Question

7 m We model pumping from spherical containers the way we do from other containers, with the axis of integration along the vertical axis of the sphere. Use the figure shown to the right to find how much work it takes to empty a full hemispherical water reservoir of radius 6 m by pumping the water to a height of 7 m above the top of the reservoir Water weighs 9800 N/m? y = - 36-Y? The amount of work required to pump the water to a height of 7 m above the top of the reservoir is approximately (Do not round until the final answer. Then round to the nearest hundredth as needed.)

Answer 1

Given a hemisphere with a slice thickness

his

lets call it dy for simplicity

We will find the volume dv, mass dm of the slice then the work dw need to lift that slice out of tank and then 7 m above it

dv = A dy

where A = area of that slice which is in form of circle of radius

36 - j 2

$dv=\pi (\sqrt{36-y^2})^2dy$

$dv=\pi (36-y^2)dy$

now dm= $\rho$ dv

$\rho$=9800 N / m³ is the density of water.

so $dm=\rho \pi (36-y^2)dy$

As per the figure,

the slice below the x-axis is at a distance y meters.

This slice must be pumped 7 m above the hemisphere,which means that it has to be displaced y + 7 meters.

so height of lift =y +7

dw =dm * g * (height of lift) ; g=9.8ms^-2

$dw=\rho g\pi (y+7)(36-y^2)dy$

The total work done could be found out by integrating all the slices starting from y=-6 till y=7 is

$W=\int_{-6}^{7}dw$

$W=\int_{-6}^{7}\rho g\pi (y+7)(36-y^2)dy$

$W=\rho g\pi \int_{-6}^{7}(36y-y^3+252-7y^2)dy$

$W=\rho g\pi (\frac{36y^2}{2}-\frac{y^4}{4}+252y-\frac{7y^3}{3})_{-6}^{7}$

$W=\rho g\pi (\frac{36(7)^2}{2}-\frac{(7)^4}{4}+252(7)-\frac{7(7)^3}{3}-\frac{36(-6)^2}{2}+\frac{(-6)^4}{4}-252(-6)+\frac{7(-6)^3}{3}$

$W=\rho g\pi (882-600.25+1764-114.33-648+324+1512-504)$

$W=\rho g\pi (2615.42)$

$W=78,91,20,752Nm^2s^{-2}$

$W=7.8X10^8Nm^2s^{-2}$

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