Question

A power cycle operates between a reservoir at temperature \(T\) and a lower- temperature reservoir at \(280 \mathrm{~K}\). At steady state, the cycle develops \(40 \mathrm{~kW}\) of power while rejecting 1000 \(\mathrm{kJ} / \mathrm{min}\) of energy by heat transfer to the cold reservoir. Determine the minimum theoretical value for \(T\), in \(\mathrm{K}\).

Short answer

The minimum theoretical value for \( T \) is approximately 949 K.

Step-by-step solution

- Understand the problem

A power cycle operates between a high-temperature reservoir at temperature \( T \) and a low-temperature reservoir at 280 K. The cycle develops 40 kW of power while rejecting 1000 kJ/min to the cold reservoir. The task is to determine the minimum theoretical value for \( T \).

- Convert power and heat transfer units

Convert the units of heat transfer from kJ/min to kW. Since 1 minute is 60 seconds, \( 1000 \mathrm{~kJ/min} = \frac{1000}{60} \mathrm{~kJ/s} = 16.67 \mathrm{~kW} \).

- Apply the First Law of Thermodynamics

The First Law of Thermodynamics for a power cycle states that the net work output \( W_{net} \) is equal to the difference between the heat added to the cycle \( Q_H \) and the heat rejected \( Q_L \).

- Relate work output to heat transfer

Given that the cycle develops 40 kW of power, we have \( W_{net} = Q_H - Q_L \). Substituting the given values: \( 40 = Q_H - 16.67 \), which simplifies to \( Q_H = 56.67 \mathrm{~kW} \).

- Use Carnot Efficiency Formula

The efficiency \( \eta \) of a Carnot cycle is given by \( \eta = 1 - \frac{T_L}{T_H} \). Rearrange this to find the high temperature reservoir \( T_H \): \( T_H = \frac{T_L}{1 - \eta} \).

- Calculate Carnot Efficiency

The efficiency can also be found using the work output and heat input: \( \eta = \frac{W_{net}}{Q_H} = \frac{40}{56.67} = 0.705 \).

- Calculate Minimum Temperature

Substitute the values into the rearranged Carnot efficiency formula: \( T_H = \frac{280}{1 - 0.705} \approx 949 \mathrm{~K} \).

Key concepts

Carnot Cycle

The Carnot cycle is an idealized thermodynamic cycle proposed by Nicolas Léonard Sadi Carnot. It offers a benchmark for the maximum possible efficiency that a heat engine can achieve. The cycle consists of four stages:

• Isothermal expansion
• Adiabatic expansion
• Isothermal compression
• Adiabatic compression

During the isothermal stages, the system exchanges heat with the surroundings while maintaining constant temperature. Adiabatic stages involve no heat transfer, with temperature changes occurring due to work done on or by the system. By using only these reversible processes, the Carnot cycle ensures no entropy is generated, making it theoretically the most efficient cycle. The mathematical representation of Carnot efficiency is given by
\( \text{Efficiency} = 1 - \frac{T_L}{T_H} \)
Where \( T_L \) is the temperature of the cold reservoir and \( T_H \) is the temperature of the hot reservoir. The closer the operating conditions are to these ideal stages, the closer a real engine's efficiency will approach that of the Carnot cycle.

First Law of Thermodynamics

The First Law of Thermodynamics, also known as the Law of Energy Conservation, asserts that energy cannot be created or destroyed, only transformed from one form to another. For a thermodynamic cycle, this law can be succinctly summarized as:
\( Q_{in} - Q_{out} = W_{net} \)
Here,

• \( Q_{in} \) is the heat energy supplied to the system
• \( Q_{out} \) is the heat energy rejected by the system
• \( W_{net} \) is the net work done by the system

In context of our power cycle problem, the energy supplied by heat to the system is partially converted to work (output), and the remainder is rejected as waste heat. Understanding how these quantities relate helps identify how the energy balance is maintained. With given rejection rate and work output, the First Law can be used to find the input heat energy.

Heat Transfer

Heat transfer, a central concept in thermodynamics, refers to the movement of thermal energy from a higher temperature region to a lower temperature one. There are three primary modes of heat transfer:

• Conduction: Occurs through direct contact between molecules.
• Convection: Involves the movement of fluid masses.
• Radiation: Transfer via electromagnetic waves.

In thermodynamic cycles, heat is transferred between the working fluid of the system and the external reservoirs.
In our problem, the power cycle rejects 1000 kJ per minute to the cold reservoir (280 K). The conversion from kJ/min to kW (division by 60) helps in analyzing the system using power (kW), which simplifies calculations. Heat transfer governs the efficiency of the cycle and directly influences the high-temperature reservoir required for given parameters.