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| <h2>Solutions of Savchenko Problems in Physics</h2> | | <h2>Solutions of Savchenko Problems in Physics</h2> |
| <p class="author"> | | <p class="author"> |
| Aliaksandr Melnichenka <br/> | | Aliaksandr Melnichenka <br/> |
| October 2023 | | October 2023 |
| </p> | | </p> |
| </header> | | </header> |
| | | |
| <h3 id="back-link"><a href="../#5.9">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../#5.9">$\leftarrow$Back</a></h3> |
| | | |
| <h3> Statement </h3> | | <h3> Statement </h3> |
| | | |
| <p> | | <p> |
| $5.9.10^*$ | | $5.9.10^*$ |
| Find the efficience of cycles shown in following figures, if as power agent it is taken a monoatomic ideal gas. | | Find the efficience of cycles shown in following figures, if as power agent it is taken a monoatomic ideal gas. |
| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="statement.png" | | <img src="statement.png" |
| loading="lazy" alt="5.9.10" width="350" /> | | loading="lazy" alt="5.9.10" width="350" /> |
| <figcaption> | | <figcaption> |
| For problem 5.9.10 | | For problem 5.9.10 |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| | | |
| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
| a) Let enumerate the states from 1 to 4 in clockwise sense starting in the coordinate $(P_1,V_1)$. The efficience is defined as | | a) Let enumerate the states from 1 to 4 in clockwise sense starting in the coordinate $(P_1,V_1)$. The efficience is defined as |
| $$\eta = 1-\frac{|Q_c|}{Q_h}=1-\frac{|Q_{34}|}{Q_{12}} \;(1)$$ | | $$\eta = 1-\frac{|Q_c|}{Q_h}=1-\frac{|Q_{34}|}{Q_{12}} \;(1)$$ |
| where $Q_{34}$ is the heat released by gas (cold reservoir) and $Q_{12}$ is the heat absorbed by gas (hot reservoir), while $Q_{23}$ and $Q_{41}$ are zero (adiabatic processes). Since $3-4$ is an isochoric process, | | where $Q_{34}$ is the heat released by gas (cold reservoir) and $Q_{12}$ is the heat absorbed by gas (hot reservoir), while $Q_{23}$ and $Q_{41}$ are zero (adiabatic processes). Since $3-4$ is an isochoric process, |
| $$Q_{34}=\Delta U_{34} = nC_v\left(T_4-T_3\right) \;(2)$$ | | $$Q_{34}=\Delta U_{34} = nC_v\left(T_4-T_3\right) \;(2)$$ |
| From Poisson Law | | From Poisson Law |
| $$P_1V_1^\gamma = P_4V_2^\gamma$$ | | $$P_1V_1^\gamma = P_4V_2^\gamma$$ |
| and the state 4 is characterized by | | and the state 4 is characterized by |
| $$P_4V_2=nRT_4$$ | | $$P_4V_2=nRT_4$$ |
| So, separating $P_4$ and putting it into above Poisson law | | So, separating $P_4$ and putting it into above Poisson law |
| $$T_4 = \frac{P_1V_1^\gamma}{nRV_2^{\gamma-1}} \;(3)$$ | | $$T_4 = \frac{P_1V_1^\gamma}{nRV_2^{\gamma-1}} \;(3)$$ |
| Proceeding similarly, $T_3$ can be found. Applying Poisson Law for process $2-3$, | | Proceeding similarly, $T_3$ can be found. Applying Poisson Law for process $2-3$, |
