Transcript

Thermodynamic Cycles

• Look at different cycles that approximate real processes

• You can categorize these processes several different ways• Power Cycles vs. Refrigeration• Gas vs. Vapor• Closed vs. open• Internal Combustion vs. External Combustion

Power Cycles

• Otto Cycle• Spark Ignition

• Diesel Cycle• Brayton Cycle

• Gas Turbine

• Rankine Cycle

η thnet

in

WQ

=

These are all heat engines. They convert heat to work, so the efficiency is:

Ideal Cycles

• We’ll be using ideal cycles to analyze real systems, so lets start with the only ideal cycle we’ve studied so far

Carnot Cycle

W Q

Q-W=0 Q=WIn addition, we know that the efficiency for a

Carnot Cycle is:

η th CarnotL

H

TT, = −1

Carnot Cycle is not a good model for most real processes

• For example• Internal combustion engine• Gas turbine

• We need to develop a new model, that is still ideal

Air-Standard Assumptions

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

• All the processes are internally reversible

• 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

Terminology for Reciprocating Devices

r VV

VV

BDC

TDC

= =maxmin

Compression Ratio

Mean Effective Pressure

∫=2

1PdVW

VPW ∆=

v

v

1-2 Isentropic Compression2-3 Constant Volume Heat Addition3-4 Isentropic Expansion4-1 Constant Volume Heat Rejection

Thermal Efficiency of the Otto Cycle

η thnet

in

net

in

in out

in

out

in

WQ

Q QQ

= = =−

= −1

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

∫ =+=+=

∆=−3

223,23,23,

2323,23,

00 PdVWWW

UWQ

bothernet

netnet

Q U

Q Q mC T Tnet

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,

=+=+=

∆=−

∫ PdVWWW

UWQ

bothernet

netnet

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

QQmC T TmC T T

,

( )( )

= −

= −−−

1

1 4 1

3 2

η th OttoT TT T

T T TT T T

,( )( )

( / )( / )

= −−−

= −−−

1

1 11

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

−−

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

kk

vvT

vv

TT

v3 = v2 and v4 = v1

TT

TT

orTT

TT

2

1

3

4

4

1

3

2

=

=

η th OttoT TT T

T T TT T T

,( )( )

( / )( / )

= −−−

= −−−

1

1 11

4 1

3 2

1 4 1

2 3 2

1

η th OttoTT, = −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 the Carnot 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

rVV

TT

Since process 1-2 is isentropic,

η th Otto kr, = − −1 11

η th OttoTT, = −1 1

2

Increasing Compression Ratio Increases the Efficiency

Typical Compression Ratios for Gasoline Engines

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 only difference is in process 2-3

Consider Process 2-3

• This is the step where heat is transferred into the system

• We model it as constant pressure instead of constant volume

23,23, uuuwq outbin −=∆=−( )2323, TTChvPuq 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 TTCuqq vout −=∆=−=( )14 TTCq vout −=

As for any heat engine…

in

out

in

netth q

qqw

−== 1η

( ) ( )2314 and TTCqTTCq pinvout −=−=

( )23

14,

)(1TTCTTCn

p

vdieselth −

−−= ( )

( )23

141TTkTT−−

−=

Rearrange

( )( )1

11232

141, −

−−=

TTkTTTTn dieselth

PVT

PVT

P P

TT

VV

rc

3 3

3

2 2

23 2

3

2

3

2

= =

= =

where

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

( )( )1

112

141, −

−−=

cdieselth rkT

TTTn

PVT

PVT

V V

TT

PP

4 4

4

1 1

14 1

4

1

4

1

= =

=

where

kk

kk

VPVP

VPVP

3344

2211 and

=

=

( )( )1

112

141, −

−−=

cdieselth rkT

TTTn 14 PP

Since Process 1-2 and Process 3-4

are both isentropic

k

k

k

k

VV

PP

VV

PP

2

3

2

3

1

4

1

4 =

1 1

kc

k

rVV

=⎟⎟⎠

⎞⎜⎜⎝

⎛=

2

3

( )( )1

112

1, −

−−=

c

kc

dieselth rkTrTn

Finally, Since process 1-2 is isentropic

1

1

2

2

1

−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

k

vv

TT

The volume ratio from 1 to 2 is the compression ratio, r

1

2

1 1 −

⎟⎠⎞

⎜⎝⎛=

k

rTT

( )⎥⎦⎤

⎢⎣

⎡−−

−= − 1111 1,

c

kc

kdieselth rkr

rη

k=1.4

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

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

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