Cycles and engines

A thermodynamic cycle is an ordered sequence of processes that returns the working substance to its initial state. A heat engine absorbs energy from a hot region, converts part of it to net work, and rejects the remainder to a colder region. The closed path is not enough by itself: every leg needs a valid state and heat–work ledger.

State functions return to their starting values

This lesson defines heat entering the working substance as positive and work done by it as positive:

\[\Delta U=Q-W\]

Internal energy \(U\) is a state function. For a complete closed cycle,

\[\Delta U_{\mathrm{cycle}}=0\]

Therefore the net transfers satisfy

\[Q_{\mathrm{net}}=W_{\mathrm{net}}\]

This equality does not mean heat equals work on every leg. One process may store energy, another may release it, and only the complete cycle has zero state change.

Closure must include temperature and composition, not only matching pressure and volume. For a fixed amount of ideal gas, matching two independent variables automatically fixes the third through the equation of state. A reactive or leaking system needs additional state variables.

Net work is oriented area

For quasistatic pressure–volume work,

\[W_{\mathrm{net}}=\oint p\,dV\]

On a standard plot with volume horizontal and pressure vertical, a clockwise loop has positive work by the gas. Expansion occurs at higher average pressure than compression. A counterclockwise loop requires net work input.

The absolute geometric area is not enough; orientation determines sign. A path traversed forward and then exactly backward encloses no signed area and produces zero net work even though nonzero work occurs on each traversal.

For sampled legs, preserve order and integrate each segment:

import numpy as np

# Clockwise rectangular cycle: expansion at high pressure.
volumes = np.array([0.01, 0.02, 0.02, 0.01, 0.01])
pressures = np.array([200000, 200000, 100000, 100000, 200000])
net_work = np.trapezoid(pressures, volumes)
print(f"Net work by the gas: {net_work:.1f} J")

Sorting samples by volume destroys direction and can erase or invent loop area.

Heat-engine energy flow

Let \(Q_h\) be the positive magnitude of heat absorbed from a hot reservoir, \(Q_c\) the positive magnitude rejected to a cold reservoir, and \(W_{\mathrm{out}}\) the positive useful work output per cycle. Energy conservation gives

\[W_{\mathrm{out}}=Q_h-Q_c\]

Thermal efficiency is

\[\eta=\frac{W_{\mathrm{out}}}{Q_h}=1-\frac{Q_c}{Q_h}\]

Efficiency is dimensionless and lies between zero and one for an engine operating as defined. It is not output work divided by net heat, because over a cycle net heat already equals work and that ratio would trivially be one. The relevant energy cost is heat drawn from the hot reservoir.

The second-law limit

No cyclic heat engine can convert all heat drawn from a single reservoir into work while leaving no other change. Some energy must be rejected or another part of the environment must change.

For a reversible engine operating between reservoirs at absolute temperatures \(T_h\) and \(T_c\), Carnot efficiency is

\[\eta_{\mathrm{Carnot}}=1-\frac{T_c}{T_h}\]

Temperatures must be in kelvin. This is the maximum efficiency for any engine operating between those two reservoir temperatures:

\[\eta\le\eta_{\mathrm{Carnot}}\]

The equality requires a reversible cycle. Finite temperature differences, friction, viscosity, electrical resistance, turbulence, leakage, and other irreversible processes lower real efficiency. The Carnot formula is not the efficiency of every cycle whose highest and lowest working-fluid temperatures happen to be \(T_h\) and \(T_c\); it refers to reservoir temperatures and reversible exchange.

The reversible Carnot cycle

An ideal Carnot cycle contains four reversible legs:

Isothermal expansion at \(T_h\), absorbing heat.
Adiabatic expansion from \(T_h\) to \(T_c\).
Isothermal compression at \(T_c\), rejecting heat.
Adiabatic compression back to \(T_h\).

