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

The speed at which the water leaves the hole
`V_(2)` = 4.21
`(m)/(s)`

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

Explanation:

Given data

Flow rate = 2.5 ×
`10^(-3)`
`(m^(3) )/(min)` = 0.0416 ×
`10^(-3)`
`(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

`(P_(1) )/(\rho g) + (V_(1) ^(2) )/(2g) + Z_(1) = (P_(2) )/(\rho g) + (V_(2) ^(2) )/(2g) + Z_(2)` ------ (1)

Since
`P_(1)` =
`P_(2)` ,
`V_(1) = 0` &
`Z_(1) - Z_(2) = h`

Equation (1) becomes

`(V_(2)^(2) )/(2 g) = h`

`V_(2)` =
`√(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)` =
`√(2 g h)` =
`√(17.71)`

`V_(2)` = 4.21
`(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 = (Q)/(V)`

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

A =
`(0.0416)/(4.21)` ×
`10^(-3)`

A = 9.88 ×
`10^(-6)`

We know that area A =
`(\pi)/(4) d^(2)`

⇒
`d^(2)` =
`(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.