UNIT 1 GAS & STEAM POWER CYCLES 1.1 PREREQUISITE … · 1.2 POWER CYCLES Ideal Cycles, Internal...

1.1 PREREQUISITE DISCUSSIONS

Discussion of this gas power cycles will involve the study of those heat engines in whichthe working fluid remains in the gaseous state throughout the cycle. We often study the idealcycle in which internal irreversibilities and complexities (the actual intake of air and fuel, theactual combustion process, and the exhaust of products of combustion among others) areremoved. We will be concerned with how the major parameters of the cycle affect theperformance of heat engines. The performance is often measured in terms of the cycleefficiency.

The cycle is defined as the repeated series of operation or processes performed on asystem, so that the system attains its original state.

The cycle which uses air as the working fluid is known as Gas power cycles. In the gas power cycles, air in the cylinder may be subjected to a series of operations

which causes the air to attain to its original position. The source of heat supply and the sink for heat rejection are assumed to be external to the

air. The cycle can be represented usually on p-V and T-S diagrams.

1.2 POWER CYCLES

Ideal Cycles, Internal Combustion

Otto cycle, spark ignition

Diesel cycle, compression ignition

Sterling & Ericsson cycles

Brayton cycles

Jet-propulsion cycle

Ideal Cycles, External Combustion

Rankine cycle

1.3 IDEAL CYCLES

Idealizations & Simplifications

Cycle does not involve any friction

All expansion and compression processes are quasi-equilibrium processes

UNIT – 1 GAS & STEAM POWER CYCLES

AIR STANDARD CYCLES

Pipes connecting components have no heat loss

Neglecting changes in kinetic and potential energy (except in nozzles & diffusers)

1.4 GAS POWER CYCLES

Working fluid remains a gas for the entire cycle

Examples:

Spark-ignition engines

Diesel engines

Gas turbines

Air-Standard Assumptions

Air is the working fluid, circulated in a closed loop, is an ideal gas

All cycles, processes are internally reversible

Combustion process replaced by heat-addition from external source

Exhaust is replaced by heat rejection process which restores working fluid to initial state

1.5 ENGINE TERMS

Top dead center

Bottom dead center

Bore

Stroke

Clearance volume

Displacement volume

Compression ratio

Mean effective pressure (MEP)

CYCLES AND THEIR CONCEPTS

1.6 OTTO CYCLE

An Otto cycle is an idealized thermodynamic cycle that describes the functioning of atypical spark ignition piston engine. It is the thermodynamic cycle most commonly found inautomobile engines. The idealized diagrams of a four-stroke Otto cycle Both diagrams

Petrol and gas engines are operated on this cycle Two reversible isentropic or adiabatic processes Two constant volume process

1.7 PROCESS OF OTTO CYCLE

Ideal Otto Cycle

Four internally reversible processes

o 1-2 Isentropic compression

o 2-3 Constant-volume heat addition

o 3-4 Isentropic expansion

o 4-1 Constant-volume heat rejection

Thermal efficiency of ideal Otto cycle:

Since V2= V3 and V4 = V1

1.8 DIESEL CYCLE

The Diesel cycle is a combustion process of a reciprocating internal combustion engine.In it, fuel is ignited by heat generated during the compression of air in the combustionchamber, into which fuel is then injected.

It is assumed to have constant pressure during the initial part of the "combustion" phaseThe Diesel engine is a heat engine: it converts heat into work. During the bottom

isentropic processes (blue), energy is transferred into the system in the form of work , butby definition (isentropic) no energy is transferred into or out of the system in the form of heat.During the constant pressure (red, isobaric) process, energy enters the system as heat .During the top isentropic processes (yellow), energy is transferred out of the system in the formof , but by definition (isentropic) no energy is transferred into or out of the system in theform of heat. During the constant volume (green,isochoric) process, some of energy flows out ofthe system as heat through the right depressurizing process . The work that leaves thesystem is equal to the work that enters the system plus the difference between the heat added tothe system and the heat that leaves the system; in other words, net gain of work is equal to thedifference between the heat added to the system and the heat that leaves the system.

1.9 PROCESSES OF DIESEL CYCLE:

1-2 Isentropic compression

2-3 Constant-Pressure heat addition

3-4 Isentropic expansion

4-1 Constant-volume heat rejection


1.8 DIESEL CYCLE










1.8 DIESEL CYCLE









For ideal diesel cycle

Cut off ratio rc

Efficiency becomes

1.10 DUAL CYCLE

The dual combustion cycle (also known as the limited pressure or mixed cycle) is athermal cycle that is a combination of the Otto cycle and the Diesel cycle. Heat is added partly atconstant volume and partly at constant pressure, the advantage of which is that more time isavailable for the fuel to completely combust. Because of lagging characteristics of fuel this cycleis invariably used for diesel and hot spot ignition engines.

