GAS POWER CYCLES

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Objectives1. Evaluate the performance of gas power cycles.

2. Develop simplifying assumptions applicable to gas power cycles.

3. Review the operation of reciprocating engines.

4. Analyze both closed and open gas power cycles.

5. Solve problems based on the Otto and Diesel cycles.

6. Solve problems based on the Brayton cycle; Brayton cycle with regeneration;and Brayton cycle with intercooling, reheating, and regeneration.

7. Identify simplifying assumptions and perform second-law analysis on gaspower cycles.

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Basic Considerations In Power Cycles Analysis

The analysis of many complexprocesses can be reduced to amanageable level by utilizingsome idealizations.

Most power-producing devices operate on cycles.

Ideal cycle: A cycle that resembles the actual cycleclosely but is made up totally of internally reversibleprocesses is called an ideal cycle.

Recall: Thermal efficiency of heat engines

Reversible cycles such as Carnot cycle have thehighest thermal efficiency of all heat engines operatingbetween the same temperature levels.

Unlike ideal cycles, they are totally reversible, andunsuitable as a realistic model.

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1. The cycle does not involve any friction. Therefore,the working fluid does not experience any pressuredrop as it flows in pipes or heat exchangers.

2. All expansion and compression processes takeplace in a quasi-equilibrium manner.

3. The pipes connecting the various components of asystem are well insulated, so heat transferthrough them is negligible.

On both P-v and T-s diagrams, the area enclosed by the process curve represents the net work of the cycle.

On a T-s diagram, the ratio of the areaenclosed by the cyclic curve to the areaunder the heat-addition process curverepresents the thermal efficiency of thecycle.

Idealizations (simplifications) in the analysis of power cycles

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Carnot Cycle - Its Value In Engineering

P-v and T-s diagrams of a Carnot cycle.Example: A steady-flow Carnot engine.

The Carnot cycle is composed of 4 totally reversible processes: isothermal heat addition, isentropic expansion, isothermal heat rejection, and isentropic compression.

For both ideal and actual cycles: Thermal efficiency increases with an increase in the average temperature at which heat is supplied to the system or with a decrease in the average temperature at which heat is rejected from the system.

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Air-standard Assumptions

The combustion process is replaced by a heat-addition process in ideal cycles.

1. The working fluid is air, which continuously circulates in a closed loop and always behaves as an ideal gas.

2. All the processes that make up the cycle are internally reversible.

3. The combustion process is replaced by a heat-addition process from an external source.

4. The exhaust process is replaced by a heat-rejection process that restores the working fluid to its initial state.

Cold-air-standard assumptions: When the working fluid is considered to be air with constant specific heats at room temperature (25°C). Air-standard cycle: A cycle for which the air-standard assumptions are applicable.

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Overview of Reciprocating EnginesThe reciprocating engine (basically a piston–cylinder device) is an invention that has proved to be very versatile and has a wide range of applications.

Reciprocating engine is thepowerhouse of the vast majority ofautomobiles, trucks, light aircraft,ships, electric power generators,and many other devices.

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Basic Components

Compression ratio:

The piston reciprocates in the cylinder between two fixed positions called the top deadcentre (TDC) - the position that forms the smallest volume in the cylinder - and the bottomdead centre (BDC) - position that forms the largest volume in the cylinder.

The distance between TDC and BDC is called the stroke ofthe engine. The diameter of the piston is called the bore.

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Performance Characteristics

Classifications of IC Engines:1. Spark-ignition (SI) or Petrol engines2. Compression-ignition (CI) or Diesel

engines

Mean effective pressure (MEP):A fictitious pressure that, if it is acted on the pistonduring the entire power stroke, would produce thesame amount of net work as that produced during theactual cycle.

Net work output per cycle:

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Otto Cycle

Sequence of processes:

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Thermal Efficiency of Otto CycleThe heat supplied to the working fluid during constant-volume heating (combustion),

The heat rejected from the working fluid during constant-volume cooling (exhaust),

Thermal efficiency,

Temperature-volume relation,

Compression ratio,

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GeneralFormulas required to solve problem based of Gas Power Cycle

1

2

1

1

1

2

1

2

VV

PP

TT

mRTPV

For S=constant process

Ideal gas equation

For V=constant process TCvqTmCvQ

For P=constant process TCpq

TmCpQ

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Formulas required to solve problem based of OTTO Cycle

1

2

1

1

1

2

1

2

VV

PP

TT

mRTPV

For S=constant process

Ideal gas equation

For V=constant process TCvqTmCvQ

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ExampleAn Otto cycle having a compression ratio of 9:1 uses air as the working fluid. Initially P1 = 95 kPa, T1 = 17oC, and V1 = 3.8 liters. During the heat addition process, 7.5 kJ of heat are added. Determine all T's, P's, th, the back work ratio, and the mean effective pressure.

