Road map to EPS 5 Lectures5: Pressure, barometric law, buoyancy water air fluid moves Fig. 7.6: Pressure in the atmosphere (compressible) and ocean (incompressible)

Road map to EPS 5 Lectures5: Pressure, barometric law, buoyancy

waterair

fluid moves

Fig. 7.6: Pressure in the atmosphere (compressible) and ocean (incompressible).

Lecture 5. EPS 5: 08 February 2010

Review discussion of the perfect gas law from Lecture 2.

1. Review pressure concepts: weight of overlying fluid ("hydrostatic"), force of molecules bouncing off of an object.

2. Further discuss the concept of density ρ ;

3. Review the concept of kinetic energy of molecules in a gas. Look at some demos showing how pressure is manifest on the molecular and hydrostatic scales.

4. Work through the concept of the barometric law (hydrostatic balance: each layer of atmosphere must support the weight of the overlying column mass of atmosphere). How does the atmosphere bring the force exerted by molecular motions into balance with the weight of overlying atmosphere?

5. Look at the distribution of pressure with altitude in the atmosphere, or depth in the ocean.

6. Discuss how a barometer works ("dry demo").

7. Introduce buoyancy.

waterair

fluid will start to move

Water columns have the same height: Pressures equal on both sides.

Water columns higher on the left: Pressure higher on the left.

Cylinder volume = h x A = h x π r2. Mass = ρ h π r2. Weight = Mass x g = ρ h π r2

Mass of water = volume x density; Which has the greater volume?

Pressure = Mass x g / A = h ρ g

Perfect gas law (a.k.a. Boyle's and Charles' Laws) PV = NkT where P is pressure, V volume, N the number of molecules in the volume, and T the absolute temperature (Kelvin; T(K)=T(C)+273.15); k is Boltzmann's constant (1.38 x 10-23 Joules/Kelvin).

Boyle’s law P1V1 = P2 V2 ; Charles’ Law P1/ T 1 = P2/ T2

P = nkT,

where n (= N/V, the number density) is the number of molecules per unit volume.

The Perfect Gas Law relates pressure to temperature (the kinetic energy of the molecules) and "number density".

The density ( ρ) is defined as the mass per unit volume. If m is the mass of one molecule, then ρm n . The pressure, density and temperature of air are

therefore related by: P = ρ (k/m) T = ρ R T , an important form of the perfect gas law. The constant (k/m) is called “R” (the gas constant), "R" = 287.5/M J kg-1 K-1.



2. Discuss the concept of density ρ ;






Kinetic energy and molecular motion

E = 1/2 m v2 = 3/2 k T

k = 1.38 10-23 Joules/Kelvin; T = 300 K (room temperature)

E = 6.21 10-21 Joules/molecule;

for one mole, N0 (6.02 1023 molecules): E0 = 3738.42 Joules/mole.

Thus the molecules in only 29 grams of air (1 mole) contain 3.78 kJ of kinetic energy.

Since 1 Watt = 1 Joule/s, this amount of energy fires up the electrical appliances in an average house for 1 second!

How fast do molecules move?

3/2 kT = 1/2 mv2

mair = 29/N0 = 4.83 10-26 kg per molecule

v = (3 kT/ m )1/2 = 500 m s-1 at T=300 K

A more exact treatment gives (8 kT/(πm))1/2 = 467 m s-1 .

This is the speed of sound! (why is that?)









Atmospheric pressure and temperatureDistribution of pressure with altitude: the barometric law.

Changes in pressure with altitude in the atmosphere (left) and depth in the ocean (right). Pressure always increases as the observer moves downward because the weight of the overlying column of fluid (air or water) increases.

Note: Altitude is conventionally measured increasing upwards from the surface of the earth, and depth increasing downwards. Therefore pressure decreases with increasing altitude in the atmosphere and pressure increases with increasing depth in the ocean.

"air is compressible" density depends on pressure

ρ = P/ [ (k/m) T ]

atmosphere ocean

Z2

Z1

D2

D1P1 P

1

P2

P2

Z (altitude) increases upward

P1> P 2 at Z 1 < Z 2

D (depth) increases downward

P1> P 2 at D 1 > D2

Z2

Z1 P

1

P2

By how much is P1 > P2? The weight of the slab of fluid between Z1 and Z2 is given by the density, ρ, multiplied by volume of the slab) and g

weight of slab = ρ(area height) g.

Set the area of the column to 1 m2, the weight isρ g (Z2 -Z1): If the atmosphere is not being accelerated, there must be a difference in pressure (P2 - P1) across the slab that exactly balances the force of gravity (weight of the slab).

Relationship between density, pressure and altitude

ρggravity

Net P1 – P2

- (P2 - P1) = weight = ρ g (Z2 -Z1).

Pressure increases as we descend in the atmosphere because the air at each level must hold up the weight of all the air above it. (Note the minus sign, pressure is lower at P2.)

But the atmosphere is compressible, meaning the density ρ depends on the pressure itself! Use the Perfect Gas Law P = ρ (k/m) T to account for this fact.

