Question

An air-standard cycle is executed in a closed system with \(0.5 \mathrm{kg}\) of air and consists of the following three processes: \(1-2 \quad\) Isentropic compression from 100 kPa and \(27^{\circ} \mathrm{C}\) to \(1 \mathrm{MPa}\) \(2-3 \quad P=\) constant heat addition in the amount of \(416 \mathrm{kJ}\) \(3-1 \quad P=c_{1} v+c_{2}\) heat rejection to initial state \(c_{1}+c_{2}\) constants) (a) Show the cycle on \(P\) -v and \(T\) -s diagrams. (b) Calculate the heat rejected. (c) Determine the thermal efficiency. Assume constant specific heats at room temperature.

Short answer

Based on the given air-standard cycle, determine the following: (a) Show the cycle on P-v and T-s diagrams. (b) Calculate the heat rejected. (c) Calculate the thermal efficiency.

Step-by-step solution

STEP 1: Identify the initial state

State 1 is given in the problem statement: pressure \(P_1 = 100\,kPa\), temperature \(T_1 = 27^{\circ} C\), and mass of air \(m = 0.5\, kg\). Using the ideal gas law, \(PV=mRT\), we can find the specific volume \(v_1\) at state 1: \(v_1 = \frac{RT_1}{P_1} = \frac{0.287\,\frac{kJ}{kg\,K} \times (27+273.15)\,K}{100\,kPa} = 0.8147\,\frac{m^3}{kg}\).

STEP 2: Find state properties after isentropic compression

In process 1-2, the air undergoes isentropic compression from 100 kPa to 1 MPa. Using the isentropic relation \(\frac{P_2}{P_1}=\left(\frac{v_1}{v_2}\right)^k\), we can find the specific volume \(v_2\) at state 2: \(v_2=\frac{v_1}{\left(\frac{P_2}{P_1}\right)^{\frac{1}{k}}} = \frac{0.8147}{\left(\frac{1000}{100}\right)^{\frac{1}{1.4}}} = 0.1691\,\frac{m^3}{kg}\). Now we can find the temperature \(T_2\) at state 2 using the ideal gas law: \(T_2 = \frac{P_2v_2}{R} = \frac{1000\,kPa \times 0.1691\,\frac{m^3}{kg}}{0.287\,\frac{kJ}{kg\,K}} = 591.5\, K = 318.35^{\circ} C\)

STEP 3: Find state properties after constant pressure heat addition

In process 2-3, the heat addition is constant at \(Q_{23} = 416\, kJ\). Using the specific heat at constant pressure, \(C_p\), we can find the temperature change \(\Delta T_{23}\): \(\Delta T_{23} = \frac{Q_{23}}{mC_p} = \frac{416}{0.5 \times 1.005} = 824.88\, K\). Now we can find the temperature \(T_3\) at state 3: \(T_3 = T_2 + \Delta T_{23} = 591.5\,K + 824.88\,K = 1416.38\, K = 1143.23^{\circ}\,C\). Since the process is constant pressure, we have \(P_3 = P_2 = 1\,MPa\). Using the ideal gas law, we can find the specific volume \(v_3\) at state 3: \(v_3 = \frac{RT_3}{P_3} = \frac{0.287\,\frac{kJ}{kg\,K} \times 1416.38\,K}{1000\,kPa} = 0.4074\,\frac{m^3}{kg}\).

STEP 4: Find state properties after heat rejection

In process 3-1, the heat is rejected to return the air to its initial state (state 1). We already know the specific volume and pressure at the initial state, \(v_1\) and \(P_1\).

STEP 5: Plot P-v and T-s diagrams

With the states and cycle processes defined, we can now plot the P-v and T-s diagrams, showing isentropic compression (1-2), constant pressure heat addition (2-3), and heat rejection following \(P=c_1 v + c_2\) (3-1).

STEP 6: Calculate heat rejected

Let's denote the heat rejected as \(Q_{31}\). We know that the heat rejected is during process 3-1, and \(Q_{31} = mC_v (T_3 - T_1)\): \(Q_{31} = mC_v (T_3 - T_1) = 0.5 \times 0.717 \times (1143.23 - 27) = 400.41\,kJ\).

STEP 7: Determine thermal efficiency

The thermal efficiency of the cycle, \(\eta_C\), is defined as the ratio of the work output to the heat input: \(\eta_C = \frac{Q_{out} - Q_{in}}{Q_{in}}\). In this cycle, the work output is given by the difference in heat input and heat rejected, so the thermal efficiency is: \(\eta_C = \frac{Q_{23} - Q_{31}}{Q_{23}} = \frac{416 - 400.41}{416} = 0.0375 = 3.75\%\). The solution to the exercise is: (a) The P-v and T-s diagrams are plotted with the data calculated for each state and process. (b) The heat rejected, \(Q_{31}\), is \(400.41\,kJ\). (c) The thermal efficiency, \(\eta_C\), is \(3.75\%\).

Key concepts

Isentropic Compression

Understanding isentropic compression is crucial when analyzing air-standard cycles. This process, which is adiabatic and reversible, involves compressing a gas without an exchange of heat with its surroundings, which is indicated by the fact that entropy—a measure of disorder—remains constant.

During isentropic compression, the pressure of the gas increases significantly while the volume decreases. This aligns with the ideal gas law which relates pressure, temperature, and volume. For air, regarded as an ideal gas in this context, we use specific heat ratios to understand the changes rigorously. Mathematically, the relationship between the final and initial pressures and volumes is given by the formula \(\frac{P_2}{P_1}=\left(\frac{v_1}{v_2}\right)^k\), where \(k\) is the specific heat ratio. This principle helps in determining various properties of the air during the compression stage of the cycle, such as temperature and volume at different states.

Constant Pressure Heat Addition

Following isentropic compression, we often explore constant pressure heat addition in an air-standard cycle. This particular process involves adding heat to the system while maintaining a stable pressure—which is typical in the combustion phase of an engine's operation.

During this stage, the temperature of the air rises due to the energy added into the system, while the pressure remains unchanged. The temperature increase is found using the specific heat at constant pressure \(C_p\) and the amount of heat added \(Q_{23}\), through the equation \(\Delta T_{23} = \frac{Q_{23}}{mC_p}\). Knowing the initial temperature, we apply the increase to find the final temperature after heat addition. Importantly, the constant pressure heat addition is what primarily contributes to the work that can be extracted from the engine cycle and is fundamental to enhancing the cycle's efficiency.

Thermal Efficiency

The concept of thermal efficiency is pivotal in the assessment of any thermodynamic cycle since it indicates the portion of heat input that is converted into useful work. For an air-standard cycle, thermal efficiency can be derived from the ratio of the net work done during the cycle to the heat added to the system.

To calculate it, we use the equation \(\eta_C = \frac{Q_{23} - Q_{31}}{Q_{23}}\), where \(Q_{23}\) is the heat added to the system and \(Q_{31}\) is the heat rejected to close the cycle. This helps us understand how much of the energy put into the cycle is used effectively and how much is lost. Typically, a higher thermal efficiency indicates a more effective engine or cycle. In practice, thermal efficiencies are often far less than 100%, due to real-world factors like friction and incomplete combustion, reflecting the challenges in converting heat to work. The concept of thermal efficiency not only governs the economic aspects of power generation but also guides improvements in design and operation to minimize energy waste.