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IZhO 2026, Теория — Задача 2. Steam Engine

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XXII International Zhautykov Olympiad/Theoretical Competition · 11 януари 2026 г. · 10 т.

THEORETICAL COMPETITION

January 11, 2026

Please read this first:

The time available for the theoretical competition is 4 hours. There are three questions.
You can use your own calculator for numerical calculations. If you don’t have one, please ask for it from Olympiad organizers.
You are provided with Writing sheet and additional paper. You can use the additional paper for drafts of your solutions but these papers will not be checked. Your final solutions which will be evaluated should be on the Writing sheets. Please use as little text as possible. You should mostly use equations, numbers, figures and plots.
Use only the front side of Writing sheets. Write only inside the boxed area.
Each task should be solved starting from a new page of the Writing sheets
At the top of each sheet of paper your country (Country) and your student code (Student Code) are printed, you only need to fill in the progressive number of each sheet (Page Number), and the total number of Writing sheets used (Total Number of Pages). If you use some blank Writing sheets for notes that you do not wish to be evaluated, put a large X across the entire sheet and do not include it in your numbering.

Условие

In this problem, we consider a simplified theoretical model of one type of steam engine. Note that at present such an engine is used only as a technical toy.

Schematic diagram of a steam engine: boiler 1 (P₀, T₀) with regulator 2 connected by a pipe with valves 3 and 4 to cylinder 5 with piston 6 moving with velocity v; rod system 7 drives flywheel 8 rotating with angular velocity ω, belt-connected to Watt governor 9 with angular velocity ω₁.

A schematic diagram of the engine under consideration is shown in the figure above. There is a steam generator 1 (a boiler with water and a heater, as well as a system for additional superheating of the steam). The generator may be regarded as a large vessel containing water vapor (without air), in which the pressure $P_0$ and temperature $T_0$ are maintained constant. The heated steam is supplied through a pipe to the working cylinder 5 with a movable piston 6. The pipe is equipped with a steam supply regulator 2 controlled by a Watt centrifugal governor 9. The working cylinder has two valves: an inlet valve 3 and an outlet valve 4. These valves open and close at certain positions of the piston. The movable piston is connected to a system of rods 7 that converts the translational motion of the piston into rotational motion of a massive flywheel 8 mounted on the engine shaft. By means of a belt drive, this wheel is connected to the Watt governor, whose angular velocity $\omega$ is equal to the angular velocity of the flywheel. The amount of steam passing through the steam supply regulator 2 depends on the rotational frequency of the governor. The flywheel shaft is connected to the working device (for whose sake all engines are created).

In all parts of the problem, steady operating regimes are considered, in which the angular velocities of rotation of the shaft and the Watt governor remain constant. Of course, the transition to such regimes is possible only in the presence of friction; however, in calculations of the steady motion, friction may be neglected.

Assume that the adiabatic index of water is $\gamma = 4/3$ , the molar mass of water is $M = 18.0 \cdot 10^{-3}\ \mathrm{kg/mole}$ .

Part 1. Steam Engine without a Governor The ideal cycle of the engine under consideration in the limit of infinitely slow motion is shown on the $P-V$ diagram. Here $V$ is the volume of the part of the cylinder between the cylinder end face and the piston that is occupied by steam (hereafter referred to as the working volume); $V_0 = 4.00\ \mathrm{L}$ is the maximum working volume; and $P$ is the pressure of the steam in this volume.

P–V diagram of the ideal cycle: isobaric segment 1–2 at pressure P₀ up to volume ηV₀, adiabatic expansion 2–3 to volume V₀ at pressure P_A, isobar 3–4 at P_A back to zero volume, and vertical segment 4–1.

Three sketches of the cylinder showing the piston positions for segments 1–2 (steam entering), 2–3 (both valves closed, expansion to ηV₀) and 3–4 (outlet open, exhaust to volume V₀).

In state 1, the piston touches the cylinder wall and the working volume is zero. The inlet valve opens, and steam begins to enter the cylinder. Along segment 1–2 of the cycle, the piston moves slowly to the right; in this process, the steam pressure in the cylinder may be assumed to be equal at all times to the pressure of the steam in the generator, $P_0 = 10.0 \cdot 10^5\ \mathrm{Pa}$ .