The adiabatic legs exchange no heat; they connect the two isotherms. For reversible heat exchange,

\[\frac{Q_h}{T_h}=\frac{Q_c}{T_c}\]

which leads to the Carnot efficiency. A finite-rate engine cannot exchange heat through an infinitesimal temperature difference while also delivering finite power, illustrating why maximum efficiency and maximum power are different goals.

Refrigerators and heat pumps

Running an engine-like cycle with net work input can move heat from cold to hot. A refrigerator's desired output is heat removed from the cold region. Its coefficient of performance is

\[\mathrm{COP}_{\mathrm R}=\frac{Q_c}{W_{\mathrm{in}}}\]

A heat pump's desired output is heat delivered to the hot region:

\[\mathrm{COP}_{\mathrm{HP}}=\frac{Q_h}{W_{\mathrm{in}}}=\mathrm{COP}_{\mathrm R}+1\]

A coefficient of performance can exceed one because it is not a heat-engine efficiency. The device moves heat in addition to converting work. For reversible devices between two reservoirs,

\[\mathrm{COP}_{\mathrm R,Carnot}=\frac{T_c}{T_h-T_c},\qquad \mathrm{COP}_{\mathrm{HP,Carnot}}=\frac{T_h}{T_h-T_c}\]

A rectangular ideal-gas cycle

Consider a clockwise rectangle with low and high pressures \(p_L\) and \(p_H\), and low and high volumes \(V_L\) and \(V_H\). Its net work is the rectangle's oriented area:

\[W_{\mathrm{net}}=(p_H-p_L)(V_H-V_L)\]

Vertical isochoric legs do no pressure–volume work. Horizontal isobaric legs contribute opposite signed work, with larger magnitude on the high-pressure expansion.

To calculate efficiency, area is not sufficient. Determine every corner temperature from \(pV=nRT\), calculate each leg's internal-energy change from a stated heat capacity, and use \(Q_i=\Delta U_i+W_i\). Add only positive heat-input legs to obtain \(Q_h\).

Cycle execution and closure

internal_energy = 1.5 * pressures * volumes  # Monatomic ideal gas.
leg_work = (pressures[1:] + pressures[:-1]) / 2 * np.diff(volumes)
leg_heat = np.diff(internal_energy) + leg_work
assert np.isclose(internal_energy[-1], internal_energy[0])
assert np.isclose(leg_heat.sum(), net_work)
heat_absorbed = leg_heat[leg_heat > 0].sum()
print("Cycle efficiency:", net_work / heat_absorbed)

For an exact closed cycle, \(\Delta U_{\mathrm{cycle}}=0\), so the last quantity should also be zero within numerical tolerance. With measured data, propagate uncertainty rather than demanding machine precision.

Verification

Test a zero-area out-and-back path, a known rectangle, and a cycle with one deliberately omitted leg. Rotating which state is called the start should not change net heat or work. Reversing the complete path should reverse net work and net heat while leaving the set of states unchanged.

Refine curved path sampling and check that work stabilizes. Require the final pressure, volume, temperature, internal energy, and composition to match their initial values within defined tolerances. Compare calculated efficiency with the Carnot bound using reservoir temperatures in kelvin. A reported efficiency above that bound indicates inconsistent heat signs, missing transfers, wrong reservoirs, or a non-closed cycle—not a superior engine.

Real engines operate in time and have power limits, heat exchangers, friction, combustion, flow losses, material constraints, and environmental effects. An ideal cycle is a benchmark for accounting and limits, not a complete machine design.

A worked Python notebook

Compare the reversible heat-engine efficiency limit with a hot reservoir at 600 kelvin. The question for this calculation is precise: how does Carnot efficiency limit change when we change cold-reservoir temperature, while keeping the other choices fixed? The earlier sections explain the physical relationship. Here we turn one part of that relationship into a calculation that you can run, inspect, and change in Google Colab.

