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