Upload
pranav-kulkarni
View
224
Download
0
Embed Size (px)
Citation preview
7/25/2019 Class 10 Handout
1/31
Fluid Mechanics AS102
Class Note No: 10
Tuesday. August 21, 2007
7/25/2019 Class 10 Handout
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)
7/25/2019 Class 10 Handout
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,
7/25/2019 Class 10 Handout
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
7/25/2019 Class 10 Handout
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
7/25/2019 Class 10 Handout
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).
7/25/2019 Class 10 Handout
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
7/25/2019 Class 10 Handout
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)
7/25/2019 Class 10 Handout
9/31
Hydrostatics & Aerostatics
todays topic:
more applications of the pressure equation bouyancy differential manometers uniform rotation
7/25/2019 Class 10 Handout
10/31
Hydrostatics & Aerostatics
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
7/25/2019 Class 10 Handout
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
7/25/2019 Class 10 Handout
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
7/25/2019 Class 10 Handout
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)
7/25/2019 Class 10 Handout
14/31
Hydrostatics & Aerostatics
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
7/25/2019 Class 10 Handout
15/31
Hydrostatics & Aerostaticsbuoyancy - static:
what about apartiallysubmerged body ?
dF2
g
Patm
Bx1
x2
h=x2
Pn
n
Figure:buoyancy
P=
Patm, x2
7/25/2019 Class 10 Handout
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
7/25/2019 Class 10 Handout
17/31
Hydrostatics & Aerostatics
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
7/25/2019 Class 10 Handout
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
7/25/2019 Class 10 Handout
19/31
Hydrostatics & Aerostaticsuniform rotation
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
7/25/2019 Class 10 Handout
20/31
Hydrostatics & Aerostatics
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 ...
Hydrostatics & Aerostatics
7/25/2019 Class 10 Handout
21/31
Hydrostatics & Aerostaticsuniform rotation
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
Hydrostatics & Aerostatics
7/25/2019 Class 10 Handout
22/31
Hydrostatics & Aerostatics
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
Figure:uniform rotation case 1
case 1 restrictions on:
there is liquid atr =0 xf3(0) > 0
< 1 (14)
no spilling xf(R) H
rHHw
1 (15)
Hydrostatics & Aerostatics
7/25/2019 Class 10 Handout
23/31
Hydrostatics & Aerostatics
uniform rotation
case 2:
- liquidoccupyingr [0, R]& there isspilling
r
x3H
P0
m
M
Figure:uniform rotation case 2
shape of the free surface:
gxf3 =1
2 2(r)2 P0+ C, r
[0, R], C=? (16)
7/25/2019 Class 10 Handout
24/31
Hydrostatics & Aerostatics
7/25/2019 Class 10 Handout
25/31
Hydrostatics & Aerostatics
uniform rotation
case 2 restrictions on:
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
7/25/2019 Class 10 Handout
26/31
Hydrostatics & Aerostaticsuniform rotation
case 3:
-no liquid occupyingr
[0, Rc)&no spilling
r
x3 H
P0
m
M
Rm
Figure:uniform rotation case 3
shape of the free surface:
gxf3=1
2 2(r)2 P0+ C, r
[Rm, R]
Rm=? C=? (19)
Hydrostatics & Aerostatics
7/25/2019 Class 10 Handout
27/31
Hydrostatics & Aerostatics
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
Hydrostatics & Aerostatics
7/25/2019 Class 10 Handout
28/31
Hydrostatics & Aerostatics
uniform rotation
case 3 restrictions on:
Rmreal
R2gHw 1(?) (20)
no spilling xf3(R) H
R2gHw
2
+ R2gHw
HHw(?) (21)
Hydrostatics & Aerostatics
7/25/2019 Class 10 Handout
29/31
Hydrostatics & Aerostaticsuniform rotation
case 4:
-no liquid occupyingr
[0, Rc)& withspilling
r
x3H
P0
m
M
Rm
Figure:uniform rotation case 4
shape of the free surface:
gxf3= 1
2 2(r)2 P0+ C, r
[Rm, R],
Rm=? C=? (22)
Hydrostatics & Aerostatics
7/25/2019 Class 10 Handout
30/31
Hydrostatics & Aerostatics
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
Hydrostatics & Aerostatics
7/25/2019 Class 10 Handout
31/31
Hydrostatics & Aerostatics
uniform rotation
case 4 restrictions on:
there is spilling
R2Hw R
Rm
xf32rdr >0
R2gH >
1
2
H
Hw(?) (23)
Rmreal
R2gH 1(?) (24)