IPhO 1977 · Theory problem 1

Otto cycle of a four-stroke combustion engine

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The compression ratio of a four-stroke internal combustion engine is \(\varepsilon = 9.5\). The engine draws in air and gaseous fuel at a temperature \(27^\circ\mathrm{C}\) at a pressure \(1\ \mathrm{atm} = 100\ \mathrm{kPa}\). Compression follows an adiabatic process from point 1 to point 2, see Fig. 1. The pressure in the cylinder is doubled during the mixture ignition (2–3). The hot exhaust gas expands adiabatically to the volume \(V_2\) pushing the piston downwards (3–4). Then the exhaust valve opens and the pressure gets back to the initial value of \(1\ \mathrm{atm}\). All processes in the cylinder are supposed to be ideal. The Poisson constant (i.e. the ratio of specific heats \(C_p/C_V\)) for the mixture and exhaust gas is \(\kappa = 1.40\). (The compression ratio is the ratio of the volume of the cylinder when the piston is at the bottom to the volume when the piston is at the top.)

Figure 1

Figure 1:

a) Which processes run between the points 0–1, 2–3, 4–1, 1–0?

b) Determine the pressure and the temperature in the states 1, 2, 3 and 4.

c) Find the thermal efficiency of the cycle.

d) Discuss obtained results. Are they realistic?

Official English text of the problem.

Answers

  • a) \(0\to1\): isobaric intake; \(2\to3\): constant-volume combustion and heat addition; \(4\to1\): constant-volume pressure decrease as the exhaust valve opens, represented by heat rejection in the ideal Otto cycle; \(1\to0\): isobaric exhaust.

  • b) Using \(27^\circ\mathrm C\simeq300\ \mathrm K\):

State Pressure Temperature
1 \(100\ \mathrm{kPa}\) \(300\ \mathrm K\)
2 \(2.34\times10^3\ \mathrm{kPa}\) \(738\ \mathrm K\)
3 \(4.68\times10^3\ \mathrm{kPa}\) \(1.48\times10^3\ \mathrm K\)
4 \(200\ \mathrm{kPa}\) \(600\ \mathrm K\)

  • c) \(\displaystyle \eta=1-\varepsilon^{\,1-\kappa}=0.594\approx59.4\%\).

  • d) The pressures and temperatures are plausible orders of magnitude for a combustion engine. The efficiency is an optimistic ideal value; heat transfer, variable heat capacities, finite combustion and gas-exchange losses reduce actual efficiency, and friction further reduces shaft output.

Solution

a) Identification of the processes

Let the minimum and maximum cylinder volumes be \[ V_{\min}=V_1,\qquad V_{\max}=V_2,\qquad \frac{V_{\max}}{V_{\min}}=\varepsilon=9.5, \] where \(V_1,V_2\) are the volume labels in the figure. States 2 and 3 occur at \(V_{\min}\); states 1 and 4 occur at \(V_{\max}\).

  • \(0\to1\): Isobaric intake. The piston moves downwards, drawing in fresh air and gaseous fuel at the external pressure \(p_0=p_1\).
  • \(2\to3\): Isochoric combustion. Idealized instantaneous combustion raises the pressure and temperature while the piston remains at its upper position. The gas does no boundary work because its volume is constant.
  • \(4\to1\): Isochoric pressure decrease. The exhaust valve opens at the lower piston position, and gas escapes until the cylinder pressure reaches the external pressure. This rapid exhaust release is represented by constant-volume heat rejection in the ideal closed Otto cycle.
  • \(1\to0\): Isobaric exhaust. The piston moves upwards, expelling exhaust gas at the external pressure.

Intake and exhaust involve matter crossing the cylinder boundary. Their idealized pressure–volume paths coincide.

b) Pressures and temperatures

Treat the working gas as ideal, with constant heat capacities and \(\kappa=C_p/C_V=1.40\). During each ideal adiabatic stroke, \[ pV^\kappa=\text{constant}, \qquad TV^{\kappa-1}=\text{constant}. \] These relations follow from the first law, \[ nC_V\,dT=-p\,dV, \] together with \(pV=nRT\) and \(R/C_V=\kappa-1\), where \(C_V\) is the molar heat capacity.

