A tank has the shape of a right circular cylinder with a radius of 4 feet and a height of 6 feet. if the tank is full, find the work required to empty the tank by pumping the water over the top of the tank. use 62.4 as the specific weight of water
+37084 From a similar posted problem posted here:
http://math.stackexchange.com/questions/1185952/how-much-work-is-done-in-pumping-water-out-over-the-top-edge-in-order-to-empty
Specific weight of water (given) = 62.4 lbf/ft^3
The force required to move the water will be the specific weight x the volume
Weight = mg = F = 62.4 x pi x r^2 x Δy = 3136.57 Δy where y = height
Work is F x d or 3136.57 (6-y) Δy where y varies from the top of the tank (6) to the bottom (o)
so integrate this with respect to Δy from 0 to 6
\(\int_{0}^{6}\) 3136.57 (6-y) Δy = |6==>0 18819.42y - 1568.29 y^2 = 56458.08 j
(IF I did this correctly..There are no promises given......I hope someone can 2nd it !)
+37084 Sorry.....wrong units in final answer ...should be ft-lb instead of j
Was dinkin' around with this.....and thought maybe since we are pumping from the top (6ft) to the bottom....we could just use the average of 3 ft to calculate work:
Total volume of water = pi x r^2 x 6 = 301.592 ft^3
Total weight of water = 301.592 x 62.4 = 18.819.396 lb
Total work to move the water 3 ft = 18,819.396 x 3 = 56458.189 ft-lb Which basically agrees with the previous answer