Actual engines have lower efficiency; both temperatures are absolute temperatures. Keep that sentence beside the results. A formula can return a number outside its useful range, and a plotting library will usually draw that number without asking whether the physical assumptions still hold. The selected interval is from 200 to 550 K. It gives us a specific domain to discuss instead of leaving the graph's range to a software default.

Use the notebook link above the lesson, run the setup cell, and then run the cells in order. The code below is also included there. NumPy handles the arrays of numerical values, SymPy keeps the equation symbolic, and Matplotlib draws the results. If these libraries or notebook cells are unfamiliar, start with Your first physics notebook.

State the relationship before choosing a graph

In this small example, q stands for cold-reservoir temperature, expressed in K. The output is Carnot efficiency limit, expressed in dimensionless. The equation below uses the numerical values in those units. Any fixed coefficients therefore carry the units needed to make the relation consistent. A value supplied in another unit must be converted before it enters this expression.

\[f(q) = 1 - \frac{q}{600}\]

This is a function of one changing input. It is not a statement that the physical system has only one variable. We deliberately hold the other inputs fixed so that one cause of variation is visible. Changing several inputs at once would make it harder to tell which change produced the output difference. Write down the held-fixed choices before trying a different experiment.

At 300 K, the expression gives 0.5 dimensionless. This is our reference calculation. It is a prediction of the stated example, not a new measurement. Its displayed digits help us compare implementations. They should not be read as a claim that physical inputs are known to six significant figures.

import numpy as np
import sympy as sp
import matplotlib.pyplot as plt

q = sp.Symbol("q", real=True)
expression = sp.sympify("1-q/600", locals={"q": q})
reference_input = 300.0
reference_output = float(expression.subs(q, reference_input))
print("Input:", reference_input, "K")
print("Predicted Carnot efficiency limit:", reference_output, "dimensionless")

sp.Symbol declares a mathematical symbol. It does not assign a numerical measurement to that symbol. subs then replaces the symbol with our chosen input, and float converts the resulting symbolic number into a standard Python numerical value. Keeping these steps separate lets us inspect the equation before numerical substitution. It also prevents a later plot from becoming the only surviving record of the model.

Compare a few cases you can explain

Before drawing hundreds of points, calculate three. The lower endpoint gives 0.666667 dimensionless. The midpoint, at 375 K, gives 0.375 dimensionless. The upper endpoint gives 0.0833333 dimensionless. Read these together with the assumptions: they describe different choices of cold-reservoir temperature within the same example, with the other inputs held fixed.

These three values are a useful first check, but they do not establish the shape between them. Two endpoints can be equal even when the curve bends or oscillates in between. Likewise, a large change between endpoints does not tell us whether the relationship is linear. We need the equation and intermediate samples to answer those questions.

evaluate = sp.lambdify(q, expression, modules="numpy")
inputs = np.array([200.0, 375.0, 550.0])
outputs = np.asarray(evaluate(inputs), dtype=float)
outputs = np.broadcast_to(outputs, inputs.shape)
for input_value, output_value in zip(inputs, outputs):
    print(f"{input_value:.6g} K -> {output_value:.6g} dimensionless")
assert np.all(np.isfinite(outputs))

lambdify turns the symbolic expression into a numerical function. We explicitly select NumPy as its numerical backend so it can accept an array. The same expression now has two useful forms: one for algebra and one for calculating many cases. The assertion checks that the chosen inputs produce finite outputs. It would catch an infinity or an undefined numerical result, but a finite value can still be physically inappropriate.

