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TRANSCRIPT
Fluid MechanicsLecture #5
Summary of the previous lecture
•We showed that the pressure at a point is independent of the direction.
•We derived a pressure equation by applying the Newton’s 2nd law to a infinitesimal fluid element without a shear stress.
TODAY’S TOPIC“PRESSURE DISTRIBUTION”
For fluids at rest
• p is only a function of the single variable, z: p=p(z)
(1) (2) (3)
•As a result, we can change the partial derivative to the total derivative
•For fluids at rest, the pressure equation is
( ) p pdp z dx dyx y
p dp pdzz dz z
dpdz
g
Physical Meaning
• p varies only in the vertical direction.• p decreases as one goes up.• p increases as one goes down.
( )dp zdz
g
Integration of the pressure equationdp gdz
dp gdz dp gdz
Is “ρg” a constant with respect to z?
• g can be assumed to be constant.• Incompressible fluids: ρ is constant.• Compressible fluids: ρ is not constant.
Incompressible/Compressible Fluids
• Incompressible fluid (liquid)• The volume of the fluid does not change under the change of temperature, velocity field, etc.
• Compressible fluid (gas)• The volume changes
Pressure Distribution in Water
Water
p=p0
h
p=?
z
z=0
z=‐h(A)
(B)
0 : atmospheric pressurep
( )dp g dz
0 ( )p p g h
0 0
Integrate from (A) to (B) : p
p h
dp g dz
( constant for incompressible fluids)dp g dz
0 0p p gh p h
Pressure at z=-10m
0p p gh 3 2
0 (1000 kg/m )(9.8 m/s )(10 m)p
3 20 (1000 kg/m )(9.8 m/s )(10 m)p
0 98 kPap
101 kPa 98 kPa
200 kPa = 2 atm2 atm: absolute pressure1 atm: gauge pressure
•The hydrostatic pressure increases by about 1 atm for every 10 meter increase in depth.
•We often convert the force into length using the “Pressure Head”
•10 meter pressure head is equivalent to 1 atm.
Pressure Distribution in Two Fluids
Water
p=p0
h1
p=?
z
z=0
z=‐h1‐h2(A)
(B)
h2
Air Temperature
Ozone Layer
•Pressure in compressible fluids (gas) changes with the temperature.
•Idea gas law: p=ρRT
•T is no longer constant; density also varies over z.
•T=T(z), p=p(T(z))
Pressure in the Air
• We cannot simply take the density term out of the integral.
( )g dz g dz
Case 1: Constant Temperature
• For z<1km, T is assumed to be constant.
pRT
const ( )T p
( )dp p gdz
0
pdp gdzRT
0
1 gdp dzp RT
2 2
1 10
1p z
p z
gdp dzp RT
2
2
1
10
lnz
pp
z
gp dzRT
22 1
1 0ln ( )p g z z
p RT
2 10
( )2
1
g z zRTp e
p
Case 2: Variable Temperature
• For z>>1km, T is no longer a constant.
• We need to know the temperature distribution, T(z)
pRT
( , )p T
( , ( )) ( , )p T z p z
Temperature Distribution
• For z<11km
0( )T z T z
0 : Temperature at the groundT
( , )dp p z gdz
( )pdp gdz
RT z
0( )pdp gdz
R T z
0
1 1( )
dp gdzp R T z
2 2
1 10 0
1 1( / ) 1
p z
p z
gdp dzp RT T z
0 0
0
Let ( / ) 1, then ( )( / )
/X T z dX T dzdz T dX
2 2 2
1 1 1
0
0 0 0
1 1 1( / ) 1
p z X
p z X
Tg gdp dz dXp RT T z RT X
2 2
1 1
1 1p X
p X
gdp dXp R X
2 1 2 1ln ln (ln ln )gp p X XR
2 2
1 1
ln lnp Xgp R X
2 2
1 1
gRp X
p X
202
01
10
1Since ( / ) 1,
1
gR
zTpX T z
p zT
1 1
22
0
If we set 0, then ,
1
agR
a
z p p
p zp T