ρ = Pav/( ( k/m )T )

Obtain the barometric law:

ΔP = -Pav [ mg / (kT) ] ΔZ.

The quantity kT/mg = H has units of length. It is a property of the atmosphere, and it is

the most important length in atmosphere, given a special name, scale height.

ΔP = -Pav ΔZ/H.or

ΔP/P = − ΔZ/H

If we had an atmosphere where the temperature did not change with altitude, the barometric law would have a very simple form in terms of the exponential function exp(), which appears on most hand calculators, exp(x) = ex, where e=2.718282….,

P(Z) = Po e-Z/H.

Po is the pressure at the ground (100,000 N/m2 or 1 atm = 1000 mb).

This is the simplest form of the Barometric Law describing the change of pressure with height in the

atmosphere.

How big is H: H = kT/mg= 1.38 e-23 x 273 /(4.8e-26 * 9.8)

H ~ 8000 ( 8 km)

The scale height H is the measure of depth in the atmosphere.

total #mol/m2 N = Hno H x local n/m-3

M = H x ρ





4. Work through the concept of the barometric law (hydrostatic balance: each layer of atmosphere must support the weight of the overlying column mass of atmosphere). How does the atmosphere bring the force exerted by molecular motions into balance with the weight of overlying atmosphere? The distribution of masses (densities) in the atmosphere will adjust until this balance is reached, because, if there are unbalanced forces, masses of air will accelerate air parcels towards the balanced distribution.




Pressure vs depth in the ocean

The weight of a slab of ocean with unit area (1 m2) is [mass of the slab]g = ρ g (D1 - D2)which gives the pressure difference between the top and bottom of the slab,

ΔP = ρ g ΔD.

Since ρ is essentially constant for water (water is incompressible), the change in pressure across the slab is proportional to the thickness of the slab but not proportional to pressure itself (contrast to atmosphere). The pressure changes by the same increment for a given depth change, and pressure increases linearly, not exponentially, with depth in the ocean,

P = Po + ρw g D

where ρw is the mean mass density of seawater.

Since the mass density of liquid water is about 1000 times greater than the density of air, the pressure becomes very large in the deep ocean.

ocean

D2

D1 P

1

P2

ocean atmosphere

1 bar = 105 N/m2

Pressure in the atmosphere (compressible) and ocean (incompressible) [Fig. 7.6 (McElroy)].





4. Work through the concept of the barometric law (hydrostatic balance: each layer of atmosphere must support the weight of the overlying column mass of atmosphere).




Densities (1 atm pressure, 300K)

water: 1000 kg/m3

air: 1.16 kg/m3

Hg (mercury): 13,600 kg/m3

sealed

“vacuum” (vapor pressure)

Diagram of a barometer:

measures atmospheric pressure

BuoyancyBuoyancy is the tendency for less dense fluids to be forced upwards by more dense fluids under the influence of gravity. Buoyancy arises when the pressure forces on an object are not perfectly balanced. Buoyancy is extremely significant as a driving force for motions in the atmosphere and oceans, and hence we will examine the concept very carefully here.

The mass density of air ρ is given by mn, where m is the mean mass of an air molecule (4.8110-26 kg molecule-1 for dry air), and n is the number density of air (n =2.69 1025 molecules m-3 at T=0o C, or 273.15 K). Therefore the density of dry air at 0 C is ρ = 1.29 kg m-3. If we raise the temperature to 10 C (285.15 K), the density is about 4% less, or 1.24 kg m-3. This seemingly small difference in density would cause air to move in the atmosphere, i.e. to cause winds.

Buoyancy force: Forces on a solid body immersed in a tank of water. The solid is assumed less dense than water and to have an area A (m2 ) on all sides. P1 is the fluid pressure at level 1, and P1x is the downward pressure exerted by the weight of overlying atmosphere, plus fluid between the top of the tank and level 2, plus the object. The buoyancy force is P1 – P1x (up ↑) per unit area of the submerged block.

P1x

Buoyancy force: Forces on a solid body immersed in a tank of water. The solid is assumed less dense than water and to have an area A (m2 ) on all sides. P1 is the fluid pressure at level 1, and P1x is the downward pressure exerted by the weight of overlying atmosphere, plus fluid between the top of the tank and level 2, plus the object. The buoyancy force is P1 – P1x (up ↑) per unit area of the submerged block.

P1x

Net Force (Net pressure forces – Gravity)





4. Work through the concept of the barometric law (hydrostatic balance: each layer of atmosphere must support the weight of the overlying column mass of atmosphere).




Road map to EPS 5 Lectures5: Pressure, barometric law, buoyancy

waterair

fluid moves

Fig. 7.6: Pressure in the atmosphere (compressible) and ocean (incompressible).

Documents

Road map to EPS 5 Lectures5: Pressure, barometric law, buoyancy water air fluid moves Fig. 7.6: Pressure in the atmosphere (compressible) and ocean (incompressible)