Chapter 4 Problem Definition Situation Fuel Tank Rotated Zero gravity

4.97: PROBLEM DEFINITION

Situation:

Stirring a liquid in a cup.

Find:

Report on the contour of the surface. Provide an explanation for the observed

shape.

SOLUTION

Stirring the cup of liquid creates a surface depressed at the center and higher at the

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4.98: PROBLEM DEFINITION

Situation:

A closed tank ﬁlled with water is rotated about a vertical axis.

D=4ft,ω=10rad/s.

Find:

Pressure at bottom center of tank (psig).

Properties:

ρ=62.4lbm/ft3=1.94 slug/ft3,γ=62.4lbf/ft3.

PLAN

Apply the equation for pressure variation equation- rotating ﬂow.

SOLUTION

Pressure variation equation- rotating ﬂow

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4.99: PROBLEM DEFINITION

Situation:

A tank of liquid is rotated on an arm.

S=0.80,D=1ft.

h=1ft,r=2ft.

VA=20ft/s,pA=25psf.

Find:

Pressure at B (psf).

Properties:

ρ=62.4lbm/ft3=1.94 slug/ft3,γ=62.4lbf/ft3.

PLAN

Apply the pressure variation equation- rotating ﬂow from point Ato point B.

SOLUTION

Pressure variation equation- rotating ﬂow

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4.100: PROBLEM DEFINITION

Situation:

A cream separator is in operation.

D=20cm,f=9000rpm.

Find:

Centripetal acceleration (m/s2).

RCF.

SOLUTION

The centripetal acceleration is

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4.101: PROBLEM DEFINITION

Situation:

A closed tank with liquid is rotated about the vertical axis.

ω=10rad/s, rB=0.5m,az=4m/s2.

Find:

Diﬀerence in pressure between points Aand B(kPa).

Properties:

ρ=1000kg/m3,S=1.2.

PLAN

Apply the pressure variation equation for rotating ﬂow between points B&C.Let

point Cbe at the center bottom of the tank.

SOLUTION

Pressure variation equation- rotating ﬂow

Then

4.102: PROBLEM DEFINITION

r1=0.5m,z1=0.5m.

z2=0m,r

2=0m.

Find:

Maximum rotational speed so that no liquid escapes from the leg on the left side

(rad/s).

PLAN

Since the ﬂuid is in rigid body rotation, apply the pressure variation equation for

rotating ﬂow. At the condition of imminent spilling, the liquid will be to the top of

theleftlegandatthebottomoftherightleg. Thus,locatepoint1beattopofthe

left (outside) leg. Locate point 2 at the bottom of the right (inside) leg.

SOLUTION

Pressure variation equation- rotating ﬂow

Term-by-term analysis

Substitute values into Eq. 1.

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4.103: PROBLEM DEFINITION

Situation:

A stagnation tube in a tank is rotated.

ω=100rad/ s,r=20cm,γ= 10000 N/m3.

Find:

Location of liquid surface in central tube.

PLAN

Pressure variation equation for rotating ﬂow from pt. 1 to pt. 2 where pt. 1 is at

liquid surface in vertical part of tube and pt. 2 is just inside the open end of the

Pitot tube.

SOLUTION

Pressure variation equation- rotating ﬂow

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4.104: PROBLEM DEFINITION

Situation:

A U-tube partially full of liquid is rotating about one leg.

f=50rpm,S=3.0,r1=1ft.

Find:

Speciﬁc gravity of other ﬂuid.

PLAN

Apply the pressure variation equation for rotating ﬂow between points 1 & 2.

SOLUTION

Pressure variation equation- rotating ﬂow

Also, by hydrostatics, because there is no acceleration in the vertical direction

(2)

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4.105: PROBLEM DEFINITION

Situation:

A manometer is rotated about one leg.

∆z=20cm,r=10cm,S=0.8.

Find:

Acceleration in g’s in leg with greatest amount of oil.

PLAN

Apply the pressure variation equation for rotating ﬂow between the liquid surfaces of

1&2Letleg1bethelegontheaxisofrotation. Letleg2betheotherlegofthe

manometer.

