017 Heat Engines

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Phys 160

Thermodynamics and Statistical

Physics

Lecture 17

Heat Engines


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4.1 Heat Engines

A heat engine is any device that absorbs heat

and converts part of that energy into work. An

important example is the steam turbine, used

to generate electricity in most of today's

power plants. The familiar internal combus-

tion engine used in automobiles does not

actually absorb heat, but we can pretend that

the thermal energy comes from outside rather

than inside and treat it, also, as a heat engine.

Unfortunately, only part of the energy

absorbed heat can be converted to work


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by a heat engine. The reason is that the heat, as

it flows in, brings along entropy, which must

somehow be disposed of before the cycle canstart over. To get rid of the entropy, every heat

engine must dump some waste heat into its

environment.The work produced by the engineis the difference between the heat absorbed

and the waste heat expelled.

How much of the heat absorbed by an engine

can be converted into work? Amazingly,

we can say a great deal without knowing

an thin about how the en ine actuall works.


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The heat absorbed by the engine comes from

a place called the hot reservoir, while the

waste heat is dumped into the cold reservoir.The temperatures of these reservoirs, Thand Tc are assumed fixed. (In general, a reser-

voir in thermodynamics is anything that's solarge that its temperature doesn't change

noticeably when heat enters or leaves. For a

steam engine, the hot reservoir is the placewhere the fuel is burned and the cold

reservoir is the surrounding environment.)


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the symbol Qh for the heat absorbed from the

hot reservoir in some given time period, and

Qc for the heat expelled to the cold reservoir.The net work done by the engine during this

time will be W. All three of these symbols will

represent positive quantities; in this chapterthe earlier sign conventions for heat and work

is not used. The benefit of operating a heat

engine is the work produced, W. The cost ofoperation is the heat absorbed, Qh. The

efficiency of an engine, e, as the benefit/cost

ratio:


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For given values of Th and Tc, what is the

maximum possible efficiency? To answer this

question, all we need are the first and

second laws of thermodynamics, plus the

assumption that the engine operates in

cycles, returning to its original state at the

end of each cycle of operation.


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The first law of thermodynamics tells us that

energy is conserved. Since the state of the

engine must be unchanged at the end of acycle, the energy it absorbs must be precisely

equal to the energy it expels. In our notation,

Thus the efficiency cannot be greater than 1,

and can equal 1 only if Qc = 0.


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To proceed further we must also invoke the

second law, which tells us that the total entropy

of the engine plus its surroundings can increasebut not decrease. Since the state of the engine

must be unchanged at the end of a cycle, the

entropy it expels must be at least as much asthe entropy it absorbs. (In this context, as

in Section 3.2, I like to imagine entropy as a

fluid that can be created but never destroyed.)Now the entropy extracted from the hot reser-

voir is just Qh/Th, while the entropy expelled to

the cold reservoir is Qc/Tc.


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So the second law tells us

In general, for the greatest maximum effici-

ency, you should make the cold reservoir very

cold, or the hot reservoir very hot, or both.The smaller the ratio Tc/Th, the more efficient

your engine can be.


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lt's easy to make an engine that's less efficient

than the limit 1 - Tc/Th, simply by producing

additional entropy during the operation.Then to dispose of this entropy you must dump

extra heat into the cold reservoir, leaving less

energy to convert to work. The most obviousway of producing new entropy is in the heat

transfer processes themselves. For instance,

when heat Qh leaves the hot reservoir, theentropy lost by that reservoir is Qh/Th; but if

the engine temperature at this time is less than

Th, then as the heat enters the engine its


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associated entropy will be greater than Qh/Th.

The first law : the efiiciency cant be greater than

1 (You cant win). The efficiency e = 1 can not beachieved unless Th = and Tc = 0, impossible

conditions. The second law: you cant even

break even..


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Problem 4.1. Recall Problem 1.34, which

concerned an ideal diatomic gas taken around

a rectangular cycle on a PV diagram.Supposenow that this system is used as a heat engine,

to convert the heat added into mechanical

work. (a) Evaluate the efficiency of this enginefor the case V2 = 3V1, P2 = 2P1.