At point 2, when the working volume reaches the value $\eta V_0$ , the inlet valve closes, and the piston continues to move to the right under the action of the steam pressure. Along segment 2–3, the expansion process is adiabatic.

At point 3, the piston reaches its extreme position, the working volume is maximal, and the pressure drops to the atmospheric value $P_A = 1.00 \cdot 10^5\ \mathrm{Pa}$ . After that, the outlet valve opens, and the piston expels the steam into the atmosphere.

When the piston returns to the extreme left position, the outlet valve closes and the inlet valve opens again (segment 4–1).

2.1 Determine the value of the coefficient $\eta$ .

2.2 To prevent steam condensation, its minimum temperature must be higher than the condensation temperature at atmospheric pressure, $t_s = 100\,{}^\circ\mathrm{C}$ .

Determine the required value of the steam temperature $T_0$ (in degrees Celsius) such that during the expansion process the temperature does not drop below the condensation temperature $t_s$ .

In what follows, assume that in the considered model the parameters $T_0$ and $\eta$ are equal to the values found above.

2.3 Calculate the mass of steam $m_0$ entering the cylinder during one cycle.

2.4 Calculate the work $A_0$ performed by the engine during one cycle.

2.5 Next, consider the operation of the engine in a real regime, taking into account that the piston moves with some finite velocity. Assume that the flywheel rotates with a constant angular velocity $\omega$ . Strictly speaking, in this case the motion of the piston is not uniform. However, as an approximation, the piston velocity in one direction may be taken to be constant and equal to its average velocity. In this problem, it is convenient to consider the rate of change of the working volume, $v = dV/dt$ .

Express the average rate of change of the working volume of the cylinder $v$ in terms of the maximum working volume $V_0$ and the angular velocity $\omega$ of the flywheel.

2.6 During the piston motion, the steam pressure $P$ in the cylinder during stage 1–2 of the cycle differs from the steam pressure in the generator $P_0$ . Assume that the rate of steam inflow is proportional to the pressure difference across the steam regulator 2, i.e.

\frac{dm}{dt} = K(P_0 - P),

where $K$ is a constant coefficient determined by regulator 2. In this part, take it to be constant and equal to $K = K_0 = 4.20 \cdot 10^{-7}\ \mathrm{kg/(Pa \cdot s)}$ .
Assume that the steam expansion process at all stages of the cycle is adiabatic.

Derive a differential equation describing the change in the mass of gas in the cylinder while the inlet valve is open. Besides the unknown function $m(t)$ , this equation should contain only known quantities.

2.7 Show that at a constant piston velocity the steam pressure $P$ in the cylinder remains constant.

2.8 Obtain an algebraic equation for determining the steam pressure $P$ in the cylinder during stage 1–2 of the cycle.

To approximately solve the resulting equation (i.e., to find the pressure $P$ ), make the following mathematical approximation: assume (only in this part of the problem) that the adiabatic index of water vapor is $\gamma \approx 1$ . In addition, when calculating the work performed by the steam, neglect atmospheric pressure.

2.9 Using the above approximation, express the steam pressure $P$ in the cylinder during stage 1–2 in terms of the pressure $P_0$ and the other known parameters of the problem.

2.10 Calculate the numerical value of the pressure $P$ at an angular velocity of the shaft $\omega = 10.0\ \mathrm{s}^{-1}$ .

2.11 Show that the formula for calculating the work $A$ performed by the engine during one cycle when the flywheel rotates with a constant angular velocity $\omega$ can be written in the form $A = \frac{A_0}{1 + \beta\frac{\omega}{K}}$ . Calculate the numerical values of the parameters $A_0$ and $\beta$ in this formula.

2.12 Assume that, in addition to the torque $M$ exerted by the piston, a constant torque $M_0$ from the working device acts on the engine shaft.

Express the average steady-state angular velocity $\omega$ of the wheel in terms of the torque $M_0$ , the coefficient $K$ , and the parameters $A_0$ and $\beta$ .