Heat addition takes place at constant volume and constant pressure process . Combination of Otto and Diesel cycle. Mixed cycle or limited pressure cycle

1.11 PROCESS OF DUAL CYCLE

Isentropic compression

Constant-volume heat rejection

Constant-pressure heat addition

Isentropic expansion

Constant-volume heat rejection

The cycle is the equivalent air cycle for reciprocating high speed compression ignitionengines. The P-V and T-s diagrams are shown in Figs.6 and 7. In the cycle, compressionand expansion processes are isentropic; heat addition is partly at constant volume andpartly at constant pressure while heat rejection is at constant volume as in the case of theOtto and Diesel cycles.

1.12 BRAYTON CYCLE

The Brayton cycle is a thermodynamic cycle that describes the workings of a constantpressure heat engine. Gas turbine engines and airbreathing jet engines use the Brayton Cycle.Although the Brayton cycle is usually run as an open system (and indeed must be run as such ifinternal combustion is used), it is conventionally assumed for the purposesof thermodynamic analysis that the exhaust gases are reused in the intake, enabling analysis as aclosed system. The Ericsson cycle is similar to the Brayton cycle but uses external heat andincorporates the use of a regenerator.

Gas turbine cycle

Open vs closed system model

With cold-air-standard assumptions

Since processes 1-2 and 3-4 are isentropic, P2 = P3 and P4 = P1

Pressure ratio is

Efficiency of Brayton cycle is

1.13 REAL TIME APPLICATIONS

PETROL ENGINES

Datsun Go Hyundai Xcent Maruti Suzuki Celerio Volkswagen Vento Nissan Terrano

DIESEL ENGINES

Isuzu Diesel Cars Datsun Diesel Cars Ashok Leyland Diesel Cars

GAS TURBINES

Indraprastha (Delhi) CCGT Power Station India Kovilkalappal (Thirumakotai) Gas CCGT Power Station India Lanco Tanjore (Karuppur) CCGT Power Plant India

1.14 TECHNICAL TERMS

TDC: Top Dead Center: Position of the piston where it forms the smallest volume BDC: Bottom Dead Center: Position of the piston where it forms the largest volume Stroke: Distance between TDC and BDC Bore : Diameter of the piston (internal diameter of the cylinder) Clearance volume: ratio of maximum volume to minimum volume VBDC/VTDC Engine displacement : (no of cylinders) x (stroke length) x (bore area) (usually given in

cc or liters) MEP: mean effective pressure: A const. theoretical pressure that if acts on piston

produces work same as that during an actual cycle Gas Power Cycles: Working fluid remains in the gaseous state through the cycle.

Sometimes useful to study an idealised cycle in which internal irreversibilities andcomplexities are removed. Such cycles are called:Air Standard Cycles

The mean effective pressure (MEP): A fictitious pressure that, if it were applied tothe piston during the power stroke, would produce the same amount of net work as thatproduced during the actual cycle.

Thermodynamics: Thermodynamics is the science of the relations between heat ,workand the properties of system

Boundary: System is a fixed and identifiable collection of matter enclosed by a real orimaginary surface which is impermeable to matter but which may change its shape orvolume. The surface is called the boundary

Surroundings: Everything outside the system which has a direct bearing on thesystem's behavior.

Extensive Property: Extensive properties are those whose value is the sum of thevalues for each subdivision of the system, eg mass, volume.

Intensive Property: Properties are those which have a finite value as the size of thesystem approaches zero, eg pressure, temperature, etc.

Equilibrium: A system is in thermodynamic equilibrium if no tendency towardsspontaneous change exists within the system. Energy transfers across the system disturbthe equilibrium state of the system but may not shift the system significantly from itsequilibrium state if carried out at low rates of change. I mentioned earlier that to definethe properties of a system, they have to be uniform throughout the system.Therefore to define the state of system, the system must be in equilibrium.

Inequilibrium of course implies non-uniformity of one or more properties). Isentropic process: Isentropic process is one in which for purposes of engineering

analysis and calculation, one may assume that the process takes place from initiationto completion without an increase or decrease in the entropy of the system, i.e., theentropy of the system remains constant.

Isentropic flow: An isentropic flow is a flow that is both adiabatic and reversible.That is, no heat is added to the flow, and no energy transformations occur due tofriction or dissipative effects. For an isentropic flow of a perfect gas, several relationscan be derived to define the pressure, density and temperature along a streamline.

Adiabatic heating: Adiabatic heating occurs when the pressure of a gas is increasedfrom work done on it by its surroundings, e.g. a piston. Diesel engines rely onadiabatic heating during their compression.

Adiabatic cooling: Adiabatic cooling occurs when the pressure of a substance isdecreased as it does work on its surroundings. Adiabatic cooling occurs in the Earth'satmosphere with orographic lifting and lee waves, When the pressure applied on aparcel of air decreases, the air in the parcel is allowed to expand; as the volumeincreases, the temperature falls and internal energy decreases.

1.15 SOLVED PROBLEMS

1. In an Otto cycle air at 1bar and 290K is compressed isentropic ally until the pressure is15bar The heat is added at constant volume until the pressure rises to 40bar. Calculate theair standard efficiency and mean effective pressure for the cycle. Take Cv=0.717 KJ/Kg Kand Runiv = 8.314KJ/Kg K.

