Question

Consider a Carnot cycle executed in a closed system with \(0.6 \mathrm{kg}\) of air. The temperature limits of the cycle are 300 and \(1100 \mathrm{K},\) and the minimum and maximum pressures that occur during the cycle are 20 and 3000 kPa. Assuming constant specific heats, determine the net work output per cycle.

Short answer

Answer: The net work output per cycle of the Carnot engine is approximately 352.25 kJ.

Step-by-step solution

Write down the given information and properties of air at room temperature

We are given the following: - Mass of air: m = \(0.6\:kg\) - Temperature limits: T1 = 300 K, T2 = \(1100\:K\) - Pressure limits: P1 = \(20\:kPa\), P2 = \(3000\:kPa\) Properties of air at room temperature (approximately 300 K): - Constant volume specific heat (Cv): \(0.718\:kJ/(kg\:K)\) - Constant pressure specific heat (Cp): \(1.005\:kJ/(kg\:K)\) - Gas constant (R): \(0.287\:kJ/(kg\:K)\)

Calculate the Carnot efficiency

The Carnot efficiency (\(\eta_{Carnot}\)) is given by the formula: $$\eta_{Carnot} = 1 - \frac{T_{low}}{T_{high}}$$ Substitute the temperature limits into the formula: $$\eta_{Carnot} = 1 - \frac{300}{1100} = 0.7273$$

Determine the heat added during the process

During the isothermal process of the Carnot cycle, the heat added (Qin) can be calculated using the following equation: $$Q_{in} = m \cdot C_p \cdot (T_{high} - T_{low})$$ Substitute the given values: $$Q_{in} = 0.6 \cdot 1.005 \cdot (1100 - 300) = 484.53 \:kJ$$

Calculate the net work output

The net work output (Wout) is given by the product of Carnot efficiency and heat added: $$W_{out} = \eta_{Carnot} \cdot Q_{in}$$ Substitute the values obtained in the previous steps: $$W_{out} = 0.7273 \cdot 484.53 = 352.25 \:kJ$$ So, the net work output per cycle of the Carnot engine is approximately \(352.25\:kJ\).

Key concepts

Carnot Efficiency

Carnot efficiency is a fundamental concept in thermodynamics, representing the highest possible efficiency that any heat engine operating between two temperatures can achieve. The formula for Carnot efficiency is expressed as \(\eta_{Carnot} = 1 - \frac{T_{low}}{T_{high}}\), where \(T_{low}\) is the absolute temperature of the cold reservoir and \(T_{high}\) is the absolute temperature of the hot reservoir, both measured in Kelvin. This equation can be derived from the second law of thermodynamics, which implies no engine can be 100% efficient if it operates on a cycle.

For example, in the provided exercise, the Carnot efficiency calculation is critical to determining the net work output of a hypothetical engine. An engine with perfect insulation and no friction would reach the efficiency calculated by this formula. However, real-world engines will always have lower efficiencies due to various inefficiencies and losses. Understanding Carnot efficiency helps us gauge the maximum work we can extract from a heat engine and sets a benchmark for comparing actual engines against an idealized version.

Thermodynamics Closed System

A thermodynamics closed system refers to a physical system that does not allow mass to cross its boundaries, although energy in the form of heat and work can transfer into and out of the system. In classical thermodynamics, we often analyze closed systems to simplify the problem and apply the fundamental laws of conservation of mass and energy.

In the context of the Carnot cycle problem, we consider the system to be closed since the mass of air, \(0.6\:kg\), remains constant throughout the cycle. This assumption allows us to apply the first law of thermodynamics, which in a closed system can be simplified to \(\Delta U = Q - W\), with the change in internal energy \(\Delta U\) being solely dependent on the heat added \(Q\) and the work done \(W\) by the system. Working with a closed system ensures that our calculations for the Carnot efficiency and the net work output per cycle do not need to account for any mass entering or leaving the system.

Specific Heat Capacity

Specific heat capacity is a property of a material that indicates how much heat energy is required to raise the temperature of one kilogram of the material by one degree Kelvin or Celsius. In equations, it is commonly denoted as \(C\) with units of \(kJ/(kg\:K)\) or \(J/(g\:\degree C)\). Specific heat capacities can be measured at constant volume \(C_v\) and constant pressure \(C_p\), which are relevant for different thermodynamic processes.

Understanding the specific heat capacity is crucial for solving thermodynamic problems, like calculating the heat added or removed during a process, which can directly affect the energy transfer within a system. In the case of the textbook exercise, the use of constant pressure specific heat \(C_p\) aligns with the isothermal expansion process of the Carnot cycle, corresponding to the boundary work done by the system while maintaining constant pressure. This specification impacts the amount of heat input \(Q_{in}\) and consequently the work output \(W_{out}\) of the cycle.