Heat Engines: the Carnot Cycle (2024)

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Applet here!

Michael Fowler

The Ultimate in Fuel Efficiency for a Heat Engine

All standard heat engines (steam, gasoline, diesel) work bysupplying heat to a gas, the gas then expands in a cylinder and pushes a pistonto do its work. So it’s easy to see howto turn heat into work, but that’s a one shot deal. We need it to keeprepeating to have a useful engine. Theheat and/or the gas must therefore be dumped out of the cylinder before thenext cycle begins, otherwise all the work the gas delivered on expanding willbe used up compressing it back!

Our aim in this lecture is to figure out just how efficientsuch a heat engine can be: what’s themost work we can possibly get for a given amount of fuel in a cyclical process? We’ll examine here the model stripped to itsessentials: an ideal gas is enclosed ina cylinder, with external thermal connections to supply and take away heat, anda frictionless piston for the gas to perform (and if necessary absorb)mechanical work:

This simplest heat engine is called the Carnot engine, for which one complete heating/cooling,expanding/contracting cycle back to the original gas volume and temperature isa Carnot cycle, named afterSadi Carnot who in 1820 derived the correctformula for the maximum possible efficiency of such a heat engine in terms ofthe maximum and minimum gas temperatures during the cycle.

Carnot's result was that if the maximum hot temperaturereached by the gas is $T_{H},$ and the coldest temperature during the cycleis $T_{C},$ (degrees kelvin, or rather just kelvin, of course) the fraction of heat energy input that comesout as mechanical work , called the efficiency, is

Efficiency = $\frac{T_{H} - T_{C}}{T_{H}} .$

This was an amazing result, because it was exactly correct,despite being based on a complete misunderstanding of the nature of heat!

How Understanding Waterwheel Efficiency was the Key to Understanding theHeat Engine

Carnot believed that heat, like electricity, was a fluid thatflowed from hot things to cold things (and somehow through space as radiation).

What motivated Carnot to attempt to calculate steam energyefficiency in 1820? Well, it was the timeof the Industrial Revolution, and the efficiency of your power supplydetermined your profit margin.

Big engineswere primarily used in mass production of cloth, in factories called mills. Upto the late 1700's these mills were located by fast flowing rivers, the powersource was a large waterwheel, it turned a long rotating rod that stretched thelength of the factory. Ropes took powerfrom pulleys on this rod to turn individual looms, which were operated bysemiskilled laborers, often children. The picture here is much later (1914),and steam-driven, but shows the poweringscheme.

Thesteam engine offered an attractive alternative: it didn't need to be close to ariver. But it needed coal or wood for fuel, unlike the watermill.

Since the main source of industrial power up to the late1700's was the waterwheel, a lot of thought went into making it as efficient aspossible, and since Carnot thought heat was a fluid, he used waterwheelthinking in analyzing the steam engine.So, how do you make a waterwheel as efficient as possible?

The water loses potential energy as it is carried down bythe wheel, so the most energy possible is $m g h$ watts, where $m$ is the mass of water flowing per second. (We're ignoring possible kinetic energycontributions from the ingoing water coming in fast $—$ this is avery small effect, and doesn't apply to Carnot's heat engine analysis.)

How is energy wasted?Obviously, we need as little friction in the wheel as possible. There must be smooth flow:no water splashing around.

The water must flow into and off the wheel without dropping any significant height,or it loses that much potential energy without producing work.

A perfect waterwheel would be reversible: it could be used to drive a copy of itselfbackwards, to lift up the same amount of water per second that fell.

Aside: A Modern Water Wheel inVirginia

There is in Virginia a pretty efficient water wheel: it’s about 80% efficient $—$ the BathCounty hydroelectric pumped storage station. This is a water wheel, actually a turbine,but that amounts to the same thing better designed, that works both ways. Water from an upper lake falls through a pipeto a turbine and the lower lake, generating electrical power. Alternatively, electric power can be suppliedto pump the water back up. Whybother? Because demand for electricityvaries, and it’s better to avoid if possible building power stations that areonly running during peak demand. It’scheaper to store power at times of low demand.

The drop h isabout 1200 feet, 380 meters. The flow rate is about a thousand tons a second. The plant generates about 3 Gigawatts, substantiallymore than a two-unit nuclear plant, such as North Anna.

Carnot’s Idea: a “Water Wheel” forHeat

Carnot's belief that heat was a fluid (we still picture it flowingthat way in thinking of heat conduction, or, say, cooking) led him to analyze the steam engine inparallel to a water wheel. In the water wheel, the water falls through agravitational potential difference and that potential energy is transformedinto work by the wheel. The"electric fluid" we now see as a fluid losing electrical potentialenergy and producing work or heat. So,what about the heat "caloric fluid" (as it was called)? Obviously, the analogy to gravitationalpotential is just temperature! As thegas in the cylinder expands, it does work, but its temperature goes down.

