A tank is is half full of oil that has a density of 900kg/m3. Find the work W re

Question

A tank is is half full of oil that has a density of 900kg/m3. Find the work W required to pump the oil out of the spout. (Use 9.8 for g and 3.14 for . Round your answer to three significant digits.)

r=6.6

h=3.3

Explanation / Answer

A method without integration: The center of gravity of the lower half sphere is 3/8 r below the center of the sphere = 4,5 m below the enter of the sphere. From there it is 4,5+12+6 m to the opening. The volume of the half sphere is 2/3*pi*r^3 = 2/3*3,14*12^3 = 3617,28 m^3. The mass is 3617,28*900 = 3255552 kg. The work is W = mgh = 3255552*9,81*22,5 = 718.581.715,2 Nm. --------- with integration: lay the half sphere and the axis of symmetry on the y axis axis, the bottom of the sphere at the origin, the top of the spout at y = 30. the equation of the half sphere is y -12 = - v(12^2-x^2) --> x^2 = 12^2 - (12-y)^2 The volume of a horizontal disc with thickness ?y is V = pi*x^2*?y = pi(12^2-(12-y)^2)*?y The weight of this disc is 900*9,81pi(12^2-(12-y)^2*?y the work to move this disc to y = 30 is W = mgx = 900*9,81pi(12^2-(12-y)^2*?y*(30-y) the total work is 900*9,81pi?(12^2-(12-y)^2*(30-y)dy from y = 0 to 12 W = 7,189*10^8 J

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