Question

Argon gas enters an adiabatic compressor at \(120 \mathrm{kPa}\) and \(30^{\circ} \mathrm{C}\) with a velocity of \(20 \mathrm{m} / \mathrm{s}\) and exits at \(1.2 \mathrm{MPa}\) \(530^{\circ} \mathrm{C},\) and \(80 \mathrm{m} / \mathrm{s}\). The inlet area of the compressor is \(130 \mathrm{cm}^{2} .\) Assuming the surroundings to be at \(25^{\circ} \mathrm{C}\), determine the reversible power input and exergy destroyed.

Short answer

The main goal of the exercise is to determine the reversible power input and exergy destroyed during the process.

Step-by-step solution

Find the mass flow rate of argon gas

First, we need to find the mass flow rate of the argon gas entering the compressor. To do this, we can use the equation: \(\dot{m} = \rho AV\), where \(\rho\) is the density of argon gas, \(A\) is the area of the inlet, and \(V\) is the velocity of the gas. However, we know that \(PV=mRT\). Therefore, \(\rho = \frac{m}{V}=\frac{P}{RT}\), where \(P\) is the pressure, \(R\) is the specific gas constant for argon (\(R_{Ar} = 0.2081 \,kJ/kg \cdot K\)), and \(T\) is the temperature in Kelvin. Given the pressure, temperature, and velocity at the inlet, firstly we can find \(\rho_{1}\), then calculate the mass flow rate \(\dot{m}\).

Apply the First Law of Thermodynamics

Next, we will apply the First Law of Thermodynamics to the compressor to find the work input. The equation for an adiabatic process is: \(\Delta h + \frac{1}{2}\Delta (V^2) = w\), where \(\Delta h\) is the change in specific enthalpy of argon, \(\Delta (V^2)\) is the change in the velocity squared, and \(w\) is the work input. Since the process is adiabatic, there is no heat transfer involved. Substituting the given values, we can find the work input.

Apply the Second Law of Thermodynamics

Now we will apply the Second Law of Thermodynamics to find the exergy destroyed. The exergy balance equation for an adiabatic compressor is: \(\dot{m}(\Delta e_x) = -\dot{W} - T_0\dot{m}(\Delta s)\), where \(\Delta e_x\) is the change in specific exergy, \(\dot{W}\) is the power input, and \(\Delta s\) is the change in specific entropy. To solve for the exergy destroyed, we need to calculate the change in specific exergy. This requires knowing the specific enthalpy and entropy changes as well as the temperature and pressure changes. The exergy change is given by the equation: \(\Delta e_x = \Delta h - T_0\Delta s + \frac{1}{2}\Delta(V^2)\) Using the known values, we can calculate the exergy destroyed during the process.