Question

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.

Short answer

Based on the given information about a two-phase flow of air and engine oil in a tube, we determined the mass flow rate of air to be approximately \(1.894 \times 10^{-3} \ \mathrm{kg/s}\). Furthermore, the calculated superficial velocities of air and engine oil are approximately \(5.029 \times 10^{-3} \ \mathrm{m/s}\) and \(2.037 \times 10^{-3} \ \mathrm{m/s}\), respectively.

Step-by-step solution

Determine the mass flow rate of the mixture

To determine the mass flow rate of the mixture, we need to use the flow quality given by $$ x = \frac{\dot{m}_\text{air}}{\dot{m}_\text{mixture}} $$ where \(x\) is the flow quality (dimensionless), \(\dot{m}_\text{air}\) is the mass flow rate of air (\(\mathrm{~kg/s}\)), and \(\dot{m}_\text{mixture}\) is the mass flow rate of the mixture (\(\mathrm{~kg/s}\)). We are given the mass flow rate of engine oil \(\dot{m}_\text{oil} = 0.9 \mathrm{~kg/s}\) and flow quality \(x = 2.1 \times 10^{-3}\), so we can find the mass flow rate of the mixture as $$ \dot{m}_\text{mixture} = \frac{\dot{m}_\text{oil}}{1 - x} $$ Substituting the given values, we get $$ \dot{m}_\text{mixture} = \frac{0.9}{1 - 2.1 \times 10^{-3}} $$ Calculating the mass flow rate of the mixture: $$ \dot{m}_\text{mixture} \approx \ 0.9019 \ \mathrm{~kg/s} $$

Determine the mass flow rate of air

Using the mass flow rate of the mixture obtained in step 1, we can find the mass flow rate of air as $$ \dot{m}_\text{air} = x \times \dot{m}_\text{mixture} $$ Substituting the values, we get $$ \dot{m}_\text{air} = 2.1 \times 10^{-3} \times 0.9019 $$ Calculating the mass flow rate of air: $$ \dot{m}_\text{air} \approx \ 1.89399 \times 10^{-3} \ \mathrm{~kg/s} $$

Determine the superficial velocities of air and engine oil

Superficial velocity is defined as the velocity of a phase if it was flowing alone in the tube. We can find the superficial velocities of air (\(v_\text{air}\)) and engine oil (\(v_\text{oil}\)) using the mass flow rates and the cross-sectional area of the tube, as $$ v_\text{air} = \frac{\dot{m}_\text{air}}{\rho_\text{air} \times A} \quad \text{and} \quad v_\text{oil} = \frac{\dot{m}_\text{oil}}{\rho_\text{oil} \times A} $$ where \(\rho_\text{air}\) and \(\rho_\text{oil}\) are the air and engine oil densities (assumed constant) and \(A\) is the cross-sectional area of the tube. The tube has a diameter of \(25 \ \mathrm{mm} = 0.025 \ \mathrm{m}\), so the area of the tube is $$ A = \pi \times \left(\frac{0.025}{2}\right)^2 $$ Calculating the area of the tube: $$ A \approx \ 4.91 \times 10^{-4} \ \mathrm{m^2} $$ For air, using the Ideal Gas Law, we can estimate the density at the given temperature (\(140 ^\circ C \ = 413.15 \ \mathrm{K}\)) and assuming standard atmospheric pressure: $$ \rho_\text{air} = \frac{P}{R_\text{air} \times T_\text{air}} $$ Using the specific gas constant for air, \(R_\text{air} = 287 \ \mathrm{J/(kg\,K)}\), and atmospheric pressure \(P \approx 101325 \ \mathrm{Pa}\), we can estimate the density of air: $$ \rho_\text{air} \approx \ \frac{101325}{287 \times 413.15} \approx 0.859 \ \mathrm{kg/m^3} $$ For engine oil, we can use a typical density value of \(\rho_\text{oil} = 900 \ \mathrm{kg/m^3}\). Using these density values and the mass flow rates obtained in previous steps, we can find the superficial velocities for air and engine oil: $$ v_\text{air} = \frac{1.89399 \times 10^{-3}}{0.859 \times 4.91 \times 10^{-4}} \quad \text{and} \quad v_\text{oil} = \frac{0.9}{900 \times 4.91 \times 10^{-4}} $$ Calculating the superficial velocities: $$ v_\text{air} \approx \ 5.029 \times 10^{-3} \ \mathrm{m/s} \quad \text{and} \quad v_\text{oil} \approx \ 2.037 \times 10^{-3} \ \mathrm{m/s} $$ Thus, the mass flow rate of air is approximately \(1.894 \times 10^{-3} \ \mathrm{kg/s}\), and the superficial velocities of air and engine oil are approximately \(5.029 \times 10^{-3} \ \mathrm{m/s}\) and \(2.037 \times 10^{-3} \ \mathrm{m/s}\), respectively.