Question

Air is compressed steadily by a compressor from \(100 \mathrm{kPa}\) and \(20^{\circ} \mathrm{C}\) to \(1200 \mathrm{kPa}\) and \(300^{\circ} \mathrm{C}\) at a rate of \(0.4 \mathrm{kg} / \mathrm{s} .\) The compressor is intentionally cooled by utilizing fins on the surface of the compressor and heat is lost from the compressor at a rate of \(15 \mathrm{kW}\) to the surroundings at \(20^{\circ} \mathrm{C}\) Using constant specific heats at room temperature, determine \((a)\) the power input to the compressor, \((b)\) the isothermal efficiency, and ( \(c\) ) the entropy generation during this process.

Short answer

Given the initial and final conditions of the air in a compressor, the mass flow rate, and the rate of heat loss, we have calculated the power input to the compressor to be \(102.3\,\text{kW}\), the isothermal efficiency to be \(38\%\), and the entropy generation during this process to be \(27.46\,\text{J/K.s}\).

Step-by-step solution

Write down the given information

We are given the following: - Initial pressure: \(P_1 = 100 \,\text{kPa}\) - Initial temperature: \(T_1 = 20^{\circ}\text{C}\) - Final pressure: \(P_2 = 1200 \,\text{kPa}\) - Final temperature: \(T_2 = 300^{\circ}\text{C}\) - Mass flow rate: \(\dot{m} = 0.4 \,\text{kg/s}\) - Heat loss rate: \(Q_{out} = 15 \,\text{kW}\) - Room temperature: \(T_0 = 20^{\circ}\text{C}\) At room temperature, the specific heats for air are: - \(c_p = 1005 \,\text{J/kg.K}\) - \(c_v = 717.5 \,\text{J/kg.K}\)

Calculate the power input to the compressor

To find the power input to the compressor, we can use the first law of thermodynamics for a steady flow process: $$\dot{W}_\text{in} = \dot{m} ((h_2 - h_1) - Q_{out})$$ For an ideal gas, the enthalpy change is related to the temperature change: $$\Delta h = h_2 - h_1 = c_p (T_2 - T_1)$$ Plugging this into the power input equation and solving for \(\dot{W}_\text{in}\): $$\dot{W}_\text{in} = \dot{m} (c_p (T_2 - T_1) - Q_{out})$$ $$\dot{W}_\text{in} = 0.4 \,\text{kg/s} (1005 \,\text{J/kg.K} (300\,^\circ\text{C} - 20\,^\circ\text{C}) - 15\,000 \,\text{W})$$ $$\dot{W}_\text{in} \approx 102300 \,\text{W} = 102.3 \,\text{kW}$$ So the power input to the compressor is \(102.3 \,\text{kW}\).

Calculate the isothermal efficiency

To find the isothermal efficiency, we can use the formula: $$\eta_\text{isoth} = \frac{\dot{W}_\text{in,ideal}}{\dot{W}_\text{in}}$$ For an ideal isothermal process, the work done is: $$\dot{W}_\text{in,ideal} = \dot{m} R T_1 \ln \left(\frac{P_2}{P_1}\right)$$ Substituting the values and calculating the ideal power input: $$\dot{W}_\text{in,ideal} = 0.4 \,\text{kg/s} (287 \,\text{J/kg.K} (293.15 \,\text{K})) \ln \left(\frac{1200\, \text{kPa}}{100\, \text{kPa}}\right)$$ $$\dot{W}_\text{in,ideal} \approx 38892 \,\text{W} = 38.89\,\text{kW}$$ Now, we can find the isothermal efficiency: $$\eta_\text{isoth} = \frac{38.89 \,\text{kW}}{102.3 \,\text{kW}} \approx 0.38 = 38\%$$ The isothermal efficiency of the compressor is \(38\%\).

