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Fluid StaticsComments

Fluid Statics

Density (ρ)mass / volume
(kg/m3)
Characteristic property of a pure substance.

Below air density values (ρ0) at various temperatures:

Air at 25° C1.22521 kg m-3
Air at 0° C1.29 kg m-3 [This is used by Giancoli]
Specific Gravity (SG)unitless( ρ of substance ) / ( ρ of water at 4° C)
PressurePressure = ( Force on fluid ) / ( Surface area of fluid )
Standard Pressure (P0)atmStandard pressure (atmospheric pressure) is the pressure the atmosphere exerts on a surface. 1 atm = 1.013 · 105 Pa = 14.69 psi
Water Pressure at Depth

P = F / A = ρ A h g / A = ρ g h

The pressure at a given depth is not dependent on area, just on the depth; generally, the pressure at a given depth of a liquid will be uniform. Note that this is just the water pressure. If the container is open to the atmosphere, then total pressure will be:

P = P0 + ρ g h
Pressure on Surface½ ρ g l ( y22 - y12 )The pressure a vertical surface such as an aquarium viewing window is determined by the above equation, where y2 is the upper edge's depth below water and y1 is the lower edge's depth below water. Note that because the surface is vertical, atmospheric pressure does not play a role.
Pressure at Elevation

P = P0 e- y ρ0 g / P0

For air pressure at an elevation y above sea level.
Gauge PressureGauges measure pressure relative to atmospheric pressure. For example, a tire gauge may read off 29 psi as the tire pressure. In fact, the absolute tire pressure is atmospheric pressure P0 (14.69 psi) plus the gauge reading of pressure PG (29 psi) for a total absolute tire pressure of 43.69 psi.

P = P0 = PG

Practice Problems
A bottle has a mass of 20 g when empty and 100 g when filled with water. When filled with another fluid, the mass is 80 g. What is the specific gravity of this other fluid?

Specific Gravity = ( ρ of substance ) / ( ρ of water at 4° C) = ( mass of substance / volume of bottle ) / ( mass of water / volume of bottle ) = ( mass of substance ) / ( mass of water )

Mass of substance = 80 g - 20 g = 60 grams
Mass of water = 100 g - 20 g = 80 grams
60 grams / 80 grams = 0.75 is the specific gravity.

What is the difference in blood pressure (mm-Hg) between the top of the head and bottom of the feet of a 1.5-meter-tall person standing vertically?

Phead = Pfeet + ρ h g
h = 1.5 m
ρ = density of blood ≈ 1050 kg / m3

Phead - Pfeet = ρ h g = 1050 · 1.5 · 9.8 = 15435 Pa
15435 Pa · 760 / 101300 = 116 mm-Hg

Pascal and Archimedes' Principles
Pascal's Principle

Pout = Pin
P = F / A
Fout / Aout = Fin / Ain
Fout / Fin = Aout / Ain

Archimedes' Principle

FB = ρfluid V g = mfluid g

FB = buoyant force
ρfluid = density of fluid
mfluid = mass of fluid displaced
V = Volume of fluid displaced, equal to the volume of the fully submerged object displacing the fluid.

Buoyant Force (FB)Archimedes' Principle addresses buoyant force. This is the phenomenon whereby an object weighs less underwater because the pressure at the bottom of the object is greater than at the top of the object. Thus, the water exerts a net force upward against gravity.
Practice Problems
The maximum gauge pressure in a hydraulic lift is 15.0 atm. What is the largest-size vehicle (kg) it can lift if the diameter of the output line is 20.0 cm?

Pascal's Principle: Pin = Fout / Aout

The problem provides pressure and diameter, so it is apparent that Pascal's Principle will provide an answer.

Aout = π r2 = π · ( ½ · .20 )2 = 0.0314159265 m2
Fout = mcar · g = mcar · 9.8 m s-1
Pin = 15.0 atm · 101325 Pascals / atm = 1519875 Pascals

1519875 = mcar · 9.8 / 0.0314159265
mcar = 4872.2736 kg
4870 kg is the maximum weight that the hydraulic can lift.

A geologist finds that a Moon rock whose mass is 10.00 kg has an apparent mass of 8.00 kg when submerged in water. What is the density of the rock?