Make a graph that answers the question

The horizontal axis shows cold-reservoir temperature; the vertical axis shows Carnot efficiency limit. Every plotted point is a prediction of the same expression. We mark the reference case separately so that a reader can connect the earlier printed value to its place on the curve. The axis labels include units because a shape without a scale cannot tell us the size of a physical effect.

input_grid = np.linspace(200.0, 550.0, 301)
output_grid = np.asarray(evaluate(input_grid), dtype=float)
output_grid = np.broadcast_to(output_grid, input_grid.shape)
fig, ax = plt.subplots(figsize=(7, 4))
ax.plot(input_grid, output_grid, label="Stated model")
ax.scatter([reference_input], [reference_output], label="Reference case", zorder=3)
ax.set_xlabel("Cold-reservoir temperature (K)")
ax.set_ylabel("Carnot efficiency limit (dimensionless)")
ax.legend()
ax.grid(alpha=0.25)
fig.tight_layout()
plt.show()

Look for a sign change, a turning point, or a region where the curve becomes steep. Connect each feature to the expression rather than to the colour of the line. If the curve has a minimum or maximum, changing the input in opposite directions may have similar effects on the output near that point. If it is nearly straight over a short interval, a local linear approximation may be useful even when the complete expression is nonlinear.

Increasing the number of samples makes the line smoother and may reveal features that a coarse grid missed. It does not improve the physical inputs or enlarge the model's validity range. For quantities that can take only discrete values, use the line as a guide to the relationship and interpret only the allowed input values. A plotting function does not enforce such restrictions for you.

Measure the local sensitivity

A derivative asks a more specific question than the whole graph: how quickly does Carnot efficiency limit change near the reference input? Its units are the output unit divided by the input unit. For this calculation, the derivative at 300 K is -0.00166667 in those ratio units. Locally, the output decreases as the input increases. This statement is local; the sign or size of the slope can be different elsewhere on the curve.

\[f'(q) = - \frac{1}{600}\]

For a small input change, multiply the derivative by that change to estimate the output change. Then compare that estimate with a fresh evaluation of the full expression. The distinction between these two calculations matters: the derivative gives a tangent approximation, while substitution evaluates the chosen expression at the new point.

slope_expression = sp.diff(expression, q)
slope_at_reference = float(slope_expression.subs(q, reference_input))
changed_input = 301.75
linear_prediction = reference_output + slope_at_reference * (
    changed_input - reference_input
)
full_prediction = float(evaluate(changed_input))
print("Symbolic derivative:", slope_expression)
print("Tangent prediction:", linear_prediction, "dimensionless")
print("Full expression:", full_prediction, "dimensionless")
print("Approximation difference:", full_prediction - linear_prediction, "dimensionless")

Here the changed input is 301.75 K. The tangent estimate is 0.497083 dimensionless, while the full expression gives 0.497083 dimensionless. If the two values are close, it means the chosen input change is small enough for this local approximation at this point. It does not mean the entire model is linear. If the slope is zero, a quadratic or higher-order term may supply the leading change instead.

This is also the beginning of uncertainty propagation. If the input has a small stated standard uncertainty, the magnitude of the slope times that uncertainty estimates one contribution to output uncertainty. Do not automatically use the same rule for a wide interval, a discontinuity, a turning point with zero first derivative, or several correlated uncertain inputs. Those situations need more of the model than one slope.

Check the derivative with numerical differences

We can estimate the same slope from values on either side of the reference point. Subtract the lower output from the upper output and divide by twice the input step. Parentheses are essential: the entire output difference must be divided by the entire input interval. Without them, Python can execute a different mathematical expression while still producing a plausible number.

The calculation below uses three decreasing step sizes. Each estimate is compared with the symbolic derivative, so the reference is independent of the numerical difference formula. Near an endpoint, it is acceptable here to evaluate a small extension only because the displayed mathematical expression is smooth there; such a check does not assert that extrapolated values describe additional physical cases.

input_span = 350.0
for fraction in [1e-2, 1e-3, 1e-4]:
    step = input_span * fraction
    upper = float(evaluate(reference_input + step))
    lower = float(evaluate(reference_input - step))
    numerical_slope = (upper - lower) / (2 * step)
    difference = numerical_slope - slope_at_reference
    print(f"step={step:.6g}: slope={numerical_slope:.6g}, difference={difference:.6g}")

For a smooth function, reducing the step often reduces the approximation error at first. Very small steps can eventually make subtraction of nearly equal floating-point values less reliable. Therefore “smaller” is not an unlimited guarantee of “better.” Compare the observed behaviour over a useful range of steps instead of choosing the smallest number the computer can represent.