Given Data:

Pressure (P1) = 1bar = 100KN/m2

Temperature(T1) = 290K



Cv = 0.717 KJ/KgKRuniv = 8.314 KJ/Kg K

To Find:

i) Air Standard Efficiency (ηotto)

ii) Mean Effective Pressure (Pm)

Solution:

Here it is given Runiv = 8.314 KJ/Kg KWe know that ,

(Here Cp is unknown)

Runiv = M R

Since For air (O2) molecular weight (M) = 28.97

8.314=28.97 R∴ R = 0.2869

(Since gas constant R = Cp-Cv )

0.2869 = Cp – 0.717∴ Cp= 1.0039 KJ/Kg K

η

Here ‘r’ is unknown.

We know that,

r= 6.919

ηotto

To Find:



Solution:



Runiv = M R


8.314=28.97 R∴ R = 0.2869


0.2869 = Cp – 0.717∴ Cp= 1.0039 KJ/Kg K

η


We know that,

r= 6.919

ηotto

To Find:



Solution:



Runiv = M R


8.314=28.97 R∴ R = 0.2869


0.2869 = Cp – 0.717∴ Cp= 1.0039 KJ/Kg K

η


We know that,

r= 6.919

ηotto

∴ ηotto = 3.87%

Mean Effective Pressure (Pm) =

Pm =

Pm = 569.92 KN/m²

2. Estimate the lose in air standard efficiency for the diesel engine for the compressionratio 14 and the cutoff changes from 6% to 13% of the stroke.

Given Data

Case (i)

Case (i)

Compression ratio (r) = 14 compression ratio (r) =14

ρ = 6% Vs ρ = 13%Vs

To Find

Lose in air standard efficiency.

SolutionCompression ratio (r) =

Case (i):

Cutoff ratio (ρ) =V3/V2

∴ ηotto = 3.87%


Pm =

Pm = 569.92 KN/m²


Given Data

Case (i)

Case (i)


ρ = 6% Vs ρ = 13%Vs

To Find



Case (i):


∴ ηotto = 3.87%


Pm =

Pm = 569.92 KN/m²


Given Data

Case (i)

Case (i)


ρ = 6% Vs ρ = 13%Vs

To Find



Case (i):


ρ =

ρ = 1.78

We know that,

ηdiesel

= 0.6043x100%

ηdiesel =60.43%

case (ii):

cutoff ratio (ρ)

=1+(0.13) (13)

ρ = 2.69

ηdiesel

= 1- (0.24855) (1.7729)= 0.5593 100%

=55.93%

Lose in air standard efficiency = (ηdiesel CASE(i) ) - (ηdiesel CASE(i) )

= 0.6043-0.5593

ρ =

ρ = 1.78

We know that,

ηdiesel

= 0.6043x100%

ηdiesel =60.43%

case (ii):

cutoff ratio (ρ)

=1+(0.13) (13)

ρ = 2.69

ηdiesel

= 1- (0.24855) (1.7729)= 0.5593 100%

=55.93%


= 0.6043-0.5593

ρ =

ρ = 1.78

We know that,

ηdiesel

= 0.6043x100%

ηdiesel =60.43%

case (ii):

cutoff ratio (ρ)

=1+(0.13) (13)

ρ = 2.69

ηdiesel

= 1- (0.24855) (1.7729)= 0.5593 100%

=55.93%


= 0.6043-0.5593

= 0.0449

= 4.49%

3. The compression ratio of an air standard dual cycle is 12 and the maximum pressure onthe cycle is limited to 70bar. The pressure and temperature of the cycle at the beginning ofcompression process are 1bar and 300K. Calculate the thermal efficiency and Mean EffectivePressure. Assume cylinder bore = 250mm, Stroke length = 300mm, Cp=1.005KJ/Kg K,Cv=0.718KJ/Kg K.

Given data:

Assume Qs1 = Qs2

Compression ratio (r) = 12

Maximum pressure (P3) = (P4) = 7000 KN/m2

Temperature (T1) = 300KDiameter (d) = 0.25mStroke length (l) = 0.3m

To find:Dual cycle efficiency (ηdual)Mean Effective Pressure (Pm)

Solution:

By Process 1-2:

= [r]γ-1

300[12

T2 = 810.58K

P2 = 3242.3KN/m2

= 0.0449

= 4.49%


Given data:

Assume Qs1 = Qs2





Solution:

By Process 1-2:

= [r]γ-1

300[12

T2 = 810.58K

P2 = 3242.3KN/m2

= 0.0449

= 4.49%


Given data:

Assume Qs1 = Qs2





Solution:

By Process 1-2:

= [r]γ-1

300[12

T2 = 810.58K

P2 = 3242.3KN/m2

By process 2-3:

T3 = 1750K

Assuming Qs1 = Qs2

mCv[T3-T2] = mCp[T4-T3]