FB = ρfluid V g

FB = ( 10.00 - 8.00 ) · g = 2.00 · g
ρfluid = 1000 kg m-3
ρobject = ?
V = ?

2.00 · g = 1000 · V · g
2.00 = 1000 · V
V = 2 · 10-3 m3 → volume of water displaced, and of object.

Density of object → 10.00 kg / ( 2 · 10-3 m3 ) = 5.00 · 103 kg m-3 = 5000 kg m-3

A cube of side length 10.0 cm and made of unknown material floats at the surface between water and oil. The oil has a density of 900 kg m-3. If the cube floats so that it is 80& in the water and 20& in the oil, what is the mass of the cube? What is the buoyant force on the cube?

FB = ρfluid V g

ρoil = 900 kg m-3
ρwater = 1000 kg m-3
Vcube = .1003 m3 = 10-3 m3
Vcube in oil = 10-3 m3 · 20% = 10-3 m3 · .20 = .0002
Vcube in water = 10-3 m3 · 80% = 10-3 m3 · .80 = .0008

FB in water = 1000 · .0008 · 9.8 = 7.84
FB in oil = 900 · .0002 · 9.8 = 1.764
FB total on cube = 7.84 + 1.764 = 9.604 Newtons is buoyant force on the cube

Mass of cube = 9.604 N / 9.8 m s-2 = 0.98 kg is mass of the cube

A Indonesian tribe makes a raft out of 10 coconut palm logs lashed together. Each log is 50 cm in diameter and has a length of 4 meters. How many people can the raft hold before they start getting their feet wet, assuming that an average tribemember has a mass of 55 kg? The specific gravity of coconut timber is approximately 0.50.

FB = ρfluid V g

ρfluid = 1000 kg m-3
FB = ?
V = ?
SGwood = 0.50
ρwood = SGwood · 1000 kg m-3 = 500 kg m-3

Notice that the tribemembers are just getting their feet wet.
Therefore, the raft must be entirely submerged.
Volume of log ≈ Volume for a cylinder = π r2 l = π · .252 · 4 = 3.14159265 m3

Recall that the upward (buoyant) force must break even with the raft's and the peoples' mass.
FB = ( mraft + mtroops ) · g = ( Vraft · ρwood + N · 40 ) · g = ( 1570.79633 + N · 40 ) g
Where N is the number of people on the raft.

( 1570.79633 + N · 40 ) g = 1000 · 3.14159265 · g
1570.79633 + N · 40 = 3141.59265
N · 55 = 1570.79633
N = 28.5599333 → The raft can support 28 people.

An open-tube mercury manometer is used to measure the pressure in an oxygen tank. When the atmospheric pressure is 1013.25 mbar, what is the absolute pressure (in Pa) in the tank if the height of the mercury in the open tube is 30.0 cm higher?

This problem is worded in a complicated manner, and with a variety of different units.
The first step is to take the material and make it simple, in SI units.

Atmospheric pressure = 1013.25 mbar = 101325 Pa
Tank pressure reading = 30.0 cm → In mm-Hg, this would simply be 30.0 cm · 10 mm cm-1 = 300 mm-Hg → Converting Torr (mm-Hg) to Pascals = 39996.7105 Pa

The problem posed is to find the absolute pressure in the tank.
Absolute pressure is gauge (or relative) pressure + atmospheric pressure.

101325 Pa + 39996.7105 Pa = 141321.71 Pa → 141000 Pa

An open-tube mercury manometer is used to measure the pressure in an oxygen tank. When the atmospheric pressure is 1013.25 mbar, what is the absolute pressure (in Pa) in the tank if the height of the mercury in the open tube is 5 cm lower, than the mercury in the tube connected to the tank?

This problem is simple, but the presentation is complicated due to wordiness and varying units.
Take the material and simplify it all into SI units.

Atmospheric pressure = 1013.25 mbar = 101325 Pa
Tank pressure reading = -7.0 cm → In mm-Hg, this would simply be -7.0 cm · 10 mm cm-1 = -70 mm-Hg → Converting Torr (mm-Hg) to Pascals = -9332.56579 Pa

With the above information, it is straightforward to figure out the absolute pressure in the tank.
Absolute pressure is gauge (or relative) pressure + atmospheric pressure.

101325 Pa + ( -9332.56579 Pa ) = 91992.4342 Pa → 92000 Pa