egr 334 thermodynamics chapter 5: sections 10-11 lecture 22: carnot cycle quiz today?

EGR 334 ThermodynamicsChapter 5: Sections 10-11

Lecture 22: Carnot Cycle

Quiz Today?

Today’s main concepts:• State what processes make up a Carnot Cycle.• Be able to calculate the efficiency of a Carnot Cycle• Be able to give the Classius Inequality• Be able to apply the Classisus Inequality to determine if a

cycle is reversible, irreversible, or impossible as predicted by the 2nd Law.

Reading Assignment:

Homework Assignment:

Read Chapter 6, Sections 1-5

Problems from Chap 5: 64, 79, 81,86

3

Energy Balance:

Recall from last time:

syssys sys

dEQ W

dt

sysdE

dt

Q W

Entropy Balance:

sys gen

QS

T

Entropy Rate Balance:

sysgen

dS Q

dt T

sys sys sysE Q W

Energy Rate Balance:

sysdS

dt

Q

T

gen

Carnot Cycle

►The Carnot cycle provides a specific example of a reversible cycle that operates between two thermal reservoirs. Other examples covered in Chapter 9 are the Ericsson and Stirling cycles.

►In a Carnot cycle, the system executing the cycle undergoes a series of four internally reversible processes:

two adiabatic processes (Q = 0) alternated with two isothermal processes ( T = constant)

T

v

2 3

1 4

p

v

2

3

14

Carnot Power Cycles

The p-v diagram and schematic of a gas in a piston-cylinder assembly executing a Carnot cycle are shown below:

Carnot Power Cycles The p-v diagram and schematic of water executing a Carnot cycle through four interconnected components are shown below:

In each of these cases the thermal efficiency is given by

H

Cmax 1

T

T

7Sec 5.10 : The Carnot Cycle

Gas only cycle

The Carnot cycle:

T

v

2 3

1 4QH

QC

Area = Work

QH

QC

Process 1-2 : Adiabatic Compression.Process 2 -3 : Isothermal Expansion

receiving QH.

Process 3 – 4 : Adiabatic Expansion.Process 4 – 1 : Isothermal

Compression, rejecting QC.


Analyzing the Carnot cycle:

Energy Balance: WUQ

VU c T VQ c T pdV

Process 2 -3 : Isothermal Expansion receiving QH.

First look at the two isothermal processes

23 VQ c T pdV 3 3 3

323 2 2 2

2

lnHH H

VRT dVQ pdV pdV dV RT RT

V V V

Process 4 – 1 : Isothermal Compression, rejecting QC.

4

141 ln

V

VRTQ C

and

41

23

41

23

41

23

ln

ln

ln

ln

VVT

VVT

VVRT

VVRT

Q

Q

Q

Q

C

H

C

H

C

H

W pdV



Energy Balance:

Then look at the two adiabatic processes (Q = 0)

12 0 VQ c dT pdV Process 1-2 : Adiabatic Compression.

Process 3 – 4 : Adiabatic Expansion.

V

RTc dT pdV dV

V

1

22

1ln

V

V

V

dV

T

dT

R

cH

C

T

T

V

3

44

3lnV

V

V

dV

T

dT

R

cC

H

T

T

V

The term

T

dT

R

c

T

dT

R

c C

H

H

C

T

T

VT

T

V

Thus,

3

4

2

1

1

2 lnlnlnV

V

V

V

V

V

1 4 1 2

2 3 4 3

V V V Vor

V V V V

VQ c dT pdV



With

3

2

4

1

V

V

V

V and

41

23

ln

ln

VVT

VVT

Q

Q

C

H

C

H

Therefore,

C

H

C

H

T

T

Q

Q

We have now proven

max 1 1C C

H H

Q T

Q T

11The Carnot Model of a Hurricane

Warm moist air

Warm air rises

Adiabatic cooling

Cools & Expands as P

As T to dew point, vapor condenses, releasing hfg and warming air

Added heat causes further rising.

12The Carnot Model of a Hurricane

A

B

C

D

vcore>>v outer

Adiabatic

Adiabatic

Isothermal Expansion

Isothermal Compression

13

Example: (5.76) One-half pound of water executes a Carnot power cycle. During the isothermal expansion, the water is heated at 600°F from a saturated liquid to a saturated vapor. The vapor then expands adiabatically to a temperature of 90°F and a quality of 64.3%(a) Sketch the cycle on a P-v diagram.(b) Evaluate the heat and work for each process in BTU(c) Evaluate the thermal efficiency.

