PHY 2311

PHYSICS 231Lecture 31: Engines and fridges

Remco ZegersQuestion hours: Thursday 12:00-13:00 & 17:15-

18:15Helproom

PHY 2312

Metabolism

U=Q+W

Change in internal energy: Must be increased: Food!

Heat transfer: Negativebody temperature < room temperature

Work done (negative)

PHY 2313

Metabolic rate

U = Q + Wt t t

Metabolic rate: rate in which food and oxygen are transformed into internal energy (to balance losses dueto heat loss and work done).

|W/t| Body’s efficiency: |U/t|

PHY 2314

Body’s efficiency

U/t~oxygen use ratecan be measured

W/t can be measured

PHY 2315

Types of processes

A: Isovolumetric V=0B: Adiabatic Q=0C: Isothermal T=0D: Isobaric P=0

PV/T=constant

First law of thermo-Dynamics:U=Q+W

PHY 2316

Isovolumetric processes (line A)

V=0W=0 (area under the curve is zero)U=Q (Use U=W+Q, with W=0)In case of ideal gas:U=3/2nRT•if P then T (PV/T=constant)

so U=negative Q=negative(Heat is extracted from the gas)

•if P then T (PV/T=constant)so U=positive Q=positive

(Heat is added to the gas)v

p

PHY 2317

Adiabatic process (line B)

Q=0No heat is added/extracted from thesystem.U=W (Use U=W+Q, with Q=0)In case of ideal gas:U=3/2nRT•if T

U=negative W=negative(The gas has done work)

•if TU=positive W=positive

(Work is done on the gas)

v

p

PHY 2318

isothermal processes

v

pT=0The temperature is not changedQ=-W (Use U=W+Q, with U=0)•if V

W=positive Q=negative(Work is done on the gas and energy extracted)

•if VW=negative Q=positive

(Work is done by the gas and energy added)

PHY 2319

isobaric process

v

p

P=0Use U=W+QIn case of ideal gas:W=-PV & U=3/2nRT•if V then T (PV/T=constant)

W: positive (work done on gas)U: negative Q: negative (heat extracted)

•if V then T (PV/T=constant)W: negative (work done by gas)

U: positive Q: positive (heat added)

PHY 23110

Cyclic processes

The system returns to itsoriginal state. Therefore,the internal energy mustbe the same after completionof the cycle (U=0)

PHY 23111

Cyclic Process, step by step 1Process A-B.Negative work is done on the gas:(the gas is doing positive work).

W=-Area under P-V diagram

=-[(50-10)*10-3]*[(1.0-0.0)*105]-½[(50-10)*10-3]*[(5.0-1.0)]*105==4000+8000W=-12000 J

U=3/2nRT=3/2(PBVB-PAVA)= = 1.5*[(1E+5)(50E-03)-(5E+5)(10E-03)]=0The internal energy has not changedU=Q+W so Q=U-W=12000 J: Heat that was added to thesystem was used to do the work!

PHY 23112

Cyclic process, step by step 2

Process B-CW=-Area under P-V diagram

=-[(10-50)*10-3*(1.0-0.0)*105]=W=4000 JWork was done on the gas

U=3/2nRT=3/2(PcVc-PbVb)= =1.5[(1E+5)(10E-3)-(1E+5)(50E-3)]=-6000 JThe internal energy has decreased by 6000 JU=Q+W so Q=U-W=-6000-4000 J=-10000 J10000 J of energy has been transferred out of the system.

PHY 23113

Cyclic process, step by step 3

Process C-AW=-Area under P-V diagramW=0 JNo work was done on/by the gas.

U=3/2nRT=3/2(PcVc-PbVb)= =1.5[(5E+5)(10E-3)-(1E+5)(10E-3)]=+6000 JThe internal energy has increased by 6000 JU=Q+W so Q=U-W=6000-0 J=6000 J6000 J of energy has been transferred into the system.

PHY 23114

Summary of the processQuantityProcess

Work(W) Heat(Q) U

A-B -12000 J 12000 J 0

B-C 4000 J -10000 J -6000

C-A 0 J 6000 J 6000

SUM -8000 J 8000 J 0

-AREA

A-B B-C C-A

PHY 23115

What did we do?

The gas performed net work (8000 J)while heat was supplied (8000 J):We have built an engine!

What if the process was done inthe reverse way?Net work was performed on thegas and heat extracted from the gas.We have built a heat pump!(A fridge)

PHY 23116

More general engine

heat reservoir Th

cold reservoir Tc

engine work

Qh

Qc

W=|Qh|-|Qc|efficiency: W/|Qh|e=1-|Qh|/|Qc|

W

The efficiency is determinedby how much of the heat yousupply to the engine is turnedinto work instead of being lostas waste.

turns water to steam

the steam moves the piston work is done

the steam is condensed

PHY 23117

Reverse direction: the fridge

heat reservoir Th

cold reservoir Tc

engine work

Qh

Qc

W

heat is expelled to outside

a piston compresses the coolantwork is done

the fridge is cooled

PHY 23118

The 2nd law of thermodynamics

1st law: U=Q+W In a cyclic process (U=0) Q=W: we cannot do more workthan the amount of energy (heat) that we put inside

2nd law: It is impossible to construct an engine that, operating in a cycle produces no other effect than theabsorption of energy from a reservoir and the performanceof an equal amount of work: we cannot get 100% efficiency

What is the most efficient engine we can makegiven a heat and a cold reservoir?

PHY 23119

Carnot engineAB isothermal expansion

DA adiabatic compression

W-, T-

Thot

CD isothermal compressionTcold

W+, Q-

BC adiabatic expansion

W-, T-

W-, Q+

PHY 23120

Carnot cycle

Work done by engine: Weng

Weng=Qhot-Qcold

efficiency: 1-Tcold/Thot

inverse Carnot cycle

A heat engine or a fridge!By doing work we cantransport heat

PHY 23121

Next lecture

Entropy and examples