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Consider a non-boiling gas-liquid two-phase flow in a 102-mm diameter tube, where the superficial gas velocity is one-third that of the liquid. If the densities of the gas and liquid are rg 5 8.5 kg/m3 and rl 5 855 kg/m3, respectively, determine the flow quality and the mass flow rates of the gas and the liquid when the gas superficial velocity is 0.8 m/s.

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DayyanKhan

Answer:

The answer for this question can be calculated using the equations attached along with the answer easily.

Explanation:

Ver imagen DayyanKhan

Answer:

mass flow rate of gas = 0.39 kg/s

mass flow rate of liquid = 115.8 kg/s

x = 0.0034 or 0.34 %

Explanation:

Outlining the parameters

d = 102 mm = 0.102m, r = d/2 = 0.051m, A = πr^2 = 0.00824m^2,

Vg = 1/3Vl = 0.8 m/s, V(liquid) = 3 * V(gas) = 2.4 m/s, rho(gas) = 58.5 kg/m3,

rho(liquid) = 5855 kg/m3

Applying the formula mass flow rate = superficial velocity * density * Area

Mathematically, m = V * rho * A

For gas phase

m(gas) = V(gas) * rho(gas) * A  

m(gas) = 0.8 * 58.5 * 0.00824

m(gas) = 0.39 kg/s

For liquid phase

m(liquid) = V(liquid) * rho(liquid) * A

m(liquid) = 2.4 * 5855 * 0.00824

m(liquid) = 115.8 kg/s

Applying the formula flow quality = mass flow rate of gas ÷ (mass flow rate of gas + mass flow rate of liquid)

Mathematically,  

x = [tex]\frac{m(gas))}{m(gas) + m(liquid)}[/tex]

x = [tex]\frac{0.39}{0.39 + 115.8}[/tex]

x = 0.0034 or 0.34 %

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