class 10 handout

7/25/2019 Class 10 Handout

1/31

Fluid Mechanics AS102

Class Note No: 10

Tuesday. August 21, 2007


2/31

Review of Last Lecture: Hydrostatics & Aerostaticsstatic pressure Eq ininertial frame

x1

x2

P 12Px1

dx1

P 12Px2

dx2

P+ 12Px1

dx1

P+ 12Px2

dx2

r

g

dx1

dx2

Figure:control volumedx1dx2dx3 in inertial frame(xi)

dx1dx2dx3 very smallsuch that

V

dV dx1dx2dx3 ...

applyNewtons 2 lawto the fluid in the C.V.

P

xi gi=0 (1)


3/31

Review of Last Lecture: Hydrostatics & Aerostaticsstatic pressure Eq intranslational frame

x1

x2

x1

x2

P

1

2

P

x

1dx

1

P 12Px

2dx2

P+ 12Px

1dx1

P+ 12Px

2dx2

r r

g

b

dx1

dx2

Figure:control volumedx1 dx

2 dx

3 in translational frame(x

i )

applyNewtons 2 lawto the fluid in the C.V.

(bi g

i) + P

xi=0 ,

d

dtbi

rest in(xm)= 0 (2)

b= biii=bj ij, g= giii=gj ij,


4/31

Review of Last Lecture: Hydrostatics & Aerostatics

IN GENERAL, the static pressure equation

- for a fluidstationary in a non-inertial frame (xm)undergoingRBM

# the component form in(xm):

kx

k

i

k

kx

i + a

i g

i

+

P

xi =0 (3)

where

=ki

k, r =xk i

k, a:=b= aki

k, g= g

ki

k

r measured in the non-inertial(xm)

memorize the above equation


5/31

Review of Last Lecture: Hydrostatics & Aerostaticshydrostatic pressure distribution - static

a body of liquid, say, water at rest in an inertial frame

M

g h

O

x3

Patm

Figure:hydrostatic pressure distribution

assumptions: v= 0 in an inertial frame a constantPatmon the body from above the gravityg acts in the body downward

( a top plane surface) the liquid is incompressible with


6/31

Review of Last Lecture: Hydrostatics & Aerostatics

hydrostatic pressure distribution - static

Eq(3) and the b.c. reduce to

dP

dx3= g , P(0) =Patm

P=Patm+g x3 =Patm+ g h,

P=Patm+ g h (4)

called thehydrostatic pressure distribution(linear variation with the depth).


7/31

Hydrostatics & Aerostatics

forces on a submerged plane static

M

A

x1

x2 x1

x2

P

Cf

dA

g h

O

Patm

Patm

Patm

Figure:forces on a plane submerged: liquid on one side; air on theother side. C- plane centroid,f- pressure center


8/31

Hydrostatics & Aerostaticsforces on a submerged plane static

assumptions: v

=0 in an inertial frame

a constantPatmon the body from above the gravityg acts in the body downward

( a top plane surface) the liquid is incompressible with

P=Patm+ g h theresultant forceatMondA,normal tothe plane, is

dFR=P dA Patm dA= g h d A= gsinx2 dA

FR

=gsin A

x2 dA= gsinxC

2 A= P

CA

xC2 the plane centroid C; PC the gage pressure at C

thecenter of pressure,f with(xf1, xf2):

x

f

1 FR=

A x1 dFR, x

f

2 FR=

A x2 dFR ... (5)


9/31


todays topic:

more applications of the pressure equation bouyancy differential manometers uniform rotation


10/31


buoyancy - static:

acting on a solid body of arbitrary shape completely submerged

in a homogeneous incompressible liquid

dF2

dF2

gdA2

Patm

B

x1

x2

h=x2

Pnn

Figure:buoyancy


11/31

Hydrostatics & Aerostaticsbuoyancy - static:

dF2

dF2

gdA2

Patm

B

x1

x2

h=x2

Pn

n

assumptions: v= 0 in an inertial frame a constantPatmon the body from above the gravityg acts in the body downward

( a top plane surface)

the liquid is incompressible with


12/31

Hydrostatics & Aerostaticsbuoyancy:

set up the RCS(xm)withx2 downward and the origin on the

top plane the pressure force acting on a smalldAof the body surface

dF= Pn dA, P=Patm+ g x2 from H.P.D.

the total pressure force acting on the surface of the body

F=

V

Pn dA= Patm

V

ndA g

V

x2 ndA

V

ndAD. Thm.

= V

1 dV =0,V

x2 ndAD. Thm.

