PHY 2311

PHYSICS 231Engines and fridges

Remco Zegers

Heat Engines

PHY 2312

First Law of thermodynamics

∆U=Uf-Ui=Q+W

∆U=change in internal energyQ=energy transfer through heat (+ if heat is

transferred to the system)W=energy transfer through work (+ if work is

done on the system)if P: constant then W=-P∆V (area under

P-V diagram

This law is a general rule for conservation of energy

PHY 2313

Types of processes

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

PV/T=constantFirst law of thermo-Dynamics:∆U=Q+W

PHY 2314

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=3/2nR∆T•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

Cv=(3/2)R so ∆U=Cvn ∆T

Cv: molar specific heat at const. vol.

PHY 2315

isobaric process

v

p∆P=0Use ∆U=W+QIn case of ideal gas:W=-P∆V & ∆U=3/2nR∆T•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)

Q=∆U-W=3/2nR∆T+ ∆(PV) = 3/2nR∆T+ nR∆T= 5/2nR∆T(used ideal gas law) Q=nCp∆T with Cp=(5/2)RCp=molar heat capacity at constant pressure

PHY 2316

isothermal processes

v

p∆T=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 V↑W=negative Q=positive(Work is done by the gas and energy added)

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=3/2nR∆T•if T↓

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

•if T↑∆U=positive W=positive

(Work is done on the gas)

v

p

PHY 2318

Cyclic processes

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

PHY 2319

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+8000 (work done by gas)W=-12000 J (work done on gas)

∆U=3/2nR∆T=3/2(PBVB-PAVA)== 1.5*[(1E+5)(50E-03)-(5E+5)(10E-03)]=0

The internal energy has not changed∆U=Q+W so Q=∆U-W=12000 J: Heat that was added to thesystem was used to do the work!

PHY 23110

Cyclic process, step by step 2Process 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=3/2nR∆T=3/2(PcVc-PbVb)==1.5[(1E+5)(10E-3)-(1E+5)(50E-3)]=-6000 J

The internal energy has decreased by 6000 J∆U=Q+W so Q=∆U-W=-6000-4000 J=-10000 J10000 J of energy has been transferred out of the system.

PHY 23111

Cyclic process, step by step 3Process C-AW=-Area under P-V diagramW=0 JNo work was done on/by the gas.

∆U=3/2nR∆T=3/2(PaVa-PcVc)==1.5[(5E+5)(10E-3)-(1E+5)(10E-3)]=+6000 J

The internal energy has increased by 6000 J∆U=Q+W so Q=∆U-W=6000-0 J=6000 J6000 J of energy has been transferred into the system.

PHY 23112

Summary of the process

08000 J-8000 JSUM

60006000 J0 JC-A

-6000-10000 J4000 JB-C

012000 J-12000 JA-B

∆UHeat(Q)Work(W)QuantityProcess

-AREA

A-B B-C C-A

PHY 23113

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 23114

exampleA gas goes from initial state I tofinal state F, given the parametersin the figure. What is the work doneon the gas and the net energy transferby heat to the gas for:a) path IBF b) path IF c) path IAF(Ui=91 J Uf=182 J)

a) work done: area under graph: W=-(0.8-0.3)10-3*2.0*105=-100 J∆U=W+Q 91=-100+Q so Q=191 J

b) W=-[(0.8-0.3)10-3*1.5*105 + ½(0.8-0.3)10-3*0.5*105]=-87.5 J∆U=W+Q 91=-87.5+Q so Q=178.5 J

c) W=-[(0.8-0.3)10-3*1.5*105]=-75 J∆U=W+Q 91=-75+Q so Q=166 J

PHY 23115

question

V (m3)

P (Pa)

1 3

3x105

1x105

Consider this cyclicprocess. Which of the followingis true?

a) This is a heat engine and the W done by the gas is +4x105 Jb) This is a fridge and the W done on the gas is +4x105 Jc) This is a heat engine and the W done by the gas is +6x105 Jd) This is a fridge and the W done on the gas is +6x105 Je) This is a heat engine and the W done by the gas is –4x105J

clockwise: work done by the gas, so engineW by gas=area enclosed=(3-1)x(3x105-1x105)=4x105 J

PHY 23116

More general engine

heat reservoir Th

cold reservoir Tc

engine work

Qh

Qc

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

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 coolant work 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 engineA→B isothermal expansion

D→A adiabatic compression

W+, ∆T+

Thot

C→D isothermal compressionTcold

W+, Q-

B→C adiabatic expansion

W-, ∆T-

W-, Q+

PHY 23120

Carnot cycle

Work done by engine: WengWeng=Qhot-Qcoldefficiency: 1-Tcold/Thot

inverse Carnot cycle

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

PHY 23121

exampleThe efficiency of a Carnot engine is 30%. The engine absorbs800 J of energy per cycle by heat from a hot reservoir at500 K. Determine a) the energy expelled per cycle and b)the temperature of the cold reservoir. c) How much workdoes the engine do per cycle?a) Generally for an engine: efficiency: 1-|Qcold|/|Qhot|

0.3=1-|Qcold|/800, so |Qcold|=-(0.3-1)*800=560 J

b) for a Carnot engine: efficiency: 1-Tcold/Thot0.3=1-Tcold/500, so Tcold=-(0.3-1)*500=350 K

c) W=|Qhot|-|Qcold|=800-560=240 J

PHY 23122

An engine is operated between a hot and a cold reservoir with Qhot=400J and Qcold=300J. a) what is the efficiencyof the engine?The engine is modified and becomes a carnot engine. As a result the efficiency is doubled. b) what is the ratio Tcold/Thot. c) what is the maximum efficiency of this engine?

a) efficiency=1-Qcold/Qhot=1-300/400=0.25b) new efficiency: 0.5=1-Tcold/Thot Tcold/Thot=0.5c) 0.5 (Carnot engine has maximum efficiency).

PHY 23123

A new powerplant

A new powerplant is designed that makes use of the temperature difference between sea water at 0m (200) andat 1-km depth (50). A) what would be the maximum efficiencyof such a plant? B) If the powerplant produces 75 MW, howmuch energy is absorbed per hour? C) Is this a good idea?

a) maximum efficiency=carnot efficiency=1-Tcold/Thot=1-278/293=0.051 efficiency=5.1%

b) P=75*106 J/s W=P*t=75*106*3600=2.7x1011 Jefficiency=1-|Qcold|/|Qhot|=(|Qhot|-|Qcold|)/|Qhot|=W/|Qhot| so |Qhot|=W/efficiency=5.3x1012 J

c) Yes! Very Cheap!! but… |Qcold|= |Qhot|-W=5.0x1012 Jevery hour 5E+12 J of waste heat is produced:Q=cm∆T 5E+12=4186*m*1 m=1E+9 kg of water is heated by 10C…perhaps not!

PHY 23124

Story of Hawaiian deep water project

• Keahole sits at a point where underwater land slopes sharply down into the sea, it was a place where warm water can be piped from the surface of the sea and cold water can be piped from depths of about a half-mile.

• A process called ocean thermal energy conversion, or OTEC, used the temperature difference between hot and cold sea water to produce 50 KW of electricity at Keahole in 1993. The process worked but it was uneconomical.

• KAILUA, HAWAI'I — Koyo USA Corp., a company selling deep-sea water from Keahole Hawai'i, is expanding its plant and has applied to sell the water in the United States

• The company is producing more than 200,000 bottles a day and says it can't keep up with demand in Japan, where it sells 1.5 liter bottles of itsMaHaLo brand for $4 to $6 each.