| $$P_2V_1^\gamma=P_3V_2^\gamma$$ | | $$P_2V_1^\gamma=P_3V_2^\gamma$$ |
| State equation for state 3 is | | State equation for state 3 is |
| $$P_3V_2=nRT_3$$ | | $$P_3V_2=nRT_3$$ |
| Separating $P_3$ and substituting into respective Poisson Law | | Separating $P_3$ and substituting into respective Poisson Law |
| $$T_3 = \frac{P_2V_1^\gamma}{nRV_2^{\gamma-1}} \;(4)$$ | | $$T_3 = \frac{P_2V_1^\gamma}{nRV_2^{\gamma-1}} \;(4)$$ |
| Substituting $(3)$ and $(4)$ into $(2)$, and as $C_v=iR/2$ | | Substituting $(3)$ and $(4)$ into $(2)$, and as $C_v=iR/2$ |
| $$Q_{34} = \frac{i}{2}\frac{V_1^\gamma}{V_2^{\gamma-1}}\left(P_1-P_2\right) \;(5)$$ | | $$Q_{34} = \frac{i}{2}\frac{V_1^\gamma}{V_2^{\gamma-1}}\left(P_1-P_2\right) \;(5)$$ |
| For the process $1-2$ | | For the process $1-2$ |
| $$Q_{12}=\Delta U_{12} = nC_v\left(T_2-T_1\right)$$ | | $$Q_{12}=\Delta U_{12} = nC_v\left(T_2-T_1\right)$$ |
| Taking account the state equations for states 1 and 2, temperatures can be found, | | Taking account the state equations for states 1 and 2, temperatures can be found, |
| $$Q_{12}=nC_v\left(\frac{P_2V_1}{nR}-\frac{P_1V_1}{nR}\right)$$ | | $$Q_{12}=nC_v\left(\frac{P_2V_1}{nR}-\frac{P_1V_1}{nR}\right)$$ |
| $$Q_{12}=\frac{i}{2}V_1\left(P_2-P_1\right) \;(6)$$ | | $$Q_{12}=\frac{i}{2}V_1\left(P_2-P_1\right) \;(6)$$ |
| Putting $(5)$ and $(6)$ into $(1)$ | | Putting $(5)$ and $(6)$ into $(1)$ |
| $$\eta = 1-\left(\frac{V_1}{V_2}\right)^{\gamma-1}$$ | | $$\eta = 1-\left(\frac{V_1}{V_2}\right)^{\gamma-1}$$ |
| Since $\gamma=C_p/C_v=1+2/i$ and $i=3$ (monoatomic gas), | | Since $\gamma=C_p/C_v=1+2/i$ and $i=3$ (monoatomic gas), |
| $$\boxed{\eta = 1-\left(\frac{V_1}{V_2}\right)^{\frac{2}{3}}}$$ | | $$\boxed{\eta = 1-\left(\frac{V_1}{V_2}\right)^{\frac{2}{3}}}$$ |
| b) Let enumerate again the states following a clockwise way, this time starting in coordinate $(P_1,T_1)$. The efficience is | | b) Let enumerate again the states following a clockwise way, this time starting in coordinate $(P_1,T_1)$. The efficience is |
| $$\eta = 1-\frac{|Q_c|}{Q_h} \;(I)$$ | | $$\eta = 1-\frac{|Q_c|}{Q_h} \;(I)$$ |
| For identifying correctly the reservoirs, let's see the following graph corresponding to the cycle in a $P$ vs $V$ plot. | | For identifying correctly the reservoirs, let's see the following graph corresponding to the cycle in a $P$ vs $V$ plot. |
| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="PVb.png" | | <img src="PVb.png" |
| loading="lazy" alt="5.9.10" width="300" /> | | loading="lazy" alt="5.9.10" width="300" /> |
| <figcaption> | | <figcaption> |
| $P$ vs $V$ plot | | $P$ vs $V$ plot |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| The hot reservoir is defined as | | The hot reservoir is defined as |
| $$Q_h = Q_{23} + Q_{34} \;(II)$$ | | $$Q_h = Q_{23} + Q_{34} \;(II)$$ |
| and cold one | | and cold one |
| $$Q_c = Q_{12} + Q_{41} \;(III)$$ | | $$Q_c = Q_{12} + Q_{41} \;(III)$$ |
