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    Work done to pump water out of tank

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    Would like second opinion or other way to solve problem, hopefully using Disk method - OTA #103642 answered last time. Perhaps OTA #103642 could send me email regarding this formula?

    Problem - A tank is in the shape of an inverted cone (pointy at the top) 6 feet high and 8 feet across at the base. The tank is filled to a depth of 3 feet. How much work is done in emptying tank through a hole at the top? (weight density of water is 62.4 lb/ft^3). Use force=density*area and disk method if possible - need simplicity and step by step understanding - Thanks!

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    Solution Preview

    In my last response I forgotten to multiply by a factor 32 ft/s^2(==g)
    So the correct answer will be:
    m*g = d* integral[0 to 3]{dV}*32 = d*(4*pi/9)*63*32
    where, d = wight density of water = 62.4 lb/ft^3
    work done = 62.4*(4*pi/9)*63*32*1.17857 = 207007 lb-ft^2/s^2 --Answer
    hence in SI ...

    Solution Summary

    The work done to pump water out of tank are analyzed.