Process Diagrams:

Review the P-v and T-s diagrams given above for the Otto cycle.

Assume constant specific heats with Cv = 0.718 kJ/kg K, k = 1.4.

Process 1-2 is isentropic; therefore, recalling that r = V1/V2 = 9,

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Q mC T Tin v ( )3 2

Let qin = Qin / m and m = V1/v1

v RTP

kJkg K

K

kPam kPa

kJmkg

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1

3

3

0 287 290

95

0 875

. ( )

.

q Qm

Q vV

kJ

mkgm

kJkg

inin

in

1

13

3 37 50 875

38 10

1727

..

.

16

T T qC

K

kJkgkJ

kg KK

in

v3 2

698 41727

0 718

31037

..

.

Using the combined gas law (V3 = V2) P P TT

MPa3 23

2

9 15 .

Process 3-4 is isentropic; therefore, 1 1 1.4 1

34 3 3

4

1 1(3103.7)9

1288.8

k kVT T T KV r

K

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Process 4-1 is constant volume. So the first law for the closed system gives, on a mass basis,

Q mC T T

q Qm

C T T

kJkg K

K

kJkg

out v

outout

v

( )

( )

. ( . )

.

4 1

4 1

0 718 1288 8 290

717 1

The first law applied to the cycle gives (Recall ucycle = 0)

w q q qkJkg

kJkg

net net in out

( . )

.

1727 717 4

1009 6

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The thermal efficiency is

th Ottonet

in

wq

kJkg

kJkg

or

,

.

. .

1009 6

1727

0 585 58 5%The mean effective pressure is

max min max min

1 2 1 2 1 1

3

3

(1 / ) (1 1/ )

1009.61298

10.875 (1 )9

net net

net net net

W wMEPV V v v

w w wv v v v v v r

kJm kPakg kPa

m kJkg

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ProblemOtto Cycle

9–34 An ideal Otto cycle has a compression ratio of 8. At the beginning of thecompression process, air is at 95 kPa and 27°C, and 750 kJ/kg of heat istransferred to air during the constant-volume heat-addition process. Assumingthat the specific heats are constant with temperature, determine:

a) the pressure & temperature at the end of heat addition process, b) the net work output, c) the thermal efficiency, and d) the mean effective pressure for the cycle.

Answers: (a) 3898 kPa, 1539 K, (b) 392.4 kJ/kg, (c) 52.3 percent, (d ) 495 kPa

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Premature ignition of the fuel produces audible noise called engine knock. It hurtsperformance and causes engine damage.

Autoignition places upper limit on compression ratios that can be used in SI engines.Specific heat ratio, k affects the thermal efficiency of the Otto cycle.

Engine Knock (Autoignition)

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Diesel Cycle: Ideal Cycle for CI Engines

The combustion process takes place over alonger interval - fuel injection starts whenthe piston approaches TDC and continuesduring the first part of power stroke.Hence, combustion process in the idealDiesel cycle is approximated as a constant-pressure heat-addition process.

In diesel engines, only air is compressed during the compression stroke, eliminatingthe possibility of autoignition. These engines can be designed to operate at highercompression ratios, typically between 12 and 24.

Fuels that are less refined (thus less expensive) can be used in diesel engines.

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1-2 Isentropic compression2-3 Constant-pressure heat addition3-4 Isentropic expansion4-1 Constant-volume heat rejection.

Sequence of processes:

Note: Petrol and diesel engines differ only in the manner the heat addition (or combustion) process takes place.

It is approximated as a constant volume process in the petrol engine cycle and as a constant pressure process in the Diesel engine cycle.

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Cutoff ratio,

Thermal Efficiency of Diesel CycleHeat supplied to the working fluid during the constant-pressure heating (combustion),

Heat rejected from the working fluid during the constant-volume cooling (exhaust),

Thermal efficiency of Diesel cycle (general),

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Formulas required to solve problem based of DIESEL Cycle

1

2

1

1

1

2

1

2

VV

PP

TT

mRTPV

For S=constant

Ideal gas equation

For V=constant TmCvQ

For P=constant TmCQ p

Cutoff ratio,

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ProblemDiesel Cycle

9–51 An ideal diesel engine has a compression ratio of 20 and uses air as the workingfluid. The state of air at the beginning of the compression process is 95 kPa and20°C. If the maximum temperature in the cycle is not to exceed 2200 K,determine:

a) the thermal efficiency, and b) the mean effective pressure.

