Question

An aircraft on the ground is to be cooled by a gas refrigeration cycle operating with air on an open cycle. Air enters the compressor at \(30^{\circ} \mathrm{C}\) and \(100 \mathrm{kPa}\) and is compressed to \(250 \mathrm{kPa}\). Air is cooled to \(70^{\circ} \mathrm{C}\) before it enters the turbine. Assuming both the turbine and the compressor to be isentropic, determine the temperature of the air leaving the turbine and entering the cabin.

Short answer

Answer: The temperature of the air leaving the turbine and entering the cabin is approximately 18°C.

Step-by-step solution

Determine the state at the compressor exit

Using the given data, we know that air enters the compressor at \(30^{\circ} \mathrm{C}\) and \(100 \mathrm{kPa}\). Let's denote this state as state 1. Since the compressor is isentropic, we can find the relationship between the temperature and pressure in an isentropic process: \(s_1 = s_2\) Using the ideal gas equation and assuming constant specific heats, we have: \(\frac{T_2}{T_1} = (\frac{P_2}{P_1})^{\frac{k-1}{k}}\) where \(T_1 = 30+273.15 = 303.15 \mathrm{K}\) and \(P_1 = 100 \mathrm{kPa}\), \(P_2 = 250 \mathrm{kPa}\), and the ratio of specific heats for air \(k=1.4\). Plugging in the given values and solving for \(T_2\), we get: \(T_2 = T_1 (\frac{P_2}{P_1})^{\frac{k-1}{k}}\) \(T_2 = 303.15 (\frac{250}{100})^{(\frac{1.4-1}{1.4})}\) \(T_2 = 354.57 \mathrm{K}\) So, at state 2, the temperature of the air leaving the compressor is \(354.57 \mathrm{K}\).

Determine the state at the turbine inlet

The air is then cooled before entering the turbine, and we are given that it is cooled to a temperature of \(70^{\circ} \mathrm{C}\). Therefore, the temperature of the air entering the turbine (state 3) is: \(T_3 = 70 + 273.15 = 343.15 \mathrm{K}\)

Determine the state at the turbine exit

Since the turbine is isentropic, we can again find the relation between the temperature and pressure for an isentropic process as follows: \(s_3 = s_4\) Again, using the ideal gas equation and assuming constant specific heats, we have: \(\frac{T_4}{T_3} = (\frac{P_4}{P_3})^{\frac{k-1}{k}}\) We know that the pressures at the inlet and outlet of the turbine are \(P_3 = 250 \mathrm{kPa}\) and \(P_4 = 100 \mathrm{kPa}\). Plugging in the given values and solving for \(T_4\), we get: \(T_4 = T_3 (\frac{P_4}{P_3})^{\frac{k-1}{k}}\) \(T_4 = 343.15 (\frac{100}{250})^{(\frac{1.4-1}{1.4})}\) \(T_4 = 291.15 \mathrm{K}\) Hence, the temperature of the air leaving the turbine and entering the cabin is \(291.15 \mathrm{K}\), or approximately \(18^{\circ} \mathrm{C}\).

Key concepts

Isentropic Process

Understanding an isentropic process is crucial when studying thermodynamics, especially in applications like gas refrigeration cycles used in aircraft cooling, as in our exercise. An isentropic process is a thermodynamic process that is both adiabatic and reversible. This means that the system does not exchange heat with its surroundings (\textbf{adiabatic}) and there are no internal inefficiencies within the system (\textbf{reversible}).

For an ideal gas undergoing an isentropic process, there's an essential relationship between the pressure and temperature of the gas that can be expressed using the variable k, the ratio of specific heats, sometimes denoted as \textbf{gamma}. The equation is known as the Poisson's equation:
\[\frac{T_2}{T_1} = \bigg(\frac{P_2}{P_1}\bigg)^{\frac{k-1}{k}}\]
where T represents the temperature and P the pressure, with subscripts 1 and 2 denoting initial and final states respectively. In the exercise, both the compressor and the turbine are assumed to be isentropic, allowing us to use this relationship to find the temperatures at the compressor exit and turbine exit.

Ideal Gas Equation

The ideal gas equation is a fundamental equation that serves as the backbone for many calculations in gas refrigeration cycles. It relates pressure, volume, and temperature of an ideal gas in the equation:\[PV = nRT\]
Here, P stands for pressure, V is volume, n represents the amount of substance in moles, R is the ideal gas constant, and T the temperature in Kelvin. In our exercise, we assume air behaves as an ideal gas, which is a good approximation under normal conditions.

Why does the ideal gas law matter in this context? It facilitates the calculation of changes in temperature or pressure during isentropic processes within the gas refrigeration cycle. When combined with the assumption of constant specific heats and accompanied by Poisson's equation, as in the solution to our exercise, it allows us to determine the final temperature of the air without needing to know its volume or the amount of substance.

Specific Heats

The term specific heats may not be intuitive, but it's a concept that affects many everyday processes, including how efficiently an aircraft is cooled. Specific heats refer to the amount of heat required to raise the temperature of a unit mass of a substance by one degree. There are two kinds of specific heats: at constant pressure (denoted as C_p) and at constant volume (denoted as C_v).

The ratio of these values is represented by the variable k or gamma, which we have encountered in the isentropic process equation. For air and several other gases under standard conditions, this ratio is approximately 1.4. Assuming constant specific heats simplifies the analysis of thermodynamic cycles, as we did in our exercise, by making it possible to directly calculate changes in temperature related to changes in pressure, through the use of the isentropic process equation.