2.13 Calculate the maximum torque $M_0$ at which the engine can operate.

2.14 Sketch a schematic graph of the dependence of the angular velocity of rotation on the torque $M_0$ .

Part 2. Governor without the Engine The angular velocity of rotation of the engine flywheel depends on the torque transmitted to the working device. When this torque decreases, the angular velocity may increase sharply, which can lead to engine failure and its further destruction. To avoid such situations, various automatic control systems are used, one of which is the Watt centrifugal governor.

Drawing of a Watt centrifugal governor: vertical shaft with two hinged arms carrying balls, a sleeve, and a lever connecting to a throttle valve of a steam pipe.

The principle of its operation is quite simple. The rotation of the engine shaft is transmitted through a drive to the shaft of the governor. During the rotation of the governor shaft, the balls deviate from the axis under the action of the centrifugal force; the faster the shaft rotates, the farther apart the balls move. The levers interact with a sleeve and move it along the shaft axis. Through a system of levers, the displacement of the sleeve is transmitted to the steam supply throttle in such a way that when the rotational speed of the shaft increases, the steam supply decreases, and when it decreases, the steam supply increases.

Idealized governor model: a rhombic hinged frame of side l on a vertical shaft rotating with angular velocity ω₁, with two balls deviating by angle θ from the vertical and a sliding lower sleeve (arrow up).

For solving this problem, the technical details of the specific device are not essential. Let us consider an idealized schematic model of the governor. A hinged rhombic frame is mounted on a vertical shaft rotating with a constant angular velocity $\omega$ . The upper vertex of the frame is fixed, while the lower vertex can freely slide along the shaft. Two massive balls are attached at the lateral vertices of the frame. It may be assumed that in the absence of shaft rotation $\theta = 0$ . When the shaft rotates, the balls rise, deviating by an angle $\theta$ from the vertical. This leads to a reduction in the steam supply to the working cylinder. We assume that the coefficient $K$ , which characterizes the steam flow rate, is related to the deflection angle by the simple relation

K = K_0 \cos\theta,

where $K_0$ is the coefficient defined earlier.

The length of a side of the frame is $l = 20\ \mathrm{cm}$ .

2.15 Find the dependence of the deflection angle $\theta$ of the balls from the vertical on the constant angular velocity $\omega$ of the governor.

2.16 Sketch a schematic graph of the dependence of the coefficient $K$ on the angular velocity $\omega$ of the governor.

2.17 Part 3. Engine with a Governor

Find the dependence of the steady-state angular velocity of rotation of the engine shaft on the torque $M_0$ for the governor system described above.

2.18 Sketch a schematic graph of the resulting dependence.

Решение

Problem 2. Steam Engine (10.0 points)

Part 1. Steam Engine without a Governor

2.1 Let us write the equation of the adiabatic process

PV^\gamma = const. \qquad (1)

Applying it to process 2–3, we obtain

\eta = \left(\frac{P_A}{P_0}\right)^{\frac{1}{\gamma}} = 0.177. \qquad (2)

2.2 To determine the initial temperature, the adiabatic equation should be written in the $(T \cdot V)$ variables:

TV^{\gamma-1} = const, \qquad (3)

which, when applied again to process 2–3, leads to the following expression:

T_0 = \frac{T_S}{\eta^{\gamma-1}} = 390\ {}^\circ\mathrm{C}. \qquad (4)

2.3 The ideal gas equation of state for point 2 has the form

P_0\eta V_0 = \frac{m_0}{M}RT_0, \qquad (5)

which yields

m_0 = M\frac{P_0\eta V_0}{RT_0} = 2.30\ \mathrm{g}, \qquad (6)

2.4 On segment 1–2 the work performed by the steam is

A_{1-2} = P_0\eta V_0, \qquad (7)

while on segment 2–3 the work along the adiabatic is

A_{2-3} = P_0 V_0\frac{\eta - \eta^\gamma}{\gamma-1}. \qquad (8)

On segment 3–4 the work is negative and equal to

A_{3-4} = -P_A V_0, \qquad (9)

therefore, the total work is

A_0 = P_0 V_0\left(\eta + \frac{\eta-\eta^\gamma}{\gamma-1}\right) - P_A V_0 = 1.25\cdot 10^3\ \mathrm{J}. \qquad (10)

