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Fluids & Thermal Physics

Thermodynamic Processes & Engines

Four processes, one energy ledger that always closes, and a ceiling on efficiency that no engineer will ever lift.

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Thermodynamic Processes & Engines

Four processes, one energy ledger that always closes, and a ceiling on efficiency that no engineer will ever lift.

23 min4 checkpoints5 simulation tasks

Introduction

Warming without heat

Pump up a bicycle tyre quickly and the pump's barrel gets hot. Friction is the tempting culprit, yet a perfectly lubricated pump still gets hot.

Predict

A gas is sealed in a perfectly insulated cylinder: no heat can cross the walls. You push the piston in, halving the volume. What happens to the gas's temperature?

Try it

No heat can cross these walls. Compress the gas until it ends hotter than 700 K.

GoalThe gas ending hotter than 700 K, with no heat in.

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Everything in this lesson is the first law of thermodynamics: conservation of energy, with the two ways energy crosses a boundary written out separately.

ΔU=Q−W\Delta U = Q - W

QQ is heat into the gas; WW is work done by the gas.

One law, four processes

For an ideal gas the internal energy depends on temperature alone, U=f2 nRTU = \tfrac{f}{2}\,n R T, with f=3f = 3 for a monatomic gas and f=5f = 5 for a diatomic one. The work is an area:

W=∫V1V2P dV\kin{W} = \int_{V_1}^{V_2} \frc{P}\, d\pos{V}
Read the graph

The gas is heated at constant pressure, 300 K to 600 K. From the work it does and its ΔU, find the heat that went in.

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J

Run the simulation first, then read the value off it.


Predict

A gas is taken round a closed cycle and returned to its starting state: same pressure, volume and temperature. What can you say about its net work and net heat?

The area is the work

Try it

One loop of this Carnot cycle delivers 138.3 J of work. Make one loop deliver twice that. Run it and watch the enclosed area fill in.

GoalOne loop delivering twice 138.3 J.

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Going right along any path, the strip under the curve down to P=0P = 0 is positive work, done by the gas. Coming back left along a lower path, the strip under that one is work done on the gas, drawn fainter with a dashed edge. What survives is the area enclosed by the loop:

Wnet=∮P dVW_{\text{net}} = \oint P\,dV

So an engine has to come back along a lower-pressure path, which means a lower temperature, which means dumping heat somewhere colder. The cold reservoir is the return path.

Predict

Your engine takes in 276 J of heat per cycle and delivers 138 J of work. A better designer works on it for a decade. What is the ceiling?

The bound, and what falls short of it

A Carnot cycle reaches the bound because all its heat is exchanged at the two extreme temperatures. Any cycle that takes in heat at some intermediate temperature buys that heat at a worse rate, and falls short.


Read the graph

A Stirling cycle between the same reservoirs. Run one loop. How much heat does it draw from the hot reservoir per cycle?

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J

Run the simulation first, then read the value off it.

Try it

Switch this Stirling engine to a diatomic gas and run a loop. Watch the net work and the efficiency.

GoalThe Stirling engine running a loop on a diatomic gas.

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Practice

During the isothermal expansion leg of the Carnot cycle, the temperature readout does not change at all. What is the heat flow during that leg?

What comes next

You heated a gas with no heat, doubled the work of a loop, and found the ceiling no engine can pass. The first law says an engine cannot make energy. It does not say why heat has to go down the temperature scale, or why no engine can beat 1−Tc/Th1 - T_c/T_h. That takes a second quantity, one which, unlike energy, is not conserved and only ever increases.

Ideas this lesson took on

Each of these is a common intuition the lesson put to the test. If one still feels right, the section that tests it is worth another look.

  • “Temperature can only change if heat flows in or out”
  • “A gas at 500 K contains a certain amount of heat”
  • “Adiabatic and isothermal are two words for the same situation”
  • “A good enough engineer could build a 100 %-efficient engine”

Finished with this lesson?

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Research behind this lesson

  • Christensen et al. 2009: Across four offerings of the same introductory calculus-based course, no more than 6 % of students gave completely correct answers about entropy changes of a system and its surroundings before instruction, and nearly two-thirds showed clear conservation-type reasoning about entropy: outcomes that changed little after ordinary instruction.
  • Loverude et al. 2002: Asked what happens when masses are added to a piston sealing an insulated cylinder, few students recognised that the first law was relevant at all, and fewer still could use work to account for the temperature change; most reasoned instead from a misinterpretation of the ideal gas law, and many did not distinguish heat, temperature, work and internal energy.
  • Meltzer 2004: Across written responses from 653 students and interviews with 32 more, students correctly learned that internal energy is a state function and then improperly over-generalised the idea to work and heat: most of those interviewed held that the net work done or net heat absorbed in a cyclic process must be zero, and no more than about 20 % applied the first law effectively even after instruction.
  • Smith et al. 2015: In upper-division thermal physics courses, surveys, homework and videotaped tutorial observations showed students struggling to connect the Carnot cycle to the second law: with particular difficulty reasoning about situations that are physically impossible, and with failures to distinguish a differential change in a state property from its net change around a cycle.