ISEGORIA / MATH ENCYCLOPEDIA
Stirling engines: heat into motion
A Stirling engine burns nothing inside itself. A fixed charge of gas, usually helium or hydrogen, is sealed between two spaces, one kept hot by a heater and one kept cold by a cooler, and two pistons shuffle it back and forth between them. Gas that expands while it is mostly hot pushes harder than the same gas compressed while it is mostly cold, and the difference turns the crank. On its way between the spaces the gas passes through the regenerator, a stack of fine wire screens that takes its heat on the way to the cold side and gives it back on the way to the hot side, and that is what lets the ideal cycle reach the Carnot efficiency. Robert Stirling patented the engine in 1816; it runs today in submarine air-independent propulsion, in solar dish generators, in cryocoolers run backwards, and in NASA’s radioisotope power studies. These experiments run an alpha engine with the exact isothermal (Schmidt) analysis, map how the work depends on the phase between the pistons and on the volume ratio, and measure what an imperfect regenerator costs.
Before you begin: Thermodynamics (ideal gas, isothermal work, the Carnot limit), trigonometry, piston engines
Predict, manipulate, then test yourself: every experiment ends with practice questions that are generated fresh at three levels. Graphs illustrate the mathematics; they do not replace a proof.
1. An alpha engine running
Two cylinders sit at 90° on one crank: the compression space, finned and cooled, and the expansion space, heated. Their volumes vary sinusoidally, the hot one leading by the phase angle \(\alpha\). Because the hot volume grows a quarter turn before the cold one, most of the gas is already on the hot side when the total volume is largest, and on the cold side when it is smallest. Schmidt’s analysis of 1871 keeps each space at its wall temperature and the pressure uniform, and that is enough to solve the engine exactly: the pressure is a shifted cosine in the crank angle, and the work per cycle has a closed form. Run the engine, change \(\alpha\) and the temperatures, and watch the loop on the pressure–volume diagram open and close.
Worked example. Take a 60 mm bore and a 40 mm stroke in both cylinders, so each sweeps 113.1 cm³, with clearances of 0.1, a cooler of 0.15, a regenerator of 0.3 and a heater of 0.15 swept volumes, helium at a mean of 10 bar, \(T_h=900\) K, \(T_k=300\) K and \(\alpha=90^\circ\). Then \(b=0.452\), \(\beta=18.4^\circ\), and the pressure swings between 6.14 and 16.29 bar. The expansion space does \(W_e=80.6\) J per cycle and the compression space takes back \(-26.9\) J, so \(W=53.7\) J, which at 1500 rpm is 1.34 kW. With isothermal spaces \(Q_e=W_e\), so the efficiency is \(53.7/80.6=66.7\%\), exactly Carnot. The ideal Stirling cycle of the same charge between the same volumes, 123.6 to 283.5 cm³, would do 195 J: the Schmidt loop has 28% of its area.
Watch out. The Schmidt analysis is an ideal: real expansion and compression spaces are closer to adiabatic than isothermal, the heat exchangers have finite area and pressure losses, and the regenerator is imperfect, so real engines reach perhaps half to two thirds of the Carnot efficiency and less work than computed here. The regenerator gas is taken at the log-mean temperature of a linear profile. The 3D engine is drawn to proportion but not to scale, and the gas dots are spread evenly through each space.
Why does the engine do no work when the pistons move in phase?
With \(\alpha=0\) the two volumes rise and fall together, so the share of gas on the hot side is the same whenever the total volume is the same, on the way out and on the way back. Pressure is then a function of volume alone, and the loop collapses to a line traced back and forth with zero area. In the formula \(\beta-\alpha\) and \(\beta\) are both zero and \(W_e=W_c=0\).
2. The Schmidt map: phase and volume ratio
A designer chooses the phase angle \(\alpha\), the ratio \(\kappa=V_{swc}/V_{swe}\) of the swept volumes, and how much dead space the heat exchangers need. The Schmidt work formula turns these into a surface. Divided by a reference pressure times the total swept volume, the work depends only on \(\alpha\), \(\kappa\), the temperature ratio \(\tau=T_k/T_h\) and the dead-volume ratio \(X\), so one map serves every engine of that shape. It vanishes at \(\alpha=0\) and \(180^\circ\) and has a single summit between them. Which summit depends on what is held fixed: the mean pressure, which sets the charge, or the maximum pressure, which sets the walls. The power is then simply \(W f\).
