Question

9–12. Consider the cylindrical tank in Example 4 that has a height of 10 m and a radius of 5 m. Recall that if the tank is full of water, then ∫₀¹⁰ 25 π ρg(15−y) dy equals the work required to pump all the water out of the tank, through an outflow pipe that is 15 m above the bottom of the tank. Revise this work integral for the following scenarios. (Do not evaluate the integrals.)

The work required to empty the tank through an outflow pipe at the top of the tank

Accepted Answer

Identify the physical setup: The tank is a cylinder with height 10 m and radius 5 m, filled with water. The density of water is $$ ho $$, and gravitational acceleration is $$ g . $$The work done to pump water depends on the distance each water slice must be lifted to the outflow point. Recall the original integral for pumping water to a pipe 15 m above the bottom: $$ \int_0^{10} 25 \pi ho g (15 - y) \, dy $$, where $$ y  is $$the vertical coordinate from the bottom, and $$ 25 \pi  is $$the cross-sectional area (since radius = 5 m, area = $$ \pi 	imes 5^2 = 25 \pi ). $$For the new scenario, the outflow pipe is at the top of the tank, which is 10 m above the bottom. Therefore, the distance each water slice at height $$ y $$ must be lifted is $$ 10 - y . $$Set up the revised integral for work as $$ W = \int_0^{10} 25 \pi ho g (10 - y) \, dy . $$This integral sums the work to lift each thin horizontal slice of water (with volume $$ 25 \pi \, dy ) $$from height $$ y  to $$the top at 10 m. This integral represents the total work required to pump all the water out through the pipe at the top of the tank. No further evaluation is needed as per the problem instructions.

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Key Concepts

Work as an Integral in Fluid Mechanics

Volume and Weight of a Cylindrical Slice

Distance Function in Work Integral