Question

A hole is punched in the side of a tank below the surface of the fluid in it. The fluid is coming out at a speed of 7 m s−1. (a) How far below the surface of the fluid was the hole punched? (b) If the volume flow rate of the fluid coming out of the tank is 0.5 L min−1, what is the radius of the hole?

The answer key shows that the numbers should be a) 2.45 m b) 0.62 mm

Can you please provide all steps to this question? Thank you so much for your time!

Answer 1

v = 7 m/s
Let the distance be H.
Velcoity of flow is given as, v = C_v (2 g H)^1/2
Where , C_v = velocity coefficient (water 0.98)
Substituing Values,
7 = 0.97 * sqrt(2 * 9.8 * H)
H = 2.60 m

I am assumed the fluid to be water and taken velocity coefficient = 0.98 , As there is no mention in the question !!

(b)
V = 0.5 l/min = 8.33 * 10^-6 m^3/s
Volume Flow rate is given as,
V = C_d A sqrt(2 g H)
8.33 * 10^-6 = C_{d *} π*r^2 * sqrt(2*9.8*2.60)
C_{d *} r^2 = 3.71 * 10^-7
Value of Cd (Discharge coefficient) is not mentioned in the question !!
Substitute and get answer !!