2.5 The working volume reaches its maximum value over half a revolution of the flywheel; therefore, the average rate of change of the volume is

v = \frac{V_0}{\pi/\omega} = \frac{V_0\omega}{\pi}. \qquad (11)

2.6 Let us establish the relation between the pressure in the cylinder and the mass of steam inside it. For this purpose, we write the adiabatic equation

P_0 V_{in}^\gamma = PV^\gamma, \qquad (12)

where $V_{in}$ is the volume occupied by the steam in the generator before it enters the working cylinder. For this volume, the equation of state is

P_0 V_{in} = \frac{m}{M}RT_0. \qquad (13)

Then the pressure in the cylinder can be expressed in terms of the steam mass using the adiabatic equation (12) and the equation of state (13) as

P = P_0\left(\frac{RT_0}{MP_0}\frac{m}{vt}\right)^\gamma, \qquad (14)

where $V = vt$ is the time-dependent working volume.
Taking into account the equation given in the problem statement, we obtain the required equation for the steam mass in the cylinder:

\frac{dm}{dt} = K(P_0 - P) = KP_0\left(1 - \left(\frac{RT_0}{MP_0}\frac{m}{vt}\right)^\gamma\right). \qquad (15)

2.7 An obvious solution of Eq. (15) is a linear function

m \propto t, \qquad (16)

and therefore, according to Eq. (14), the steam pressure in the cylinder remains constant.
2.8 In accordance with Eq. (15), the gas mass increases linearly

m = K(P_0 - P)t, \qquad (17)

and thus, from Eq. (15) and the relation $V = vt$ , we obtain that the gas pressure must satisfy

P = P_0\left(\frac{RT_0}{MP_0}\frac{\pi K(P_0-P)}{V_0\omega}\right)^\gamma. \qquad (18)

2.9 Now we use the approximation $\gamma \approx 1$ , which allows us to obtain an explicit expression for the pressure:

P = \frac{P_0}{1 + \frac{MV_0\omega}{\pi KRT_0}}. \qquad (19)

2.10 The numerical value of the pressure for the given parameters is

P = 9.10\cdot 10^5\ \mathrm{Pa}. \qquad (20)

2.11 The work performed by the engine during one cycle can be calculated using Eq. (10), where $P_0$ should be replaced by the pressure $P$ given by Eq. (19). This leads to

A = P_0 V_0\frac{\eta + \frac{\eta-\eta^\gamma}{\gamma-1}}{1 + \frac{MV_0\omega}{\pi KRT_0}}. \qquad (21)

Thus, the parameters of this formula are

A_0 = P_0 V_0\left(\eta + \frac{\eta-\eta^\gamma}{\gamma-1}\right) = 1.44\cdot 10^3\ \mathrm{J} \qquad (22)

\beta = \frac{MV_0}{\pi RT_0} = 4.16\cdot 10^{-9}\ \mathrm{m\cdot s^2}. \qquad (23)

2.12 In the steady-state regime, the work performed by the steam per cycle is equal to the work done on the load:

\frac{A_0}{1 + \beta\frac{\omega}{K}} = 2\pi M_0. \qquad (24)

Hence, the mean angular velocity in the steady-state regime is

\omega = \frac{K}{\beta}\left(\frac{A_0}{2\pi M_0} - 1\right). \qquad (25)

2.13 From Eq. (25) it follows that the maximum torque is

M_{0max} = \frac{A_0}{2\pi} = 230\ \mathrm{N\cdot m}. \qquad (26)

2.14 A schematic plot of the corresponding dependence is shown in the figure below.

Graph of the angular velocity $\omega$ versus $2\pi M_0/A_0$ — a steeply decreasing curve reaching zero at 1.

Part 2. Governor without the Engine

2.15 It is more convenient to solve this part of the problem in a rotating reference frame. In the equilibrium state, the torque due to gravity is balanced by the torque due to the centrifugal force:

mgl\sin\theta = m\omega^2 l^2\sin\theta\cos\theta. \qquad (27)

It follows that the deflection angle is determined by

\cos\theta = 1, \quad \omega < \sqrt{\frac{g}{l}}, \qquad (28)

\cos\theta = \frac{g}{\omega^2 l}, \quad \omega \geq \sqrt{\frac{g}{l}}. \qquad (29)

Schematic of the centrifugal governor: vertical shaft, rod of length $l$ deflected by angle $\theta$, ball, horizontal vector $m\omega^2 r$ and downward vector $mg$.