0.718 [1750-810.58] = 1.005 [T4-1750]T4 = 2421.15K

By process 4-5:

We know that, = 1.38

T5 = 1019.3K

By process 2-3:

T3 = 1750K

Assuming Qs1 = Qs2


0.718 [1750-810.58] = 1.005 [T4-1750]T4 = 2421.15K

By process 4-5:


T5 = 1019.3K

By process 2-3:

T3 = 1750K

Assuming Qs1 = Qs2


0.718 [1750-810.58] = 1.005 [T4-1750]T4 = 2421.15K

By process 4-5:


T5 = 1019.3K

Heat supplied Qs = 2

Qs = 1349KJ/Kg

Heat rejected T1]Qr

= 516.45 KJ/Kg

ηdual

ηdual = 61.72%

Stroke volume (Vs) =

Vs = 0.0147m3

Mean Effective Pressure (Pm)= 832.58/0.0147

Pm = 56535 KN/m2

4. A diesel engine operating an air standard diesel cycle has 20cm bore and30cmstroke.the clearance volume is 420cm3.if the fuel is injected at 5% of the stroke,findthe air standard efficiency.

Given Data:-

Bore diameter (d) =20cm=0.2mk

Stroke, (l) =30cm=0.3m

Clearance volume, (v2 ) =420cm3=420/1003= m3

To Find:-


Qs = 1349KJ/Kg

Heat rejected T1]Qr

= 516.45 KJ/Kg

ηdual

ηdual = 61.72%


Vs = 0.0147m3


Pm = 56535 KN/m2


Given Data:-




To Find:-


Qs = 1349KJ/Kg

Heat rejected T1]Qr

= 516.45 KJ/Kg

ηdual

ηdual = 61.72%


Vs = 0.0147m3


Pm = 56535 KN/m2


Given Data:-




To Find:-

Air standard efficiency,(diesel) Solution:-

Compression ratio, r = v1/v2

= (vc+vs)/vc

We knowthat,

Stroke volume, vs=area*length

=

= )

Vs=

Therefore,

Compression ratio, (r) =

r = 23.42

Cut off ratio, /+5% ) /

We know the equation,



= (vc+vs)/vc

We knowthat,


=

= )

Vs=

Therefore,


r = 23.42





= (vc+vs)/vc

We knowthat,


=

= )

Vs=

Therefore,


r = 23.42



the pressure corresponding to state 2 and the cycle is completed. Such a cycle is

known as Rankine Cycle.

Processes:

1-2 → Isentropic (reversible adiabatic) compression in pump.

2-3 → Constant Pressure heat addition in boiler.

3-4 → Isentropic expansion in turbine.

4-1 → Constant pressure heat removal in condenser.

s

T

6 5

4

32’

2

1

P2

P1

1.16 RANKINE CYCLE The simplest way of overcoming the inherent practical difficulties of Carnot Cycle

without deviating too far from it is to keep processes 2-3 and 3-4 of Carnot Cycle

unchanged and to continue the process 4-1 in the condenser until all vapour is

converted to liquid water. Water is then pumped into Boiler till its pressure is raised to

If changes in kinetic and potential energies are neglected, the area under the curve 2-3

i.e. area 2-2’-3-5-6-2 represents the heat transfer to the working fluid in Boiler, which is

equal to (h3 – h2) and area under the curve 1-2 i.e. area 1-4-5-6-1 represents the heat

transferred from the working fluid in condenser, which is equal to (h4 – h1). The

difference between the two areas, namely area 1-2-2’-3-4-1, represents the work

obtained from the cycle. The thermal efficiency of the cycle is given by

( ) ( )( )

( ) ( )( ) H

PT

23

1243

23

1423

H

netth

QWW

hhhhhh

hhhhhh

2653'22area143'221area

QW

−=

−−−−

=

−−−−

=

−−−−−−−−−−

==η

where WT and WP are the turbine work and pump work respectively per kg of steam flow

through the cycle and h1, h2, h3, h4 are the specific enthalpies of the working fluid.

We know that the efficiency of the Carnot cycle depends only on the temperature levels

of high and low temperature reservoirs. Efficiency o the Rankine cycle similarly depends

on the average temperature at which the heat is transferred to and from the working

fluid. Any change that increases the average temperature at which heat is transferred to

the working fluid will increase the efficiency of the Rankine cycle. Similarly, any change

that decreases the average temperature at which heat is transferred from the working

fluid will increase the efficiency of the Rankine cycle.

An advantage of the Rankine cycle over all other power cycles is its low back work ratio,

which is expressed as the ratio of the pump work to the turbine work, i.e.

