Class 09 Handout

7/25/2019 Class 09 Handout

1/18

Fluid Mechanics AS102

Class Note No: 09

Monday. August 20, 2007


2/18

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)


3/18


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 ...


4/18

Hydrostatics & Aerostatics

todays topic:

rederive the static pressure equation

applications of the pressure equation


5/18

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)


6/18


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)


7/18

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)


8/18

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)


9/18

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


2 dx


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)


10/18


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


2 dx


i )

inx1 direction,

the force=

P 12

Px1

dx1

dx2dx

3

P+

1

2

P

x1dx1

dx2dx

3

+ g1 dx

1dx

2dx

3 (11)


11/18


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,


12/18


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


13/18


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.


14/18

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


15/18


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


16/18


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


17/18

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


18/18


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 ...

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