An open-top 20.0[m] tall cylindrical tank is filled with water as shown below. A

Question

An open-top 20.0[m] tall cylindrical tank is filled with water as shown below. A small puncture on the side of the tank, is located just 15.0[m] above ground. The flow rate out of the tank is 1.00 [lt/min]. a) How fast is the water flowing through the hole? b) If the radius of the cylindrical tank is 10.0[m], what is the size of the hole? c) How far from the edge of the tank is the puddle? d) If we want punch another hole, closer to the ground, and get the new stream to land in the same puddle, how height should we do it?

Explanation / Answer

Here ,

height of water surface above the water is h

a)

h = 20 - 15

h = 5 m

velocity of the water = sqrt(2 * g * h)

velocity of the water = sqrt(2 * 9.8 * 5)

velocity of the water = 9.9 m/s

the velocity of the water is 9.9 m/s

b)

let the area of hole is A

A * 9.9 = volume flow rate

A * 9.9 = 1 *10^-3/60

A = 1.684 *10^-6 m^2

the area of puncture is 1.684 *10^-6 m^2

c)

time of flight , t= sqrt(2 * h /g)

t = sqrt(2 * 15/9.8)

t = 1.75 s

distance of puddle = 1.75 * 9.9

distance of puddle = 17.32 m

the distance of puddle is 17.32 m

d)

let the height is h

distance = speed of flow * time of flight

17.32 = sqrt(2 * 9.8 * (20 - h)) * sqrt(2*h/9.8)

solving for h

h = 5 m

the height of the hole must be 5 m

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