Gas Cycles Otto,Diesel,Dual cycles

8/10/2019 Gas Cycles Otto,Diesel,Dual cycles

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Gas Cycles

Assumptions of air standard cycle Analyze three cycles

Otto

Diesel

Brayton


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Assumptions of Air Standard Cycle

Working fluid is air

Air is ideal gas

Combustion process is replaced by heat addition

process Heat rejection is used to restore the fluid to its initial

state and complete the cycle

All processes are internally reversible

Constant or variable specific heats can be used


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Gas cycles have many

engineering applications

Internal combustion engine Otto cycle

Diesel cycle

Gas turbines

Brayton cycle

Refrigeration

Reversed 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

TDC

BDC

V

V=

CV

CVDVr

+=


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Mean Effective Pressure

Mean Effective Pressure (MEP) is a fictitiouspressure, that if acted on the piston during the

entire power stroke, would produce the same

amount of net work.

minmax VV

W

MEP

net

=


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The net work output ofa cycle is equivalent tothe product of the meaneffect pressure and thedisplacement volume


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Real and Idealized Cycle


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Idealized Otto Cycle


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Idealized Otto Cycle

1-2 - Adiabatic Compression (Isentropic)

2-3 - Constant Volume Heat Addition

3-4 - Adiabatic Expansion (Isentropic)

4-1 - Constant Volume Heat Rejection


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Performance of Cycle

in

net

q

w

=Efficiency:

Lets start by getting heat input:

23in uuq =


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Cycle Performance

Get net work from energy balance of cycle:

outinnet qqw =Substituting for qin and qout:

=netw )uu()uu( 1423

Efficiency is then:

in

net

q

w=


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Cycle Performance

Substituting for net work and heat input:

)u-(u

)u-(u-)u-(u

23

1423=

We can simplify the above expression:

)u-(u)u-(u1

23

14=


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Cold Air Standard Cycle

cp, cv, and k are constant at ambienttemperature (

70 F) values.

Assumption will allow us to get a quick first

cutapproximation of performance of cycle.


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Cycle performance with cold air

cycle assumptions

If we assume constant specific heats:

)T-(Tc

)T-(Tc

)u-(u

)u-(u

1 23v

14v

23

14

==

)T-(T

)T-(T1

23

14=


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Cycle performance with cold air

cycle assumptions

1k

2

1

r11

TT1 ==

This looks like the Carnot efficiency, but it is

not! T1 and T2 are not constant.

What are the limitations for this expression?


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Differences between Otto and

Carnot Cycles

T

s

1

2

3

4

T

s

1

2

3

4

2

3


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Effect of compression ratio on

Otto cycle efficiency


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Sample ProblemThe 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

P

v

1

2

3

4

T1 = 299 K

P1 = 0.95 bar

r = V1 /V2 = V4 /V3 = 9

Q23 = 8.6 kJ


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Major assumptions

Kinetic and potential energies are zero Closed system

1 is start of compression

Ideal cycle: 1-2 isentropic compression, 2-3

constant volume heat addition, etc.

Cold cycle constant properties


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Carry through with solutionCalculate mass of air:

kg10x29.6RT

VPm 3-

1

11 ==

Compression occurs from 1 to 2:

ncompressioisentropicV

VTT

1

2

112

=

k

( ) ( ) 11.42 9K27322T +=

K705.6T2= But we need T3!


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Get T3 with first law:

( )23v23 TTmcpe)keu(mWQ =++=Solve for T3:

2

v

3 Tc

qT += K705.6

kgkJ0.855

kg6.29x10kJ8.6

3

=

K2304.7T3=


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Thermal Efficiency

11.41k 911

r11 ==

585.0=


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Lets take a look at the Diesel cycle.


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Idealized Diesel cycle1-2 - Adiabatic Compression (Isentropic)

2-3 - Constant Pressure Heat Addition

3-4 - Adiabatic Expansion (Isentropic)

4-1 - Constant Volume Heat Rejection


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Performance of cycle

in

net

q

w

=Efficiency:

Heat input occurs from 2 to 3 inconstant pressure process:

23in hhq = Why enthalpies?


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Diesel Cycle Performance

Get net work from energy balance of cycle:

outinnet qqw =Substituting for qin and qout:

=netw )uu()hh( 1423

As we did with the Otto cycle:

in

net

q

w=


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Diesel Cycle Performance

Substituting for net work and heat input:

)h-(h)u-(u-)h-(h

23

1423=

We can simplify the above expression:

)h-(h)u-(u1

23

14=


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For cold cycle analysis

)T-(Tc

)T-(Tc1

23p

14v= )T-k(T)T-(T

123

14=

Its possible to rewrite this in a simpler form if we

define a new term:

ratiocutoffV

Vr

2

3c ==


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Cold cycle efficiency

( )

= 1

1111

c

k

c

krk

r

r

Efficiency is dependent on compressionratio and cutoff ratio.


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k = 1.4

Th th l ffi i f th id l Di l l


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The thermal efficiency of the ideal Diesel cycle as afunction of compression and cutoff rates (k=1.4)


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Comparison between Otto and

Diesel cyclesFor same compression ratio

OTTO > DIESELFor same combustion temperature

DIESEL > OTTO


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Sample ProblemA Diesel air-standard cycle has a compression ratio

of 15:1. The pressure and temperature at thebeginning of the compression are 100 kPa and 17 C,

respectively. If the maximum temperature of the

cycle is 2250 K, determine:

1. the cutoff ratio2. the thermal efficiency

3. the mean effective pressure

Draw cycle


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Draw cycle

Apply assumption as before with


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pp y assu pt o as be o e w t

Otto cycle

Cold cycle assumptions

Kinetic and potential energies neglected

Ideal cycle


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More analysis

Cutoff ratio is then:

62.27.8562250 ==cr

Thermal efficiency:

( )

= 111

1 1c

k

c

krk

r

r


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More analysis

Plug numbers into thermal efficiency:

( )

=

162.24.1

162.2

15

11

4.1

4.0

574.0=

Gas Cycles Otto,Diesel,Dual cycles

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