Question

The data listed below are claimed for a power cycle operating between reservoirs at \(527^{\circ} \mathrm{C}\) and \(27^{\circ} \mathrm{C}\). For each case, determine if any principles of thermodynamics would be violated. (a) \(Q_{\mathrm{H}}=700 \mathrm{~kJ}, W_{\text {cycle }}=400 \mathrm{~kJ}, Q_{\mathrm{C}}=300 \mathrm{~kJ}\). (b) \(Q_{\mathrm{H}}=640 \mathrm{~kJ}, W_{\text {cycle }}=400 \mathrm{~kJ}, Q_{\mathrm{C}}=240 \mathrm{~kJ}\). (c) \(Q_{\mathrm{H}}=640 \mathrm{~kJ}, W_{\text {cycle }}=400 \mathrm{~kJ}, Q_{\mathrm{C}}=200 \mathrm{~kJ}\)

Short answer

Case (a) does not violate thermodynamics principles, while cases (b) and (c) do violate.

Step-by-step solution

Title - Calculate Carnot Efficiency

First, determine the Carnot efficiency for the cycle operating between the given temperatures. Use the formula \text{Carnot efficiency}, \ \ \( \text{Carnot Efficiency} = 1 - \frac{T_C}{T_H} \), where \( T_C \) and \( T_H \) are the absolute temperatures of the cold and hot reservoirs, respectively. Convert the given temperatures from Celsius to Kelvin: \ \( T_H = 527^{\circ} \mathrm{C} + 273.15 = 800.15 \mathrm{~K} \) \ \( T_C = 27^{\circ} \mathrm{C} + 273.15 = 300.15 \mathrm{~K} \) \ Then, calculate the Carnot efficiency: \ \( \text{Carnot Efficiency} = 1 - \frac{300.15}{800.15} = 0.6247 = 62.47\% \)

Title - Compare Claimed Efficiency with Carnot Efficiency

Next, calculate the claimed efficiencies for each case and compare them with the Carnot efficiency calculated in Step 1 to ensure none of them violate the second law of thermodynamics (no real engine can be more efficient than a Carnot engine). Use the claimed efficiencies formula: \ \( \text{Efficiency}_i = \frac{W_{\text{cycle }}}{Q_{\text{H}}} \)

Title - Evaluate Case (a)

Calculate the efficiency: \ \( \text{Efficiency}_a = \frac{400}{700} = 0.571 = 57.14\% \) \ Compare this with the Carnot efficiency (62.47\%) to determine if it violates any principles. The claimed efficiency is less than Carnot efficiency, so it does NOT violate any principles.

Title - Evaluate Case (b)

Calculate the efficiency: \ \( \text{Efficiency}_b = \frac{400}{640} = 0.625 = 62.5\% \) \ Compare this with the Carnot efficiency (62.47\%) to determine if it violates any principles. The claimed efficiency is slightly higher than the Carnot efficiency, so it DOES violate the second law of thermodynamics.

Title - Evaluate Case (c)

Calculate the efficiency: \ \( \text{Efficiency}_c = \frac{400}{640} = 0.625 = 62.5\% \) \ Compare this with the Carnot efficiency (62.47\%) to determine if it violates any principles. The claimed efficiency is slightly higher than the Carnot efficiency, so it DOES violate the second law of thermodynamics.

Key concepts

Carnot cycle

The Carnot cycle is a theoretical thermodynamic cycle that provides the upper limit on efficiency for any heat engine. It is reversible, consisting of four processes: two isothermal (constant temperature) and two adiabatic (no heat exchange). During the isothermal expansion, a system absorbs heat from the hot reservoir at a constant high temperature. Next, in an adiabatic process, the system expands further without exchanging heat, dropping in temperature. This is followed by an isothermal compression, where heat is released to the cold reservoir, and finally, another adiabatic compression back to the starting state. Importantly, the Carnot cycle helps us understand the maximum possible efficiency of energy conversion between heat and work.

second law of thermodynamics

The second law of thermodynamics is crucial for understanding the limitations of energy conversion processes. It states that in any cyclic process, the entropy (disorder) of an isolated system will either increase or remain constant, meaning that energy transformations are inherently irreversible. An implication of this law is that no real engine can be 100% efficient, as there's always some energy lost as waste heat. This law also helps us understand why the Carnot efficiency represents a theoretical maximum since it assumes a perfectly reversible cycle, which is impossible in real-world engines.

energy conversion

Energy conversion in thermodynamic cycles involves transforming heat energy into work. In a power cycle, heat is absorbed from a high-temperature source, part of it is converted into useful work, and the remainder is expelled to a low-temperature sink. The efficiency of this conversion depends on how effectively the cycle can convert the heat into work. The Carnot cycle provides an idealized model for understanding the limits of this conversion. It indicates the highest theoretical efficiency that a heat engine can achieve, serving as a benchmark for real machines.

efficiency calculation

Efficiency calculation in thermodynamics is essential for evaluating the performance of heat engines. It is determined by the ratio of the work output to the heat input. The efficiency formula is given by: \( \text{Efficiency} = \frac{W_{\text{cycle}}}{Q_{\text{H}}} \), where \( W_{\text{cycle}} \) is the work done by the cycle and \( Q_{\text{H}} \) is the heat absorbed from the hot reservoir. The efficiency cannot exceed the Carnot efficiency, which is \( 1 - \frac{T_C}{T_H} \), where \( T_C \) and \( T_H \) are the absolute temperatures of the cold and hot reservoirs. This calculation helps in comparing the practical efficiency of engines to the theoretical limit.

absolute temperature

Absolute temperature is a temperature measurement on a scale where 0 is the lowest possible temperature, known as absolute zero. The Kelvin scale is commonly used for this purpose in thermodynamic calculations. Converting temperatures from Celsius to Kelvin is essential in efficiency calculations, as the formulas depend on absolute temperature values. For instance, to convert Celsius to Kelvin, add 273.15 to the Celsius temperature. This ensures accuracy when applying equations like the Carnot efficiency formula, which relies on absolute temperatures to determine theoretical maximum efficiency.