Carnot assumed thatthe steam engine was nothing but a water wheel for this caloric fluid, so themost efficient engine would have minimal friction, but also, in analogy withthe water entering and leaving the wheel gently with no intermediate loss ofheight, the heat would enter and leave the gas in the engine isothermally(remember the temperature is analogous to the gravitational potential, thus the height). Therefore, by analogy with $g h,$ the drop in temperature $T_{H} - T_{C}$ measures the potential energy given up by aunit amount of the “heat fluid”.

The most efficient steam engine would have isothermal heatexchange (negligible temperature differences in heat exchange), like the mostefficient water wheel (only a tiny drop as the water enters and leaves thewheel). Of course, this is thetheoretical limit: some drop is necessary for operation. But the importantpoint is that in the limit of perfect efficiency, both engine and waterwheelare reversible $—$ ifsupplied with work, they could transform it into the same amount of heat theywould need to generate that work in the first place.

But how does that relate to the energy expended producingthe heat in the first place? Well, Carnotknew something else: there was an absolute zero of temperature. Therefore, he reasoned, if you cooled thefluid down to absolute zero, it would give up all its heat energy. So, the maximum possible amount of energy youcan extract by cooling it from $T_{H}$ to $T_{C}$ is, what fraction is that of cooling it toabsolute zero?

It’s just $\frac{T_{H} - T_{C}}{T_{H}}!$

Of course, the caloric fluid picture isn’t right, but this result is! This is the maximum efficiency of aperfect engine: and remember, this engine is reversible. We'll see how to use that important fact later.

Getting Work Out of a Hot Gas Efficiently:Isothermal and Adiabatic Flows

Now let’s turn to the details of getting the most work outof a heated gas. We want the process tobe as close to reversible as possible:there are two ways to move the piston reversibly: isothermally, meaning heat gradually flows in or out, from areservoir at a temperature infinitesimally different from that of the gas inthe piston, and adiabatically, inwhich there is no heat exchange at all, the gas just acts like a spring.

So, as the heat is supplied and the gas expands, thetemperature of the gas must stay the same as that of the heat supply (the “heatreservoir”): the gas is expanding isothermally.Similarly, it must contract isothermally later in the cycle as it shedsheat.

To figure out the efficiency, we need to track the enginethrough a complete cycle, finding out how much work it does, how much heat istaken in from the fuel, and how much heat is dumped in getting ready for thenext cycle. You might want to look atthe applet to get thepicture at this point: the cycle has four steps, an isothermal expansion asheat is absorbed, followed by an adiabatic expansion, then an isothermalcontraction as heat is shed, finally an adiabatic contraction to the originalconfiguration. We’ll take it one step ata time.

Step 1: Isothermal Expansion

So the first question is: How much heat is supplied, and how much work is done, as the gas expands isothermally? Taking the temperature of the heat reservoirto be $T_{H}$ ( $H$ for hot), the expanding gas follows theisothermal path $P V = n R T_{H}$ in the $(P, V)$ plane.

The work done by the gas in a small volume expansion $Δ V$ is just $P Δ V,$ the area under the curve (as we proved in thelast lecture).

Hence the work done in expanding isothermally from volume $V_{a}$ to $V_{b}$ is the total area under the curve betweenthose values,

$workdoneisothermally = \int_{V_{a}}^{V_{b}} P d V = \int_{V_{a}}^{V_{b}} \frac{n R T_{H}}{V} d V = n R T_{H} \ln \frac{V_{b}}{V_{a}} .$

There is no change in its internal energyduring this expansion, so the total heat supplied must be $n R T_{H} \ln \frac{V_{b}}{V_{a}},$ the same as the external work the gas hasdone.

In fact, this isothermal expansion is only the first step:the gas is at the temperature of the heat reservoir, hotter than its othersurroundings, and will be able to continue expanding even if the heat supply iscut off. To ensure that this furtherexpansion is also reversible, the gas must not be losing heat to thesurroundings. That is, after the heatsupply is cut off, there must be no further heat exchange with thesurroundings, the expansion must be adiabatic.

Step 2: Adiabatic Expansion

By definition, no heatis supplied in adiabatic expansion, but work is done.

The work the gas does in adiabatic expansion is like that ofa compressed spring expanding against a force $—$ equal tothe work needed to compress it in the first place, for an ideal (and perfectlyinsulated) gas. So adiabatic expansionis reversible.

Steps 3 and 4: Completing the Cycle

We’ve looked in detail at the work a gas does in expandingas heat is supplied (isothermally) and when there is no heat exchange(adiabatically). These are the twoinitial steps in a heat engine, but it is necessary for the engine to get backto where it began, for the next cycle.The general idea is that the piston drives a wheel (as in the diagram atthe beginning of this lecture), which continues to turn and pushes the gas backto the original volume.

But it is also essential for the gas to be as cold aspossible on this return leg, because the wheel is now having to expend work on the gas, and we want that to be as little work as possible $—$ it’scosting us. The colder the gas, the lesspressure the wheel is pushing against.