Worked example. At \(\tau=1/3\) with \(X=0.5\) and equal cylinders, the work at \(\alpha=90^\circ\) is \(0.293\,p_{\mathrm{mean}}V_T\), and the best phase for \(\kappa=1\) is \(86.7^\circ\), though it does only 0.2% more work than the 90° that crankshafts make easy. The summit, at \(\alpha=86.6^\circ\) and \(\kappa=1.14\), is barely higher, at 0.295. With no dead space at all the map is sharper: the summit moves to \(\alpha=76.9^\circ\), \(\kappa=0.66\) and \(0.514\,p_{\mathrm{mean}}V_T\). Judged by the peak pressure instead, the dead-space-free optimum is \(\alpha=113^\circ\), \(\kappa=0.46\). An engine on the map with a 113 cm³ hot cylinder at 10 bar mean does 66.3 J a cycle, 1.66 kW at 1500 rpm.
Watch out. The map inherits every assumption of the Schmidt analysis, and real optima differ: adiabatic spaces, flow friction in the heat exchangers and the cost of a larger regenerator all push the best design. The dead volume here is split 1:2:1 between cooler, regenerator and heater, and the clearance volumes are zero. The surface height is to scale within one normalisation, not between the two.
Why does dead volume move the best phase angle towards 90°?
Dead volume is gas that never changes volume, so it damps the pressure swing: \(b\) falls and the pressure becomes nearly \(p_{\mathrm{mean}}(1-b\cos(\theta+\beta))\). In that small-\(b\) limit the work is about \(\pi p_{\mathrm{mean}}b\) times the swept volumes times sines of the phase, and the expansion work \(\propto\sin(\alpha-\beta)\) is largest when the two volume waves are in quadrature, so the optimum tends to 90° as the swing shrinks.
3. The regenerator: the heart of the cycle
In the ideal Stirling cycle the gas is heated at constant volume from \(T_c\) to \(T_h\) and later cooled back again. That heat, \(c_v(T_h-T_c)\) per kilogram, is larger than the heat taken in along the hot isotherm, and if the heater had to supply it the efficiency would collapse. The regenerator stores it instead: a matrix of fine screens whose temperature runs from cold to hot along the flow. Gas going to the cold side leaves its heat in the screens; gas coming back picks it up. Its effectiveness \(\varepsilon\) is the fraction returned; the heater must make up the rest. Drag the thermometer and watch the efficiency climb towards Carnot only as \(\varepsilon\to1\).
Worked example. Helium (\(R=2077\) J/kg·K, \(\gamma=5/3\), so \(c_v=3116\)) between 300 K and 900 K with a volume ratio of 2 does \(W=2077\times600\times\ln2=864\) kJ per kilogram and takes in 1296 kJ on the hot isotherm. The isochoric heating is \(3116\times600=1869\) kJ, 1.44 times as much. With no regenerator \(\eta=864/(1296+1869)=27.3\%\); with \(\varepsilon=0.9\) the heater supplies only 187 kJ of it and \(\eta=58.3\%\); at 0.95 it is 62.2%, and only at \(\varepsilon=1\) does it reach Carnot, 66.7%. At \(\varepsilon=0.9\) the gas reaches the heater at 840 K.
Watch out. This is the ideal Stirling cycle, two isotherms and two isochores, with every loss except the regenerator’s ignored. The relation \(\varepsilon=\mathrm{NTU}/(1+\mathrm{NTU})\) holds for a balanced counterflow regenerator with linear temperature profiles; a real one also loses heat along its length by conduction and costs pumping work, which is why its screens are fine but not endless. The parcel motion in the figure is schematic; their temperatures follow the linear profiles.
Why does a regenerator not change the work, only the efficiency?
The work is the area of the cycle, and the two isochores contribute none: no volume change, no work. The regenerator only decides where the isochoric heat comes from, the matrix or the heater. With a perfect regenerator the only heat bought from outside is \(RT_h\ln r\) on the hot isotherm, and \(W/Q=1-T_c/T_h\).