Back work ratio = T

P

WW

( ) ( )( ) H

PT

23

1243th Q

WWhh

hhhh −=

−−−−

=η

METHODS OF IMPROVING THE EFFICIENCY OF RANKINE CYCLE The Rankine cycle efficiency is given by

Since the pumping work is very small compared to the turbine work, it may be

neglected. Hence, the efficiency of the Rankine cycle can be approximated as

( )( )

( )( )23

s

23

43th hh

hhhhh

−Δ

=−−

=η

It can be seen that the Rankine efficiency depends on three values, h2, h3 and turbine

expansion work (Δh)s. The enthalpy of the steam entering the turbine h3 is determined

by the pressure and temperature of the steam entering the turbine. The enthalpy of feed

water h2 is determined by the condenser pressure (as in this case h2 = h1 since the

pump work is negligible). The isentropic heat drop (Δh)s in the turbine is determined by

the pressure and temperature at the entrance of steam turbine and the pressure at the

end of expansion in the turbine. That means the Rankine efficiency depends on

pressures P1 ( i.e. P4 ) , P2 (i.e. P3 ) and temperature T3.

a) Effect of Lowering Condenser pressure By lowering the condenser pressure we can

achieve higher efficiency as the enthalpy of

the steam leaving the turbine decreases

thereby increasing heat drop (Δh)s in the

turbine. However, there is a limit to which

the condenser pressure can be lowered and

this limit is the saturated pressure Psat

corresponding to the condenser

temperature.

b) Effect of Increasing Steam pressure in the Boiler

In this analysis the maximum temperature of the steam T3 as well as the exhaust

pressure P4 are held constant.

P2, P2’, P2’’ are the pressures of steam at the entrance of the turbine at temperature T3

such that P2’’ > P2’ > P2 . x4, x4’,x4’’ are the qualities of steam at he exhaust pressure P4

of the turbine, where x4>x4’>4’’ .It is evident that as pressure n the Boiler increases, the

isentropic heat drop Δh)s increases with the result that the Rankine cycle efficiency

increases.

An adverse effect resulting from increasing the steam pressure in the Boiler is the

greater amount of moisture in the steam at the end of expansion in the turbine. If the

moisture content in the turbine exceeds ~10%, the turbine blades also get eroded which

leads to serious wear of the turbine blades.

c) Effect of Superheating of Steam in the Boiler

The moisture in the steam at the

end of the expansion can be

reduced by superheating and

x4’

x4”

x4 (Δhs)″

(Δhs)′

(Δhs)

h

s

P2’’ P2’

T3 P4

P2

increasing the superheat temperature of steam, T3’. Hence, it is natural to avoid the

erosion of the turbine blades by an increase of boiler pressure accompanied by

superheating at a higher temperature. By superheating to a higher temperature, the

heat drop in the turbine is increased from (h3 – h4) to (h3’ – h4’), thereby increasing the

efficiency of the Rankine cycle. However, the maximum temperature to which the steam

can be superheated is limited by materials.

d) Effect of Superheating &Reheating of Steam in the Boiler We have noted earlier that the efficiency of the Rankine cycle can be increased by

increasing the steam pressure in the boiler. But this increases the moisture content of

the steam in the lower stages of the turbine, which may lead to erosion of the turbine

blades. The reheat cycle has been developed to take advantage of the increased

pressure of boiler, avoiding the excessive moisture of the steam in the low pressure

stage.

In the reheat cycle, the high pressure superheated steam after expansion in the high

pressure turbine is reheated at constant pressure, usually to the entrance temperature

of the steam in the high pressure turbine. After this, it expands in the low pressure

s

P3 P4

P6

4

3 5

6

Constant temperature h

turbine to the exhaust pressure. Reheating has a two-fold advantage. Firstly, it reduces

excessive moisture in the low pressure stages of turbine, and secondly, a large amount

of work may be obtained at the cost of additional consumption of heat required for

reheating the steam. The net effect is an improvement in the thermal efficiency of the

cycle. Thus with reheat cycle, the efficiency of the cycle is increased without increase in

the maximum pressure or maximum temperature of the cycle.

The efficiency of the reheat cycle is given by

( ) ( ) ( )( ) ( )4523

126543

H

PTth hhhh

hhhhhhQ

WW−+−

−−−+−=

−=η

a) Turbine and pump losses

1-2s & 3-4s → Isentropic drop.

1-2 & 3-4 → irreversible drop.

b) Pump loses

a-b: friction loss → entropy increases

b-c: heat loss → entropy decreases.

1.17 REAL RANKINE CYCLE: DEVIATION FROM IDEAL CYCLES

Example CRITERIA FOR THE COMPARISON OF CYCLES The choice of power plant for a given purpose is determined largely by considerations of

operating cost and capital cost. The former is primarily a function of overall efficiency

of the plant, while the latter depends mainly on its size and complexity.

The Second Law tells us that even in the best power cycle, some heat must be rejected.

The best form of cycle is one in which (i) all the heat supplied is transferred while the

working fluid is at constant temperature TH, and all the heat rejected from the working

fluid is at constant temperature TL; (ii) all processes are reversible. The efficiency of the

such cycle is H

LH

TTT − , which is known as ideal cycle efficiency or Carnot efficiency.

However, all real processes are irreversible and irreversibility reduces cycle efficiency.

