Work done to pump water out of tank
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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Hi,
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
hence,
work done = 62.4*(4*pi/9)*63*32*1.17857 = 207007 lbft^2/s^2 Answer
hence in SI ...
Solution Summary
The work done to pump water out of tank are analyzed.
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