Answer to Question #264036 in Civil and Environmental Engineering for hydr

Question

An open cylindrical tank 4 ft in diameter and 6 ft deep is filled with water and rotated about its axis at 60 rpm. How much liquid is spilled, and how deep is the water at the axis? Ans. 15.3 ft , 3.55 ft , give the solution.

Answer 1

Concept:

For no spilling of water, rise above the original water level is equal to fall below the original water level

Height of paraboloid of revolution, 

where, 

ω = Angular velocity

R = Radius of cylinder

Calculations:

Given Data:

Diameter = 2 m

Height of Tank = 4 m

Depth of water = 1.5 m

So, Radius, R = 1 m

Rise above the original water level for no spilling = 4 - 1.5 = 2.5 m

As, Rise above level = Fall below level

So, Fall = 2.5 m

So, the total height of paraboloid = Rise + Fall = 2.5 + 2.5 = 5 m

Putting all that data in above equation, we get,

$5 = ω ^2 × 1 ^2 /2 × 9.81$

ω = 9.9 rad/sec