Sec 5.10 : The Carnot Cycle

p

v

14



15

state 1 2 3 4

T (°F/R) 600 600 90 90

p (psi) 1541 1541 0.6988 0.6988

x 0 1 0.643

v (ft3/lb) 0.02363 0.2677 300.74

u (Btu/lb) 609.9 1090.0 687.8



state 1 2 3 4

T (°F/R) 600 600 90 90

p (psi)

x 0 1 0.643

v (ft3/lb)

u (Btu/lb)

Isothermal IsothermalAdiabatic Adiabatic

Using Table A-23

3 0.01610 0.643(467.7 0.01610) 300.74 / mv ft lb

3 58.07 0.643(1040.2 58.07) 687.8 / mu Btu lb

16

Example: (5.76) (b) Evaluate the heat and work for each process in BTU



state 1 2 3 4

T (°F/R) 600 600 90 90

p (psi) 1541 1541 0.6988

0.6988

x 0 1 0.643

v (ft3/lb) 0.02363

0.2677

300.74

u (Btu/lb) 609.9 1090.0

687.8

Process

U Q W

1 - 2 240 274.8

34.81

2 - 3 0

3 - 4

4 - 1 0

12 2 1 0.5 1090 609.9 / 240m mU m u u lb Btu lb Btu

12 2 1 2 1 2 1( ) ( )W pdV p V V p mv mv mp v v

22 3

2

144 1(1541 / ) 0.5 0.2677 0.02363 /

1 778f m mf

in Btulb in lb ft lb

ft ft lb

Process 1-2

34.81 Btu

Process

U Q W

1 - 2

2 - 3 0

3 - 4

4 - 1 0

12 12 12 240 34.81 274.8Q U W Btu

17

Example: (5.76) (b) Evaluate the heat and work for each process in BTU



state 1 2 3 4

T (°F/R) 600 600 90 90

p (psi) 1541 1541 0.6988

0.6988

x 0 1 0.643

v (ft3/lb) 0.02363

0.2677

300.74

u (Btu/lb) 609.9 1090.0

687.8

Process

U Q W

1 - 2 240 274.8

34.81

2 - 3 -201

0 201

3 - 4

4 - 1 0Process 2-3

23 2 1 0.5 687.8 1090 / 201.1m mU m u u lb Btu lb Btu

23 23 23 0 ( 201.1) 201.1W Q U Btu

23 0Q (adiabatic process)

Process

U Q W

1 - 2 240 274.8

34.81

2 - 3 0

3 - 4

4 - 1 0

18

Example: (5.76) (b) Evaluate the heat and work for each process in Btu



state 1 2 3 4

T (°F/R) 600 600 90 90

p (psi) 1541 1541 0.6988

0.6988

x 0 1 0.643

v (ft3/lb) 0.02363

0.2677

300.74

u (Btu/lb) 609.9 1090.0

687.8

Process

U Q W

1 - 2 240 274.85

34.81

2 - 3 -201

0 201

3 - 4 -142.6

4 - 1 0H H

C C

Q T

Q T

For Process 3 – 4: for the Carnot cycle:

34

550274.85 142.6

1060C

C HH

TQ Q Q Btu Btu

T

where QH = Q12 = 274.8 Btu

34 3 4 3 3 4 3( ) ( )W pdV p V V p m v v 34 4 3( )U m u u

19




state 1 2 3 4

T (°F/R) 600 600 90 90

p (psi) 1541 1541 0.6988

0.6988

x 0 1 0.643

v (ft3/lb) 0.02363

0.2677

300.74

u (Btu/lb) 609.9 1090.0

687.8

Process

U Q W

1 - 2 240 274.85

34.81

2 - 3 -201

0 201

3 - 4 -142.6

4 - 1 0continuing for Process 3 – 4:

recalling h = u + pv

4 3 34 4 4 3 3( ) ( )m u u Q m p v p v 34 34 34U Q W

34 4 4 4 3 3 3( ) ( )Q m u p v m u p v 34

4 4 4 4 3 3 3( ) ( )Q

h u p v u p vm

2

2 34 2

142.6 144 1(687.8 / (0.6988 / )(300.74 / ))

0.5 1 778m f mm f

Btu in Btuh Btu lb lb in ft lb

lb ft lb ft

441.5 / mBtu lb

20




state 1 2 3 4

T (°F/R) 600 600 90 90

p (psi) 1541 1541 0.6988

0.6988

x 0 1 0.643 0.368

v (ft3/lb) 0.02363

0.2677

300.74

172.1

u (Btu/lb) 609.9 1090.0

687.8 419.5

Process

U Q W

1 - 2 240 274.85

34.81

2 - 3 -201

0 201

3 - 4 -142.6

4 - 1 0continuing for Process 3 – 4:

Then using Table A2 at h4 = 441.5 Btu/lbm and T4 = 90 deg.

4 44

4

441.5 58.070.368

1042.7f

fg

h hx

h

which let the state 4 intensive properties be found:3

4 0.01610 0.368(467.7 0.01610) 172.1 / mv ft lb

4 58.07 0.368(1040.2 58.07) 419.5 / mu Btu lb

state 1 2 3 4

T (°F/R) 600 600 90 90

p (psi) 1541 1541 0.6988

0.6988

x 0 1 0.643

v (ft3/lb) 0.02363

0.2677

300.74

u (Btu/lb) 609.9 1090.0

687.8

21

Example: (5.76)(b) Evaluate the heat and work for each process in BTU


state 1 2 3 4

T (°F/R) 600 600 90 90

p (psi) 1541 1541 0.6988

0.6988

x 0 1 0.643 0.368

v (ft3/lb) 0.02363

0.2677

300.74

172.1

u (Btu/lb) 609.9 1090.0

687.8 419.5

Process

U Q W

1 - 2 240 274.85

34.81

2 - 3 -201 0 201

3 - 4 -134.2

-142.6 -8.32

4 - 1 0


34 4 3 0.5 (419.5 687.8) / 134.2m mU m u u lb Btu lb Btu and for Process 4-1:

34 3 4 3 3 4 3W pdV p V V p m v v

22 3

34 2

144 1(0.6988 / )(0.5 ) 172.1 300.74 /

1 778f m mf

in BtuW lb in lb ft lb

ft ft lb

8.32 Btu

Process

U Q W

1 - 2 240 274.85

34.81

2 - 3 -201 0 201

3 - 4 -142.6

4 - 1 0

22

Example: (5.76)(b) Evaluate the heat and work for each process in BTU


state 1 2 3 4

T (°F/R) 600 600 90 90

p (psi) 1541 1541 0.6988

0.6988

x 0 1 0.643 0.368

v (ft3/lb) 0.02363

0.2677

300.74

172.1

u (Btu/lb) 609.9 1090.0

687.8 419.5

Process

U Q W

1 - 2 240 274.85

34.81

2 - 3 -201 0 201

3 - 4 -134.2

-142.6 -8.32

4 - 1 95.2 0 -95.2


41 1 4 0.5 (609.9 419.5) / 95.2m mU m u u lb Btu lb Btu

41 41 41 0 95.2 95.2W Q U Btu

Finally, process 4 – 1:

Process

U Q W

1 - 2 240 274.85

34.81

2 - 3 -201 0 201

3 - 4 -134.2

-142.6 -8.32

4 - 1 0

23

Example: (5.76)


state 1 2 3 4

T (°F/R) 600 600 90 90

p (psi) 1541 1541 0.6988

0.6988

x 0 1 0.643 0.368

v (ft3/lb) 0.02363

0.2677

300.74

172.1

u (Btu/lb) 609.9 1090.0

687.8 419.5

Process

U Q W

1 - 2 240 274.85

34.81

2 - 3 -201 0 201

3 - 4 -134.2

-142.6 -8.32

4 - 1 95.2 0 -95.2


Thermal efficiency of the cycle:

(c) Evaluate the thermal efficiency.

max

5501 1 0.481 48%

1060C

H

T

T

34.81 201 8.32 95.2 131.990.480 48%

274.85 274.85Cycle

in

W

Q

Clausius Inequality

►The Clausius inequality is developed from the Kelvin-Planck statement of the second law and can be expressed as:

cycleb

Q

T

The nature of the cycle executed is indicated by the value of scycle:

scycle = 0 no irreversibilities present within the system

scycle > 0 irreversibilities present within the system

scycle < 0 impossible

25Sec 5.11 : The Clausius Inequality

We have shown that: and thusC

H

C

H

T

T

Q

Q

For an ideal/reversible process

C

C

H

H

T

Q

T

Q

Therefore:0

C

C

H

H

T

Q

T

Q

p

v

Now consider a general process

Each part of the cycle is divided into an infinitesimally small process

0C

C

H

H

T

dQ

T

dQ

Then sum (integrate) the entire process

0

bT

Q dQ = heat transfer at boundary

T = absolute T at that part of the cycle.

26Sec 5.11 : The Clausius Inequality

For a real process, Qreal > Qreversible

Therefore:

P

v

We can then define σ, where

0

bT

Q

cyclebT

Q

and σcycle = 0 reversible process

σcycle > 0 irreversible process

σcycle < 0 impossible process

We now also have the mathematical definition of enthalpy.

TdSdQdST

dQ

More on this in Chapter 6

27

Example: (5.81) A system executes a power cycle while receiving heat transfer at a temperature of 500 K and discharging 1000 kJ by heat transfer at a temperature of 300 K. There are no other heat transfers. Appling the Clausius Inequality, determine σcycle if the thermal efficiency is (a) 60% (b) 40% (c) 20 %. In each case is the cycle reversible, or impossible?

Sec 5.11 : The Clausius Inequality

W=?

TH= 500 K

TH= 300 K

Q = 50 kJ

28



bcycle T

Q

HCH

C QQQ

Q 11

and

Therefore:

CHH

C

C

H

Hcycle TT

QT

Q

T

Q 11

K

kJ

KKkJcycle 667.0

300

6.01

500

11000

Since σcycle is negative, the cycle is impossible.

W=?

TH= 500 K

TH= 300 K

Q = 50 kJ

29



CHH

C

C

H

Hcycle TT

QT

Q

T

Q 11

K

kJ

KKkJcycle 0

300

4.01

500

11000

Since σcycle is zero, the cycle is internally reversible.

W=?

TH= 500 K

TH= 300 K

Q = 50 kJ

30



CHH

C

C

H

Hcycle TT

QT

Q

T

Q 11

K

kJ

KKkJcycle 667.0

300

2.01

500

11000

Since σcycle is positive, the cycle is irreversible.

W=?

TH= 500 K

TH= 300 K

Q = 50 kJ

31

End of Lecture 21 Slides

egr 334 thermodynamics chapter 5: sections 10-11 lecture 22: carnot cycle quiz today?

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