SOLUTION

Pressure variation equation- rotating ﬂow

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4.106: PROBLEM DEFINITION

Situation:

A fuel tank rotated in zero-gravity environment.

f=3rpm, r1=1.5m,zA=1m.

Find:

Pressure at exit (Pa).

Properties:

ρ=800kg/m3,p1=0.1kPa.

PLAN

Apply the pressure variation equation for rotating ﬂow from liquid surface to point

A. Call the liquid surface point 1.

SOLUTION

Pressure variation equation- rotating ﬂow

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4.107: PROBLEM DEFINITION

Situation:

A rotating set of tubes has liquid in the bottom of it.

D1=2d,D2=d.

r2=,z2=4.

Find:

Derive a formula for the angular speed when the water will begin to spill.

PLAN

Start with pressure variation equation for rotating ﬂow. Let point 1 be at the liquid

surface in the large tube and point 2 be at the liquid surface in the small tube.

SOLUTION

Pressure variation equation- rotating ﬂow

Thechangeinvolumeinleg1hastobethesameasleg2.So

The elevation diﬀerence between 1 and 2 will be

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4.108: PROBLEM DEFINITION

Situation:

Water ﬁlls a tube that is closed at one end.

D=1cm,r=40cm,ω=50rad/s.

Find:

Force exerted on closed end (N).

Properties:

ρ=1000kg/m3

PLAN

Apply the pressure variation equation for rotating ﬂow from the open end of the tube

to the closed end.

SOLUTION

Pressure variation equation- rotating ﬂow

Then

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4.109: PROBLEM DEFINITION

Situation:

Water sits in a U-tube that is closed at one end.

D=1cm,=2cm.

Find:

Rotational speed when water will begin to spill from open tube (rad/s).

Properties:

ρ=1000kg/m3,γ=9810N/m3.

PLAN

Apply the pressure variation equation for rotating ﬂow between water surface in leg

A-A to water surface in open leg after rotation.

SOLUTION

When the water is on the verge of spilling from the open tube, the air volume in the

closed part of the tube will have doubled. Therefore, we can get the pressure in the

Pressure variation equation- rotating ﬂow

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4.110: PROBLEM DEFINITION

Situation:

Water is pumped from a reservoir by a centrifugal pump consisting of a disk with

radial ports.

r=5cm,f=3000rpm, z1=0m.

Find:

Maximum operational height (m).

PLAN

Apply the pressure variation equation for rotating ﬂow

Locate point 1 at the liquid surface where z=0.

Locate point 2 at the outer edge of the rotating disk.

SOLUTION

Pressure variation equation

Rotational Rate

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4.111: PROBLEM DEFINITION

Situation:

A tank rotated about the horizontal axis and water in tank rotates as a solid body.

V=rω,z=−1,0,+1 m,ω=5rad/ s.

Find:

Pressure gradient each value of z (kPa/m).

Properties:

ρ=1000kg/m3.

PLAN

Apply the pressure variation equation for rotating ﬂow.

SOLUTION

Pressure variation equation- rotating ﬂow.

when z=−1m

when z=+1m

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4.112: PROBLEM DEFINITION

Situation:

A tank 4 ft in diameter and 12 feet long rotated about horizontal axis and water

in tank rotates as a solid body. Maximum velocity is 25 ft/s.

V=rω,Vmax =25 ft/s.

D=4ft,L=12ft.

Find:

Maximum pressure diﬀerence in tank (psf).

Point of minimum pressure (ft).

Properties:

ρ=62.4lbm/ft3=1.94 slug/ft3,γ=62.4lbf/ft3.

PLAN

SOLUTION

Below the axis both gravity and acceleration cause pressure to increase with decrease

in elevation; therefore, the maximum pressure will occur at the bottom of the cylin-

Solving: r=γ/ρω2;pmin occurs at zmin =+g/ω2.Using the equation for pressure

variation in rotating ﬂows between the tank bottom where the pressure is a maximum

(zmax =−r0)and the point of minimum pressure.

SOLUTION

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From the analysis section above,

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