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    Fluid Volume Calculation

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    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.

    © BrainMass Inc. brainmass.com October 3, 2022, 12:27 am ad1c9bdddf
    https://brainmass.com/math/integrals/fluid-volume-calculation-424158

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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

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com October 3, 2022, 12:27 am ad1c9bdddf>
    https://brainmass.com/math/integrals/fluid-volume-calculation-424158

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