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Post

Efficiency of a Thermodynamic Cycle

Created by @nathanedwards

at November 1st 2023, 3:05:03 am.

AP_Physics_2-Questions

#efficiency

#thermodynamics

#heat-engine

Question:

An engine operates in a thermodynamic cycle where the working fluid absorbs heat, performs work, and releases waste heat. The cycle consists of two isothermal processes and two adiabatic processes. The temperature of the hot reservoir is 800 K and the temperature of the cold reservoir is 300 K. Calculate the efficiency of this heat engine.

Solution:

The efficiency of a heat engine can be calculated using the formula:

\text{{Efficiency}} = \frac{{\text{{Work done}}}}{{\text{{Heat absorbed}}}}

The work done by the engine can be calculated as the difference between the heat absorbed and the waste heat released. The heat absorbed can be calculated using the formula:

\text{{Heat absorbed}} = Q_h - Q_c

where $Q_h$ is the heat absorbed from the hot reservoir and $Q_c$ is the heat released to the cold reservoir.

Given that the hot reservoir temperature is 800 K and the cold reservoir temperature is 300 K, we can express these temperatures in Kelvin:

$T_h = 800 \, \text{K}$

$T_c = 300 \, \text{K}$

The efficiency of the engine can now be calculated:

\text{{Efficiency}} = \frac{{\text{{Work done}}}}{{\text{{Heat absorbed}}}} = \frac{{Q_h - Q_c}}{{Q_h}}

Now, let's express the heat absorbed in terms of the temperatures:

Q_h = T_h \Delta S_{\text{hot}}

Q_c = T_c \Delta S_{\text{cold}}

where $\Delta S_{\text{hot}}$ and $\Delta S_{\text{cold}}$ represent the changes in entropy for the hot and cold reservoirs, respectively. Since the process is isothermal, the changes in entropy can be calculated using the equation:

\Delta S = nR \ln{\left(\frac{{V_f}}{{V_i}}\right)}

where $n$ represents the number of moles of the working fluid, $R$ is the gas constant, and $V_f$ and $V_i$ represent the final and initial volumes of the system, respectively.

For the isothermal processes, the formula simplifies to:

\Delta S = nR \ln{\left(\frac{{V_f}}{{V_i}}\right)} = nR \ln{\left(\frac{{V_i}}{{V_f}}\right)}

Since the engine undergoes two isothermal and two adiabatic processes, the change in volume can be related to the change in temperature using the equations for the adiabatic processes:

\frac{{T_1 \cdot V_1^{(\gamma - 1)}}}{{T_2 \cdot V_2^{(\gamma - 1)}}} = 1

\frac{{T_3 \cdot V_3^{(\gamma - 1)}}}{{T_4 \cdot V_4^{(\gamma - 1)}}} = 1

where $T_1, T_2, T_3,$ and $T_4$ represent the temperatures at the corresponding states, and $V_1, V_2, V_3,$ and $V_4$ represent the volumes at the corresponding states. $\gamma$ is the adiabatic constant, which depends on the working fluid (e.g., 1.4 for diatomic ideal gases).

Now, by substituting the equations for entropy and using the relations for the adiabatic processes, we can simplify the expression for the efficiency:

\text{{Efficiency}} = \frac{{Q_h - Q_c}}{{Q_h}} = \frac{{T_h \Delta S_{\text{hot}} - T_c \Delta S_{\text{cold}}}}{{T_h \Delta S_{\text{hot}}}}

\text{{Efficiency}} = 1 - \frac{{T_c}}{{T_h}} \cdot \frac{{\Delta S_{\text{cold}}}}{{\Delta S_{\text{hot}}}}

Now, we can substitute the expressions for $\Delta S_{\text{cold}}$ and $\Delta S_{\text{hot}}$ and solve for the efficiency:

\text{{Efficiency}} = 1 - \frac{{T_c}}{{T_h}} \cdot \frac{{nR \ln{\left(\frac{{V_4}}{{V_3}}\right)}}}{{nR \ln{\left(\frac{{V_2}}{{V_1}}\right)}}}

\text{{Efficiency}} = 1 - \frac{{T_c}}{{T_h}} \cdot \frac{{\ln{\left(\frac{{V_4}}{{V_3}}\right)}}}{{\ln{\left(\frac{{V_2}}{{V_1}}\right)}}}

Finally, we can use the relations for adiabatic processes to simplify the expression further:

\text{{Efficiency}} = 1 - \frac{{T_c}}{{T_h}} \cdot \frac{{\ln{\left(\frac{{V_1}}{{V_2}}\right)}}}{{\ln{\left(\frac{{V_1}}{{V_2}}\right)}}}

\text{{Efficiency}} = 1 - \frac{{T_c}}{{T_h}}

Let's substitute the given values and calculate the efficiency:

\text{{Efficiency}} = 1 - \frac{{300 \, \text{K}}}{{800 \, \text{K}}} = 1 - 0.375 = 0.625

Therefore, the efficiency of this heat engine is 0.625 or 62.5%.