The larger impurities and cloudy sediments in the water of an aquarium can be cl
ID: 1852141 • Letter: T
Question
The larger impurities and cloudy sediments in the water of an aquarium can be cleared by means of a filter and the water can be aerated by using an aerator, which is driven by an air pump, as shown in the figure. For filtering, the water in the aquarium must be recirculated through the filter at a volumetric flow rate of 0.4 times 10-4 m3/s. The velocity of the water-air mixture in the glass tube is 0.45 m/s. The cross-sectional areas of the glass tube and the plastic tube are 1 times 10-4 m2 and 0.2 times 10-4 m2, respectively. Assume that the densities of water, air and water-air mixture are 1000 kg/m3, 1.2 kg/m3 and 900 kg/m3, respectively. Determine the mass flow rate and velocity of the air at the exit of the air pump. The volumetric flow rate through the aerator is 3 times 10-4 m3/s.Explanation / Answer
Mass flow rate of water through glass tube, Mw = density of water * volumetric flow rate of water through glass tube Mw = 1000 * 0.4 * 10^-4 = 0.04 kg/s Now, Volumetric flow rate of (air + water mixture) through glass tube = area * velocity = 1 * 10^-4 * 0.45 = 0.45 * 10^-4 m^3/s Mass flow rate of (air + water mixture) through glass tube, Maw = density of (air + water mixture) * volumetric flow rate of (air + water mixture) through glass tube Maw = 900 * 0.45 * 10^-4 = 0.0405 kg/s Now, mass flow rate of air through the glass tube must be the difference of the two... So, Ma = 0.0405 - 0.0400 = 0.0005 kg/s Volumetric flow rate through aerator, V1 = 3 * 10^-4 m^3/s So, mass flow rate through aerator, M1 = density of air * volumetric flow rate through aerator M1 = 1.2 * 3 * 10^-4 =3.6 * 10^-4 kg / s So, total mass flow rate of air through the pump must be the sum of these two mass flow rates of air... M = M1 + Ma = 0.0005 + 0.00036 = 0.00086 kg/s Hence velocity = mass flow rate / density of air * area of plastic tube = 0.00086 / (1.2 * 0.2 * 10^-4) = 35.833 m/s Answers : Mass flow rate = 0.00086 kg/s Velocity = 35.833 m/s
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.