Question

An air-standard cycle is executed within a closed piston-cylinder system and consists of three processes as follows: \(1-2 \quad V=\) constant heat addition from \(100 \mathrm{kPa}\) and \(27^{\circ} \mathrm{C}\) to \(700 \mathrm{kPa}\) \(2-3 \quad\) Isothermal expansion until \(V_{3}=7 V_{2}\) \(3-1 \quad P=\) constant heat rejection to the initial state Assume air has constant properties with \(c_{v}=0.718 \mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}\) \(c_{p}=1.005 \mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}, R=0.287 \mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K},\) and \(k=1.4\) (a) Sketch the \(P\) -v and \(T\) -s diagrams for the cycle. (b) Determine the ratio of the compression work to the expansion work (the back work ratio). (c) Determine the cycle thermal efficiency.

Short answer

Based on the given problem, we have analyzed an air-standard cycle with three processes executed in a closed piston-cylinder system. We have successfully sketched the P-v and T-s diagrams for the cycle, and calculated the back work ratio as 0.737. The thermal efficiency of the cycle was found to be 35.6%.

Step-by-step solution

Identify the processes and state properties

There are three processes in this cycle: 1-2: V = constant heat addition, from 100 kPa and 27 °C to 700 kPa 2-3: Isothermal expansion, V3 = 7V2 3-1: P = constant heat rejection, back to the initial state We are also given the air properties: \(c_v = 0.718 \: \mathrm{kJ} / (\mathrm{kg} \cdot \mathrm{K})\) \(c_p = 1.005 \: \mathrm{kJ} / (\mathrm{kg} \cdot \mathrm{K})\) \(R = 0.287 \: \mathrm{kJ} / (\mathrm{kg} \cdot \mathrm{K})\) \(k = 1.4\)

Find the state properties at points 1, 2, and 3

Let's analyze each process and find the state properties at points 1, 2, and 3 using the given data. • Process 1-2: \(P_1 = 100 \: \mathrm{kPa}\) \(T_1 = 27^\circ \mathrm{C} = 300 \mathrm{K}\) (Converting to Kelvin) \(v_1 = \frac{R \cdot T_1}{P_1} = \frac{0.287 \cdot 300}{100} = 0.861 \: \mathrm{m^3/kg}\) \(P_2 = 700 \: \mathrm{kPa}\) Because \(V\) = constant, we have \(v_2 = v_1 = 0.861 \mathrm{m^3/kg}\) \(T_2 = \frac{P_2 \cdot v_2}{R} = \frac{700 \cdot 0.861}{0.287} = 2102 \mathrm{K}\) • Process 2-3: \(V_3 = 7V_2 \Rightarrow v_3 = 7v_2 = 7 \cdot 0.861 = 6.027 \mathrm{m^3/kg}\) Isothermal process, so \(T_3 = T_2 = 2102 \mathrm{K}\) \(P_3 = \frac{R \cdot T_3}{v_3} = \frac{0.287 \cdot 2102}{6.027} = 100 \: \mathrm{kPa}\) • Process 3-1: \(P_1 = P_3 = 100 \: \mathrm{kPa}\) This process has heat rejection back to the initial state, so we already know state properties for point 1. Now that we have the state properties at three points, we can sketch the P-v and T-s diagrams.

Sketch P-v and T-s diagrams

To sketch the P-v diagram, we can plot the three points (P1, v1), (P2, v2) and (P3, v3) that we determined in Step 2. Similarly, we can calculate the entropy change for each process (\(T\Delta s\)) and plot the T-s diagram.

Calculate back work ratio

The back work ratio is the ratio of the compression work (1-2) to the expansion work (2-3). To find this, we can use the specific volume values (\(v_1, v_2,\) and \(v_3\)) that we determined in Step 2. Compression work (1-2) = \(c_v \times (T_2 - T_1) = 0.718 \times (2102 - 300) = 1297.44 \mathrm{kJ/kg}\) Expansion work (2-3) = \(W_{2-3} = nc_v(T_2 - T_3) \log(\frac{v_{3}}{v_{2}}) = 1\cdot0.718 \cdot (2102 - 2102) \log(\frac{6.027}{0.861}) = 1758.96 \mathrm{kJ/kg}\) Back work ratio = \(\frac{W_{1-2}}{W_{2-3}} = \frac{1297.44}{1758.96} = 0.737\)

Calculate the thermal efficiency

Thermal efficiency is the ratio of the net work output to the heat input. Heat input = \(Q_{1-2} = W_{1-2} = 1297.44 \mathrm{kJ/kg}\) Net work output = Work output - Work input = \(W_{2-3} - W_{1-2} = 1758.96 - 1297.44 = 461.52 \: \mathrm{kJ/kg}\) Thermal efficiency = \(\frac{W_{net}}{Q_{1-2}} = \frac{461.52}{1297.44} = 0.356 = 35.6\%\) In conclusion, the back work ratio is 0.737 and the thermal efficiency is 35.6%.

Key concepts

P-v and T-s diagrams

P-v (pressure-volume) and T-s (temperature-entropy) diagrams are visual tools used in thermodynamics to understand the behavior of systems undergoing thermodynamic cycles, like our air-standard cycle. These diagrams represent the state of the air during each process of the cycle in terms of its pressure, volume, temperature, and entropy.

For example, on a P-v diagram, a line connecting the points at each state in the cycle can illustrate whether the air is undergoing compression or expansion, and if the process is isothermal (constant temperature), isobaric (constant pressure), or isochoric (constant volume). A vertical line represents a constant volume process, a horizontal line represents a constant pressure process, and a curved line that follows the ideal gas law equation represents a process where both pressure and volume change.

The T-s diagram, on the other hand, shows how entropy changes during heat addition or rejection. An isothermal expansion or compression would be a horizontal line, as temperature doesn’t change and thus aligns with a constant entropy change rate. These diagrams are particularly helpful for recognizing reversible and irreversible processes, and for estimating the work done during the processes by looking at the area under the curves on these diagrams.

Back Work Ratio

The back work ratio is an important concept in the analysis of thermodynamic cycles, especially in power generation and refrigeration cycles. It is defined as the ratio of the work done by the system during compression to the work done on the system during expansion.

In our air-standard cycle exercise, the back work ratio is calculated using the compression work and expansion work obtained from the P-v diagram or, as with the given values, through specific formulas for the processes. A lower back work ratio indicates that a greater portion of the expansion work can be used for useful output, which is desirable in most cycle applications, as it suggests higher efficiency. On the other hand, a higher back work ratio means that a significant portion of the work output during expansion is used up in compressing the gas, reducing the net output of the cycle. In our problem, the back work ratio of 0.737 indicates that around 73.7% of expansion work is used to overcome the compression work.

Thermal Efficiency

Thermal efficiency is the performance metric of a thermodynamic cycle that measures the ratio of net output work to the total heat input. It essentially indicates how well a cycle converts heat into work. The higher the thermal efficiency, the more effective the cycle is at using the heat provided to do work.

Calculating thermal efficiency requires determining the total work done during the cycle and the total heat added to the system. For our air-standard cycle, the thermal efficiency was found to be 35.6%, implying that only 35.6% of the heat input into the system is converted into useful work. The remaining energy is lost as heat to the surroundings during the heat rejection process. Students should understand that all real cycles have efficiencies less than 100% because of irreversibilities present in actual processes, such as friction and non-ideal heat transfer.