A tank is full of water. Find the work required to pump the water out of the spo

Question

A tank is full of water. Find the work required to pump the water out of the spout. (Use 9.8 m/s2 for g. Use 1000 kg/m3 as the density of water. Assume r = 9 m and h = 3 m

Explanation / Answer

Solution: I am supposing that the origin be at the center of the bottom of the tank. Let us consider a slice of height: dx at a height x above the origin. Also Let y = the radius of the slice => I can write: y = m x + r, here: m = (R-r)/h Now: Volume of the slice = PI * y^2 dx = PI * (m x + r)^2 dx (Weight of water in the slice) = d * PI * (m x + r)^2 dx ( d = density) As we know that the water in the slice has to be lifted by height h-x Hence the work done in lifting the water in the slice = d * PI * (m x + r)^2 * (h-x) dx Total work done W = integral(0,h) d * PI * (m x + r)^2 * (h-x) dx W = PI d integral(0,h) (m x + r)^2 * (h-x) dx W = PI d integral(0,h) (m^2 x^2 + 2mxr + r^2) * (h-x) dx W = PI d integral(0,h) (m^2 x^2 h - m^2 x^3 + 2mxrh - 2mx^2r + r^2 h - r^2 x) dx W = PI d (m^2 h^4 /3 - m^2 h^4 /4 + 2 m r h^3 /2 - 2 m r h^3 /3 + r^2 h^2 - r^2 h^2/2) W = PI d (m^2 h^4 /12 + m r h^3 /3 + r^2 h^2 /2) Substitute m = (R - r)/h = (12 - 6)/15 = 0.4 also other variables: W = 3.14 * 62.5 (0.4^2 15^4 /12 + 0.4 * 6 * 15^3 /3 + 6^2 15^2 /2) W = 4.66 * 10^7 lb-ft Ans Please rate

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