Question

A 1.90m diameter closed cylinder, 2.75m high is completely filled with oil having sp.gr. of 0.8 under a pressure of 5 kg/cm^2 at the top.

1. a. What angular speed (RPM) can be imposed on the cylinder so that the maximum pressure at .90 m from the center of the tank is 14 kg/cm^2 ?
2. b. If the cylinder is open what is the maximum angular speed (RPM) can be imposed so that the water touches the bottom of the tank?

Answer 1

Ans) Given,

Diameter = 1.90 m

Height = 2.75 m

Specific gravity of oil = 0.80  

Pressure at top = 5 kg/cm²

We know,

  ² x² /2g = y

At x1 = 0.90m , y = y1

Pressure at y1 = 14 kg/cm² + Pressure at top

= 14 + 5

= 19 kg/cm²

or 190000 kg/m²

g y1 = 190000

y1 = 190000 / (9.81 x 800)

   y1 = 48.28 m

Now,

² (0.90)² / (2 x 9.81 ) = 24.20

= 48.21 rad/sec

or 44.21 x 60 / (2 x 3.14)

= 466.44 rpm  

Ans b) Now, when cylinder is open gauge pressure = 0

  ² x² /2g = y

Here, y = H

= 2gH/x²

= 2 x 9.81 x 2.75 / 0.95²

= 8.3 rad/sec

or 8.3 x 60 / (2x3.14) rpm

or 79.6 rpm