (state 4) in the turbine and produce power.

• The exhaust gases leave the turbine.

Brayton cycleBy using the air-standard assumptions, replacing the combustion process by a constant pressure heat addition process, and replacing the exhaust discharging process by a constant pressure heat rejection process, the open cycle described above can be modeled as a closed cycle, called ideal Brayton cycle.

25

Open Cycle Gas Turbine

Air

GCompressor

Turbine

CombustorFuel

Generator

50 – 70 % of turbine power

Pressure ratio: usually about 15, but up to 40 and moreTurbine inlet temperature (TIT): 900° - 1700°CTurbine exit temperature (TET): 400° - 600°CPower: 100 kW – 300 MW

Exhaust

26

Closed Cycle Gas TurbineHeat Source

G

Compressor TurbineGenerator

Condensate from Process

Steam to Process

Heat Exchanger

2

1

3

4

Working fluid circulates in a closed circuit and does not cause corrosion or erosion

Any fuel, nuclear or solar energy can be used

The ideal Brayton cycle is made up of four internally reversible processes.

1-2 Isentropic compression

2-3 Constant pressure heat addition

3- 4 Isentropic expansion

4-1 Constant pressure heat rejection

Brayton cycle

Steady Flow Energy Equation

Efficiency of Brayton Cycle

( )232323 TTchhqq pin

( ) ( )414141 TTchhqq pout

23

41

23

4123 1

qq

qqq

addedheatworknet

th

1

111

2

3

1

4

2

1

23

14

TTTT

TT

TTTT

th 1

Efficiency of Brayton CycleConsider process 1 – 2, Isentropic compression

Cp

TCpTC

pTpCpV

11

1

1

2

1

2

1 1

prpp

TT

ratiopressurerpp

p 1

2

Consider process 3 – 4, Isentropic expansion

1

1

3

4

3

4 1

prpp

TT

2

3

Efficiency of Brayton CycleFrom equations 2 and 3:

Substituting equations 2, 3 and 4 in equation 1

2

3

1

4

3

4

2

1 TT

TT

TT

TT

4

1

2

3

2

3

1

11 1

111

pp

th

rTTTT

r 1

11

p

th

r5

31

Work ratio

)()()(

outputWork Net Work

34

3412

34

3412

TTTTTT

wwwrw

( ) )1(

3

11

pw rTTr

32

Equation shows that the work ratio increases in direct proportion to the ratio T3 /T1 and inversely with a power of the pressure ratio.

On the other hand, thermal efficiency equation shows that thermal efficiency increases with increased pressure ratio.

33

Compressor work:w12 = - (h2 – h1 ) = -Cp(T2 – T1)

Heat supplied during the cycle:q23 = (h3 – h2) = Cp(T3 – T2)

23

3412

inputHeat Net Work

qww

Summary of Equations

)1(

11

pr

Turbine work:w34 = (h3 – h4) = Cp(T3 – T4)

34

3412

outputWork Net Work

wwwrw

( )

)1(

3

11

pw rTT

r

Work ratio

Efficiency

34

Improving cycle efficiency and work ratio

According to the equation (5) above, the efficiency of Brayton cycle depends only on the pressure ratio. To maximize the efficiency the pressure ratio has to be increased. This is done by compressing air isentropically from 1 to 2 to the maximum possible pressure. When pressure is increased the temperature also rises proportionately to the order of (-1)/.

The maximum temperature (T3) is fixed by the metallurgical properties, ability to withstand the high temperatures by the turbine materials. In practice the minimum temperature (T1) is limited to the atmospheric temperature.

35

Isentropic compression process to reach the maximum temperature T3 is shown by 1-2. What is shown in the figure is the compression close to T3. The temperature T3 is obtained by heat addition form 2 to 3. If we reach T3 only by compression the points 2 and 3 will coincide. Under this condition the compression work and turbine work becomes equal giving zero net work. Therefore the maximum theoretical pressure ratio is obtained when work ratio is zero

36

1

1

3

1

2max)(

TT

pp

rp

Therefore, for zero net work

( ) )1(

3

110

prTT

It is clear from the cycle 1-2’-3’-4’ of figures, the net work also becomes zero when pressure ratio is unity i.e. p2 = p1. Here also the figures show a pressure ratio very close to 1 for illustration. Here the maximum temperature T3 is achieved by heat supply at the same pressure.

Therefore, the pressure ratio (rp) has to be in between (rp)max and 1.

37

Pressure ratio for maximum net work

The net work out put per unit mass is given by the following equation.

)()( 3412 TTcTTcw pp

Since

1/)1(

/)1(

1

212 Tr

pp

TT p

3/)1(4

1 Tr

Tp

and

111 /)1(3/)1(

1

pppp rTcrTcw

01111/)12(3/11

pp

pp

p rTc

rTc

drdw

38

)1(2/

1

3

TT

rp

( ))1/(

1

3max

TT

rp

Since

max)( pp rr

The maximum net work is obtained when the pressure ratio equals the square root of maximum theoretical pressure ratio.

