Question

Water is stored in an open stored in an open storage tank. The absolute pressure in the tank above the liquid is 1.0 atm. The water level is 8m above the base of the tank. What is the instantaneous velocity of a fluid jet when a 0.5 cm diameter orifice is open at point A. The opening is 0.8m above the base. consider v1=0 at the instant the opening is made (1)

Answer 1

`V_(A)= 11.88 m/s`

Step-by-step explanation:

given data:

water level at point A = 8 m

diameter of orifice = 0.5 cm

velocity at point A = 0

h1 =0.8 m

h2 = 8 - h1 = 8 - 0.8 = 7.2 m

Applying Bernoulli theorem between point A and B

`P_(o)+\rho _(water)gh_(2)+(1)/(2)\rho v_(B)^(2)+\rho _(water)gh_(1)=P_(o) +(1)/(2)\rho v_(A)^(2)+\rho _(water)gh_(1)`

`V_(A)=\sqrt{2gh_(2)}`

`V_(A)=√(2*9.81*7.2)`

`V_(A)= 11.88 m/s`