2.16 A schematic plot of this dependence is shown in the figure below.

Graph of $K/K_0$ versus $\omega\sqrt{l/g}$: constant at 1 up to 1, then decreasing to about 0.12 at 2.8; grid with ticks 0–3 on the horizontal axis and 0–1.2 on the vertical axis.

Part 3. Engine with a Governor

2.17 Taking into account that $K = K_0\cos\theta$ and using Eq. (19), substitution into Eq. (25) yields

\omega = \sqrt[3]{\frac{K_0 g}{\beta l}\left(\frac{A_0}{2\pi M_0} - 1\right)}. \qquad (30)

2.18 A schematic plot of the corresponding dependence is shown in the figure below.

Two graphs of $\omega$ versus $2\pi M_0/A_0$: a steeply decreasing curve (without governor) and a slowly decreasing curve starting from a small value (with governor).

	Content	Points
2.1	Formula (1): $PV^{\gamma} = const$	0.2	0.6
	Formula (2): $\eta = \left(\frac{P_A}{P_0}\right)^{\frac{1}{\gamma}}$	0.2
	Numerical value in formula (2): $\eta = 0.177$	0.2
2.2	Formula (3): $TV^{\gamma-1} = const$	0.2	0.6
	Formula (4): $T_0 = \frac{T_S}{\eta^{\gamma-1}}$	0.2
	Numerical value in formula (4): $T_0 = 390\ {}^{\circ}\mathrm{C}$	0.2
2.3	Formula (6): $m_0 = M\frac{P_0\eta V_0}{RT_S}$	0.2	0.4
	Numerical value in formula (6): $m_0 = 2.30\ \mathrm{g}$	0.2
2.4	Formula (7): $A_{1-2} = P_0\eta V_0$	0.2	1.0
	Formula (8): $A_{2-3} = P_0 V_0 \frac{\eta - \eta^{\gamma}}{\gamma - 1}$	0.2
	Formula (9): $A_{3-4} = -P_A V_0$	0.2
	Formula (10): $A_0 = P_0 V_0 \left(\eta + \frac{\eta - \eta^{\gamma}}{\gamma - 1}\right) - P_A V_0$	0.2
	Numerical value in formula (10): $A_0 = 1.25 \cdot 10^3\ \mathrm{J}$	0.2
2.5	Formula (11): $v = \frac{V_0\omega}{\pi}$	0.2	0.2
2.6	Formula (12): $P_0 V_{in}^{\gamma} = PV^{\gamma}$	0.2	1.0
	Formula (13): $P_0 V_{in} = \frac{m}{M}RT_0$	0.2
	Formula (14): $P = P_0 \left(\frac{RT_0}{MP_0}\frac{m}{vt}\right)^{\gamma}$	0.2
	Formula (15): $\frac{dm}{dt} = KP_0\left(1 - \left(\frac{RT_0}{MP_0}\frac{m}{vt}\right)^{\gamma}\right)$	0.4
2.7	Formula (16): $m \propto t$	0.4	0.6
	Reference to formula (14) or (15)	0.2
2.8	Formula (17): $m = K(P_0 - P)t$	0.2	0.6
	Formula (18): $P = P_0\left(\frac{RT_0}{MP_0}\frac{\pi K(P_0 - P)}{V_0\omega}\right)^{\gamma}$	0.4
2.9	Formula (19): $P = \frac{P_0}{1 + \frac{MV_0\omega}{\pi KRT_0}}$	0.2	0.2
2.10	Numerical value in formula (20): $P = 9.10 \cdot 10^5\ \mathrm{Pa}$	0.2	0.2
2.11	Formula (21): $A = P_0 V_0 \frac{\eta + \frac{\eta - \eta^{\gamma}}{\gamma - 1}}{1 + \frac{MV_0\omega}{\pi KRT_0}}$	0.2	1.0
	Formula (22): $A_0 = P_0 V_0\left(\eta + \frac{\eta - \eta^{\gamma}}{\gamma - 1}\right)$	0.2
	Numerical value in formula (22): $A_0 = 1.44 \cdot 10^3\ \mathrm{J}$	0.2
	Formula (23): $\beta = \frac{MV_0}{\pi RT_0}$	0.2
	Numerical value in formula (23): $\beta = 4.16 \cdot 10^{-9}\ \mathrm{m} \cdot \mathrm{s}^2$	0.2
2.12	Formula (24): $\frac{A_0}{1 + \beta\frac{\omega}{K}} = 2\pi M_0$	0.4	0.6
	Formula (25): $\omega = \frac{K}{\beta}\left(\frac{A_0}{2\pi M_0} - 1\right)$	0.2
2.13	Formula (26): $M_{0max} = \frac{A_0}{2\pi}$	0.4	0.6
	Numerical value in formula (26): $M_{0max} = 230\ \mathrm{N} \cdot \mathrm{m}$	0.2
2.14	Graph: turns zero after the maximum torque is achieved	0.1	0.4
	maximum torque is pointed	0.1
	monotonic increase when the torque goes to zero	0.2
2.15	Formula (27): $mgl\sin\theta = m\omega^2 l^2 \sin\theta\cos\theta$	0.2	1.0
	Formula (28): $\cos\theta = 1$ , $\omega < \sqrt{\frac{g}{l}}$	0.4
	Formula (29): $\cos\theta = \frac{g}{\omega^2 l}$ , $\omega \ge \sqrt{\frac{g}{l}}$	0.4
2.16	Graph: horizontal segment is present	0.1	0.4
	the value of $K_0$ is pointed for the horizontal segment	0.1
	monotonically goes to zero when the angular velocity increases	0.2
2.17	Formula (30): $\omega = \sqrt[3]{\frac{K_0 g}{\beta l}\left(\frac{A_0}{2\pi M_0} - 1\right)}$	0.4	0.4
2.18	Graph: fracture point is present	0.1	0.2
	the left segment decrease faster than the right segment	0.1
Total			10.0

