Question

A heat engine cycle is executed with steam in the saturation dome between the pressure limits of 7 and \(2 \mathrm{MPa}\). If heat is supplied to the heat engine at a rate of \(150 \mathrm{kJ} / \mathrm{s}\), the maximum power output of this heat engine is \((a) 8.1 \mathrm{kW}\) (b) \(19.7 \mathrm{kW}\) \((c) 38.6 \mathrm{kW}\) \((d) 107 \mathrm{kW}\) \((e) 130 \mathrm{kW}\)

Short answer

Answer: The maximum power output of the heat engine is approximately 19.7 kW.

Step-by-step solution

State the Clausius Inequality to Find Maximum Power Output

The Clausius inequality states that: \(\int_{\text{cycle}}\frac{\delta Q}{T} \le 0\). The equality holds for a reversible process, and since we want to find the maximum power output, we'll consider a reversible heat engine. For a reversible cycle, the Clausius theorem implies that \(\int_{\text{cycle}}\frac{\delta Q}{T} = 0\).

Calculate the Change in Entropy

For a reversible cycle, the change in entropy is zero, which means: \(\Delta S = \int_{1}^{2}\frac{\delta Q_{H}}{T_H} + \int_{1}^{2}\frac{\delta Q_{L}}{T_L} = 0\), where \(\delta Q_{H}\) and \(\delta Q_{L}\) are the heat transfer at the high and low temperatures, \(T_H\) and \(T_L\) respectively.

Obtain the Reversible Work Output

For a reversible cycle, the work output is equal to the difference in heat transfer at high and low temperatures: \(W = \delta Q_{H} - \delta Q_{L} = \delta Q_{H} (1 - \frac{T_L}{T_H})\).

Determine the Efficiency

The efficiency of a reversible heat engine is given by \(\eta = \frac{W}{\delta Q_H} = 1 - \frac{T_L}{T_H}\).

Apply the Carnot Limit

The temperature ratio of the steam within the saturation dome is \(T_H = \frac{T_{sat(7MPa)}}{T_{sat(2MPa)}}\) where \(T_{sat(7MPa)}\) and \(T_{sat(2MPa)}\) are the saturation temperatures corresponding to the 7 MPa and 2 MPa pressure limits. Using the steam tables, we get: \(T_{sat(7MPa)} = 285.830^{\circ}\mathrm{C} = 558.98\,\mathrm{K}\) \(T_{sat(2MPa)} = 212.377^{\circ}\mathrm{C} = 485.527\,\mathrm{K}\) Hence, we have \(T_L = 485.527\,\mathrm{K}\) and \(T_H = 558.98\,\mathrm{K}\).

Calculate the Efficiency

Using the previously obtained efficiency expression: \(\eta = 1 - \frac{T_L}{T_H} = 1 - \frac{485.527}{558.98} = 0.1314\). The efficiency of the heat engine is \(13.14\%\).

Determine the Maximum Power Output

Now, we can determine the maximum power output of the heat engine by multiplying the efficiency by the rate of heat transfer: \(P = \eta \times \frac{\delta Q}{\delta t} = 0.1314 \times 150\,\mathrm{kJ/s} = 19.71\,\mathrm{kW}\). Therefore, the maximum power output of the heat engine is approximately \(19.7\,\mathrm{kW}\). The correct answer is (b) \(19.7\,\mathrm{kW}\).

Key concepts

Clausius Inequality

The Clausius inequality is a fundamental concept in thermodynamics, especially relating to the second law. It's expressed as \(\int_{\text{cycle}}\frac{\delta Q}{T} \le 0\) for any thermodynamic cycle a system undergoes. In simple terms, it stipulates that for any spontaneous process, the total entropy of the system and surroundings together will not decrease. In the context of heat engines, equality holds for reversible processes, which means that there are no inefficiencies or internal friction within the system.

In practical terms, when we consider a heat engine, we often think about its efficiency and power output. The Clausius inequality helps us understand the best possible scenario for a heat engine—a reversible process. Such an ideal system could, theoretically, convert heat into work with maximum efficiency. However, in the real world, all processes have some degree of irreversibility, leading to the generation of entropy, or disorder, and a consequent loss of potential work.

Entropy Change

Entropy can be thought of as a measure of disorder or randomness in a system, with the concept closely linked to the second law of thermodynamics. The change in entropy \(\Delta S\) during a process quantifies how much disorder is increased or decreased. For a reversible process within a closed system, the total entropy change is zero, which is represented by \(\Delta S = 0\).

It is crucial to understand that during a reversible heat engine cycle, such as the ideal Carnot cycle, the engine absorbs heat \(\delta Q_H\) at high temperature \(T_H\) and releases heat \(\delta Q_L\) at a lower temperature \(T_L\). The entropy change at each point in the cycle sums to zero since the increase in entropy from the heat addition at high temperature is exactly balanced by the decrease in entropy from the heat rejection at low temperature. This concept is essential because it sets the fundamental limit on the maximum work output and efficiency for any heat engine.

Carnot Efficiency

The Carnot efficiency \(\eta\) is the theoretical maximum efficiency that a heat engine can achieve when operating between two temperatures, and it's an essential benchmark for all real engines. The formula for Carnot efficiency is \(\eta = 1 - \frac{T_L}{T_H}\) where \(T_H\) and \(T_L\) are the absolute temperatures at which the heat is absorbed and rejected, respectively.

The remarkable insight behind the Carnot efficiency is that no engine operating between these two temperatures can be more efficient than a Carnot engine. This limit is based on the assumption of reversible processes and no creation of entropy. In the original exercise, the actual efficiency of the heat engine was obtained by applying this limit using the calculated saturation temperatures at the given pressures. The efficiency gives us a peek at what percentage of the heat supplied to the engine is converted into useful work. In reality, all engines have efficiency values below this ideal due to practical limitations and irreversibilities in processes.

Saturation Temperature

The saturation temperature of a substance is the temperature at which a phase change occurs at a given pressure. For water and steam—a common working fluid in heat engines—this is the temperature at which water boils into steam or steam condenses into water at a particular pressure. It is vital to heat engine cycles, especially when dealing with steam, as it defines the upper and lower temperature limits for processes such as steam generators and condensers.

In the heat engine cycle problem, the working fluid, steam, operates within the saturation dome characterized by the pressure and temperature limits of \(2\mathrm{MPa}\) and \(7\mathrm{MPa}\). Using steam tables, one can determine the corresponding saturation temperatures for these pressures. By knowing the saturation temperatures, you can then calculate the temperature ratio that is necessary for finding the Carnot efficiency of the heat engine. As pressure increases, so does the saturation temperature, which in turn impacts the efficiency and power output of the heat engine.