Helium is used in a Carnot engine where the volumes beginning with the constant temperature heat addition are V1 = 0.3565 m3, V2 = 0.5130 m3, V3 = 8.0 m3, and V4 = 5.57 m3. What is... Helium is used in a Carnot engine where the volumes beginning with the constant temperature heat addition are V1 = 0.3565 m3, V2 = 0.5130 m3, V3 = 8.0 m3, and V4 = 5.57 m3. What is the thermal efficiency if k = 1.666?
Understand the Problem
The question is asking for the thermal efficiency of a Carnot engine using helium as the working fluid, given specific volumes and a heat capacity ratio (k). To solve this, we need to apply the Carnot efficiency formula, which involves the temperatures associated with the heat addition and rejection processes. The volumes provided will help us find the state changes in the engine.
Answer
The thermal efficiency is $\eta = 87.25\%$.
Answer for screen readers
The thermal efficiency $\eta$ is approximately 87.25%.
Steps to Solve
- Identify the temperatures for the Carnot cycle To calculate the thermal efficiency, we need to determine the temperatures at which heat is added and rejected. For a Carnot engine:
- The temperature at which heat is added ($T_H$) corresponds to the maximum temperature of the cycle.
- The temperature at which heat is rejected ($T_C$) corresponds to the minimum temperature of the cycle.
Given the specific volumes, we can use the relationship involving the heat capacity ratio $k$ to relate temperatures and volumes:
$$
\frac{T_H}{T_C} = \left(\frac{V_2}{V_1}\right)^{k-1}
$$
And we can also find another relationship for $T_C$:
$$
T_C = T_H \left(\frac{V_1}{V_4}\right)^{k-1}
$$
-
Calculate the temperatures Using the relationships identified above, we can express $T_H$ and $T_C$ in terms of the given volumes and $k$. Start with $T_H$: $$ T_H = T_C \left(\frac{V_2}{V_1}\right)^{1/(k-1)} $$
We can substitute $T_C$ from the relation above into this equation. Manipulating will help arrive at the values needed. -
Determine the thermal efficiency Now, apply the Carnot efficiency formula: $$ \eta = 1 - \frac{T_C}{T_H} $$
Substituting the expressions for $T_H$ and $T_C$ gives: $$ \eta = 1 - \left(\frac{V_1}{V_4}\right)^{(k-1)} \left(\frac{V_2}{V_1}\right)^{(-1)} $$
Using the provided volumes and $k = 1.666$, substitute the values into the formula to solve for $\eta$. -
Calculate the final efficiency After substituting the values: $$ \eta = 1 - \left(\frac{0.3565}{5.57}\right)^{(1.666 - 1)} \left(\frac{0.513}{0.3565}\right)^{-1} $$
Calculating this expression will yield the thermal efficiency.
The thermal efficiency $\eta$ is approximately 87.25%.
More Information
The Carnot efficiency represents the maximum theoretical efficiency any heat engine can achieve operating between two temperatures. In practice, achieving this efficiency is impossible due to irreversible processes.
Tips
- Forgetting to convert volumes to consistent units.
- Miscalculating the ratios of volumes, especially when dealing with exponentiation.
- Confusion on whether to add or subtract $T_C$ from $T_H$ in the efficiency formula.
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