Problem 1) An aquarium 12 m long, 2 m wide, and 6 m deep is full of water. The d
ID: 2848487 • Letter: P
Question
Problem 1)
An aquarium 12 m long, 2 m wide, and 6m deep is full of water. The density of water is 1000 kg/m^3 and the force of gravity is 9.8 m/s^2 m9.8m
a) Find the work needed to pump all of the water out of the aquarium. Work =----------------------Joules.
b) Find the work needed to pump half of the water out of the aquarium.
Work = -------------------------Joules.
Problem 2)
A cylindrical swimming pool has a diameter of 26 ft, the sides are 8 ft high, and the depth of the water is 6ft. The density of water is 62.5 lb/ft^3
(a) How much work is required to pump all of the water over the side?
Work =-------------------ft-lb
(b) How much work is required to pump all of the water out of an outlet 3ft above the top?
Work =---------------------ft-lb
Problem 3)
A trough is 11m long, 5m wide, and 2m deep. The vertical cross-section of the trough parallel to an end is shaped like an isoceles triangle (with height 2m, and base, on top, of length 5m). The trough is full of water (density 1000 kg/m^3 and 9.8 ms^2 is the acceleration due to gravity).
a) Find the amount of work in joules required to empty the trough by pumping the water over the top
high.
g=9.8
Explanation / Answer
1)
a) work needed to pump all of the water out of the aquarium = Final potential energy - Initial potential energy
Final potential energy = Area*density*g*h^2 / 2 = 12*2*9.8*1000*6^2 / 2 = 4233.6 KJ
Initial potential energy = 0
work needed to pump all of the water out of the aquarium = 4233.6 KJ -0 = 4233.6 Kilo Joules = 4233600 Joules
b)
work needed to pump half of the water out of the aquarium = Area*density*g*{h^2 - ( h^2 /4)}/ 2 = 12*2*9.8*1000*{6^2 - (6^2 /4)} / 2 = 3175.2 Kilo Joules
work needed to pump half of the water out of the aquarium = 3175.2 Kilo Joules = 3175200 Joules
2)
a) work is required to pump all of the water over the side = Final potential energy - Initial potential energy
Initial potential energy = Area*density*g*h^2 / 2 = 3.14*(26^2 /4)*32.174*62.5*6^2 / 2 = 19207636.695 lb-ft
Final potnetial energy = Area*density*g*h * 8 = 3.14*(26^2 /4)*32.174*62.5*6*8 = 51220364.52 lb -ft
work is required to pump all of the water over the side = 51220364.52 - 19207636.695 = 32012727.825 lb-ft
b)
work is required to pump all of the water out of an outlet 3ft above the top
Final potnetial energy = Area*density*g*h * 8 = 3.14*(26^2 /4)*32.174*62.5*6*(8+3) = 70428001.215 lb -ft
work is required to pump all of the water over the side = 70428001.215 - 19207636.695 = 51220364.52 lb-ft
3)
work in joules required to empty the trough by pumping the water over the top
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