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

$dh / dt = 0.2658 mm / min$