A tank of water sits at the edge of a table of height 0.8 m. The tank springs a

Question

A tank of water sits at the edge of a table of height 0.8 m. The tank springs a very small leak at its base, and water sprays out a distance of 0.5 m from the edge of the table. What is the water level h in the tank? (Assume the tank is open to the air at the top.)
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A tank of water sits at the edge of a table of height 0.8 m. The tank springs a very small leak at its base, and water sprays out a distance of 0.5 m from the edge of the table. What is the water level h in the tank? (Assume the tank is open to the air at the top.)

Explanation / Answer

the time it takes for a differential mass element of water to reach ground from the opening is

Height Table = 1/2 g t^2

t = sqrt( 2 H / g) = sqrt (2 * 0.8 / 9.8) = 0.4 sec

Thus, the water velocity at the orifice is

V = 0.5 / 0.4 = 1.25 m/s

Now we need to calculate the velocity V. This is done by using the Bernoulli Principle on the dam, that is

V^2 / 2 + gz + Pressure / density = constant

The orifice right at the bottom of the tank it spitting out the water horizontally. From Bernoulli Principle the velocity is then given by

V^2 / 2 + Patm/density = g h + Patm/ density

or

V^2 = 2gh

and then, the water level at the tank is,

h = V^2 / (2g) = (1.25)^2 / (2 *9.8) = 0.08 m

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