Question

A 10-ft diameter by 15-ft height vertical tank is receiving water (p = 62.1 lb/cu ft) at the rate of 300 gpm and is discharging through a 6-in ID line with a constant speed of 5 fps. At a given instant, the tank is half full. Find the water level and the mass change in the tank 15 min. later.

Answer 1

Volume of the vertical tank $= \frac{Diameter^2}{4} \times \pi \times height$

$= \frac{10^2}{4} \times 3.14 \times 15 = 1178.097 \;ft^3$

Rate of water receipt in the vertical tank = 300 gpm = 40.1042 ft³/min

Velocity of water coming out = 5 fps = 300 fpm

Rate of water being removed from the tank = (cross section area of pipe) $\times$ (velocity)

$= \frac{\pi (inner \; diametr)^2}{4} \times velosity \\ = \frac{\pi (0.5)^2}{4} \times 300 \\ = 58.904 \;ft^3/min$

Rate at which water level is reducing has to be determined.

Volume of water reduced per minute $= (58.904 -40.1042) \;ft^3/min$

$= \frac{18.8}{(\frac{\pi \times 10^2}{4})} = 0.23936 \;ft/min$

Reduction in level in 15 minutes $= 0.23936 \times 15 = 3.5905 \;ft$

Reduction in mass = (volume reduction) $\times$ density

$= 18.8 \times 62.1 = 1167.48\; lb/min$

Mass reduction in 15 minutes $= 1167.48 \times 15 = 17512.2 \;lb$

Level of water remaining in the tank after 15 min:

(7.5 feet – 3.5905 feet) = 3.9095 feet