(b) Calculate the efficiency of an "ideal"

engine operating between the sametemperature extremes.


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How to make an engine that does achieve the

maximum possible efficiency for a given Th

and Tc ?Minimize the production of extra entropy

during heat transfer itself.

Every engine has a so-called "working sub-stance," which is the material that actually

absorbs heat, expels waste heat, and does

work. In many heat engines the workingsubstance is a gas. Imagine, then, that we first

want the gas to absorb some heat Q h from the

hot reservoir. In the rocess, the entro of the


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Reservoir decreases by Qh/Th, while the

entropy of the gas increases by Qh/Tgas. To

avoid making any new entropy, we would needto make Tgas = T. This isn't quite possible,

because heat won't flow between objects at

the same temperature. So let's make Tgas justslightly less than T , and keep the gas at this

temperature (by letting it expand) as it absorbs

the heat. This step of the cycle, then, requiresthat the gas expand isothermally.

.

l l d h f h l h


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Similarly, during the portion of the cycle when

the gas is dumping the waste heat into the cold

reservoir, we want its temperature to be onlyinfinitesimally greater than T, to avoid creating

any new entropy. And as the heat leaves the

gas, we need to compress it isothermally tokeep it at this temperature. So we have an

isothermal expansion at a temperature just less

than T h, and an isothermal compression at atemperature just greater than Tc. The only

remaining question is how we get the gas from

one temperature to the other and back.


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We don't want any heat to flow in or out when

the gas is at intermediate temperatures, so

these intermediate steps must be adiabatic. Theentire cycle consists of four steps; isothermal

expansion at Th, Adiabatic expansion from Th to

Tc, isothermal compression at Tc, and adiabaticcompression from Tc back up to Th. The

theoretical importance of this cycle was first

pointed out by Sadi Carnot in 1824, so the cycleis now known as the Carnot cycle.

i ibl di l f h


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It is possible to prove directly, from the

formulas for isothermal and adiabatic

processes, that an ideal gas taken around aCarnot cycle realizes the maximum possible

efficiency 1 - Tc/Th. As long as we know that no

new entropy was created during the cycle, thestrict equality must hold in equation 4.4, and

therefore the efficiency must be the maximum

allowed by equation 4.5. This conclusion holdseven if the gas isn't ideal, and, for that matter,

even if the working substance isn't a gas at al


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Although a Carnot cycle is very efficient, it's

also horribly impractical. The heat flows so

slowly during the isothermal steps that it takesforever to get a significant amount of work out

of the engine. So don't bother installing a

Carnot engine in your car; while it wouldincrease your gas mileage, you'd be passed on

the highway by pedestrians.


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Problem 4.6. To get more than an infinitesimal

amount of work out of a Carnot engine, we wouldhave to keep the temperature of its working

substance below that of the hot reser-voir and above

that of the cold reservoir by non-infinitesimal

amounts. Consider, then, a Carnot cycle in which the

working substance is at temperature Thw as it

absorbs heat from the hot reservoir, and at

temperature Tcw as it expels heat to the coldreservoir. Under most circumstances the rates of heat

transfer will be directly proportional to the temp-

erature differences:


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Assume here for simplicity that the constants

of proportionality (K) are the same for both of

these processes. Let us also assume that bothprocesses take the same amount of time, so

the t's are the same in both of these

equations. *(a) Assuming that no new entropy is created

during the cycle except during the two heat

transfer processes, derive an equation thatrelates the four temperatures Th, T c, Thw, and

Tcw.


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(b) Assuming that the time required for the


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(b) Assuming that the time required for the

two adiabatic steps is negligible, write down an

expression for the power (work per unit time)

output of this engine. Use the first and second

laws to write the power entirely in terms of the

four temperatures (and the constant K), theneliminate Tcw using the result of part (a).


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017 Heat Engines

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