Mathematical hint for the theoretical competition

The following formulas may be useful:

$\int x^n dx = \frac{x^{n+1}}{n+1} + C$ , where $n \neq -1$ is a fixed number, $C$ refers to an arbitrary constant;

$\int \frac{dx}{x} = \ln|x| + C$ , where $C$ stands for an arbitrary constant;

$\int \frac{dx}{(a^2+x^2)^{3/2}} = \frac{x}{a^2\sqrt{a^2+x^2}} + C$ , where $a$ – constant, $C$ – arbitrary constant;

$(1+x)^{\gamma} \approx 1 + \gamma x + \frac{\gamma(\gamma-1)}{2}x^2$ , for $|x| \ll 1$ and any value of $\gamma$ ;

$\tan x \approx \sin x \approx x$ , for $|x| \ll 1$ ;

$\ln(1+x) \approx x$ , for $|x| \ll 1$ .

List of physical constants

Acceleration of gravity $g = 9.80\ \mathrm{m/s^2}$ ;

Universal gas constant $R = 8.31\ \mathrm{J/(mole\cdot K)}$ ;

Speed of light in vacuum $c = 3.00 \cdot 10^{8}\ \mathrm{m/s}$ ;

Reduced Planck constant $\hbar = 1.05 \cdot 10^{-34}\ \mathrm{J\cdot s}$ ;

Magnetic constant $\mu_0 = 4\pi \cdot 10^{-7}\ \mathrm{H/m}$ ;

Electric constant $\varepsilon_0 = 8.85 \cdot 10^{-12}\ \mathrm{F/m}$ ;

Stefan-Boltzmann constant $\sigma = 5.67 \cdot 10^{-8}\ W/(m^2 \cdot K^4)$ ;

Elementary charge $e = 1.60 \cdot 10^{-19}\ \mathrm{C}$ ;

Electron mass $m_e = 9.11 \cdot 10^{-31}\ \mathrm{kg}$ .

Оригинал в Архива: IZhO-2026-Theory_eng.pdf · официални решения: IZhO-2026-Theory_eng_sol.pdf