IOM 2019, Теория — Задача 2. The Stirling Cycle and Regeneration

• 1.1. $T'_H = T'_C = \frac{T_H + T_C}{2}$

• 1.2. $r_{\max} = \frac{\delta - \eta_0}{2\delta}$ ≈ $12.5$ %

• 1.3. $\eta_e = \frac{7\delta\eta_0}{4(\delta + \eta_0)}$ ≈ $21$ %

• 2.1. $V \cdot T^{-2} = \mathrm{const}$

• 2.2. $U(V, T) = \frac{\alpha T^4}{4V}$

• 2.3. $p(V, T) = -\frac{\alpha T^4}{12V^2}$

• 2.4. $U(p, V) = -3pV;\quad pV^{2/3} = \mathrm{const}$

• 3.1. 473 %

  Correct range is from 472% to 474% (answer table).

• 3.2. 21.5 %

Possible Solution

1.1. Let's consider the cycle of WB1. The amount of heat transferred from WB1 to WB2 during isochoric cooling can be related to temperature change $Q_R = C_V(T_H - T'_H)$, where $C_V$ is the specific heat of the working body at constant volume. According to the statement of the problem, the same amount of heat is transferred by WB1 to WB2 during the isochoric heating $Q_R = C_V(T'_C - T_C)$. From these equations we obtain $T_H - T'_H = T'_C - T_C$. Therefore, $T_H + T_C = T'_H + T'_C$. The total amount of heat received by each working body in one cycle is $Q_+ = Q_V + Q_T = C_V(T_H - T_C) + A_H$ because in the isothermal process the internal energy of ideal gas does not change, so $Q_T$ is equal to work $A_H$ done on the "hot" isotherm. Therefore, the regeneration factor $r \equiv \dfrac{Q_R}{Q_+} = \dfrac{C_V(T_H - T'_H)}{C_V(T_H - T_C) + A_H}$. Note that WB1 transmits heat to WB2 by means of heat exchange while cooling down to temperature $T'_H$, at the same time WB2 is heating to temperature $T'_C$. Therefore, in order to prevent the reverse heat flow, the requirement $T'_C \le T'_H$ must be satisfied. Since $T_H + T_C = T'_H + T'_C$, this requirement means that $T'_H \ge \dfrac{T_H + T_C}{2}$. Therefore, the maximum value of $r$ corresponds to $T'_H = T'_C = \dfrac{T_H + T_C}{2}$.

1.2. This quantity is equal to $r_{\max} = \dfrac{C_V(T_H - T_C)}{2C_V(T_H - T_C) + 2A_H}$. The absolute value of isothermal work is equal to the area under the process curve in pressure-volume coordinates, which is $|A| = \displaystyle\int_{V_{\min}}^{V_{\max}} p(V)\, dV$. According to the Clapeyron-Mendeleev equation, the process equation for a constant amount $\nu$ of ideal gas is $p(V) = \dfrac{\nu RT}{V}$. Thus, even without calculating the integral, it is clear that $\dfrac{|A_H|}{|A_C|} = \dfrac{T_H}{T_C}$. For an engine without regeneration, gas work in one cycle is equal to $A = |A_H| - |A_C| = \delta \cdot A_H$. It means that $\eta_0 = \dfrac{A}{Q_+} = \dfrac{\delta \cdot A_H}{C_V(T_H - T_C) + A_H}$. From this ratio we obtain $C_V(T_H - T_C) = \dfrac{\delta - \eta_0}{\eta_0} A_H$. Thus, the maximum value of the regeneration factor in the proposed scheme is $r_{\max} = \dfrac{\delta - \eta_0}{2\delta} = 0{,}125 = 12{,}5\,\%$. One may note that the problem is correct only if $\eta_0 < \delta$, which is natural because $\delta$ is equal to the efficiency of the Carnot cycle with the same temperatures of the heater and the cooler.

