The radius of the hemisphere is 5ft, the curved portion of the hemisphere is on the ground so that it opens upward. How much work is required to pump water out of the top of the tank. Weight of water is 62.5lb/ft^3.
2007-03-25 08:32:18 · 3 answers · asked by sprintdawg007 3 in Science & Mathematics ➔ Engineering
Work = Force x Distance
Force = weight of liquid in the tank = Volume of hemisphere (cu.ft.) x 62.5 Lb per cubic foot = 3.1416 x diameter cubed/12 x 62.5 = 16,362.5 Lb.
Distance = distance in feet from the bottom of the tank to the center of gravity of the tank = 2.88feet
Work = 16, 362.5 Lb x 2.88 Ft = 47,122.56 Ft-Lb
This does not take into account pressure losses in the piping and pump efficiencies.
2007-03-25 15:48:33 · answer #1 · answered by gatorbait 7 · 0⤊ 0⤋
The evaporation enthalpy for liquid water at 20 C is 2454,a million KJ/kg. it is the ability required to first warmth the water to the boiling element, then make it evaporate. 500 m^3 = 500 000 kg water we desire 500 000*=a million 227 050 MJ 1227050/250=4908,2 seconds. or 80 one,803 minutes.
2016-10-19 21:42:29 · answer #2 · answered by ? 4 · 0⤊ 0⤋
ever hear of a siphon???? Pump a column up to the top of the tank (weight * height = work) and let the atmosphere do the work draining the rest of the tank
2007-03-25 08:37:29 · answer #3 · answered by Anonymous · 0⤊ 1⤋