Hence the ratio of actual cycle efficiency to ideal cycle efficiency, i.e. the efficiency ratio

is one measure of comparison. Some cycles are more sensitive to irreversibilities than

others. That is, two cycles may have the same ideal cycle efficiencies, but allowing for

the process efficiencies, their actual cycle efficiencies may be markedly different.

Hence, Work Ratio rw is a criterion, which indicates the cycle sensitiveness to the

irreversibilities. Any cycle consists of both positive (turbine work) and negative (pump

work) work. The work ratio rw is defined as the ratio of net work to the positive work

done in the cycle. That is

T

PTw W

wWr −=

If rw is near unity, then the effect of irreversibility on the cycle efficiency is less.

However, if rw is slightly greater than zero, quite a small amount of component

inefficiencies is sufficient to reduce the network output to zero thereby reducing the

actual cycle efficiency to zero.

Hence, we can say that a high ideal cycle efficiency together with high work ratio

provides a reliable indication that a real power plant will have a good overall efficiency.

The next consideration is some criterion which will indicate the relative size of plant for a

given power output. In general, the size of component depends on the amount of

working fluid, which has to be passed through them. A direct indication of relative sizes

of steam power plant is therefore provided by the Specific Steam Consumption (ssc)

i.e. mass flow of steam required per unit power output. If W is the net work output in

kJ/kg, then ssc can be found from

⎥⎦⎤

⎢⎣⎡=⎥⎦

⎤⎢⎣⎡

⎥⎦⎤

⎢⎣⎡=

kWhkg

W3600

hs3600x

kWskgor

kJkg

W1ssc

For small power plants, gas is ideally preferred as the working fluid. The gasoline

engines, diesel engines and gas turbines are common examples.

The analysis of the air-standard cycle is based on the assumptions that are far from

real. In actual internal combustion (IC) engines, chemical reaction occurs inside the

engine cylinder as a result of combustion of air-fuel mixture and. The IC engines are

actually operated on Open cycles in which the working fluid does not go through a

cycle. The accurate analysis of IC engine is very complicated. However, it is

advantageous to analyse the performance of an ideal closed cycle that closely

approximates the real cycle. One such approach is air-standard cycle, which is based

on certain assumptions. The assumptions for idealized air-standard cycles are:

1) The working fluid, air, is assumed to be an ideal gas. The equation of state is

given by the equation pv = RT and the specific heats are assumed to be

constant.

2) All processes that constitute the cycle are reversible.

3) No chemical reaction occurs during the cycle. Heat is supplied from a high

temperature reservoir (instead of chemical reaction) and some heat is rejected to

the low temperature reservoir during the cycle.

1.18 Gas Power / Air Standard Cycles.

4) The mass of air within the system remains constant throughout the cycle.

5) Heat losses from the system to the atmosphere are assumed to be zero.

In this we shall discuss about the Brayton cycle, Otto cycle and Diesel cycle.

1.19 BRAYTON CYCLE The Brayton cycle is widely used as the basis for the operation of Gas turbine.

A schematic diagram of a simple gas turbine (open cycle) and the corresponding p-v

and T-s diagrams are shown below.

Air is drawn from he atmosphere into compressor, where it is compressed reversibly

and adiabatically. The relative high pressure air is then used in burning the fuel in the

combustion chamber. The air-fuel ratio quite high (about 60:1) to limit the temperature

burnt gases entering the turbine. The gases then expand isentropically in the turbine. A

portion of the work obtained from the turbine is utilised to drive the compressor and the

auxiliary drive and the rest of the power output is the net power of the gas turbine plant.

Simple gas turbine Brayton cycle with closed cycle consists of

1 – 2 Isentropic compression in the compressor.

2 – 3 Constant pressure heat addition.

v

p

4

32

1

s

T

4

3

2

1

3 – 4 Isentropic expansion in the turbine.

4 – 1 Constant pressure heat rejection.

Assuming constant specific heats, the

thermal efficiency of the cycle

H

Lth Q

Q1−=η

or 23

14th hh

hh1−−

−=η

( )( )23P

14P

TTCTTC1

−−

−=

The thermal efficiency can be written as

⎥⎦

⎤⎢⎣

⎡−

⎥⎦

⎤⎢⎣

⎡−

−=

−−

−=

1TTT

1TTT

1

TTTT1

2

32

1

41

23

14thη

Now γγ 1

1

2

1

2

pp

TT

−

⎟⎟⎠

⎞⎜⎜⎝

⎛= and

γγ 1

4

3

4

3

pp

TT

−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

Since, p2 = p3 and p1 = p4, it follows

2

3

1

4

4

3

1

2

TT

TTor

TT

TT

==

Hence, γγ

γγη

1

p1

1

22

1th r

11

pp

11TT1

−

− ⎟⎟⎠

⎞⎜⎜⎝

⎛−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−=−=

Where rp is the pressure ratio.