=

V

x2 dV = vol(V) i2

F= [ vol(V) g] i2, F1 =F3 =0

liquid weight displaced&upward or oppositetog


13/31

Hydrostatics & Aerostaticsbuoyancy:

thecenter of buoyancyB

xB1 F2 =

F on V

x1dF2=

V

x1 P n2 dA

= Patm

V

x1 n2 dA g

V

x1 x2n2 dA

D.Thm= PatmV

x1x2

dV gV

(x1 x2)x2

dV

= g xC1 vol(V), xC1 thecentroidof V (6)

xB3 F2 =

F on V x

3dF2=

Vx3 P n2 dA

= g xC3 vol(V), xC3 thecentroidof V (7)

xB1 =x

C1, x

B3 =x

C3 (8)


14/31


buoyancy:

the Archimedes principle:the buoyant force on a submerged body is equal to the

weightof liquid it hasdisplaced, and acts through the

centroidof the displaced volume


15/31

Hydrostatics & Aerostaticsbuoyancy - static:

what about apartiallysubmerged body ?

dF2

g

Patm

Bx1

x2

h=x2

Pn

n

Figure:buoyancy

P=

Patm, x2


16/31

Hydrostatics & Aerostaticsdifferential manometers :

PA, APB, B

C

g

h1

h2

h3

c d e

f g

Figure:differential manometer

assumptions: v= 0 in an inertial frame the gravityg acts in the body downward the liquids are incompressible

PB PA =?

H d i & A i


17/31


differential manometers :

Pc=PA+A g h1, Pc=Pd=Pe,

Pf =Pe+Cg h2, Pf =Pg, Pg=PB+Bg h3

PB PA=

PB+Bg h3 =PA+A g h1+Cg h2 (9)

H d t ti & A t ti


18/31

Hydrostatics & Aerostaticsuniform rotation

a cylindrical container of water rotating at a constant

around its axis, the water is at rest relative to the container

Hw

gg

R

x1

x3

x

1

x3H

P0 P0

n

m

M

Figure:uniform rotation

set(xm)in the inertial frame; attach(xm)to the rotating

cylinder (see the sketch).

H d t ti & A t ti


19/31


assumptions:

incompressible liquid like water with; b=0; = i3 = i

3; ( given and constant) g= gi3; a gas above the liquid with constantP0,

neglect the surface tension effecteq. (3)

P

x1= 2x1 ,

P

x2= 2x2 ,

P

x3= g (10)

P= 1

2 2 (r)2 g x3 + C, (11)

(r)2 := (x1 )2 + (x2 )

2,

C=? (12)

H drostatics & Aerostatics


20/31


uniform rotation

how to findC? acombinationof the factors as follows

b.c.:P=P0 on the free surface (fromt= P0n, t=

Tn)

volume preservation, spilling or no spilling ?

the effect of ,R,Hw &H ...



21/31


case 1: low - liquidoccupyingr [0, R]&no spilling

r

x3 H

P0

m

M

Figure:uniform rotation case 1

shape of the free surface:

gxf3= 1

2 2(r)2 P0+ C, r

[0, R] (13)

R

0

xf32rdr

vol p.= R2Hw



22/31


C P0=gHw

1()2

, :=

||R

2gHw g xf3 = 12

2(r)2 + g Hw[1()2], r [0, R]

r

x3 HP0

m

M


case 1 restrictions on:

there is liquid atr =0 xf3(0) > 0

< 1 (14)

no spilling xf(R) H

rHHw

1 (15)



23/31


uniform rotation

case 2:

- liquidoccupyingr [0, R]& there isspilling

r

x3H

P0

m

M



gxf3 =1

2 2(r)2 P0+ C, r

[0, R], C=? (16)


24/31



25/31


uniform rotation


there is spilling

R2Hw R

0

xf32rdr =... >0

R

2

gHw

2>

H

Hw 1 (17)

there is liquid at r =0 xf3(0)>0

R

2gH

2


26/31


case 3:

-no liquid occupyingr

[0, Rc)&no spilling

r

x3 H

P0

m

M

Rm



gxf3=1

2 2(r)2 P0+ C, r

[Rm, R]

Rm=? C=? (19)



27/31


uniform rotation

case 3 continue:how to findRm&C?

volume preservation

R

Rm

xf32rdr =R2Hw

C P0 = gR2Hw

R2 (Rm)2

1

42(R2 + (Rm)

2)

xf3(Rm) =0

Rm

R =

1

R

2

gHw

1



28/31


uniform rotation


Rmreal

R2gHw 1(?) (20)

no spilling xf3(R) H

R2gHw

2

+ R2gHw

HHw(?) (21)



29/31


case 4:

-no liquid occupyingr

[0, Rc)& withspilling

r

x3H

P0

m

M

Rm



gxf3= 1

2 2(r)2 P0+ C, r

[Rm, R],

Rm=? C=? (22)



30/31


uniform rotation

case 4 continue:how to findRm&C?

spilling xf3(R) =H

C P0=gH1 R

2gH

2 xf3(Rm) =0

Rm

R =

1 R2gH

2



31/31


uniform rotation


there is spilling

R2Hw R

Rm

xf32rdr >0

R2gH >

1

2

H

Hw(?) (23)

Rmreal

R2gH 1(?) (24)

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