39

Variation of efficiency and the net work out put with rp and T3

We will see how the efficiency and net work output vary with the pressure ratio and the maximum temperature.

Let us assume the following data are available

Min. temperature, T1 = 30oC = 303K

Isentropic eff. for compressor, comp = 0.85

Isentropic eff. for the turbine, Tub = 0.90

40

Isentropic efficiency

Performance of turbines/compressors are measured by isentropic efficiencies.

The actual work input to the compressor is more and the actual work output from the turbine is more because of irreversibility.

Isentropic efficiencies involve a comparison between the actual performance of a device and the performance that would be achieved under idealized circumstances for the same inlet state and the same exit pressure.

Isentropic efficiency - Turbine

The desired output from a turbine is the work output. Hence,

the definition of isentropic efficiency of a turbine is the ratio of

the actual work output of the turbine to the work output of the

turbine if the turbine undergoes an isentropic process between

the same inlet and exit pressures.

WorkIsentropicWorkTurbineActual

T

The isentropic efficiency of turbine can be written as

12

12

hhhh

s

aT

h1 = enthalpy at the inlet

h2a = enthalpy of actual process at the exit

h2s = enthalpy of isentropic process at the exit 

12

12

TTTT

s

aT

Isentropic efficiency - compressor

The isentropic efficiency of a compressor or pump is defined as

the ratio of the work input to an isentropic process, to the work

input to the actual process between the same inlet and exit

pressures.

WorkActualWorkCompressorIsentropic

C

The isentropic efficiency of compressor can be written as

12

12

hhhh

a

sC

h1 = enthalpy at the inlet

h2a = enthalpy of actual process at the exit

h2s = enthalpy of isentropic process at the exit 

12

12

TTTT

a

sC

45

The Back Work Ratio

Therefore, the turbine used in gas-turbine power plants are larger than those used in steam power plants of the same net power output, P. Usually more than half of the turbine work output is used to drive the compressor.

turbine

comp

WW

ratioworkBack

46

Deviation of Actual Gas-Turbine Cycles from Ideal One

Pressure drop Isentropic efficiency

Example 1A four stroke SI engine has the compression ratio of 6 and swept volume of 0.15m3. Pressure and temperature at the beginning of compression are 98kPa and 60oC respectively. Heat supplied in

the cycle is 150kJ. cp = 1kJ/kgK, cv = 0.71kJ/kgK

Determine

(i) the pressure , volume and temperature at all main state points

(ii) Efficiency

(iii) Mean effective pressure

Example 2

An ideal diesel cycle using air as working fluid has a compression ratio of 16 and a cut off ratio of 2. The intake conditions are 100kPa, 20oC, and 2000cm3.Determine (a) Temperature and pressure at the end of each process(b) Net work output(c) Thermal efficiency(d) Mean effective pressure

cp = 1.0045kJ/kgK, cv 0.7175kJ/kgK

Example 3

In an air standard Brayton cycle the minimum and maximum temperature are 300K and 1200K respectively. The pressure ratio is 10.

(i) Find out temperatures after compression and expansion

(ii) Calculate the compressor and turbine work, each in kJ/kg of air, and thermal efficiency of the cycle.

Example 4

A gas turbine receives air at 1bar, 300K and compresses it adiabatically to 6.2bar. The isentropic efficiency of compressor is 0.88. The fuel has a heating value of 44186kJ/kg and the fuel –air ratio is 0.017kg fuel/kg of air. The turbine efficiency is 0.9. Calculate the work of turbine and compressor per kg of air compressed and the thermal efficiency.For products of combustion cp = 1.147kJ/kgK, g = 1.33.

For air cp = 1.005kJ/kgK, g = 1.4.

51

The ideal air-standard Brayton cycle operates with air entering the compressor at 95 kPa, 22oC. The pressure ratio rp is 6:1 and the air leaves the heat addition process at 1100 K. Determine • the compressor work • the turbine work per unit mass flow, • the cycle efficiency,• the back work ratio, and compare the compressor exit

temperature to the turbine exit temperature.

Assume constant properties.

Example 5

Example 6 In a gas turbine plant, working on the Brayton cycle, helium at 30° C and 22 bar is compressed to a pressure of 64 bar and then heated to a temperature of 1200 °C. After expansion in the turbine, the gas is cooled to initial pressure and temperature.Assume the following:Isentropic efficiency of the compressor – 0.85Isentropic efficiency of the turbine – 0.8Pressure loss in the combustion chamber – 1.2 barPressure loss in the cooler – 0.5 barSpecific heat (Cp) of the products of combustion is the same as that of helium and it is equal to 5.1926 kJ/kg K. Ratio of specific heats of helium – 1.667Determine the following;• Temperature at the end of compression and expansion.• Heat supplied, heat rejected and the net work per kg of helium.• Thermal efficiency of the plant• Flow rate of helium required to give an output of 12 MW.