It can be seen that increasing the pressure ratio can increase the efficiency of the

Brayton cycle.

s

T 3

2

1

4

The highest temperature of the

cycle occurs at the end of

combustion process (state 3) and

the maximum temperature that the

turbine blades can withstand limits

it. This also limits the pressure ratio

that can be used for the cycle. In

gas turbine cycle, T1 is the

temperature of the atmosphere and

T3 is the temperature of the burnt

gases entering the turbine. Both

these temperatures are fixed, first

by ambient conditions and second by metallurgical conditions. Between these two

extreme values of temperatures, there exists an optimum pressure ratio for which

the net work output of the cycle is maximum.

The net work output of the turbine is given by

( ) ( )12P43Pn TTCTTCW −−−=

⎥⎦

⎤⎢⎣

⎡−−⎥

⎦

⎤⎢⎣

⎡−−= 1

TTTC1

TTTC

1

21P

3

43P

Now 2

1

3

4

2

3

1

4

TT

TTor

TT

TT

==

Pressure ratio, rp

1.20 Effect of Pressure Ratio

As mentioned above, the thermal efficiency of the Brayton cycle depends on the

pressure ratio and the ratio of specific heat. For air, γ = 1.4 and the efficiency vs.

pressure ratio plot is shown below.

( )( ) ⎥⎦

⎤⎢⎣⎡ −−

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡−−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−⎟⎟

⎠

⎞⎜⎜⎝

⎛−

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

−

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

⎥⎦

⎤⎢⎣

⎡−−⎥

⎦

⎤⎢⎣

⎡−−=∴

−

−

−

−

1rTC1r

1TC

1ppTC1

pp

1TC

1TTTC1

TTTCW

1

P1P1

P

3P

1

1

21P1

1

2

3P

1

21P

2

13Pn

γγ

γγ

γγ

γγ

The optimum pressure ratio is obtained by differentiating the net work with respect to

pressure ratio, rP and putting the derivative as zero.

Let n1=

−γ

γ

( )

( )[ ]1rTC1r1TCW n

P1PnP

3Pn −−⎥⎦

⎤⎢⎣

⎡−−=

or [ ] [ ]1nP1P

1nP3P

P

n rnTCrnTCdr

dW −−− −−−=

or 1nP1P

1nP3P rnTCrnTC0 −−− −=

or 1nP1

1nP3 rnTrnT −−− =

or ( ) ( )

3

11n1nP T

Tr =−−−−

or 3

1n2P T

Tr =−

or 1

3n2P T

Tr =

i.e. ( )n21

1

3optimumP T

Tr ⎟⎟⎠

⎞⎜⎜⎝

⎛=

or ( )( )12

1

3optimumP T

Tr−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

γγ

The maximum work is given by

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−⎟⎟

⎠

⎞⎜⎜⎝

⎛−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−⎟⎟

⎠

⎞⎜⎜⎝

⎛−= 1

TTTC1

TTTCW

21

1

31P

21

3

13Pmax

( )

[ ]213P

121

313P

TTC

TTT2TC

−=

⎥⎦⎤

⎢⎣⎡ +−=

i.e. [ ]2minmaxPmax TTCW −=

Effect of Reversibility In an ideal gas turbine plant, the compression and expansion processes are isentropic

and there is no pressure drop in the combustion chamber and heat exchanger. But

because of irreversibilities associated with the compressor and the turbine and because

of the pressure drop in the actual flow passages and combustion chamber, the actual

gas turbine plant cycle differs from the ideal one.

Compressor efficiency

12

1'2c hh

hh−−

=η

Turbine efficiency

'43

43T hh

hh−−

=η

T

s

4’ 4

3

2’ 2

1

Air is drawn from the atmosphere into

compressor and compressed isentropically

to state 2. It is then heated at constant

pressure in the regenerator to state x by hot

burnt gases from the turbine. Since the

temperature of air is increased before it

reaches the combustion chamber, less

amount of fuel will be required to attain the

designed turbine inlet temperature of the

products of combustion. After the combustion at constant pressure in the combustion

chamber, the gas enters the turbine at state 3 and expands to state 4 isentropically. It

then enters the counter-flow regenerator, where it gives up a portion of its heat energy

to the compressed air from the compressor and leaves the regenerator at state y. In the

y

Combustion Chamber

3x

2

14

Fuel

Regenerator

Turbine Compressor

4

3

x2

y

1

1.21 Regenerative Brayton Cycle The temperature of the exhaust gases of simple gas turbine is higher than the

temperature of air after compression. If the heat energy in the exhaust gases instead of

getting dissipated in the atmosphere is used in heating air after compression, it will

reduce the energy requirement from the fuel, thereby increasing the efficiency of the

cycle.

ideal cycle, the temperature of the compressed air leaving the regenerator is equal to

the temperature of the burnt gases leaving the turbine, i.e. Tx = T4. But in practice, the

temperature of the compressed air leaving the regenerator is less than Tx.

The effectiveness of the regenerator is given by the ratio of the increase in enthalpy of

the working fluid flowing through the regenerator to the maximum available enthalpy

difference.

