How do you calculate velocity of water free falling under gravity from outlet at the botom of open tank?

Answer 1

The same as any other free-falling object There is nothing significantly unique about water in this regard. It will accelerate at 9.8 meters per second squared until it achieves terminal velocity. Not sure what that is for a water droplet or a water stream. For calculating free fall from a tank the velocity at the exit (1)and flow quantity in the pipe(2) must be calculated. the velocity at the exit ===> U2 = 2xZxG---------(1) where; Velocity of fluid at exit ---U Fluid height in the tank ---Z Gravitational acceleration coefficient ----G flow quantity in the pipe====> m=VxS-----------------(2) Where; mass flow rate in exit pipe---m exit pipe transection area -------s velocity of flowing fluid------V Calculation example:The flow quantity of water from a tank .The height of water in the tank is 1 meter and water exits the tank from 1/2 inch pipe Pipe transection area--- (1.25cm/2)2.Pi= 1.227cm2 Velocity of water at boottom of tank U2=100x980x2 U=443cm/sec this could be taken as V then m can be calculated as, m=443cm/sec x1.227cm2=543.2cm3/sec----12592gm/min

Correction: The mass flow rate equation ism/t = density*area*velocity Your units would be mass per unit time The volumetric flow rate is Q = area*velocity Your units would be volume per unit time The above is volumetric flow rate as opposed to mass flow rate.

Answer 2

ignoring air resistance, velocity (v) after 10 metre drop (from rest) at earth surface can be calculated from :

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v^2 = (u^2 ) + (2 * a * s), where :

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v = final velocity = ? (metres / second)

u = initial velocity ( 0 metres / second)

a = acceleration due to gravity at earths surface = 9.82 (m/s) / s

s = distance travelled (10 metres)

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so:

v^2 = 0 + (2 * 9.82 * 10)

v^2 = 196.4

v = 14.014 metres / second