Question

a rectangular water tank 10' wide 10' long and 6' tall is buried underground so that its top is 4' below ground level. assume water weighs 62.4 pounds per foot cube and the tank is full of water. how much work does it take to pump out 2/3 of the water in the tank to a height 2' above ground level?

Answer 1

liven length of the water tank (d)= 10' = 10 foot e = 10 x 0.3048 m l= 3.048 m width of the tank (W) = 10x0.3048m = 3.048 m the tank t = 1.219m below the 6 6x 0.3048 m d=1.8288 m depth of 1 2foot 14.post ground level to foot 16 foot 12/3x6 = 40 6 foot y = io foot 10 foot 10.500 16foot 11 늦 R - y to foot know that Since we work (W) force X distance so el. @ al = foy dy (here dy shows that distance is changing with redusing water level in tank)

Now where м 3 of displaced water J = gravitational constant = 9.8 mys2 M- mass ma density x volume given density (D)= 62.4 pounds per foot cabe 0.454 kg (0.304873 m² = 1000.45 kg /mnd D = 62.4x

and volume of displaced water v= 4 x lexio Na 1.2192 mx 3.048m x 3.048m V= 11:3267 m3 $0 1000 - 45 X 11:3267 kg 11326.74 kg . so N 11326 74 x 9.8 I'll x 105 N E = Sa work done Epoot from to displace / water to to foot-3.048m 1.8288 m 3.048 (1.Uxlos y dy 1.8288 J 3.048 trolixlos W = yo 니 18288 1:11x105 W = 9.29 – 3.3447 2 W = 1.11 X 105 X 5.94 I 2 W = 3.297 x 105 Joules so work done to displace water is 3297X16 J.