Calculate the entropy generation

To find the entropy generation, we can use the entropy equation: $$\Delta S_\text{system} = \dot{m}(s_2 - s_1) - \frac{Q_{out}}{T_0}$$ For an ideal gas, the entropy change is related to the temperature and pressure change: $$\Delta S = c_p \ln \left(\frac{T_2}{T_1}\right) - R \ln \left(\frac{P_2}{P_1}\right)$$ Plugging this into the entropy generation equation and solving for \(\Delta S_\text{system}\): $$\Delta S_\text{system} = 0.4 \,\text{kg/s} (1005 \,\text{J/kg.K} \ln \left(\frac{573.15 \,\text{K}}{293.15 \,\text{K}}\right) - 287 \,\text{J/kg.K} \ln \left(\frac{1200\, \text{kPa}}{100\, \text{kPa}}\right) - \frac{15\,000 \,\text{W}}{293.15 \,\text{K}})$$ $$\Delta S_\text{system} \approx 27.46 \,\text{J/K.s}$$ The entropy generation during this process is \(27.46 \,\text{J/K.s}\).

Key concepts

First Law of Thermodynamics

Understanding the First Law of Thermodynamics is essential when exploring energy transfer within systems like compressors. This law, often described as the law of energy conservation, states that energy cannot be created or destroyed in an isolated system. Instead, energy can only be transformed from one form to another.

In the context of our compressor exercise, the first law can be represented as an equation that equates the power input to the change in enthalpy of the air minus the heat lost to the surroundings. Mathematically, we express this as
\[\begin{equation}\text{Power input} (\text{W}_{\text{in}}) = \text{mass flow rate} (\dot{m}) \times (\text{change in enthalpy} (\text{h}_2 - \text{h}_1) - \text{heat loss to surrounds} (Q_{\text{out}}))\end{equation}\].
Enthalpy represents the total heat content of the air and is a key factor in understanding how energy is being used in compressing the air within the system. The solution to the problem illustrates how to apply this law by calculating the actual power input required to achieve the desired increase in air pressure and temperature, considering the cooling effect of the compressor fins.

Isothermal Efficiency

Isothermal efficiency is a metric used to evaluate the performance of compressors and other thermodynamic systems that work with the compression and expansion of gases. It is defined as the ratio of the work input for an ideal isothermal process to the actual work input required by the system. An isothermal process is one that occurs at a constant temperature, where the system's temperature doesn't change despite work being done on it or by it.

In our exercise, isothermal efficiency is given by\[\begin{equation}\eta_{\text{isoth}} = \frac{\text{Work input for an ideal isothermal process}}{\text{Actual Power input}}\end{equation}\].
The ideal isothermal work input is lower than the actual power input because real-life scenarios involve additional factors like friction and heat loss, which are not present in an ideal scenario. Thus, the isothermal efficiency offers insight into how close the system's performance is to an ideal model and where improvements may be made for better energy use. The calculated isothermal efficiency of the compressor indicates how effectively the system is performing against an ideal standard, providing a practical measure of system performance.

Entropy Generation

Entropy is a fundamental concept in thermodynamics that deals with the disorder or randomness within a system. In thermodynamic processes, entropy can be generated due to irreversible actions like friction, mixing, chemical reactions, or heat transfer across a finite temperature difference.

The calculation of entropy generation helps us assess the irreversibilities in a process. The entropy generation within the compressor is found using the relationship\[\begin{equation}\Delta S_{\text{system}} = \dot{m}(s_2 - s_1) - \frac{Q_{\text{out}}}{T_0}\end{equation}\],
where \(s_2 - s_1\) represents the change in the specific entropy of the air, \(Q_{\text{out}}\) is the rate of heat loss, and \(T_0\) is the surrounding temperature. This expression considers the transmission of energy due to heat flow and how it impacts the entropy, or disorder, of the system.

Understanding the entropy generation in a process like air compression allows us to grasp where the process deviates from ideal behavior. With higher entropy generation, the process is less efficient as more waste (like dissipated heat) is generated. Therefore, in our textbook example, identifying the entropy generation clarifies the nature and magnitude of inefficiencies in the compressor's operation.