Question

Consider a non-boiling gas-liquid two-phase flow in a tube, where the ratio of the mass flow rate is \(\dot{m}_{l} / \dot{m}_{g}=300\). Determine the flow quality \((x)\) of this non-boiling two-phase flow.

Short answer

Answer: The flow quality (x) of this non-boiling gas-liquid two-phase flow is \(\frac{1}{301}\).

Step-by-step solution

Write down the given information

We are given the mass flow rate ratio, \(\dot{m}_{l} / \dot{m}_{g}=300\).

Write down the flow quality formula

The flow quality (x) is defined as \(x = \frac{\dot{m}_g}{\dot{m}_l + \dot{m}_g}\).

Express the mass flow rates in terms of the given ratio

Given ratio is, \(\frac{\dot{m}_{l}}{\dot{m}_{g}}=300\). Let's denote \(\dot{m}_{g}\) as \(m_g\) and \(\dot{m}_{l}\) as \(m_l\). Then we can rewrite the given ratio as \(\frac{m_l}{m_g}=300\). From this equation, we can say that \(m_l = 300m_g\).

Substitute the mass flow rates in the flow quality formula

Now, we can substitute \(m_l = 300m_g\) into the flow quality formula: \(x = \frac{m_g}{(300m_g) + m_g}\).

Solve for the flow quality

After substitution and simplification, we get \(x = \frac{m_g}{301m_g}\). As \(m_g\) is nonzero, we can cancel them, and we get \(x = \frac{1}{301}\). So, the flow quality (x) for this non-boiling two-phase flow is \(\frac{1}{301}\).