Question

A right circular cylindrical tank of 5 metre height is filled with water .Water comes out from there through a pipe having the length of diameter 8 cm. at a speed of 225 metre / minute and the tank becomes empty after 45 minutes. Let us write by calculting, the length of diameter of the tank.​

Answer 1

$r\approx1.8\ m$ is the radius of the tank.

Step-by-step explanation:

Given:

height of the cylindrical tank, $h=5\ m$

diameter of the outlet pipe, $d_o=0.08\ m$

speed of flow at the outlet, $v=225\ m.min^{-1}$

time after which the tank becomes empty, $t=45\ min$

The area of pipe at outlet:

$A_o=\frac{\pi.d_o^2}{4}$

$A_o=\frac{\pi\times 0.08^2}{4}$

$A_o=0.005\ m^2$

Hence the volume flow rate:

$\dot V=v.A_o$

$\dot V=225\times 0.005$

$\dot V=1.131\ m^3.min^{-1}$

Now the volume of the tank drained in 45 minutes:

$V=\dot V\times t$

$V=1.131\times 45$

$V\approx50.894\ m^3$

Now the cross sectional area of the cylindrical tank:

$A=\frac{V}{h}$

$A=\frac{50.894}{5}$

$A\approx10.1788\ m^2$

We know the area of cylinder:

$\pi.r^2=10.1788$

$r\approx1.8\ m$

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TOPIC: volume flow rate

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