04/19/23 PHY 113 -- Lecture 21 1

Announcements

1. 3rd exam Redo due Thursday, Dec. 4th

Presentations for Exam 3 (or 1 or 2) ??? if ?? when?? how many??

2. Final exam, Wednesday, Dec. 10th

Extra problem solving sessions?? if ?? when?? how many??

3. Physics colloquium Thursday, Dec. 4th at 4 PM Professor Scott Wollenwebber, WFU School of Medicine --

"Positron Emission Tomography: From Basic Physics to Functional Images"

4. Today’s lecture – analysis of thermodynamic processesEfficiency of a thermodynamic processCarnot, Otto, Diesel processesNotion of entropy

04/19/23 PHY 113 -- Lecture 21 2

Review of thermodynamic ideas

“First law” of thermodynamics: Eint = Q W

For an ideal gas: PV = nRT

Special cases: Isovolumetric (V=constant) W = 0

Isobaric (P=constant)

Isothermal process (T=constant) Eint = 0

Adiabatic process (Q = 0)

V

PVV C

C-R

CTnCTR-n

E : where,1γ

; 1γ

int

i

fii

i

fV

V V

VVP

V

VnRTPdVW

f

i

lnln

1-γ1-γγγ ffiiffii VTVTVPVP

1

R

CP

04/19/23 PHY 113 -- Lecture 21 3

Extra credit:

Show that the work done by an ideal gas which has an initial pressure Pi and initial volume Vi when it expands adiabatically to a volume Vf is given by:

1γ

11γ f

iV

V

ii

V

VVPPdVW

f

i

04/19/23 PHY 113 -- Lecture 21 4

P (

1.01

3 x

105 )

Pa

Vi Vf

Pi

Pf

A

B C

D

Examples process by an ideal gas:

AB BC CD DA

Q

W 0 Pf(Vf-Vi) 0 -Pi(Vf-Vi)

Eint

1-γ

)( ifi PPV 1-γ

)(γ iff VVP 1-γ

)( iff PPV 1-γ

)(γ ifi VVP-

1-γ

)( ifi PPV 1-γ

)( iff PPV 1-γ

)( iff VVP 1-γ

)( ifi VV-P

Efficiency as an engine:

e = Wnet/ Qinput

04/19/23 PHY 113 -- Lecture 21 5

Otto cycle:

04/19/23 PHY 113 -- Lecture 21 6

P

b

c d

e

04/19/23 PHY 113 -- Lecture 21 7

P QAB=0

QCD=0

1γ)(2

BC

BC

PPVQ

1γ)(1

AD

DAPPV

Q

)()(

112

1

BC

AD

BC

DA

BC

DABC

PPVPPV

QQ

Q

QQe

21

21 ;

VPVP

VPVP

CD

BA

121 /

11 VV

e

Otto cycle

Example: r=5, =1.4

e=0.475

04/19/23 PHY 113 -- Lecture 21 8

Diesel cycle

V

A

D

QAB=0

QCD=0

1γ

)(γ

BCBBC

VVPQ

1γ)(

ADD

DAPPV

Q

BACD

BACD

VVVV

VVVVe/1

/1

/1

/1

γ1

1

Example: VD/VC=5, VA/VB=15, =1.4 e=0.558

04/19/23 PHY 113 -- Lecture 21 9

Stirling engine

P

Th

Tc

A

C

D

B

DAAB

CDAB

QQWW

e

AB BC CD DA

Q

W 0 0

Eint 0 0

A

Bh V

VnRT ln

A

Bh V

VnRT ln

D

Cc V

VnRT ln

D

Cc V

VnRT ln

1γ ch TTnR

1γ ch TTnR

1γ ch TTnR

1γ ch TTnR

Example:

Th = 3Tc VB=VC=5VA=5VD = 1.3

%7.66

%6.50

1γ/ln

/ln

Carnot

e

TTnRVVnRT

VVTTnRe

chABh

ABch

04/19/23 PHY 113 -- Lecture 21 10

Carnot process

04/19/23 PHY 113 -- Lecture 21 11

Carnot cycle

AB BC CD DA

Q 0 0

W

Eint 0 0

A

Bh V

VnRT ln

A

Bh V

VnRT ln

D

Cc V

VnRT ln

1γ ch TTnR

1γ ch TTnR

D

Cc V

VnRT ln

1γ ch TTnR

1γ ch TTnR

h

c

AB

CD

h

c

AB

CDAB

T

T

VV

VV

T

T

Q

QQe

1

/ln

/ln1

04/19/23 PHY 113 -- Lecture 21 12

Carnot cycle

QBC=0

QDA=0

A

BHAB V

VnRTQ ln

D

CCCD V

VnRTQ ln

H

C

AB

CDAB

T

T

Q

QQe

1

CT

04/19/23 PHY 113 -- Lecture 21 13

Examples

Efficiency of a Carnot engine operating between the temperatures of Tc=0oC and Th=100oC:

For a Carnot engine, it is clear that we cannot achieve e=100%; not possible to completely transform heat into work. It is possible to show that the Carnot engine is the most efficient that one can construct between the two operating temperatures Tc and Th.

%8.2615.37315.273

1 e

04/19/23 PHY 113 -- Lecture 21 14

Carnot cycle for cooling and heating

“coefficient of performance”

COPheating=|Qh/W|=Th/(Th-Tc)

COPcooling=|Qc/W|=Tc/(Th-Tc)

Example: Suppose that on a cold winter day, a heat pump has a compressor which brings outdoor air at Tc=-3oC into a room at Th=22oC. What is the COP?

COP=295.15/25=11.8

04/19/23 PHY 113 -- Lecture 21 15

More about Carnot cycle

A

Bhh V

VnRTQ ln

A

Bcc V

VnRTQ ln

0||||

: thatNotice

c

c

h

h

c

c

h

h

TQ

TQ

TQ

TQ

B

AAB T

dQS

:entropy Define

0

ln

0

ln

DA

A

BCD

BC

A

BAB

S

VV

nRS

S

VV

nRS

04/19/23 PHY 113 -- Lecture 21 16

Carnot cycle shown in a T-S diagram:

04/19/23 PHY 113 -- Lecture 21 17

Other examples of entropy calculations:

Ideal gas:

Isovolumetric process: Isobaric process:

Melting of solid having mass m and latent heat L at melting temperature TM:

A

BB

A

V

TTnR

TdTnR

S

dTnR

dTnCdQ

ln1γ1γ

1γ

A

BB

A

P

TTnR

TdTnR

S

dTnR

dTnCdQ

ln1γ

γ1γ

γ

1γγ

M

m

TLm

TLdm

S 0

04/19/23 PHY 113 -- Lecture 21 18

P (

1.01

3 x

105 )

Pa

Vi Vf

Pi

Pf

A

B C

D

Peer instruction question:

Consider the “square cycle” shown below. What can you say about the entropy change in each cycle:

(A)SABCDA= 0

(B) SABCDA> 0

(C) SABCDA< 0

04/19/23 PHY 113 -- Lecture 21 19

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04/19/23 PHY 113 -- Lecture 21 22