Gas Cycles

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Gas Cycles Carnot Cycle T2 T1 s1 s2 Work W 1 2 3 4 1-2 - ADIABATIC COMPRESSION (ISENTROPIC) 2-3 - HEAT ADDITION (ISOTHERMAL) 3-4 - ADIABATIC EXPANSION (ISENTROPIC) 4-1 - WORK (ISOTHERMAL) Heat Q

description

Gas Cycles. Carnot Cycle. 1-2 - ADIABATIC COMPRESSION (ISENTROPIC) 2-3 - HEAT ADDITION (ISOTHERMAL) 3-4 - ADIABATIC EXPANSION (ISENTROPIC) 4-1 - WORK (ISOTHERMAL). Heat Q. 3. 2. T2. Work W. 1. T1. 4. s2. s1. Carnot Cycle. - PowerPoint PPT Presentation

Transcript of Gas Cycles

Page 1: Gas Cycles

Gas Cycles

Carnot Cycle

T2

T1

s1 s2

Work W

1

2 3

4

1-2 - ADIABATIC COMPRESSION (ISENTROPIC)2-3 - HEAT ADDITION (ISOTHERMAL)3-4 - ADIABATIC EXPANSION (ISENTROPIC)4-1 - WORK (ISOTHERMAL)

Heat Q

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Carnot Cycle

Carnot cycle is the most efficient cycle that can be executed between a heat source and a heat sink.

However, isothermal heat transfer is difficult to obtain in reality--requires large heat exchangers and a lot of time.

2

1

T

T-1

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Carnot Cycle

Therefore, the very important (reversible) Carnot cycle, composed of two reversible isothermal processes and two reversible adiabatic processes, is never realized as a practical matter.

Its real value is as a standard of comparison for all other cycles.

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Gas cycles have many engineering applications

Internal combustion engineOtto cycleDiesel cycle

Gas turbines Brayton cycle

RefrigerationReversed Brayton cycle

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Some nomenclature before starting internal combustion engine cycles

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More terminology

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Terminology

Bore = d Stroke = s Displacement volume =DV = Clearance volume = CV Compression ratio = r

4

ds

2

CV

CVDVr

TDC

BDC

V

V

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Mean Effective Pressure

Mean Effective Pressure (MEP) is a fictitious pressure, such that if it acted on the piston during the entire power stroke, it would produce the same amount of net work.

minmax VV

WMEP net

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The net work output of a cycle is equivalent to the product of the mean effect pressure and the displacement volume

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Real Otto cycle

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Real and Idealized Cycle

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Otto Cycle P-V & T-s Diagrams

Pressure-Volume Temperature-Entropy

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Otto Cycle Derivation

Thermal Efficiency:

For a constant volume heat addition (and rejection) process;

Assuming constant specific heat:

Q

Q - 1 =

Q

Q - Q =

H

L

H

LHth

T C m = Q vin

1-TT

T

1 - TT

T-1 =

)T - T( C m

)T - T( C m - 1 =

2

32

1

41

23v

14vth

T C m = Q v Rej

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For an isentropic compression (and expansion) process:

where: γ = Cp/Cv

Then, by transposing,

T

T = V

V = V

V = T

T

4

3

3

4

1-

2

1

1-

1

2

T

T = T

T

1

4

2

3

Otto Cycle Derivation

T

T-1 = 2

1thLeading to

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Differences between Otto and Carnot cycles

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The compression ratio (rv) is a volume ratio

and is equal to the expansion ratio in an otto cycle engine.

Compression Ratio

V

V = V

V = r3

4

2

1v

1 + v

v = rv

v + v = volume Clearance

volume Total = r

cc

sv

cc

ccsv

where Compression ratio is defined as

Otto Cycle Derivation

Page 18: Gas Cycles

Then by substitution,

)r(

1 - 1 = )r( - 1 = 1-

v

-1vth

)r( = V

V = T

T -1v

1

2

-1

2

1

The air standard thermal efficiency of the Otto cycle then becomes:

Otto Cycle Derivation

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Summarizing

Q

Q - 1 =

Q

Q - Q =

H

L

H

LHth T C m = Q v

1-TT

T

1 - TT

T-1 =

2

32

1

41

th

)r( = V

V = T

T -1v

1

2

-1

2

1

)r(

1 - 1 = )r( - 1 = 1-

v

-1vth

T

T = T

T

1

4

2

3

2

11T

T th

where

and then

Isentropic behavior

Otto Cycle Derivation

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Heat addition (Q) is accomplished through fuel combustion

Q = Lower Heat Value (LHV) BTU/lb, kJ/kg

Q A

F m =Q

fuelain

cycle

Otto Cycle Derivation

T C m = Q vin also

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Effect of compression ratio on Otto cycle efficiency

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Sample Problem – 1The air at the beginning of the compression stroke of an air-standard Otto cycle is at 95 kPa and 22C and the cylinder volume is 5600 cm3. The compression ratio is 9 and 8.6 kJ are added during the heat addition process. Calculate:

(a) the temperature and pressure after the compression and heat addition process(b) the thermal efficiency of the cycle

Use cold air cycle assumptions.

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Draw cycle and label points

T1 = 295 K

P1 = 95 kPa

r = V1 /V2 = V4 /V3 = 9

Q23 = 8.6 kJ

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Carry through with solution

kg 10 x 29.6RT

VPm 3-

1

11

Calculate mass of air:

Compression occurs from 1 to 2:

ncompressio isentropic V

VTT

1

2

112

k

11.42 9K 27322T

K 705.6T2 But we need T3!

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Get T3 with first law:

23v23 TTmcQ Solve for T3:

2v

3 Tc

qT K705.6

kgkJ0.855

kg6.29x10kJ8.6 3

K2304.7T3

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Thermal Efficiency

11.41k 9

11

r

11

585.0

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Sample Problem – 2

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Solution

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Diesel Cycle P-V & T-s Diagrams

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Sample Problem – 3

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Gasoline vs. Diesel Engine