The efficiency of a Carnot engine working between 227°C and 27°C is

Asked in WBPSC CS Prelims 2023 · Topic: Heat and Thermodynamics (Second Law of Thermodynamics (Entropy, Heat Engines, Refrigerators, Carnot Cycle))

1. 100%
2. 50%
3. 40%
4. 20%

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• Heat rejection is at constant volume from 5 to 1.The question states "same maximum pressure and temperature". This means $P_{max}$ and $T_{max}$ are the same for all three cycles. Consider the heat rejection process: it's at constant volume for Otto and Dual, and at constant volume for Diesel (if the state 5 is reached at $V_1$). However, the heat rejection in Diesel is at constant volume from state 5 to state 1.Let's consider the heat added and rejected. Efficiency $\eta = 1 - Q_{out}/Q_{in}$. $Q_{in, Otto} = c_v (T_4 - T_2)$. $Q_{in, Diesel} = c_p (T_4 - T_2)$. $Q_{in, Dual} = c_v (T_3 - T_2) + c_p (T_4 - T_3)$.$Q_{out, Otto} = c_v (T_5 - T_1)$. $Q_{out, Diesel} = c_v (T_5 - T_1)$. $Q_{out, Dual} = c_v (T_5 - T_1)$.For the same $T_{max} (=T_4)$ and $P_{max} (=P_4)$, the initial states (after compression) $T_2, P_2$ will be different.Let's analyze the heat addition relative to the total heat capacity. For same $T_{max}$ and $P_{max}$: - Diesel cycle Heat is added at constant pressure. To reach $T_{max}$ at $P_{max} $. - Otto cycle Heat is added at constant volume. To reach $T_{max}$, pressure will increase. If $P_{max}$ is also the limit, then the constant volume heat addition cannot go beyond the point where $P=P_{max}$ and $T=T_{max}$. - Dual cycle Heat addition at constant volume then constant pressure.When $P_{max}$ and $T_{max}$ are fixed, the Otto cycle tends to have the highest efficiency. This is because the constant volume heat addition process (Otto) is more effective than constant pressure (Diesel) for raising temperature for the same amount of heat input. For the same $T_{max}$ and $P_{max}$, the Otto cycle can operate with a higher compression ratio, leading to higher efficiency.Consider the T- s diagram. Heat addition at constant volume is a curve with increasing temperature and pressure. Heat addition at constant pressure is a curve with increasing temperature and volume. Efficiency is higher if heat is added at higher temperatures and rejected at lower temperatures.For same $P_{max}$ and $T_{max}$: - The Diesel cycle will have a larger cut-off ratio compared to the Otto cycle's heat addition process duration. - The Otto cycle will have a higher compression ratio for a given $P_{max}$ and $T_{max}$ compared to the Diesel cycle when the heat added is constrained.A well-known result states that for the same $P_{max}$, $T_{max}$, and compression ratio $r$: $ \eta_{Otto} > \eta_{Dual} > \eta_{Diesel}$. However, the question states "same maximum pressure and temperature", which is a slightly different condition than "same compression ratio".Let's reconsider the statement "same maximum pressure and temperature". This means the end point of heat addition ($T_4, P_4$) is the same for all three cycles. - For Otto, heat is added at constant volume $V_2$. So, $P_2 \to P_4$ and $T_2 \to T_4$. Here $P_4 = P_{max}$ and $T_4 = T_{max}$. - For Diesel, heat is added at constant pressure $P_2$. So, $P_2 = P_{max} $. $T_2 \to T_4$. Here $T_4 = T_{max}$. So $P_2 = P_{max}$ and $T_4 = T_{max}$. - For Dual, heat is added at constant volume $V_2$ to $V_3$, then at constant pressure $P_3$ to $P_4$. Here $P_4 = P_{max}$ and $T_4 = T_{max}$.The comparison of efficiencies depends on how these cycles achieve the same $P_{max}$ and $T_{max}$. If we consider the T- s diagram: Efficiency is roughly $1 - (T_{rejection} / T_{addition})$. Heat addition in Otto is isochoric, Diesel is isobaric, Dual is mixed. Heat rejection is isochoric for all if they end at the same state 1.For same $T_{max}$ and $P_{max} $: The Otto cycle allows for a higher compression ratio, leading to higher efficiency. The Diesel cycle is limited in efficiency by its isobaric heat addition, especially at high cut-off ratios. The Dual cycle falls in between.Therefore, for the same maximum pressure and temperature, the order of efficiency is generally: $\eta_{Otto} > \eta_{Dual} > \eta_{Diesel}$.Now let's check the options: (1) $\eta_{\text{Diesel}} > \eta_{\text{Otto}}$ (False) (2) $\eta_{\text{Otto}} > \eta_{\text{Diesel}}$ (True) (3) $ \eta_{\text{Dual}} > \eta_{\text{Diesel}}$ (True) (4) $\eta_{\text{Otto}} > \eta_{\text{Dual}}$ (True)The question asks to choose the correct option. This means we need to select the option that represents the correct relationship. Option A says (1) $\eta_{\text{Diesel}} > \eta_{\text{Otto}} $. This is false. Option B says (2) $ \eta_{\text{Otto}} > \eta_{\text{Diesel}}$. This is true. Option C says (3) $\eta_{\text{Dual}} > \eta_{\text{Diesel}}$. This is true. Option D says (4) $\eta_{\text{Otto}} > \eta_{\text{Dual}}$. This is true.This is a multiple-choice question where multiple statements are true. However, usually, there's only one best answer. The question asks for "the following option will be true". This implies selecting one statement from the given options.Let's re-read the question: "If the Otto, Diesel and Dual cycles are compared for the same maximum pressure and temperature, the following option will be true :". Then options are (1), (2), (3), (4) referring to the statements.So we need to select the option that contains a true statement. Statement (1) is False. Statement (2) is True. Statement (3) is True. Statement (4) is True.This suggests there might be an issue with the question expecting a single correct answer when multiple statements are true.Let's assume the question is asking for the most comprehensive or strongest true statement, or there's a specific order that is implied. The established order is $\eta_{Otto} > \eta_{Dual} > \eta_{Diesel}$.Statement (2) $\eta_{Otto} > \eta_{Diesel}$ is true. Statement (3) $\eta_{Dual} > \eta_{Diesel}$ is true. Statement (4) $\eta_{Otto} > \eta_{Dual}$ is true.All three statements (2), (3), and (4) are true based on the established order.
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