The tank shown above has height h = y2 - y1 = 90 cm. Its top surface has area A2

Question

The tank shown above has height h = y2 - y1 = 90 cm. Its top surface has area A2 = 119 cm2, and water flows out through a hole of diameter 1.6 cm in the bottom of the side wall. Suppose the top surface of the fluid is subjected to an external gauge pressure P2 = 0.791 atm. Assume the top surface of the fluid has negligible velocity (v2 = 0) and the water flows out into atmospheric pressure. Find the velocity v1 of the water as it leaves the tank.

HELP: Use Bernoulli's principle.
HELP: Be careful about gauge pressure vs. absolute pressure.

Explanation / Answer

0.5*d*v1^2=0.5*d*v2^2+g*(y2-y1)+P2

and v1*pi*(0.8)^2=v2*(119) []by eq. of continuity]

v2=0.0169v1

putting this in the first equation we get the result

0.5*1000*(1-0.000285)v1^2=9.81*0.9 +0.791*101.325*10^3

499.8575*v1^2=80156.904

v1^2=160.34

thus v1=12.66 m/s

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