Question

3. (22 points) Consider the tank below with a cylindrical middle and hemispherical bottom and top. Assume that the tank is half full of water. Set up, but DO NOT SOLVE an integral (or integrals) that can be used to determine the work required to pump all of the water out of a spout level with the top of the tank. Note: the density of water is 1000 kg/m3. Make sure to indicate an axis, with 0 marked. In your work, make sure to demonstrate (with justification) the force(s) on cross sections and the distance(s) traveled by water at cross-sections 1mT 3m M112 Spri.. df Integral Formulas.pdfFormula Sheet for...pdf

Answer 1

The tank below with a cylindrical middle and hemispherical bottom and top. Work done to pump out = present potential energy. The water in the tank is taken floor as the down level Y^bar_1 = y/2 T^bar_1 = 2m Y^bar_2 = 4 + 8 92) y^bar_2 = 4 + 15 Y^bar_2 = 4.75m \r\n m_1 = (pi R^2 h) times l m_1 = pi times 4 times 4 times 10^3 m_1 = 5025 482461 kg m_2 = (2/3 pi R^3)l m_2 = (2/3 times pi times 12)^3) times 10^3 m_2 = 16.75516 m_2 = 16755.16 kg P.E. = P.E_1 + P.E_2 P.E = m_1 g y^bar_1 + m_2 g y^bar_2 P.E = (9.81) [(50265.48246 times 2) + (16755.16 times 4.75)] P.E. = 1766957.334 joules the work done to pump out water = 1766.957334 kj