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Fluid Mechanics AS102
Class Note No: 09
Monday. August 20, 2007
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Review of Last Lecture: Hydrostatics & Aerostatics
derivation of the static pressure equation- for a fluidstationaryin a non-inertial frame undergoingRBM
# inertial (xi) and non-inertial (x
i ), r= r + b
# control volum inside the fluid domain# Newtons 2nd law# volume integral eq to differential eq
[ ( r) + a g] + P=0 (1)
constant
and constant a:=
d2b
dt2 certain combination among
,
d2b
dt2 ,g
ff
r the position measured in(xm)
7/25/2019 Class 09 Handout
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Review of Last Lecture: Hydrostatics & Aerostatics
derivation of the static pressure equation- for a fluidstationaryin a non-inertial frame undergoing
RBM# the component form in(xm):
kx
k
i
k
kx
i + a
i g
i
+
P
xi=0 (2)
where
=ki
k, r =xk i
k, a= a
ki
k, g= g
ki
k
# the component form in(xn)obtained similarly.[note thatr is the position measured in(xm)]the link is throughr = r + b or xi=x
j i
j ii+ bii
xj i
j =xiii biii x
j =i
j ii(xi bi),
=ik
xk
=ik
xk
xi
=ii ik
xk,
=
ki
k=kik ...
k =i
kimm ...
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Hydrostatics & Aerostatics
todays topic:
rederive the static pressure equation
applications of the pressure equation
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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)
hints:
P(x1+1
2 dx1, x2, x3) =P(xi) + P
x1
1
2 dx1+ higher order terms ...
applyNewtons 2 lawto the fluid in the C.V.:
inx1 direction,
v1 =0 d
dt v1 dx1 dx2 dx3= 0 (3)
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Hydrostatics & Aerostatics
static pressure Eq in inertial frame
x1
x2
P 1
2
P
x1
dx1
P 12Px
2
dx2
P+ 12Px1
dx1
P+ 1
2
P
x2 dx2
r
g
dx1
dx2
Figure:control volume: static pressure Eq in inertial frame (xi)
2nd law
P 12Px1
dx1
dx2 dx3
P+
1
2
P
x1dx1
dx2 dx3
+ g1 dx1 dx2 dx3=0 (4)
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Hydrostatics & Aerostaticsstatic pressure Eq in inertial frame
x1
x2
P 12Px1
dx1
P 12Px2
dx2
P+ 12Px1
dx1
P+ 12Px2
dx2
r
g
dx1
dx2
Figure:control volume: static pressure Eq in inertial frame (xi)
inx1 direction,
Px1
g1 =0 (5)
inxidirection,
P
xi
gi=0 (6)
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Hydrostatics & Aerostaticsstatic pressure Eq intranslational frame
x1
x2
x1
x2
P 12
Px1 dx
1
P 12Px
2dx2
P+ 12Px
1dx1
P+ 12Px
2
dx2
r r
g
b
dx1
dx2
Figure:control volumedx1 dx
2 dx
3 in translational frame(x
i )
r=r + b, r=xiii, r =xj i
j, b=biii, i
i
pick= ii (7)
dii
dt =0, viin(xm)
= dxidt , v
i
in(xm)= dx
idt =0 (8)
xi=x
i + b
i, dx
i=dx
i , v
i= b
i (9)
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Hydrostatics & Aerostaticsstatic pressure Eq in translational frame
x1
x2
x1
x2
P 1
2
P
x
1
dx
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.:
inx1 direction,
rate of l. m.= d
dt
v1 dx1dx2dx3
=
d
dt
b1 dx
1 dx
2 dx
3 c.m.
= b1 dx
1 dx
2 dx
3 (10)
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Hydrostatics & Aerostaticsstatic pressure Eq in translational frame
x1
x2
x1
x2
P 1
2
P
x
1
dx
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 )
inx1 direction,
the force=
P 12
Px1
dx1
dx2dx
3
P+
1
2
P
x1dx1
dx2dx
3
+ g1 dx
1dx
2dx
3 (11)
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Hydrostatics & Aerostaticsstatic pressure Eq in translational frame
inx1 direction,
2nd law (b1 g1) +
P
x1=0 (12)
inxidirection,
(bi gi) + P
xi=0 (13)
generally, in a translational(xi)withi
i =ii,
(bi g
i) + P
xi=0 ,
d
dtbi
rest in(xm)= 0 (14)
where
b= biii=b
j i
j, g= giii=g
j i
j,
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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 (15)
where=ki
k, r =xk i
k, a= a
ki
k, g= g
ki
k
r the position measured in(xm) Q? can we derive the above eq from tensoral
transformations?
Not straightforward! there is a relative motion between(xm)and(xn)and /tshould be treated adequately
memorize the above equation
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Hydrostatics & Aerostatics
the static pressure equation
# incompressiblefluids like water
every quantity is known except P solve eq (15) forP
# compressiblefluids like air
= (P, T)
more equations needed to make the model determinate perfect gas:
P=RT , R=R
M, R =8.314
J
mol.oK (16)
[P] = Nm2
, [] = kgm3
, [R] = Jkg.oK
, [T] = oK
Rspecific gas constant, R universal gas constant,M
chemical molecular weight for air: M=28.97 g
mol, R=287 J
kg.oK
2eqs. &3unknowns givenT =T(x)like in ISA, etc.
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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
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Review of Last Lecture: Hydrostatics & Aerostatics
hydrostatic pressure distribution - static
fix a RCS(xm)to the body withx3 downward and the origin
in the top plane =0; b= 0; (supposedly static) g= gi3; P=Patm atx3 =0
Eq (15)reduces to
P
x1=0,
P
x2=0,
P
x3= g
P=g x3+ C b.c. P=Patm+g x3 =Patm+ g h,
P=Patm+ g h ,
Pgage :=P Patm= g h, h 0
H d i A i
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Hydrostatics & Aerostatics
forces on a submerged plane static
M
A
x1
x2 x
1
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
H d i & A i
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Hydrostatics & Aerostaticsforces on a submerged plane static
M
A
x1
x2 x1x2
P
C
f
dA
g hO
Patm
Patm
Patm
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
H d t ti & A t ti
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Hydrostatics & Aerostatics
forces on a submerged plane - static
set up a RCS as shown in the sketch the resultant force at MondA,normal tothe plane, is
dFR=P dA Patm dA= g h d A= gsinx2 dA
FR=gsin
A
x2 dA= gsinx2CA =PCA
x2C the plane centroidC; PC the gage pressure atC
thecenter of pressure,f with(x1f
, x2f
):
x1fFR=
A
x1 dFR, x2fFR=
A
x2 dFR ...