Question

An air-standard cycle with variable specific heats is executed in a closed system and is composed of the following four processes: \(1-2 \quad v=\) constant heat addition from 14.7 psia and \(80^{\circ} \mathrm{F}\) in the amount of \(300 \mathrm{Btu} / \mathrm{lbm}\) 2-3 \(P=\) constant heat addition to \(3200 \mathrm{R}\) \(3-4 \quad\) Isentropic expansion to 14.7 psia 4-1 \(P=\) constant heat rejection to initial state (a) Show the cycle on \(P\) -V and \(T\) -s diagrams. (b) Calculate the total heat input per unit mass. (c) Determine the thermal efficiency.

Short answer

Q: Explain the given air-standard cycle including its processes. The given air-standard cycle consists of four processes: 1. Constant volume heat addition from the initial state (14.7 psia and \(80^{\circ} \mathrm{F}\)) with an energy input of 300 Btu/lbm. 2. Constant pressure heat addition up to a temperature of 3200 R. 3. Isentropic expansion (back to the initial pressure of 14.7 psia). 4. Constant pressure heat rejection returning to the initial state.

Step-by-step solution

Understand the Given Air-Standard Cycle

Analyze the given cycle, consisting of four processes: 1. Process 1-2: Constant volume heat addition from the initial state (14.7 psia and \(80^{\circ} \mathrm{F}\)) with an energy input of 300 Btu/lbm. 2. Process 2-3: Constant pressure heat addition up to a temperature of 3200 R. 3. Process 3-4: Isentropic expansion (back to the initial pressure of 14.7 psia). 4. Process 4-1: Constant pressure heat rejection returning to the initial state.

Sketch the P-V and T-s Diagrams for the Given Cycle

Draw the P-V and T-s diagrams for the given air-standard cycle, labeling the states (1, 2, 3, and 4) and the processes (1-2, 2-3, 3-4, and 4-1). The P-V diagram will show a constant volume process (1-2), followed by a constant pressure process (2-3), then an isentropic expansion (3-4), and finally a constant pressure process returning to the initial state (4-1). The T-s diagram will show an increase in temperature without a change in entropy (1-2), followed by a constant pressure heat addition (2-3), then an isentropic expansion (3-4), and finally a constant pressure heat rejection returning to the initial state (4-1).

Calculate the Initial State and Properties

Calculate the enthalpy and specific volume at the initial state (1). The temperature at state 1, \(T_1=80^{\circ} \mathrm{F}+460=540\,\mathrm{R}\). Assuming air to be an ideal gas, we can use the ideal gas law to calculate the specific volume \(v_1\): \(P_1v_1 = R T_1\), where \(R=287\,\mathrm{J}/\mathrm{(kg\cdot K)}\) for air, \(R=\frac{1545\,\mathrm{ft\cdot lb_f}}{\mathrm{(lb_m\cdot R)}}\), and \(1\,\mathrm{psia}=6895\,\mathrm{Pa}\). Lastly, use the enthalpy-temperature relationship to get the enthalpy and specific volume at state 1.

Calculate the Properties at State 2

Determine the enthalpy at state 2 using the given heat input of 300 Btu/lbm (1 Btu/lbm = 2.326 kJ/kg): \(h_2=h_1+300\,\mathrm{Btu/lbm}\). Since the process is constant volume, \(v_2=v_1\). Then, use the enthalpy and specific volume to calculate the temperature and pressure at state 2.

Calculate the Properties at State 3

Using the given temperature at state 3, \(T_3=3200\,\mathrm{R}\), and the constant pressure process, \(P_3=P_2\), calculate the specific volume \(v_3\) using the ideal gas law. Then, determine the enthalpy at state 3 using the enthalpy-temperature relationship.

Calculate the Properties at State 4

Because process 3-4 is isentropic, the entropy at state 4 is equal to the entropy at state 3, \(s_4=s_3\). Using the pressure at state 4, \(P_4=P_1=14.7\,\mathrm{psia}\), the entropy, and the ideal gas assumption, calculate the temperature, enthalpy, and specific volume at state 4.

Determine Total Heat Input per Unit Mass

Calculate the total heat input \(Q_{in}\) per unit mass using the enthalpy changes in processes 1-2 and 2-3: \(Q_{in}= (h_2-h_1) + (h_3-h_2)\).

Determine Work Output per Unit Mass

Calculate the work output \(W_{out}\) per unit mass using the enthalpy change in processes 3-4: \(W_{out} = (h_3-h_4)\).

Calculate the Thermal Efficiency

Determine the thermal efficiency (\(\eta\)) of the cycle using the calculated heat input and work output values: \(\eta = \frac{W_{out}}{Q_{in}}\). Having followed the steps, we have shown the cycle on P-V and T-s diagrams, calculated the total heat input per unit mass, and determined the thermal efficiency.