Question

A piston-cylinder device contains \(0.15 \mathrm{kg}\) of air initially at \(2 \mathrm{MPa}\) and \(350^{\circ} \mathrm{C}\). The air is first expanded isothermally to \(500 \mathrm{kPa},\) then compressed polytropically with a polytropic exponent of 1.2 to the initial pressure, and finally compressed at the constant pressure to the initial state. Determine the boundary work for each process and the net work of the cycle.

Short answer

The boundary works for each process are as follows: - Isothermal expansion (process 1-2): \(W_{12} = 2925.64\, \mathrm{J}\) - Polytropic compression (process 2-3): \(W_{23} = -2495.94\, \mathrm{J}\) - Constant-pressure compression (process 3-1): \(W_{31} = -2898\, \mathrm{J}\) The net work of the cycle is approximately \(-1468.3\, \mathrm{J}\).

Step-by-step solution

Calculate the initial and final volumes

We'll start by calculating the initial and final volumes for each process. To do this, we'll use the ideal gas law: \(PV = m\cdot R\cdot T\) For the initial state (state 1): \(P_1 = 2\,\mathrm{MPa} = 2\times10^6\,\mathrm{Pa}\) \(T_1 = 350^{\circ}\mathrm{C} = 350 + 273.15 = 623.15\,\mathrm{K}\) \(m = 0.15\,\mathrm{kg}\) \(R = \frac{287\,\mathrm{J/(kg\cdot K)}}{1\,\mathrm{kg}}\) (specific gas constant for air) Now we can find the initial volume \(V_1\): \(V_1 = \frac{m\cdot R\cdot T_1}{P_1} = \frac{0.15\cdot287\cdot623.15}{2\times10^6} = 0.01347\,\mathrm{m^3}\)

Calculate volume for the final state of each process

Next, we need to compute the final volume for each process. For process 1-2 (isothermal expansion): \(P_2 = 500\,\mathrm{kPa} = 500\times10^3\,\mathrm{Pa}\) (given) Since it's an isothermal process, \(T_2 = T_1 = 623.15\,\mathrm{K}\) Now we can find the final volume for process 1-2, \(V_2\): \(V_2 = \frac{m\cdot R\cdot T_2}{P_2} = \frac{0.15\cdot287\cdot623.15}{500\times10^3} = 0.05388\,\mathrm{m^3}\) For process 2-3 (polytropic compression): \(P_3 = P_1 = 2\times10^6\,\mathrm{Pa}\) (given) We need to find the temperature at state 3 (\(T_3\)) and the final volume \(V_3\). We know that \(PV^n = \mathrm{constant}\) for a polytropic process. \(\left(\frac{P_2}{P_1}\right) = \left(\frac{V_1}{V_2}\right)^{1.2}\) Solving for \(V_3\), we have: \(V_3 = V_1\left(\frac{P_2}{P_1}\right)^{\frac{1}{1.2}} = 0.01347\left(\frac{500\times10^3}{2\times10^6}\right)^{\frac{1}{1.2}} = 0.02793\,\mathrm{m^3}\) For process 3-1 (constant-pressure compression): Since it's a constant-pressure process, \(V_1 = V_1\)

Calculate the work for each process

Now we can find the boundary work for each process: For process 1-2 (isothermal expansion): \(W_{12} = m\cdot R\cdot T_1 \cdot \ln\frac{P_1}{P_2} = 0.15\cdot287\cdot623.15\cdot\ln\frac{2\times10^6}{500\times10^3} = 2925.64\,\mathrm{J}\) For process 2-3 (polytropic compression): We first find the temperature at state 3 (\(T_3\)) using the ideal gas law: \(T_3 = \frac{P_3\cdot V_3}{m\cdot R} = \frac{2\times10^6\cdot0.02793}{0.15\cdot287} = 1036.93\,\mathrm{K}\) Now we can find the work, \(W_{23}\): \(W_{23} = \frac{m\cdot R\cdot (T_3 - T_2)}{1 - 1.2} = \frac{0.15\cdot287\cdot(1036.93 - 623.15)}{1 - 1.2} = -2495.94\,\mathrm{J}\) For process 3-1 (constant-pressure compression): \(W_{31} = P\cdot (V_3 - V_1) = 2\times10^6\cdot (0.02793 - 0.01347) = -2898\,\mathrm{J}\)

Calculate the net work of the cycle

Finally, we can find the net work of the cycle by summing up the work done in each process: \(W_{net} = W_{12} + W_{23} + W_{31} = 2925.64 - 2495.94 - 2898 = -1468.3\,\mathrm{J}\) The net work of the cycle is approximately \(-1468.3\,\mathrm{J}\).

Key concepts

Piston-Cylinder Device

Imagine a typical piston-cylinder device as a cylinder with a movable piston on top, often used in thermodynamic problems to represent engines or compressors. The space inside the cylinder is filled with a gas. As the piston moves up and down, it alters the volume of the gas inside, doing what's known as 'boundary work' on the system.

Boundary work occurs when the force exerted by the gas on the piston results in a displacement, which is essentially the energy transfer due to volume change. It's a critical concept in thermodynamics since engines' performance and efficiency are linked to the work they do during expansion and compression cycles.

Isothermal Process

An isothermal process is characterized by a constant temperature throughout the entire operation, meaning the gas's temperature doesn't change as it expands or contracts. During an isothermal expansion, the internal energy of an ideal gas remains constant, and all the heat added to the system is used to do work on the surroundings.

For an ideal gas undergoing isothermal expansion, the boundary work can be calculated using the formula:
\( W = nRT \text{ln} \left(\frac{V_2}{V_1}\right) \)
Here, \(n\) is the number of moles, \(R\) the universal gas constant, and \(V_1\) and \(V_2\) the initial and final volumes, respectively. This formula can be derived from the ideal gas law and the first law of thermodynamics.

Polytropic Process

In contrast, a polytropic process involves a situation where the pressure and volume of a gas are related by a constant exponent, designated as 'n', in the equation \(PV^n = \text{constant}\). This exponent defines the process and can represent a range of processes including isothermal (\(n=1\)), adiabatic (\(n = \text{gamma, which is the heat capacity ratio}\)), and others.

The work done during a polytropic process can be found using the equation:
\( W = \frac{P_1V_1 - P_2V_2}{1 - n} \)
where \(P_1\) and \(V_1\) are the initial pressure and volume, and \(P_2\) and \(V_2\) are the final states. Polytopic processes often occur in real-world applications like compressing or expanding in a non-ideal way.

Ideal Gas Law

The ideal gas law is a fundamental equation in thermodynamics that relates the pressure \(P\), volume \(V\), and temperature \(T\) of an ideal gas. It can be written in the form:
\( PV = nRT \)
In this equation, \(n\) represents the number of moles of the gas, and \(R\) is the ideal, or universal, gas constant. An 'ideal gas' is a hypothetical gas that follows this law exactly, without any deviation.

In real-world scenarios, gases exhibit ideal behavior under low pressure and high temperature. The ideal gas law is crucial when calculating the initial and final states of a gas in processes like isothermal and polytropic, as seen in the exercise solution provided. It allows one to determine necessary properties such as volume or temperature when other state variables are known.