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Chapter 5: Problem 24

A certain reversible power cycle has the same thermal efficiency for hot and cold reservoirs at 1000 and \(500 \mathrm{~K}\), respectively, as for hot and cold reservoirs at temperature \(T\) and \(1000 \mathrm{~K}\). Determine \(T\), in \(\mathrm{K}\).

Short Answer

Expert verified
The unknown temperature T is 2000 K.

Step by step solution

01

Understand the Given Temperatures

Identify the initial hot and cold reservoir temperatures provided. These are 1000 K (hot) and 500 K (cold) for one scenario, and T (unknown hot temperature) and 1000 K (cold) for the other scenario.
02

Understand Thermal Efficiency of Reversible Cycles

Recall the formula for the thermal efficiency of a reversible cycle: enoittal xdrofroftnecreiplaT-yreseventyC/1storbgçeness ß wosui ⇔ \[ \eta = 1 - \frac{T_C}{T_H} \], where \( T_C \) is the cold reservoir temperature and \( T_H \) is the hot reservoir temperature.
03

Apply the Thermal Efficiency Formula to Both Scenarios

Write the efficiency equation for the first pair of temperatures: \[ \eta_1 = 1 - \frac{500}{1000} = 1 - 0.5 = 0.5 \]. Write the efficiency equation for the second scenario: \[ \eta_2 = 1 - \frac{1000}{T} \]. Since the efficiencies are equal by the problem's condition, set the two equations equal: \[ 0.5 = 1 - \frac{1000}{T} \].
04

Solve for the Unknown Temperature T

From the equation \( 0.5 = 1 - \frac{1000}{T} \): 1. Subtract 1 from both sides: \[ 0.5 - 1 = -\frac{1000}{T} \]. 2. Simplify the left side: \[ -0.5 = -\frac{1000}{T} \]. 3. Remove the negative signs: \[ 0.5 = \frac{1000}{T} \]. 4. Solve for T: \[ T = \frac{1000}{0.5} \]. 5. Compute the value: \[ T = 2000 \; \text{K} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Thermodynamic Efficiency
Thermodynamic efficiency is a measure of how well a power cycle converts heat from a hot reservoir into useful work. For a reversible power cycle, which is an idealized process without energy dissipation, the efficiency can be calculated using the formula: \[ \eta = 1 - \frac{T_C}{T_H} \] where \( \eta \) is the efficiency, \( T_C \) is the temperature of the cold reservoir, and \( T_H \) is the temperature of the hot reservoir. This formula represents the maximum possible efficiency for the given temperatures and shows that efficiency increases as the temperature difference between the hot and cold reservoirs increases. If the temperatures are closer together, the efficiency decreases. Understanding this principle helps us compare different power cycles and optimize energy conversion.
Hot and Cold Reservoirs
In thermodynamics, a power cycle operates between two thermal energy reservoirs: a hot reservoir, which supplies heat to the cycle, and a cold reservoir, where the waste heat is expelled. These reservoirs are assumed to have infinite thermal capacities and maintain constant temperatures. The hot reservoir might be a high-temperature furnace or combustion chamber, while the cold reservoir could be the ambient environment like a river, ocean, or the atmosphere. The temperatures of these reservoirs play a crucial role in determining the efficiency of the power cycle. The bigger the temperature difference between them, the more efficient the cycle can potentially be.
Temperature Calculation
To solve the given exercise, we need to find the unknown temperature \( T \) for the hot reservoir in the second scenario, using the condition that the thermal efficiencies are the same for both scenarios. Given temperatures in the first scenario are 1000 K for the hot reservoir and 500 K for the cold reservoir. In the second scenario, the hot reservoir's temperature is \( T \) and the cold reservoir's temperature is 1000 K. We start by calculating the efficiency for the first scenario: \[ \eta_1 = 1 - \frac{500}{1000} = 0.5 \] Next, we write the efficiency for the second scenario: \[ \eta_2 = 1 - \frac{1000}{T} \] Since \( \eta_1 = \eta_2 \), we equate the expressions: \[ 0.5 = 1 - \frac{1000}{T} \] Solving for \( T \): \[ 0.5 = 1 - \frac{1000}{T} \] Subtracting 1 from both sides and simplifying: \[ -0.5 = -\frac{1000}{T} \] Removing the negative signs: \[ 0.5 = \frac{1000}{T} \] Solving for \( T \): \[ T = \frac{1000}{0.5} = 2000 \] Therefore, \( T \) equals 2000 K. This calculation confirms that the condition of equal efficiencies is satisfied when the hot reservoir temperature is 2000 K.

