Answer to Question #84769 in Mechanics | Relativity for Hardy

Question

Q1.The water in a tank is 1.8 m deep and over the surface is air at pressure 70 kPa (gauge). Find the flow rate from an orifice of 50 mm in the bottom of the tank if the Cd = 0.6.

Answer 1

Imagine that the water did not have gauge pressure over it of 70 kPa. In this case the flow rate would be

$Q_{im}=C_d A\sqrt{2gh},$

where

$A=\pi d^2/4$

and

$h=1.8 \text{ m}$

, wouldn't it?

But this additional influence exerted by the gauge pressure

$p_g$

can be represented as additional

$H=\frac{p_g}{\rho g},$

meters of water over 1.8 meters that are already in the tank. Thus we can replace common

$h$

for

$h'=h+H$

in the first expression in the beginning. Therefore, the real flow rate is:

$Q=C_d A\sqrt{2gh'}=C_d\frac{\pi d^2}{4} \sqrt{2g(h+\frac{p_g}{\rho g}})=$ $=0.6\cdot \frac{3.14\cdot 0.05^2}{4} \sqrt{2\cdot 9.8(1.8+\frac{70\cdot 10^3}{1000\cdot 9.8}})=0.0156 \text{ m}^3/\text{s}.$