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