A non-boiling two-phase flow of air and engine oil in a 25 -mm diameter tube has a bulk mean temperature of \(140^{\circ} \mathrm{C}\). If the flow quality is \(2.1 \times 10^{-3}\) and the mass flow rate of the engine oil is \(0.9 \mathrm{~kg} / \mathrm{s}\), determine the mass flow rate of air and the superficial velocities of air and engine oil.

To find the mass flow rate of the mixture, let's first define the flow quality (given as \(2.1 \times 10^{-3}\)): Flow quality \((x)\) = mass flow rate of air / mass flow rate of the mixture We are also given that the mass flow rate of the engine oil is \(0.9 \mathrm{~kg} / \mathrm{s}\). Let's define the mass flow rate of the air as m_air and the mass flow rate of the mixture as m_mix. Then, we can rearrange the equation and solve for m_mix: \(m_{mix} = \frac{m_{air}}{x}\)

Now, we can use the given flow quality and the mass flow rate of engine oil to find the mass flow rate of air: \(m_{air} = x * m_{mix}\) Substituting the mass flow rate of engine oil: \(m_{air} = 2.1 \times 10^{-3} * 0.9 = 0.00189 \mathrm{~kg} / \mathrm{s}\)

Using the mass flow rate of air, we can now find the mass flow rate of the mixture: \(m_{mix} = \frac{m_{air}}{x} = \frac{0.00189}{2.1 \times 10^{-3}} = 0.9 \mathrm{~kg} / \mathrm{s}\)

The superficial velocity for each phase can be calculated using the following formula: Superficial velocity = mass flow rate / (density * cross-sectional area of the tube) First, we need to find the cross-sectional area of the tube: \(A = \pi \left(\frac{D}{2}\right)^2 = \pi \left(\frac{25 \times 10^{-3}}{2}\right)^2 = 4.91 \times 10^{-4} \mathrm{~m}^2\) Now, we will need the densities of air and engine oil at \(140^{\circ} \mathrm{C}\). For this exercise, we will use the following approximate densities (in \(\mathrm{kg/m^3}\)): Density of air = 0.947 Density of engine oil = 845 Then, we can calculate the superficial velocities: Superficial velocity of air: \(v_{air} = \frac{m_{air}}{(\rho_{air} * A)} = \frac{0.00189}{(0.947 * 4.91 \times 10^{-4})} = 4.19 \mathrm{~m/s}\) Superficial velocity of engine oil: \(v_{oil} = \frac{m_{oil}}{(\rho_{oil} * A)} = \frac{0.9}{(845 * 4.91 \times 10^{-4})} = 2.17 \times 10^{-3} \mathrm{~m/s}\) So, the mass flow rate of air is approximately \(0.00189 \mathrm{~kg} / \mathrm{s}\), the superficial velocity of air is approximately \(4.19 \mathrm{~m/s}\), and the superficial velocity of engine oil is approximately \(2.17 \times 10^{-3} \mathrm{~m/s}\).