| Process $2-3$ is isobaric, then | | Process $2-3$ is isobaric, then |
| $$Q_{23}=nC_p\left(T_2-T_1\right) \;(IV)$$ | | $$Q_{23}=nC_p\left(T_2-T_1\right) \;(IV)$$ |
| Process $3-4$ is isothermal, so | | Process $3-4$ is isothermal, so |
| $$Q_{34} = A_{34} = nRT_2\ln{\frac{V_4}{V_3}}$$ | | $$Q_{34} = A_{34} = nRT_2\ln{\frac{V_4}{V_3}}$$ |
| Since $V=nRT/P$ according state equation | | Since $V=nRT/P$ according state equation |
| $$Q_{34} = nRT_2\ln{\frac{P_2}{P_1}} \;(V)$$ | | $$Q_{34} = nRT_2\ln{\frac{P_2}{P_1}} \;(V)$$ |
| Substituting $(IV)$ and $(V)$ into $(II)$and taking account that $C_p=(i+2)R/2$ | | Substituting $(IV)$ and $(V)$ into $(II)$and taking account that $C_p=(i+2)R/2$ |
| $$Q_h = \frac{i+2}{2}nR\left(T_2-T_1\right)+nRT_2\ln{\frac{P_2}{P_1}} \;(VI)$$ | | $$Q_h = \frac{i+2}{2}nR\left(T_2-T_1\right)+nRT_2\ln{\frac{P_2}{P_1}} \;(VI)$$ |
| Process $1-2$ is isothermal, so | | Process $1-2$ is isothermal, so |
| $$Q_{12}=A_{12}=nRT_1\ln{\frac{V_2}{V_1}}$$ | | $$Q_{12}=A_{12}=nRT_1\ln{\frac{V_2}{V_1}}$$ |
| and newly, taking account state equation | | and newly, taking account state equation |
| $$Q_{12}=A_{12}=nRT_1\ln{\frac{P_1}{P_2} \;(VII)}$$ | | $$Q_{12}=A_{12}=nRT_1\ln{\frac{P_1}{P_2} \;(VII)}$$ |
| Finally, process $4-1$ is isobaric | | Finally, process $4-1$ is isobaric |
| $$Q_{14} = nC_p\left(T_1-T_2\right) \;(VIII)$$ | | $$Q_{14} = nC_p\left(T_1-T_2\right) \;(VIII)$$ |
| Substituting $(VII)$ and $(VIII)$ into $(III)$ | | Substituting $(VII)$ and $(VIII)$ into $(III)$ |
| $$Q_c = nRT_1\ln{\frac{P_1}{P_2}}+\frac{i+2}{2}nR\left(T_1-T_2\right) \;(IX)$$ | | $$Q_c = nRT_1\ln{\frac{P_1}{P_2}}+\frac{i+2}{2}nR\left(T_1-T_2\right) \;(IX)$$ |
| Substituting $(VI)$ and $(IX)$ into $(I)$ and developing algebraically | | Substituting $(VI)$ and $(IX)$ into $(I)$ and developing algebraically |
| $$\eta = \frac{2\left(T_2-T_1\right)\ln{\frac{P_2}{P_1}}}{(i+2)\left(T_2-T_1\right)+2T_2\ln{\frac{P_2}{P_1}}}$$ | | $$\eta = \frac{2\left(T_2-T_1\right)\ln{\frac{P_2}{P_1}}}{(i+2)\left(T_2-T_1\right)+2T_2\ln{\frac{P_2}{P_1}}}$$ |
| Since gas is monoatomic, $i = 3$, hence | | Since gas is monoatomic, $i = 3$, hence |
| $$\boxed{\eta = \frac{2\left(T_2-T_1\right)\ln{\frac{P_2}{P_1}}}{5\left(T_2-T_1\right)+2T_2\ln{\frac{P_2}{P_1}}}}$$ | | $$\boxed{\eta = \frac{2\left(T_2-T_1\right)\ln{\frac{P_2}{P_1}}}{5\left(T_2-T_1\right)+2T_2\ln{\frac{P_2}{P_1}}}}$$ |
| </p> | | </p> |
| | | |
| <p style="text-align: right; font-style: italic; font-size: 14;"> | | <p style="text-align: right; font-style: italic; font-size: 14;"> |
| BSc. Luis Daniel Fernández Quintana<br> | | BSc. Luis Daniel Fernández Quintana<br> |
| Physics Department (FCNE)<br> | | Physics Department (FCNE)<br> |
| Universidad de Oriente, Cuba<br> | | Universidad de Oriente, Cuba<br> |
| </p> | | </p> |
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