UNIT 1 GAS & STEAM POWER CYCLES 1.1 PREREQUISITE … · 1.2 POWER CYCLES Ideal Cycles, Internal...
Transcript of 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
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
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
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
Pressure (P2) = 15bar = 1500KN/m2
Pressure (P3) = 40bar = 4000KN/m2
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:
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:
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
∴ 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%
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%
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
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%
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%
Lose in air standard efficiency = (ηdiesel CASE(i) ) - (ηdiesel CASE(i) )
= 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%
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%
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
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
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
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
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:-
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:-
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:-
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,
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,
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,
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