Question

A small circular hole 6.00 mm in diameter is cut in the side of a large water tank, 14.0 m below the tank's water level. The top of the tank is open to the air.

What is the speed of efflux?

What is the volume discharged per unit time?

Answer 1

Concepts and reason

The main concept used to solve this problem is velocity of efflux. Initially, use the expression of velocity of efflux and find its value. Then, use the expression of volume discharge per unit time to find its value.

Fundamentals

The velocity of efflux can be given as follows:

\(v=\sqrt{2 g h}\)

Here, \(\mathrm{g}\) is the acceleration due to gravity and \(\mathrm{h}\) is the height of the orifice from the top of the fluid opening. The volume discharged per unit time can be give as follows:

\(Q=A v\)

Here, \(\mathrm{A}\) is the area of the orifice and \(\mathrm{v}\) is the velocity of efflux.

 

(a) The velocity of efflux can be given as follows:

$$ v=\sqrt{2 g h} $$

Substitute \(9.8 \mathrm{~m} / \mathrm{s}^{2}\) for \(\mathrm{g}\) and \(14.0 \mathrm{~m}\) for \(\mathrm{h}\) in the above expression.

\(v=\sqrt{2\left(9.8 \mathrm{~m} / \mathrm{s}^{2}\right)(14.0 \mathrm{~m})}\)

\(=16.565 \mathrm{~m} / \mathrm{s}\)

\(=16.6 \mathrm{~m} / \mathrm{s}\)

Part a The speed of efflux is \(16.6 \mathrm{~m} / \mathrm{s}\)

The size of the orifice is negligible as compared to the height of the orifice from the top of the tank. Thus, the velocity of the efflux is same as the velocity of a freely falling object, falling from a height \(\mathrm{h}\).

 

(b) The volume discharged per unit time can be give as follows:

\(Q=A v\)

The area of the orifice can be calculated as follows:

\(A=\pi\left(\frac{d}{2}\right)^{2}\)

Here, \(\mathrm{d}\) is the diameter of the orifice.

Substitute \(\pi\left(\frac{d}{2}\right)^{2}\) for \(\mathrm{A}\) in the expression \(Q=A v\)

\(Q=\frac{\pi d^{2} v}{4}\)

Substitute \(6.00 \mathrm{~mm}\) for \(\mathrm{d}\) and \(16.565 \mathrm{~m} / \mathrm{s}\) for \(\mathrm{v}\) I the above expression.

\(Q=\frac{\pi(6.00 \mathrm{~mm})^{2}(16.565 \mathrm{~m} / \mathrm{s})}{4}\left(\frac{10^{-3} \mathrm{~m}}{1 \mathrm{~mm}}\right)^{2}\)

\(=4.68 \times 10^{-4} \mathrm{~m}^{3} / \mathrm{s}\)

Part \(b\)

The volume discharged per unit time is \(4.68 \times 10^{-4} \mathrm{~m}^{3} / \mathrm{s}\)

The orifice is in the shape of a circle such that the area of the orifice is the area of a circle. The radius is the half of

the diameter such that the \(r^{2}\) is written as \(\left(\frac{d}{2}\right)^{2}\) in the expression of area.

 

 

Part a 

The speed of efflux is \(16.6 \mathrm{~m} / \mathrm{s}\).

Part b

The volume discharged per unit time is \(4.68 \times 10^{-4} \mathrm{~m}^{3} / \mathrm{s}\)