Question

An air-standard cycle with variable specific heats is executed in a closed system and is composed of the following four processes: \(1-2 \quad\) Isentropic compression from 100 kPa and \(22^{\circ} \mathrm{C}\) to \(600 \mathrm{kPa}\) 2-3 \(\quad v=\) constant heat addition to \(1500 \mathrm{K}\) \(3-4 \quad\) Isentropic expansion to \(100 \mathrm{kPa}\) 4-1 \(P=\) constant heat rejection to initial state (a) Show the cycle on \(P\) -v and \(T\) -s diagrams. (b) Calculate the net work output per unit mass. (c) Determine the thermal efficiency.

Short answer

Question: Draw the closed system air-standard cycle on P-v and T-s diagrams, and then calculate the net work output per unit mass and the thermal efficiency of the cycle. The cycle consists of the following processes: Isentropic compression from State 1 (100 kPa, \(22^\circ\mathrm{C}\)) to State 2 (600 kPa), constant volume heat addition from State 2 to State 3 (\(1500 \mathrm{K}\)), isentropic expansion from State 3 to State 4 (100 kPa), and constant pressure heat rejection from State 4 to State 1 (initial state).

Step-by-step solution

Identify the given parameters and the process

We have a closed system undergoing a cycle, comprised of the following processes: 1. Isentropic compression: From State 1 (100 kPa, \(22^{\circ} \mathrm{C}\)) to State 2 (600 kPa). 2. Constant volume heat addition: From State 2 to State 3 (\(1500 \mathrm{K}\)). 3. Isentropic expansion: From State 3 to State 4 (100 kPa). 4. Constant pressure heat rejection: From State 4 to State 1 (initial state).

Determine properties at each state

To determine the properties at each state, we need to utilize the Ideal Gas Law and the Isentropic Relationships. State 1: Given properties are: Pressure: \(P_1 = 100 \mathrm{kPa},\) Temperature: \(T_1 = 22^{\circ} \mathrm{C} \Rightarrow 22 + 273.15 = 295.15 \mathrm{K}\) Calculate the specific volume at State 1 using the Ideal Gas Law: \(v_1 = \frac{R T_1}{P_1}\) State 2: Given property is: Pressure: \(P_2 = 600 \mathrm{kPa}\) We know that the compression from State 1 to State 2 is isentropic, so: \(\frac{P_2 v_2^{\gamma}}{P_1 v_1^{\gamma}} = 1\) State 3: Given property is: Temperature: \(T_3 = 1500 \mathrm{K}\) Since the heat addition is at constant volume, \(v_3 = v_2\) Calculate \(P_3\) using Ideal Gas Law: \(P_3 = \frac{R T_3}{v_3}\) State 4: Given property is: Pressure: \(P_4 = 100 \mathrm{kPa}\) The expansion from State 3 to State 4 is isentropic, so: \(\frac{P_4 v_4^{\gamma}}{P_3 v_3^{\gamma}} = 1\)

Calculate the Net Work Output per Unit Mass

For isentropic compression (1-2) and expansion (3-4), the work can be calculated as: \(W_{1-2} = \frac{R(T_1 - T_2)}{1-\gamma}\) \(W_{3-4} = \frac{R(T_3 - T_4)}{1-\gamma}\) For constant volume heat addition (2-3), the work is zero. For constant pressure heat rejection (4-1), the work can be calculated as: \(W_{4-1} = P_1(v_1 - v_4)\) The net work output per unit mass is the sum of these work values: \(W_{net} = W_{1-2} + W_{3-4} + W_{4-1}\)

Determine the Thermal Efficiency

The thermal efficiency can be calculated as the ratio of the net work output to the heat input: \(\eta_{th} = \frac{W_{net}}{Q_{input}}\) The heat input (Q_input) is the heat added during the constant volume heat addition (2-3), which can be calculated as: \(Q_{input} = C_v (T_3-T_2)\) Now we can calculate the thermal efficiency: \(\eta_{th} = \frac{W_{net}}{Q_{input}}\) Follow these steps to determine the properties at each state, calculate the net work output per unit mass, and determine the thermal efficiency for the given air-standard cycle.