Question

Air is compressed steadily by an 8 -kW compressor from \(100 \mathrm{kPa}\) and \(17^{\circ} \mathrm{C}\) to \(600 \mathrm{kPa}\) and \(167^{\circ} \mathrm{C}\) at a rate of \(2.1 \mathrm{kg} / \mathrm{min} .\) Neglecting the changes in kinetic and potential energies, determine ( \(a\) ) the increase in the exergy of the air and ( \(b\) ) the rate of exergy destroyed during this process. Assume the surroundings to be at \(17^{\circ} \mathrm{C}\).

Short answer

Question: Calculate (a) the increase in the exergy of the air, and (b) the rate of exergy destroyed during the process in a steady-state air compressor. Solution: (a) The increase in exergy of the air is 1.086 kg/min. (b) The rate of exergy destroyed during the process is 5.720 kW.

Step-by-step solution

Write down the given information

We are given the following information: - Power input to the compressor, \(W_{in} = 8\,\text{kW}\) - Air flow rate, \(m = 2.1\,\text{kg/min}\) - Initial pressure, \(P_1 = 100\,\text{kPa}\) - Initial temperature, \(T_1 = 17^\circ\text{C} = 290\,\text{K}\) - Final pressure, \(P_2 = 600\,\text{kPa}\) - Final temperature, \(T_2 = 167^\circ\text{C} = 440\,\text{K}\) - Surrounding temperature, \(T_0 = 17^\circ\text{C} = 290\,\text{K}\)

Calculate change in exergy

To find the increase in the exergy of the air, we can use the exergy change equation for an ideal gas: $$\Delta ex = m\left(\frac{T_2 - T_1}{T_0}\right)$$ Plugging in the values, we get: $$\Delta ex = (2.1\,\text{kg/min})\left(\frac{440\,\text{K} - 290\,\text{K}}{290\,\text{K}}\right)$$ $$\Delta ex = 2.1\,\text{kg/min} \times \frac{150\,\text{K}}{290\,\text{K}}$$ $$\Delta ex = 1.086\,\text{kg/min}$$ Therefore, the increase in exergy of the air is \(1.086\,\text{kg/min}\).

Calculate rate of exergy destroyed

The rate of exergy destroyed can be found using the steady-state exergy balance equation: $$\dot E_{destroyed} = \dot W_{in} - \dot E_{out}$$ The exergy destruction rate can be determined by multiplying the exergy change by the mass flow rate: $$\dot E_{out} = m \times \Delta ex = 2.1\,\text{kg/min} \times 1.086\,\text{kg/min} = 2.280\,\text{kW}$$ Now, we can calculate the rate of exergy destroyed: $$\dot E_{destroyed} = \dot W_{in} - \dot E_{out} = 8\,\text{kW} - 2.280\,\text{kW} = 5.720\,\text{kW}$$ Therefore, the rate of exergy destroyed during this process is \(5.720\,\text{kW}\).

Key concepts

Thermodynamics

Thermodynamics is a branch of physics that deals with heat, work, temperature, and energy. In essence, thermodynamics looks at how energy is converted from one form to another and how it affects matter. The laws of thermodynamics govern the principles of energy conservation, energy quality, and entropy.

When studying thermodynamics, especially in relation to processes such as the compression of air in a compressor, we often talk about the first and second laws of thermodynamics. The first law relates to the conservation of energy, stating that energy cannot be created or destroyed, only transformed. The second law introduces the concept of entropy and exergy and indicates that the quality of energy can degrade over time, meaning that not all input energy can be used for work due to irreversibilities in a process.

Exergy analysis, which is common in thermodynamics, assesses the quality or 'usefulness' of energy in a system and how much work could potentially be extracted. It helps engineers and scientists determine the performance of thermal systems and the maximum possible efficiency.

Steady-State Exergy Balance

In thermodynamics, particularly for systems that are operating continuously, the concept of steady-state is crucial. A steady-state exergy balance implies that on a time-averaged basis, the exergy content within a control volume stays constant.

For a system in steady-state, the exergy entering and leaving the control volume will not change with time. However, within that control volume, exergy can be destroyed due to irreversibilities like friction, mixing, chemical reactions, and heat transfer with a temperature difference.

The steady-state exergy balance is formulated in a way that the rate of exergy destruction plus the rate of exergy transfer out of the system equals the rate of exergy transfer into the system. Mathematically, the steady-state exergy balance can be expressed as \[\dot{E}_{in} - \dot{E}_{out} = \dot{E}_{destroyed}\], where \(\dot{E}_{in}\) is the rate of exergy entering, \(\dot{E}_{out}\) is the rate of exergy leaving, and \(\dot{E}_{destroyed}\) is the rate at which exergy is being destroyed within the system.

This balance is crucial for the second part of the textbook problem, where students are asked to find the rate of exergy destroyed during the air compression process.

Ideal Gas Exergy Change

The concept of exergy change for an ideal gas is significant when analyzing energy conversion processes. For an ideal gas, the change in exergy is associated with changes in temperature, pressure, and volume, often resulting from a process such as compression or expansion.

To calculate the change in exergy, the formula \[\Delta ex = m\left(\frac{T_2 - T_1}{T_0}\right)\] is used, where \(m\) is the mass flow rate, \(T_1\) and \(T_2\) are the initial and final temperatures, and \(T_0\) is the reference temperature, generally the ambient temperature.

It's important to note that this formula assumes no external work is done on or by the gas other than what's inherent in the change of state. In the textbook problem, this change in exergy helps determine how much more useful energy the air has after compression compared to before, neglecting kinetic and potential energy changes. Understanding how to calculate ideal gas exergy change is crucial for students in solving these types of thermodynamic problems.