Assume constant specific heats for air at room temperature.

Answers: (a) 63.5 percent, (b) 933 kPa

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For the same compression ratio, thermal efficiency of Otto cycle is greater than thatof the Diesel cycle.

As the cutoff ratio decreases, the thermal efficiency of the Diesel cycle increases. When rc =1, the efficiencies of the Otto and Diesel cycles are identical.

Thermal efficiencies of large diesel engines range from about 35 to 40 percent.Higher efficiency and lower fuel costsmake diesel engines attractive inapplications such as in locomotive engines,emergency power generation units, largeships, and heavy trucks.

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ProblemDiesel Cycle

9-59A six-cylinder, four-stroke, 4.5-L compression-ignition engine operates on theideal Diesel cycle with a compression ratio of 17. The air is at 95 kPa and 55°Cat the beginning of the compression process and the engine speed is 2000 rpm.The engine uses light diesel fuel with a heating value of 42,500 kJ/kg, an air–fuelratio of 24, and a combustion efficiency of 98 percent. Using constant specificheats at 850 K, determine:

a) the maximum temperature in the cycle and the cutoff ratio,b) the net work output per cycle and the thermal efficiency, c) the mean effective pressure, d ) the net power output, and e) the specific fuel consumption, in g/kWh, defined as the ratio of the

mass of the fuel consumed to the net work produced.

Answers: (a) 2383 K, 2.7 (b) 4.36 kJ, 0.543, (c) 969 kPa, (d ) 72.7 kW, (e) 159 g/kWh

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Approximating the combustion process asa constant-volume or a constant-pressureheat-addition process is overly simplisticand not quite realistic.A better approach would be to model thecombustion process in both SI and CIengines as a combination of two heat-transfer processes, one at constant volumeand the other at constant pressure.The ideal cycle based on this concept iscalled the dual cycle.

Dual Cycle: Realistic Ideal Cycle for CI Engines

Note: Both the Otto and the Diesel cycles can be obtained as special cases of the dual cycle.

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BRAYTON CYCLE: THE IDEAL CYCLE FOR GAS-TURBINE

An open-cycle gas-turbine engine. A closed-cycle gas-turbine engine.

The combustion process is replaced by a constant-pressure heat-addition processfrom an external source, and the exhaust process is replaced by a constant-pressureheat-rejection process to the ambient air.

1-2 Isentropic compression (in a compressor)2-3 Constant-pressure heat addition3-4 Isentropic expansion (in a turbine)4-1 Constant-pressure heat rejection

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T-s and P-v diagrams for the ideal Brayton cycle.

Pressure ratio

Thermal efficiency of the

ideal Brayton cycle as a

function of the pressure ratio.

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The fraction of the turbine work used to drive the compressor is called the back work ratio.

The two major application areas of gas-turbine engines are aircraft propulsion and electric power generation.

The highest temperature in the cycle islimited by the maximum temperature thatthe turbine blades can withstand. Thisalso limits the pressure ratios that can beused in the cycle.The air in gas turbines supplies thenecessary oxidant for the combustion ofthe fuel, and it serves as a coolant tokeep the temperature of variouscomponents within safe limits. An air–fuelratio of 50 or above is not uncommon.

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Effect of the pressure ratio on the net work done.

w w wC T T C T TC T T T C T T T

C Tr

C T r

net turb comp

p p

p p

pp

k k p pk k

( ) ( )( / ) ( / )

( ) ( )( )/( )/

3 4 2 1

3 4 3 1 2 1

3 1 11

1 1

1 1 1

Note that the net work is zero when

/( 1)

3

1

1k k

p pTr and rT

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For fixed T3 and T1, the pressure ratio that makes the work a maximum is obtained from:

dwdr

net

p

0

This is easier to do if we let X = rp(k-1)/k w C T

XC T Xnet p p 3 11 1 1( ) ( )

dwdX

C T X C Tnetp p

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10 1 1 0 0[ ( ) ] [ ]

Solving for X

Then, the rp that makes the work a maximum for the constant property case and fixed T3 and T1 is

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Formulas required to solve problem based of BRAYTON Cycle

1

2

1

1

1

2

1

2

VV

PP

TT

mRTPV

For S=constant process

Ideal gas equation

For P=constant process TCpq

TmCpQ

Pressure ratio