Answer:
The speed at which the water leaves the hole
= 4.21 ![\frac{m}{s}](https://tex.z-dn.net/?f=%5Cfrac%7Bm%7D%7Bs%7D)
The value of diameter of the hole d = 0.112 m = 11.2 cm
Step-by-step explanation:
Given data
Flow rate = 2.5 ×
= 0.0416 ×
![\frac{m^{3} }{sec}](https://tex.z-dn.net/?f=%5Cfrac%7Bm%5E%7B3%7D%20%7D%7Bsec%7D)
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
------ (1)
Since
=
,
& ![Z_{1} - Z_{2} = h](https://tex.z-dn.net/?f=Z_%7B1%7D%20-%20Z_%7B2%7D%20%3D%20h)
Equation (1) becomes
![\frac{V_{2}^{2} }{2 g} = h](https://tex.z-dn.net/?f=%5Cfrac%7BV_%7B2%7D%5E%7B2%7D%20%20%7D%7B2%20g%7D%20%3D%20h)
= ![\sqrt{2 g h}](https://tex.z-dn.net/?f=%5Csqrt%7B2%20g%20h%7D)
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
=
= ![\sqrt{17.71}](https://tex.z-dn.net/?f=%5Csqrt%7B17.71%7D)
= 4.21 ![\frac{m}{s}](https://tex.z-dn.net/?f=%5Cfrac%7Bm%7D%7Bs%7D)
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}](https://tex.z-dn.net/?f=A%20%3D%20%5Cfrac%7BQ%7D%7BV%7D)
Put the values of Q & V in the above formula we get
A =
× ![10^{-3}](https://tex.z-dn.net/?f=10%5E%7B-3%7D)
A = 9.88 × ![10^{-6}](https://tex.z-dn.net/?f=10%5E%7B-6%7D)
We know that area A = ![\frac{\pi}{4} d^{2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cpi%7D%7B4%7D%20d%5E%7B2%7D)
⇒
=
× 9.88 × ![10^{-6}](https://tex.z-dn.net/?f=10%5E%7B-6%7D)
⇒
= 0.01258
⇒ d = 0.112 m = 11.2 cm
this is the value of diameter of the hole.