kindly help me with this problem I provide some information for the solution needed. Use free floating decimals in all your calculations and in expressing your final answers. • Solve problem SYSTEMATICALLY and NEATLY thank you!  kindly follow this format GIVEN,REQUIRED,SOLUTION 2. What rotational speed must be maintained in operating a 10-in. diameter centrifugal bowl handling a 2-in. thick layer of liquid with an approximate density of 72.5 lb/ft3 if an absolute pressure of 47.2 psi is required at the wall of the bowl to apply hydrostatic equilibrium?

Elements Of Electromagnetics
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ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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kindly help me with this problem I provide some information for the solution needed.

Use free floating decimals in all your calculations and in expressing your final answers.
• Solve problem SYSTEMATICALLY and NEATLY thank you! 

kindly follow this format GIVEN,REQUIRED,SOLUTION

2. What rotational speed must be maintained in operating a 10-in. diameter centrifugal bowl
handling a 2-in. thick layer of liquid with an approximate density of 72.5 lb/ft3 if an absolute
pressure of 47.2 psi is required at the wall of the bowl to apply hydrostatic equilibrium?

 

Hydrostatic Equilibrium Principle:
In a stationary mass of a single static fluid, the pressure is constant in any cross section parallel
to the earth's surface but varies from height to height.
Figure 2: Pressures in vessels of various shapes filled with the same fluid (same properties)
Reference
(datum)
p+ dp
Area = S
dz
Fgovity
Fig 3: A column
of a stationary fluid
EF = 0
Fupward pressure - Fdownward pressure - Fgravity = 0
Since:
F = pressure x area; Fgravity = gravitational acceleration x mass
Then:
ps - (p + dp ) S - (g/g-)pSdz = 0
dp + (g/g.)) pdZ = 0
Transcribed Image Text:Hydrostatic Equilibrium Principle: In a stationary mass of a single static fluid, the pressure is constant in any cross section parallel to the earth's surface but varies from height to height. Figure 2: Pressures in vessels of various shapes filled with the same fluid (same properties) Reference (datum) p+ dp Area = S dz Fgovity Fig 3: A column of a stationary fluid EF = 0 Fupward pressure - Fdownward pressure - Fgravity = 0 Since: F = pressure x area; Fgravity = gravitational acceleration x mass Then: ps - (p + dp ) S - (g/g-)pSdz = 0 dp + (g/g.)) pdZ = 0
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