Question

Consider a non-boiling gas-liquid two-phase flow in a \(102-\mathrm{mm}\) diameter tube, where the superficial gas velocity is one-third that of the liquid. If the densities of the gas and liquid are $\rho_{g}=8.5 \mathrm{~kg} / \mathrm{m}^{3}\( and \)\rho_{l}=855 \mathrm{~kg} / \mathrm{m}^{3}$, respectively, determine the flow quality and the mass flow rates of the gas and the liquid when the gas superficial velocity is $0.8 \mathrm{~m} / \mathrm{s}$.

Short answer

Question: Determine the flow quality and the mass flow rates of the gas and liquid when the gas superficial velocity is \(0.8 \mathrm{~m} / \mathrm{s}\) in a non-boiling gas-liquid two-phase flow in a tube with a diameter of 102 mm, where the superficial gas velocity is one-third that of the liquid. The densities of the gas and the liquid are 8.5 kg/m³ and 855 kg/m³, respectively. Answer: The flow quality is 0.25, the mass flow rate of the gas is 0.0551 kg/s, and the mass flow rate of the liquid is 16.6 kg/s.

Step-by-step solution

Calculate Superficial Liquid Velocity

Given that the superficial gas velocity is one-third that of the liquid, let's denote the superficial liquid velocity as \(V_l\). Then, the given relationship is: \(V_g = \frac{1}{3}V_l\) We are given the superficial gas velocity \(V_g = 0.8 \mathrm{~m} / \mathrm{s}\). Now we can solve for \(V_l\): \(V_l = 3V_g = 3 ( 0.8 \mathrm{~m} / \mathrm{s}) = 2.4 \mathrm{~m} / \mathrm{s}\) So the superficial liquid velocity is \(2.4 \mathrm{~m} / \mathrm{s}\).

Calculate Flow Quality

The flow quality \(x\) is defined as the ratio of superficial gas velocity to the total superficial velocity. Let's denote the total superficial velocity as \(V_T\). Then, we can find \(x\) as follows: \(V_T = V_g + V_l = 0.8 \mathrm{~m} / \mathrm{s} + 2.4 \mathrm{~m} / \mathrm{s} = 3.2 \mathrm{~m} / \mathrm{s}\) \(x = \frac{V_g}{V_T} = \frac{0.8 \mathrm{~m} / \mathrm{s}}{3.2 \mathrm{~m} / \mathrm{s}} = 0.25\) So the flow quality is \(0.25\).

Calculate Mass Flow Rates

The mass flow rate of the gas and the liquid can be determined using their respective velocities and densities. First, let's calculate the cross-sectional area of the tube, using the diameter \(D = 102~\mathrm{mm}\): \(A = \pi (\frac{D}{2})^2 = \pi (\frac{102 \times 10^{-3} \mathrm{~m}}{2})^2 = 0.008159 \mathrm{~m}^2\) Now, we can calculate the mass flow rates as follows: \(\dot{m}_g = \rho_g V_g A = 8.5 \mathrm{~kg} / \mathrm{m}^{3} \cdot 0.8 \mathrm{~m} / \mathrm{s} \cdot 0.008159 \mathrm{~m}^2 = 0.0551 \mathrm{~kg} / \mathrm{s}\) \(\dot{m}_l = \rho_l V_l A = 855 \mathrm{~kg} / \mathrm{m}^{3} \cdot 2.4 \mathrm{~m} / \mathrm{s} \cdot 0.008159 \mathrm{~m}^2 = 16.6 \mathrm{~kg} / \mathrm{s}\) So, the mass flow rate of the gas is \(0.0551 \mathrm{~kg} / \mathrm{s}\), and the mass flow rate of the liquid is \(16.6 \mathrm{~kg} / \mathrm{s}\).