The effectiveness of regenerator is given by

2x

2xR hh

hh−−

= ′η

Assuming constant specific heat

2x

2xR TT

TT−−

= ′η

The thermal efficiency of an ideal gas turbine cycle with regenerator is

x3

1y

x3

1y

H

Lth TT

TT1

hhhh

1QQ1

−

−−=

−

−−=−=η

Since for ideal condition

Tx = T4 and Ty = T2

Therefore,

⎥⎦

⎤⎢⎣

⎡−

⎥⎦

⎤⎢⎣

⎡−

−=−−

=

3

43

1

21

43

12th

TT1T

1TTT

1TTTT

η

Since γγ

γγ

γγ 1

2

1

1

3

4

3

4

1

1

2

1

2

pp

pp

TTand

pp

TT

−−−

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

∴

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

⎟⎟⎠

⎞⎜⎜⎝

⎛−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−⎟⎟

⎠

⎞⎜⎜⎝

⎛

−=

−

−

γγ

γγ

η

1

1

2

1

1

2

3

1th

pp

11

1pp

TT1

γγ 1

1

2

3

1

pp

TT1

−

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

So the thermal efficiency of an ideal regenerative gas turbine cycle depends not only on

the pressure ratio but also on the ratio of two extreme temperatures. In this case, lower

the pressure ratio, higher the efficiency, the maximum value being 3

13

TTT − when rP = 1.

Consider the compressor work first.

The curvature of the constant

pressure lines on T-s diagram is such

that the vertical distance between

them reduces as we go towards the

left (shown by the arrow). Therefore,

further to the left the compression

process 1-2 takes place, smaller is

the work required to drive the

compressor. State 1 is determined by

the atmospheric pressure and temperature. But if the compression is carried out in two

stages, 1-3 and 4-5 with the air is being cooled at constant intermediate pressure pi

between the stages; some reduction of compression work can be achieved. The sum of

temperature rises (T3 – T1) and (T5 – T4) will be clearly less than (T2 – T1). Ideally, it is

possible to cool the air to atmospheric condition i.e. T4 = T1, and in this case

Intercooling is complete.

pi 2

3

1

5

4

T

s

This is the Carnot cycle efficiency based upon maximum and minimum temperatures of

the cycle.

1.22 Intercooling and Reheating

The addition of regenerator improves the ideal efficiency but does not improve the work

ratio. The latter may be reduced by reducing the compressor work or increasing the

turbine work.

With isentropic compression and

complete Intercooling, the

compressor work is given by

( ) ( )45P13P TTCTTCW −−−−=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧−⎟⎟

⎠

⎞⎜⎜⎝

⎛−

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧−⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

−

−

1ppTC

1ppTC

1

i

21P

1

1

i1P

γγ

γγ

The saving in work will depend on

the choice of intermediate

Intercooling pressure pi. By equating idp

dW to zero the condition for minimum work is

given by

21i ppp =

Hence, Pii

2

1

i rpp

pp

==

and P1

2Pi r

ppr ==

Thus for minimum compressor work,

the compression ratio and work inputs

for two stages are equal. The

compression work can further be

reduced by increasing the number of

stages and intercoolers. But the

additional complexity and cost make

more than three stages uneconomic.

Similarly, it can be shown that the

work output from the turbine is

increased by multi-stage expansion

HP Compressor Intercooler

43

15

Fuel

LP Compressor

HP Turbine

LP Turbine

Reheater

98

6 10

pi 6

8

7

9

10

T

s

with reheating. The figure in the left illustrates the relevant part of the cycle showing

expansion in two stages with reheating to the metallurgical limit i.e. T9 = T6.

The turbine work is increased from W6-7 to W6-8 + W9-10 which is given by

Wnet = CP(T6 – T8)+CP(T9 – T10)

It is possible to show that with isentropic expansion, the optimum intermediate pressure,

this time for maximum work, is given by

PPi76i rrorppp ==

Reheating can also be extended to more than two stages, although this is seldom done

in practice and with open cycle plant a limit is set by the oxygen available for

combustion.

Although intercooler and reheaters improve the work ratio, these devices by themselves

can lead to decrease in ideal efficiency. This is because the heat supplied is increased

as well as net work output. The full advantage is only reaped if a regenerator is also

included in the plant. The additional heat required for the colder air leaving the

compressor can be obtained from the hot exhaust gases, and there is a gain in ideal

cycle efficiency as well as work ratio.

BRAYTON CYCLE WITH INTERCOOLING, REHEATING AND REGENERATION

Regenerator 10

HP Intercoole

43

15

Fuel

LP

HP

LP Turbine

Reheater

87

6 9

The thermal efficiency is given by

( ) ( )( ) ( )7856

32110

H

Lth hhhh

hhhh1

QQ1

−+−

−+−−=−=η

T

s

7 5

9

8 6

4

3

2

1

10

UNIT 1 GAS & STEAM POWER CYCLES 1.1 PREREQUISITE … · 1.2 POWER CYCLES Ideal Cycles, Internal...

Documents

Transcript of UNIT 1 GAS & STEAM POWER CYCLES 1.1 PREREQUISITE … · 1.2 POWER CYCLES Ideal Cycles, Internal...