Question

Air at \(80 \mathrm{kPa}, 27^{\circ} \mathrm{C},\) and \(220 \mathrm{m} / \mathrm{s}\) enters a diffuser at a rate of \(2.5 \mathrm{kg} / \mathrm{s}\) and leaves at \(42^{\circ} \mathrm{C}\). The exit area of the diffuser is \(400 \mathrm{cm}^{2}\). The air is estimated to lose heat at a rate of \(18 \mathrm{kJ} / \mathrm{s}\) during this process. Determine \((a)\) the exit velocity and \((b)\) the exit pressure of the air.

Short answer

Question: The air enters a diffuser with a mass flow rate of 0.5 kg/s, a temperature of 15°C, a pressure of 100 kPa, and a velocity of 180 m/s. The exit area of the diffuser is 0.25 m², and the exit temperature is 25°C. Determine the exit velocity and exit pressure of the air, assuming there is no work transfer and that the air can be modeled as an ideal gas. Answer: To determine the exit velocity and exit pressure of the air, follow these steps: 1. Convert the inlet and outlet air temperatures to Kelvin: T1= 288.15 K, T2 = 298.15 K. 2. Use the ideal gas law to find the specific volume at the inlet and outlet of the diffuser. 3. Utilize the mass flow rate and specific volume at the exit of the diffuser to find the exit velocity, V2. 4. Calculate the enthalpy at the inlet and outlet using the specific heat at constant pressure, cp, and the temperature difference. 5. Substitute the known values of q, m, V1, V2, and h1 into the energy conservation equation and solve for h2. 6. Calculate the exit pressure, P2, using the calculated h2 and specific volume at the exit of the diffuser. Using these steps, you can calculate the exit velocity and exit pressure of the air at the end of the diffuser.

Step-by-step solution

Mass Flow Rate Continuity Equation

The mass flow rate is constant throughout the diffuser. Thus, we can use the mass conservation equation with the given mass flow rate and the exit area of the diffuser to find the exit velocity. The mass conservation equation can be expressed as: $$ \rho_{1} A_{1} V_{1} = \rho_{2} A_{2} V_{2} $$

Specific Volume and Ideal Gas Law

Since we are dealing with air, we can assume it behaves as an ideal gas. To find the specific volume, we use the ideal gas equation: $$ \rho = \frac{P}{RT} $$ Where P is the pressure, R is the specific gas constant, and T is the temperature in Kelvin. Remember to convert the Celsius temperatures to Kelvin by adding 273.15.

Energy Conservation Equation for Air

Using the specific internal energy of air denoted by u and the enthalpy h, the energy conservation equation can be written as: $$ q + m \cdot (h_1 + \frac{V_1^2}{2}) = m \cdot (h_2 + \frac{V_2^2}{2}) + w $$ Where m is the mass flow rate, q is the heat transfer rate, and w is the work rate. In a diffuser, there is no work transfer, so w = 0.

Solve for Exit Velocity

Now, we will use the mass flow rate, the ideal gas law, and the energy conservation equation to solve for the exit velocity, V2: 1. Convert both inlet and outlet air temperatures to Kelvin by adding 273.15. 2. Use the ideal gas law to find the specific volume at the inlet and outlet of the diffuser. 3. Using the mass flow rate and specific volume at the exit of the diffuser, find the exit velocity, V2.

Solve for Exit Pressure

Similarly, we will find the exit pressure, P2, by using the energy conservation equation and the specific volume calculated earlier: 1. Calculate the enthalpy at the inlet and outlet (assumed to be incompressible) using specific heat at constant pressure, cp, and the temperature difference. 2. Substitute the known values of q, m, V1, V2, and h1 into the energy conservation equation. 3. Solve for h2. 4. Calculate the exit pressure, P2, using the calculated h2 and specific volume at the exit of the diffuser. Following these steps will provide you with the exit velocity and exit pressure of the air at the end of the diffuser.