An absolute slope difference is useful when the expected slope is near zero. A relative difference would divide by a very small reference and could look enormous even when the absolute discrepancy is tiny. Away from zero, a relative difference can help compare quantities with very different scales. The tolerance should reflect the question and the numerical method, rather than a universal decimal threshold copied into every test.

Change the example and preserve the reasoning

Make one copy of the reference calculation before editing it. Choose a second value of cold-reservoir temperature inside the stated interval, predict whether Carnot efficiency limit will rise or fall, and then evaluate the expression. Place the prediction and the result in a text cell. If they disagree, identify whether the mistake was in the sign, the unit conversion, the algebra, or the assumption about how the curve behaves.

Next choose a shorter plotting interval around the reference point. Compare the local curve with its tangent approximation. Then widen the interval again. The difference shows why a useful local estimate need not describe the whole range. Keep the axis labels and numerical scale visible when comparing the plots; automatic axis limits can make two different effects appear equally large.

Finally, return to the physical limit stated at the start: Actual engines have lower efficiency; both temperatures are absolute temperatures. Describe one change to the setup that would violate that limit. State what new input, equation, or observation would be needed before trusting an answer in the changed situation. This last step connects the Python experiment to the physics instead of treating successful execution as the only criterion for a useful result.

Before saving, restart the runtime and run all cells in order. A result that depends on an earlier, now-deleted input cell is not reproducible from the saved notebook. Keep unrounded values for later calculations, round only the reported output, and label any new data as measured, assumed, or generated. The graph and printed values should remain traceable to the same expression and the same choices of inputs.

Recover the output change by integrating the slope

There is another way to check that the expression and its derivative describe the same relationship. Add up the small output changes predicted by the slope across the chosen input interval. In the continuous limit, this integral equals the output at the upper endpoint minus the output at the lower endpoint. The result has units of dimensionless: the input unit in the integration cancels the input unit in the denominator of the slope.

This statement is about the mathematics of this particular relationship. It does not assign a new physical meaning to an arbitrary area under a graph. For example, an area under force versus distance represents work, while an area under a different pair of quantities may represent something else or have no standard name. Always carry the axis units through the calculation before naming an integral.

slope_function = sp.lambdify(q, slope_expression, modules="numpy")
endpoint_change = float(evaluate(550.0) - evaluate(200.0))
for points in [101, 201, 401]:
    grid = np.linspace(200.0, 550.0, points)
    slopes = np.asarray(slope_function(grid), dtype=float)
    slopes = np.broadcast_to(slopes, grid.shape)
    integrated_change = np.trapezoid(slopes, grid)
    print(
        points,
        "integrated change:",
        integrated_change,
        "endpoint change:",
        endpoint_change,
    )

The trapezoidal rule joins neighbouring slope samples with straight segments and adds their signed areas. A negative contribution remains negative; replacing every slope by its magnitude would calculate a different quantity. Increasing the sample count provides a practical refinement check. If the values do not settle toward the endpoint difference, first inspect the interval, singularities, and array ordering.

Do not conclude that two agreeing calculations prove the model is physically complete. Both use the same selected relationship, so an omitted effect would be omitted from both. Their agreement is evidence that the numerical tools implement that relationship consistently. Comparing with an independent experiment or a more appropriate physical model is a separate test.

Keep this comparison in the notebook when you change the expression. It is particularly useful after adding a coefficient, changing a sign convention, or converting an input unit. A code cell that produces a smooth graph can conceal these mistakes; a second calculation of the same output change gives you another place to notice them.

Sources

Share to: