Answer:
a. 71884.8 π lb/ft-s²∫₀⁵(9 - y)dy
b. 23961.6 π lb/ft-s²∫₀⁵(5 - y)dy
c. 99840π lb/ft-s²∫₀⁶rdr
Step-by-step explanation:
.(a) Write down an integral for the work needed to pump all of the water to a point 4 feet above the tank.
The work done, W = ∫mgdy where m = mass of cylindrical tank = ρA([5 + 4] - y) where ρ = density of water = 62.4 lb/ft³, A = area of base of tank = πd²/4 where d = diameter of tank = 12 ft.( we add height of the tank + the height of point above the tank and subtract it from the vertical point above the base of the tank, y to get 5 + 4 - y) and g = acceleration due to gravity = 32 ft/s²
So,
W = ∫mgdy
W = ∫ρA([5 + 4] - y)gdy
W = ∫ρA(9 - y)gdy
W = ρgA∫(9 - y)dy
W = ρgπd²/4∫(9 - y)dy
we integrate W from y from 0 to 5 which is the height of the tank
W = ρgπd²/4∫₀⁵(9 - y)dy
substituting the values of the other variables into the equation, we have
W = 62.4 lb/ft³π(12 ft)² (32 ft/s²)/4∫₀⁵(9 - y)dy
W = 71884.8 π lb/ft-s²∫₀⁵(9 - y)dy
.(b) Write down an integral for the fluid force on the side of the tank
Since force, F = ∫PdA where P = pressure = ρgh where h = (5 - y) since we are moving from h = 0 to h = 5. So, P = ρg(5 - y)
The differential area on the side of the tank is given by
dA = 2πrdy
So. F = ∫PdA
F = ∫ρg(5 - y)2πrdy
Since we are integrating from y = 0 to y = 5, we have our integral as
F = ∫ρg2πr(5 - y)dy
F = ∫ρgπd(5 - y)dy since d = 2r
substituting the values of the other variables into the equation, we have
F = ∫₀⁵62.4 lb/ft³π(12 ft) × 32 ft/s²(5 - y)dy
F = 23961.6 π lb/ft-s²∫₀⁵(5 - y)dy
.(c) How would your answer to part (a) change if the tank was on its side
The work done, W = ∫mgdr where m = mass of cylindrical tank = ρAh where ρ = density of water = 62.4 lb/ft³, A = curved surface area of cylindrical tank = 2πrh where r = radius of tank, d = diameter of tank = 12 ft. and h = height of the tank = 5 ft and g = acceleration due to gravity = 32 ft/s²
So,
W = ∫mgdr
W = ∫ρAhgdr
W = ∫ρ(2πrh)hgdr
W = ∫2ρπrh²gdr
W = 2ρπh²g∫rdr
we integrate from r = 0 to r = d/2 where d = diameter of cylindrical tank = 12 ft/2 = 6 ft
So,
W = 2ρπh²g∫₀⁶rdr
substituting the values of the other variables into the equation, we have
W = 2 × 62.4 lb/ft³π(5 ft)² × 32 ft/s²∫₀⁶rdr
W = 99840π lb/ft-s²∫₀⁶rdr