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Most popular questions from this chapter

Using the Kelvin-Planck statement of the second law of thermodynamics, demonstrate the following corollaries: (a) The coefficient of performance of an irreversible refrigeration cycle is always less than the coefficient of performance of a reversible refrigeration cycle when both exchange energy by heat transfer with the same two reservoirs. (b) All reversible refrigeration cycles operating between the same two reservoirs have the same coefficient of performance. (c) The coefficient of performance of an irreversible heat pump cycle is always less than the coefficient of performance of a reversible heat pump cycle when both exchange energy by heat transfer with the same two reservoirs. (d) All reversible heat pump cycles operating between the same two reservoirs have the same coefficient of performance.

A refrigeration cycle having a coefficient of performance of 3 maintains a computer laboratory at \(18^{\circ} \mathrm{C}\) on a day when the outside temperature is \(30^{\circ} \mathrm{C}\). The thermal load at steady state consists of energy entering through the walls and windows at a rate of \(30,000 \mathrm{~kJ} / \mathrm{h}\) and from the occupants, computers, and lighting at a rate of \(6000 \mathrm{~kJ} / \mathrm{h}\). Determine the power required by this cycle and compare with the minimum theoretical power required for any refrigeration cycle operating under these conditions, each in \(\mathrm{kW}\).

Two reversible power cycles are arranged in series. The first cycle receives energy by heat transfer from a reservoir at temperature \(T_{\mathrm{H}}\) and rejects energy to a reservoir at an intermediate temperature \(T\). The second cycle receives the energy rejected by the first cycle from the reservoir at temperature \(T\). and rejects energy to a reservoir at temperature \(T_{\mathrm{C}}\) lower than \(T\). Derive an expression for the intermediate temperature \(T\) in terms of \(T_{\mathrm{H}}\) and \(T_{\mathrm{C}}\) when (a) the net work of the two power cycles is equal. (b) the thermal efficiencies of the two power cycles are equal.

A refrigeration cycle operating between two reservoirs receives energy \(Q_{\mathrm{C}}\) from a cold reservoir at \(T_{\mathrm{C}}=280 \mathrm{~K}\) and rejects energy \(Q_{\mathrm{H}}\) to a hot reservoir at \(T_{\mathrm{H}}=320 \mathrm{~K}\). For each of the following cases determine whether the cycle operates reversibly, irreversibly, or is impossible: (a) \(Q_{\mathrm{C}}=1500 \mathrm{~kJ}, W_{\text {cycle }}=150 \mathrm{~kJ}\). (b) \(Q_{\mathrm{C}}=1400 \mathrm{~kJ}, Q_{\mathrm{H}}=1600 \mathrm{~kJ}\). (c) \(Q_{\mathrm{H}}=1600 \mathrm{~kJ}, W_{\text {cycle }}=400 \mathrm{~kJ}\). (d) \(\beta=5\).

For each \(\mathrm{kW}\) of power input to an ice maker at steady state, determine the maximum rate that ice can be produced, in \(\mathrm{kg} / \mathrm{h}\), from liquid water at \(0^{\circ} \mathrm{C}\). Assume that \(333 \mathrm{~kJ} / \mathrm{kg}\) of energy must be removed by heat transfer to freeze water at \(0^{\circ} \mathrm{C}\), and that the surroundings are at \(20^{\circ} \mathrm{C}\).
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