Question

A cylindrical tank of radius R, filled to the top with a liquid, has a small hole in the side, of radius r, at distance d below the surface. Find an expression for the volume flow rate through the hole. A 4.0-mm-diameter hole is 1.0 m below the surface of a 2.0-m-diameter tank of water. What is the rate, in mm/min, at which the water level will initially drop if the water is not replenished?

Answer 1

Answer:

a)$V=\pi *r^2 * \sqrt{2gd}$

b)$dh / dt = 0.2658 mm / min$

Explanation:

From the question we are told that

Diameter of hole $d_h=4mm=>0.004m$

Depth of hole $D=0mm=>0.001m$

Diameter of tank $d_t=2mm=>0.002m$

Generally the equation for pressure is mathematically given as

$Pressure P= \rho*g*d$

$P= 1/2*\rho *v^2$

Where

$v = \sqrt {2gd}$

$V = Area*v$

$V=\pi *r^2 * \sqrt{2gd}$

Generally the level at which the water level will initially drop if the water is not replenished is mathematically given by

$dh / dt = (r/R)^2 *sqrt{2gd}\\dh / dt = (2/2000)^2 *sqrt(2*9.81*1) \\dh / dt = 4.429*10^-3 mm/s$

Therefore the level at which the water level will initially drop if the water is not replenished

$dh / dt = 0.2658 mm / min$

Answer 2

The rate, in mm/min, at which the water level will initially drop will be 1.0625 mm/min.

Given data:

The diameter of hole is, d = 4.0 mm = 0.004 m.

The depth of hole is, h = 1.0 m.

The diameter of tank is, d' = 2.0 m.

The given problem is based on the flow rate, which is defined as the flow of liquid through a given section per unit time.

Let us first obtain the equation of pressure as,

$P=\dfrac{1}{2} \times \rho \times v^{2}$

Here, v is the velocity of efflux and its value is,

$v=\sqrt{2gh} \\\\v^{2}=2gh$

And the level at which the water level will initially drop if the water is not replenished is mathematically given by,

$\dfrac{dH}{dt}=(r/R)^{2} \times v$

Here,

r is the radius of hole.

R is the radius of tank.

Solving as,

$\dfrac{dH}{dt}=((d/2) /(d'/2))^{2} \times \sqrt{2gh} \\\\\dfrac{dH}{dt}=((0.004/2) /(2/2))^{2} \times \sqrt{2 \times 9.8 \times 1}\\\\\dfrac{dH}{dt}=1.77 \times 10^{-5} \;\rm m/s\\\\\dfrac{dH}{dt}=1.77 \times 10^{-5} \times 6 \times 10^{4} \;\rm mm/min\\\\\dfrac{dH}{dt}=1.0625 \;\rm mm/min$

Thus, we can conclude that the rate, in mm/min, at which the water level will initially drop will be 1.0625 mm/min.

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