Question

A tank is full of water. Find the work required to pump the water out of the spout. Use the fact that water weighs 62.5 lb/ft3. (Assume r = 6 ft, R = 12 ft, and h = 15 ft.)

Answer 1

Answer:

Work done, $W = 464,062.5\pi$

Solution:

Consider the weight of water is $62.5 lb/ft^{3}$

Let us assume:

r = 6 ft

R = 12 ft

h = 15 ft

Now, let us consider the volume of the cross-section (horizontal) be dV, from the rim top let y be its vertical distance and dw be its weight

Now,

radius = $(r - R)\frac{y}{h} + R$

radius = $(6 - 12)\frac{y}{15} + 12$

radius = $12 - 0.4y$

dV = $\pi(radius)^{2} = \pi(12 - 0.4y)^{2}dy$

The weight of cross-section is given by:

$dw = 62.5(12 -0.4y)^{2}\pi dy$

Now, the work done is given by:

$W = \int_{a}^{b} ydw$

$W = \int_{0}^{15} 62.5(12 - 0.4y)^{2}\pi ydy$

$W = 62.5\pi \int_{0}^{15} (144y - 9.6y^{2} + 0.16y^{3})dy$

$W = 62.5\pi [\frac{144y^{2}}{2} - \frac{9.6y^{3}}{3} + \frac{0.16y^{4}}{4}]_{0}^{15}$

On solving the above equation , we get:

$W = 464,062.5\pi$

Answer 2

Work done / required to pump water out of the spout = 464,062.5 π

Given data :

weight of water = 62.5 Ib/ft³

inner radius ( r ) = 6 ft

Outer radius ( R ) = 12 ft

Height ( h ) = 15 ft

Determine the work required to pump water out of the spout

Vertical distance = y

dv = cross section volume

dw = weight

First step : determine the radius of the tank

Radius = ( r - R ) $\frac{y}{h} + R$

Radius = ( 6 - 12 ) $\frac{y}{15} + 12$

∴ Radius = 12 - 0.4y

Next step : determine  dv ( cross section volume )

Dv = $\pi r^{2}$  

     = $\pi ( 12 - 0.4y )^{2}$dy

Weight of cross-section ( dw ) = 62.5 ( 12.5 - 0.4y )² $\pi dy$

Final step:  The work done to pump water out of the spout is

Work done = $\int\limits^1_0 {y} \, dw$   =  ∫¹⁵₀ 62.5 ( 12.5 - 0.4y )² $\pi dy$

∴ Work done ( W ) = 464,062.5 $\pi$

Hence we can conclude that the Work done / required to pump water out of the spout = 464,062.5 π

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