A tank of 5m height is filled with water. calculate the velocity of efflux through a hole 3m below the surface of water.
Answers
Let’s consider a cylindrical tank with a circular hole in the bottom. The velocity of the jet exiting the hole ( Vj=2gh−−−√ ) is a function of the depth of water in the tank as shown very nicely in Vishnu Ganesh’s answer here:
The volume flow rate exiting the tank will be:
rate=Vj(Aj)=2gh−−−√Aj
(Area of the jet (hole) = Aj)
Be sure to use consistent units.
Here’s how to solve for the time to drain the tank:
Since the fluid level is decreasing as the tank drains and the rate at which the fluid level drops varies with depth, we must do some calculus.
I will examine the time (dt) it takes to drop an infinitely small distance = dh and then derive a general expression for the time it takes to drain the tank from level h1 to level h2.
Consider a very small time interval = dt
Area of the jet (hole) = Aj
Area of the tank = At
Fluid volume exiting the jet in time dt is Aj(Vj)dt
Fluid volume removed from the tank in time dt is At(dh)
The fluid volume exiting the jet in time dt is equal to the fluid volume removed from the tank in time dt:
Aj(Vj)dt=At(dh)
or
dt=AtAjvjdh
dt=AtAj2gh√dh
dt=AtAj2g√h−12dh
t=∫dt=AtAj2g√∫h1h2h−12dh
t=2AtAj2g√(h1−−√−h2−−√)
To find the time to drain the tank, simply set h2 = 0
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