To ensure that the engine is as efficient as possible, thisreturn path to the starting point $(P_{a}, V_{a})$ must also be reversible. We can’t just retrace the path taken in thefirst two legs, that would take all the work the engine did along those legs,and leave us with no net output. Now thegas cooled during the adiabatic expansion from b to c, from $T_{H}$ to $T_{C},$ say, so we can go some distance back along thereversible colder isotherm $T_{C} .$ Obviously, that can't get us all the wayback to $(P_{a}, V_{a}),$ because that’s at the hotter temperature $T_{H} .$ It's equally clear, though, that our best betis to stay as cold as possible for as long as possible, provided we can getback to the start on a reversible path (otherwise we're losingefficiency). There's really only one option:we stay on the cold isotherm until we meet the adiabat that passes through theoriginal point, then complete the cycle by going up that adiabat (remember theadiabats are steeper than the isotherms).

To picture the Carnot cycle in the (P, V) plane, recall fromthe previous lecture the graph showing two isotherms and two adiabats:

Carnot’s cycle is around that curved quadrilateral havingthese four curves as its sides.

Let us redraw this, slightly less realistically but moreconveniently:

Efficiency of the Carnot Engine

In a complete cycle of Carnot’s heat engine, the gas tracesthe path abcd. The important question is: what fraction of the heat supplied from thehot reservoir (along the red top isotherm), let’s call it $Q_{H},$ isturned into mechanical work? Thisfraction is of course the efficiencyof the engine.

Since the internal energy of the gas is the same at the endof the cycle as it was at the beginning—it’s backto the same $P$ and $V$ $—$ it must bethat the work done equals the net heat supplied,

$W = Q_{H} - Q_{c},$

$Q_{C}$ beingthe heat dumped as the gas is compressed along the cold isotherm.

The efficiency is the fraction of the heat input that is actually converted to work, so

$efficiency = \frac{W}{Q_{H}} = \frac{Q_{H} - Q_{C}}{Q_{H}} .$

This is the answer, but it’s not particularly useful:measuring heat flow, especially the waste heat, is quite difficult. In fact, it was long believed that the heatflow out was equal to that flowing in, and this seemed quite plausible becausethe efficiency of early engines was very low.

But there’s a better way to express this.

Now the heat supplied along the initial hot isothermal path ab is equal to the work done along that leg,(from the paragraph above on isothermal expansion):

$Q_{H} = n R T_{H} \ln \frac{V_{b}}{V_{a}}$

and the heat dumped into the cold reservoir along cd is

$Q_{C} = n R T_{C} \ln \frac{V_{c}}{V_{d}} .$

$Q_{H} - Q_{C}$ lookscomplicated, but actually it isn’t!

The expression can be greatly simplified using the adiabaticequations for the other two sides of the cycle:

$\begin{array}{l} T_{H} V_{b}^{γ - 1} = T_{C} V_{c}^{γ - 1} \\ T_{H} V_{a}^{γ - 1} = T_{C} V_{d}^{γ - 1} . \end{array}$

Dividing the first of these equations by the second,

$(\frac{V_{b}}{V_{a}}) = (\frac{V_{c}}{V_{d}})$

and using that in the preceding equation for $Q_{C},$

$Q_{C} = n R T_{C} \ln \frac{V_{a}}{V_{b}} = \frac{T_{C}}{T_{H}} Q_{H} .$

So for the Carnotcycle the ratio of heat supplied to heat dumped is just the ratio of the absolutetemperatures!

$\frac{Q_{H}}{Q_{C}} = \frac{T_{H}}{T_{C}}, or \frac{Q_{H}}{T_{H}} = \frac{Q_{C}}{T_{C}} .$

Remember this: it’ll be importantin developing the concept of entropy.

The work done can now be written simply:

$W = Q_{H} - Q_{C} = (1 - \frac{T_{C}}{T_{H}}) Q_{H} .$

Therefore theefficiency of the engine, defined as the fraction of the ingoing heat energy that is converted to available work, is

$efficiency = \frac{W}{Q_{H}} = 1 - \frac{T_{C}}{T_{H}} .$

These temperatures are of course in degrees Kelvin, so forexample the efficiency of a Carnot engine having a hot reservoir of boilingwater and a cold reservoir ice cold water will be $1 - (273 / 373) = 0.27$ ,just over a quarter of the heat energy is transformed into useful work. This is the very same expression Carnot foundfrom his water wheel analogy.

After all the effort to construct an efficient heat engine,making it reversible to eliminate “friction” losses, etc., it is perhapssomewhat disappointing to find this figure of 27% efficiency when operatingbetween 0℃ and 100℃. Infact, when in the early 1800's the first steam locomotives were designed, itwas found that the power/weight ratio needed to move along a track could onlybe achieved by having high pressure boilers, meaning boiling water at a fewatmospheres (up to ten of so) pressure. At 6 atmospheres pressure, for example, theboiling temperature is about 280℃, or say 550K (kelvin), so operatingbetween that and room temperature at 300K gives a theoretical efficiency ofabout 250/550, or 45%, a big improvement.

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