1.3. When regeneration is used, the cycle work is the same and the heat generated by the heater during the cycle is reduced by $Q_R = r \cdot Q_+$. Consequently, the efficiency of the engine with regeneration $\eta = \dfrac{A}{Q_+ - Q_R} = \dfrac{\eta_0}{1 - r}$, and for the maximum regeneration factor $\eta_{\max} = \dfrac{2\delta\eta_0}{\delta + \eta_0} = 0{,}24$. Taking into account mechanical losses, the engine efficiency is $\eta_e = \dfrac{7}{8}\eta_{\max} = \dfrac{7\delta\eta_0}{4(\delta + \eta_0)} = 0{,}21 = 21\,\%$.

2.1. The amount of heat $Q_p$ received by the substance during isobaric heating from temperature $T_1$ to temperature $T_2$ can be calculated using heat capacity $Q_p = \displaystyle\int_{T_1}^{T_2} C_p(T)\, dT = \dfrac{\beta(p)}{2}(T_2^2 - T_1^2)$. According to the problem statement, the substance performs work related to the volume change: $A_p = -\dfrac{\beta(p)}{4}(T_2^2 - T_1^2) = p(V_2 - V_1)$. Since $p = \mathrm{const}$, the change of volume in the isobaric process is proportional to the change of the absolute temperature squared. Taking into account that $V|_{T \to 0} \to 0$, we find the isobaric equation for this substance in the temperature-volume coordinates: $V \cdot T^{-2} = \mathrm{const}$.

2.2. It follows from the result of problem 2.1 that the pressure of the unknown substance is a function of $x \equiv \dfrac{T^2}{V}$, i.e. the thermal equation of state should have the form $p = f\left(\dfrac{T^2}{V}\right)$. In addition, the change in the internal energy in this process is $U(V_2, T_2) - U(V_1, T_1) = Q_p - A_p = \dfrac{3\beta(p)}{4}(T_2^2 - T_1^2)$. Using $U|_{T \to 0} \to 0$, we find that $U(V, T) = \dfrac{3\tilde{\beta}(T^2/V)}{4} T^2$, where $\tilde{\beta}(x) \equiv \beta(f(x))$. On the other hand, the amount of heat obtained in the isochoric process (when the work is not performed) is equal to the change of internal energy $U(V, T_2) - U(V, T_1) = Q_V = \displaystyle\int_{T_1}^{T_2} C_V(T)\, dT = \dfrac{\gamma(V)}{4}(T_2^4 - T_1^4)$. Taking into account that here too $U|_{T \to 0} \to 0$, we come to the conclusion that the caloric equation of state looks like $U(V, T) = \dfrac{\gamma(V)}{4} T^4$. Let's compare the two results obtained: since $\dfrac{3\tilde{\beta}(T^2/V)}{4} T^2 \equiv \dfrac{\gamma(V)}{4} T^4 \Rightarrow \tilde{\beta}(T^2/V) \equiv \dfrac{T^2}{3}\gamma(V)$, there must be a constant $\alpha$, such that $\gamma(V) = \dfrac{\alpha}{V}$, and then $\tilde{\beta}(x) = \dfrac{\alpha T^2}{3V}$. It means that $U(V, T) = \dfrac{\alpha T^4}{4V}$. Now it is clear from the problem statement that $\alpha > 0$.

2.3. The relationship between the caloric and thermal equations of state leads to the equation:

In other words, $2x\dfrac{df}{dx} - f = -\dfrac{\alpha}{4}x^2$. It is easy to see that this requirement is satisfied by the function $f(x) = -\dfrac{\alpha}{12}x^2$, i.e. the thermal equation of state has the form $p(V, T) = -\dfrac{\alpha T^4}{12V^2}$.

Note: Actually, it is clear that the general solution to the above differential equation is $f(x) = D \cdot \sqrt{x} - \dfrac{\alpha}{12}x^2$, where $D$ is an arbitrary constant. However, the isochore equation has a form given in the statement only if $D = 0$.

