# Fluid Volume Calculation

Please see the attachment.

The end plates (isosceles triangles) of the trough shown below were designed to withstand a fluid force of 6000 lb. How many cubic feet can the tank hold without exceeding this limitation?

Assuming the density is 62.4 lb/ft^3, the maximum volume is ? ft^3.

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

The fluid force acting on each end plate is given by

F_0 = integral_S(P(y) dx dy),

where S is the surface the water makes with an end plate and P(y) is the water pressure at height y. We have

P(y) = rho(d-y), where rho = 62.4 lb/ft^3 is the density of water. Thus we have

F_0 = integral_S(rho(d-y) dx dy).

Since rho is constant, we can bring it outside the integral. Thus

F_0 = rho integral_S((d-y) dx dy).

The next thing to do is to evaluate the surface integral. We have |x(y)| <= 3y/7, so we may write it as

F_0 = rho integral_0^d(integral_{-3y/7}^{3y/7}((d-y) dx dy))

= rho integral_0^d(6y/7(d-y) dy)

= 6/7 rho integral_0^d((yd - y^2) dy)

= 6/7 rho (1/2 y^2 d - 1/3 y^3)_{y=0}^d

= 1/7 rho d^3.

Thus we have d = (7F_0/rho)^(1/3).

Now the total volume of water in the tank is given by V = 1/2 lwd, where l = 30 ft is the length of the tank, w = 6/7 d is the width of the top surface of water in the tank. Thus we have

V = 3/7 ld^2 = (3l/7)*(7F_0/rho)^(2/3).

Plugging in the numbers, we get

V = (3)(30)/7 * ((7)(6000)/62.4))^(2/3) = 987 ft^3

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