Question

A large storage tank, open to the atmosphere at the top and fi lled with water, develops a small hole in its side at a point 16.0 m below the water level. If the rate of fl ow from the leak is 2.50 103 m3 /min, determine (a) the speed at which the water leaves the hole and (b) the diameter of the hole

Answer 1

Answer:

The speed at which the water leaves the hole $V_{2}$ = 4.21 $\frac{m}{s}$

The value of  diameter of the hole d = 0.112 m = 11.2 cm

Step-by-step explanation:

Given data

Flow rate = 2.5 × $10^{-3}$ $\frac{m^{3} }{min}$ = 0.0416 × $10^{-3}$ $\frac{m^{3} }{sec}$

Height (h) = 16 m

(a) The speed at which the water leaves the hole :-

Apply bernouli equation for the water tank at point 1 & 2

$\frac{P_{1} }{\rho g} + \frac{V_{1} ^{2} }{2g} + Z_{1} = \frac{P_{2} }{\rho g} + \frac{V_{2} ^{2} }{2g} + Z_{2}$ ------ (1)

Since $P_{1}$ = $P_{2}$ , $V_{1} = 0$ & $Z_{1} - Z_{2} = h$

Equation (1) becomes

$\frac{V_{2}^{2} }{2 g} = h$

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

This is the speed at which the water leaves the hole.

Put the values of g & h in the above formula

⇒ 2 g h = 2 × 9.81 × 16 = 17.71

$V_{2}$ = $\sqrt{2 g h}$ = $\sqrt{17.71}$

$V_{2}$ = 4.21 $\frac{m}{s}$

This is the speed at which the water leaves the hole.  

(b)Diameter of the hole :-

We know that flow rate Q = A × V

$A = \frac{Q}{V}$

Put the values of Q & V in the above formula we get

A =  $\frac{0.0416}{4.21}$ × $10^{-3}$

A = 9.88 × $10^{-6}$

We know that area A = $\frac{\pi}{4} d^{2}$

⇒ $d^{2}$ = $\frac{4}{\pi}$ × 9.88 × $10^{-6}$

⇒ $d^{2}$ = 0.01258

⇒ d = 0.112 m = 11.2 cm

this is the value of  diameter of the hole.