It should be noted that the pressure of our substance is always negative! This is very unusual for normal gases, but negative pressure may be observed in a condensed state within some range of the state parameters. Besides, the varieties of matter with similar equations of state are sometimes considered in inflationary cosmology (quintessence, generalized Chaplygin's gas).

2.4. Using the results of 2.3 and 2.2, we find that $U(p, V) = -3pV$. The adiabatic equation is obtained from the condition $p\, dV + dU = 0$, wherefrom it follows that $\dfrac{2}{3} p\, dV + V dp = 0 \Rightarrow d(pV^{2/3}) = 0$. So, the adiabatic equation is $pV^{2/3} = \mathrm{const}$.

3.1. The Stirling cycle for WB′ in pressure-volume coordinates looks as shown in the figure. The isothermal equation is $p = -\dfrac{\alpha T^4}{12V^2}$, and therefore the work done on WB′ in the process 1-2, is equal to $A'_{12} = \dfrac{\alpha T_C^4}{12} \displaystyle\int_{V_1}^{V_2} \dfrac{dV}{V^2} = \dfrac{\alpha T_C^4}{12}\left(\dfrac{1}{V_1} - \dfrac{1}{V_2}\right)$, and similarly $A'_{34} = \dfrac{\alpha T_H^4}{12}\left(\dfrac{1}{V_2} - \dfrac{1}{V_1}\right)$. Therefore, the total work done on WB′ is $A' = \dfrac{\alpha(T_H^4 - T_C^4)(V_1 - V_2)}{12V_1V_2}$. The amount of heat that WB′ received in the cycle is $Q'_X = Q_{23} + Q_{12}$. For isochoric process $Q_{23} = U_3 - U_2 = \dfrac{\alpha}{4V_2}(T_H^4 - T_C^4)$. For isothermal process

Therefore, $Q'_X = \dfrac{\alpha}{12V_1V_2}[3V_1T_H^4 - (4V_2 - V_1)T_C^4]$. Consequently, $\chi = \dfrac{Q'_X}{A'} = \dfrac{3V_1T_H^4 - (4V_2 - V_1)T_C^4}{(T_H^4 - T_C^4)(V_1 - V_2)}$. According to the problem statement, $n \equiv \dfrac{T_H}{T_C}$ and $m \equiv \dfrac{V_1}{V_2}$ are exactly the same as in the WB-cycle. That means that $\dfrac{1}{n} = 1 - \dfrac{T_H - T_C}{T_H} = 1 - \delta \Rightarrow n = \dfrac{1}{1 - \delta} = \dfrac{25}{18} \approx 1{,}3889$.

The efficiency of the Stirling cycle for three-atomic ideal gas is calculated simply: the work produced in the cycle $A = \nu R(T_H - T_C)\ln(V_1/V_2)$, while the heat of the heater $Q_H = 3\nu R(T_H - T_C) + \nu R T_H \ln(V_1/V_2)$. Therefore, the efficiency $\eta_0 = \dfrac{\delta \cdot \ln m}{3\delta + \ln m}$. Thus, $m = \exp\left(\dfrac{3\delta\eta_0}{\delta - \eta_0}\right) = e^{2{,}52} \approx 12{,}4286$. It's easy to note that $\chi = \dfrac{3mn^4 + m - 4}{(n^4 - 1)(m - 1)} \approx 4{,}7325 \approx 473\,\%$.

3.2. When using regeneration, the WB-cycle efficiency remains the same, i.e. $A + A' = \eta_0(Q_H + Q'_H)$. On the other hand, $\chi \equiv \dfrac{Q'_H}{A'} - 1$ and, therefore, $A' = \dfrac{Q'_H}{1 + \chi}$. Finally, we use the known regeneration factor: $Q'_H = \dfrac{r}{1 - r} Q_H$. From these equations we find $\eta = \dfrac{kA}{Q_H} = \dfrac{k}{1 - r}\left(\eta_0 - \dfrac{r}{1 + \chi}\right) \approx 0{,}2149 \approx 21{,}5\,\%$.

TABLE OF ANSWERS AND CRITERIA