At state 1, the supplied conditions give \[ p_1=100\ \mathrm{kPa},\qquad T_1=27^\circ\mathrm C\simeq300\ \mathrm K. \]

During adiabatic compression \(1\to2\), \[ p_2=p_1\left(\frac{V_{\max}}{V_{\min}}\right)^\kappa =p_1\varepsilon^\kappa, \] \[ T_2=T_1\left(\frac{V_{\max}}{V_{\min}}\right)^{\kappa-1} =T_1\varepsilon^{\kappa-1}. \] Numerically, \[ \varepsilon^{\kappa-1}=9.5^{0.40}=2.46087, \qquad \varepsilon^\kappa=9.5^{1.40}=23.3783, \] so \[ \boxed{p_2\simeq2.34\times10^3\ \mathrm{kPa}}, \qquad \boxed{T_2\simeq738\ \mathrm K}. \]

During constant-volume ignition \(2\to3\), the pressure doubles. Under the ideal working-gas approximation, \(p/T\) is constant at fixed volume, hence \[ p_3=2p_2,\qquad T_3=2T_2. \] Thus \[ \boxed{p_3\simeq4.68\times10^3\ \mathrm{kPa}}, \qquad \boxed{T_3\simeq1.48\times10^3\ \mathrm K}. \]

During adiabatic expansion \(3\to4\), \[ p_4=p_3\left(\frac{V_{\min}}{V_{\max}}\right)^\kappa =\frac{2p_1\varepsilon^\kappa}{\varepsilon^\kappa} =2p_1, \] \[ T_4=T_3\left(\frac{V_{\min}}{V_{\max}}\right)^{\kappa-1} =\frac{2T_1\varepsilon^{\kappa-1}}{\varepsilon^{\kappa-1}} =2T_1. \] Therefore, \[ \boxed{p_4=200\ \mathrm{kPa}}, \qquad \boxed{T_4\simeq600\ \mathrm K}. \]

c) Thermal efficiency

Take work done by the gas as positive. The intake and exhaust strokes contribute equal and opposite work: \[ W_{01}=p_1(V_{\max}-V_{\min}), \qquad W_{10}=-p_1(V_{\max}-V_{\min}). \] They therefore cancel. The constant-volume processes contribute no boundary work.

For the two adiabatic strokes, the first law gives \[ W_{12}=nC_V(T_1-T_2),\qquad W_{34}=nC_V(T_3-T_4). \] The net work per cycle is consequently \[ W_{\mathrm{net}} =nC_V\big[(T_3-T_2)-(T_4-T_1)\big]. \]

The combustion energy is represented by heat supplied at constant volume: \[ Q_{\mathrm{in}}=nC_V(T_3-T_2). \] Hence \[ \eta=\frac{W_{\mathrm{net}}}{Q_{\mathrm{in}}} =1-\frac{T_4-T_1}{T_3-T_2}. \] The quantity \(nC_V(T_4-T_1)\) is the equivalent heat rejected in the closed Otto-cycle representation.

Since \(T_3=2T_2\) and \(T_4=2T_1\), \[ \eta=1-\frac{T_1}{T_2} =1-\frac{1}{\varepsilon^{\kappa-1}}. \] Therefore, \[ \boxed{\eta=1-9.5^{-0.40}=0.59364\approx59.4\%.} \]

Within this ideal model, efficiency depends on the compression ratio and \(\kappa\). The specified pressure doubling determines the temperatures and work per charge, but cancels from the efficiency.

d) Discussion of realism

The compression temperature is approximately \(465^\circ\mathrm C\), the peak temperature approximately \(1200^\circ\mathrm C\), and the peak pressure approximately \(4.68\ \mathrm{MPa}\). These are reasonable orders of magnitude for a combustion engine.

At the end of expansion, the gas remains at approximately \(327^\circ\mathrm C\) and twice atmospheric pressure. Thus the exhaust still carries substantial energy. Even the ideal cycle rejects \[ 1-\eta\simeq40.6\% \] of the supplied energy.

An actual engine departs from the model in several ways:

  • Heat passes from the hot gas into the cylinder walls, so compression and expansion are not perfectly adiabatic.
  • Combustion takes finite time while the piston moves, so heat addition is not exactly isochoric.
  • Heat capacities depend on temperature and composition. Their increase at high temperature makes a constant \(\kappa=1.40\) an optimistic approximation.
  • Intake and exhaust require pressure differences across valves, producing pumping losses; their work contributions no longer cancel exactly.
  • Incomplete combustion and mechanical friction reduce useful output.

Moreover, actual exhaust release is an open process: escaping gas carries energy out of the cylinder, and intake supplies a fresh charge. Constant-volume cooling of a fixed gas charge is its ideal Otto-cycle representation.

Thus the calculated \(59.4\%\) is the efficiency of the stipulated ideal model; the actual efficiency and shaft output will be lower.

The solution was checked against the official solution.