Thermodynamic Cycle Presentation

8/19/2019 Thermodynamic Cycle Presentation

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Thermodynamic Cycles



• Look at different cycles that approximate realprocesses

• You can categorize these processes several

different ways• Power Cycles vs. Refrigeration

• Gas vs. Vapor

• Closed vs. open• Internal Combustion vs. External Combustion



WQ

Q-W=0 Q=W

In addition, we know that the efficiency for aCarnot Cycle is:

η th Carnot

L

H

T

T , = −

1



Carnot Cycle is not a good model for most realprocesses

• For example• Internal combustion engine

• Gas turbine

• We need to develop a new model, thatis still ideal



Air-Standard Assumptions

• Air continuously circulates in a closedloop and behaves as an ideal gas

• All the processes are internallyreversible

• Combustion is replaced by a heat-addition process from the outside

• Heat rejection replaces the exhaust

process

• Also assume a constant value for Cp,

evaluated at room temperature



erminology for Reciprocating Devices

r V

V

V

V

BDC

TDC

= =max

minCompression Ratio



Mean Effective Pressure

∫=2

1PdV W

V PW ∆=



v

v



1-2 Isentropic Compression

2-3 Constant Volume Heat Addition

3-4 Isentropic Expansion

4-1 Constant Volume Heat Rejection



hermal Efficiency of the Otto Cycle

η thnet

in

net

in

in out

in

out

in

W

Q

Q

Q

Q Q

Q

Q

Q

= = = −

= −1

Apply First Law Closed System to Process 2-3, V = Constant

∫ =+=+=∆=−

3

223,23,23,

2323,23,

00 PdV W W W

U W Q

bother net

net net

Q U

Q Q mC T T

net

net in v

,

, ( )

23 23

23 3 2

=

= = −

∆



Apply First Law Closed System to Process 4-1,V = Constant

001

441,41,41,

4141,41,

=+=+=

∆=−

∫ PdV W W W

U W Q

bother net

net net

Q U

Q Q mC T T Q mC T T mC T T

net

net out v

out v v

,

, ( )( ) ( )

41 41

41 1 4

1 4 4 1

=

= − = −= − − = −

∆



η th Ottoout

in

v

v

QQ

mC T T

mC T T

,

( )

( )

= −

= − −

−

1

1 4 1

3 2

η th OttoT T T T

T T T

T T T

,( )( )

( / )

( / )

= − −−

= − −

−

1

11

1

4 1

3 2

1 4 1

2 3 2

Recall processes 1-2 and 3-4 are isentropic, so

1

3

4

4

3

1

2

1

1

2

T and

−−

⎟⎟ ⎠ ⎞⎜⎜

⎝ ⎛ =⎟⎟

⎠ ⎞⎜⎜

⎝ ⎛ =

k k

vvT

vv

T T

v3 = v2 and v4 = v1

T T

T T

or

T T

T T

2

1

3

4

4

1

3

2

=

=



η th OttoT T T T

T T T

T T T

,( )( )

( / )

( / )

= − −−

= − −

−

1

11

1

4 1

3 2

1 4 1

2 3 2

1

η th Otto

T

T , = −1 1

2

Is this the same as the Carnot efficiency?



Efficiency of the Otto Cycle vs. Carnot Cycle

• There are only two temperatures in theCarnot cycle

• Heat is added at TH

• Heat is rejected at TL

• There are four temperatures in the Otto

cycle!!• Heat is added over a range of temperatures

• Heat is rejected over a range of temperatures



1

1

1

2

2

1 1−

−

=⎟⎟

⎠

⎞⎜⎜

⎝

⎛ =

k

k

r V

V

T

T

Since process 1-2 is isentropic,

η th Otto k r , = −

−1

1

1

η th Otto

T

T , = −1 1

2

Increasing Compression RatioIncreases the Efficiency

Typical CompressionRatios for GasolineEngines



Why not use higher compression Ratios?

• Premature Ignition• Causes “Knock”

• Reduces the Efficiency

• Mechanically need a better design



Diesel Engines

• No spark plug• Fuel is sprayed into hot compressed air



State Diagrams for the Diesel Cycle



Diesel Cycle Otto Cycle

The onlydifference

is inprocess2-3



Consider Process 2-3

• This is the step where heat is transferred into thesystem

• We model it as constant pressure instead of constant

volume23,23, uuuwq out bin −=∆=−( )2323, T T C hvPuq pin −=∆=∆+∆=

Consider Process 4-1• This is where heat is rejected

• We model this as a constant v process

• That means there is no boundary work

uwq ∆=− 4141

( )4141 T T C uqq vout −=∆=−=

( )14 T T C q vout −=



As for any heat engine…

in

out

in

net th

q

q

q

w−== 1η

( ) ( )2314 and T T C qT T C q pinvout −=−=

( )23

14,

)(1

T T C

T T C n

p

vdieselth

−

−−=

( )

( )23

141

T T k

T T

−

−−=



Rearrange

( )

( )1

11

232

141,

−

−−=

T T kT

T T T n dieselth

PV

T

PV

T P P

T

T

V

V r c

3 3

3

2 2

2

3 2

3

2

3

2

= =

= =

where

rc is called the cutoff ratio – it’s the ratio ofthe cylinder volume before and after thecombustion process



( )

( )1

11

2

141,

−

−−=

c

dieselth

r kT

T T T n

PV

T

PV

T V V

T

T

P

P

4 4

4

1 1

1

4 1

4

1

4

1

= =

=

where



k k

k k

V PV P

V PV P

3344

2211 and

=

=

( )( )111

2

141,

−

−−=

c

dieselthr kT T T T n 14

PP

Since Process 1-2 and Process 3-4

are both isentropic

k

k

k

k

V V

PP

V V

PP

2

3

2

3

1

4

1

4=

1 1

k c

k

r V V =⎟⎟

⎠ ⎞⎜⎜

⎝ ⎛ =

2

3



)( )111

2

1,

−

−−=

c

k

cdieselth

r kT r T n

Finally, Since process 1-2 is

isentropic

1

1

2

2

1

−

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ =

k

vv

T T

The volume ratiofrom 1 to 2 is the

compression ratio, r

1

2

1 1 −

⎟ ⎠

⎞⎜⎝

⎛ =

k

r T

T



( )⎥⎦⎤⎢

⎣⎡

−−−=

− 1111

1,

c

k

c

k dieselthr k

r r

η

k=1.4



The efficiency of the Otto cycle is always higher than theDiesel cycle

Why use the Diesel cycle?Because you can use higher compression ratios

Thermodynamic Cycle Presentation

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