Heat engines

Heat engines

Use heat to do work in a cyclic process: same process is repeated over and over again

Reservoirs

Keep the same temperature

Supply the working substance with energy so that its temperature remains or becomes equal to Treservoir

All processes are “slow” and reversible from the reservoir’s point of view

The Carnot cycle

Consists of four slow steps:

12: isothermal23: adiabatic34: isothermal41: adiabatic

The first step

U=0 as T=0 (nearly true for non-ideal gas)

Stotal=0, hence this step is reversible

1

221 ln

VV

nRTWQ HH

1

221 ln

VV

nRTQ

SH

Hgas

H

Hreservoir

TQ

S 21

Four questions

In the second step, Q, W, U, S are

a) positive

b) zero

c) negative

d) dependent on the

gas used

Comments on the Carnot cycle

It is customary to define QC as positive even though heat flows out of the system

Heat loss to cold reservoir necessary to bring system back to original state

Scycle=0, hence

Shot reservoir+ Scold reservoir = 0 C

C

H

H

TQ

TQ

Work and efficiency

Work done during cycle: W = QH – QC

This is independent of working substance!

HH

C

H

C

HCarnot T

TTT

QQ

QW

11

inheat outwork

efficiency

Reversibility

There is no net change in entropy: the entropy of gas and environment are unchanged

Hence the entire Carnot cycle is reversible, and it can be run as a “fridge”

“Reversible” means: we can reverse the cycle and restore the original state without any residual changes outside the system

Maximum efficiency

Imagine we have some heat engine converting heat into work, and a Carnot engine running backwards

The Carnot engine takes heat QC from the cold reservoir, work W is done on the gas and the engine gives QH to the hot reservoir

Maximum efficiency II

other=Carnot: no net work done, no net heat transfer from either reservoir

other > Carnot: Say the other engine delivers QC to cold reservoir. Then it takes in QH + , which is used to do work W + . However:

This violates second law of thermodynamics

Carnot engine is most efficient

If there was a more efficient engine, net work could be done by extracting heat from the hot reservoir without any other changes anywhere.

Thus disorderly thermal motion would be converted into orderly mechanical motion

Entropy would decrease, not allowed by 2nd law

The Otto cycle

Four steps:

12: slow adiabatic23: fast isochoric34: slow adiabatic41: fast isochoric

Temperature and heat

The temperature T2 (and T4) can be evaluated:

Heat added:41:23:

1

2

12

1 '

VV

TT

CTVCpV

H

)(

)(

2

4

CVC

HVH

TTnCQ

TTnCQ

Entropy in the Otto cycle

Step 41: gas: fast process, calculate S as if it were slow reservoir: slow process, calculate as usual

HenceH

HV

H

Hreservoir

HV

T

TV

gas

TTT

nCTQ

S

TT

nCTT

nCSH

414

414 ln

d

4

01414 reservoirgas SS

Entropy in the Otto cycle II

Likewise in step 23:

We conclude for the entire cycle

Since

The cycle is thus not reversible

03232 reservoirgas SS

0 reservoirgascycle SSS

0 ,0 reservoircyclegas SSS

PS225 – Thermal Physics topicsThe atomic hypothesisHeat and heat transferKinetic theoryThe Boltzmann factorThe First Law of ThermodynamicsSpecific HeatEntropyHeat enginesPhase transitions