Chapter

Fluid Statics

Blinn College - Physics 1401 - Terry Honan

. - Fluids and Density

There are three primary states of matter: solid, liquid and gas. Other more exotic states can occur under extreme temperatures and pressures; the most common of these is a plasma. We have intuition for these three states: A solid has a shape and a volume, although it can be deformed into a different shape. Liquids have a volume but no specific shape. A gas has no shape and will expand to fill any volume. Our intuition for solid, liquids and gases can be deceiving. For instance, at high pressures the distinction between liquids and gases often disappears. Also, things that are obviously solids to us, like glasses and plastics, can actually be very very slow-flowing liquids.

In this chapter we are studying fluids. A fluid is defined as a liquid or a gas. Lumping together these two distinct things may seem awkward at first but we will see that they share many common properties. The most important of these properties is pressure.

All matter can be ascribed a density, where density ρ is defined as a mass per unit volume.

ρ=(m/V)=(mass/volume)

Mass values in the metric system were defined based on properties of water; a cubic centimeter, which is the same as a milliliter, of water has one gram (QuantityUnit) of mass. A liter (QuantityUnit) of water, which is the same as a cubic decimeter, then has a kilogram of mass and cubic meter has a thousand kilograms of mass.

(Subscript[ρ, water])=1(((QuantityUnit))/(((QuantityUnit))^3))=1(((QuantityUnit))/((QuantityUnit)))=1000(((QuantityUnit))/(((QuantityUnit))^3))

Water is a very useful reference density, since if something floats it is less dense than water and if it sinks it is more dense than water. The specific gravity is defined as the ratio of the density of a substance to the density of water; this is the same as the numerical value of density in units of (QuantityUnit)/(((QuantityUnit))^3).

Example . - What is the Average Density of the Earth

M=5.97×(10^24)(QuantityUnit)  and  R=6.38×(10^6)(QuantityUnit)

Solution

V=(4/3)π (R^3)=1.0878×(10^21)(((QuantityUnit))^3)  ⟹  ρ=(M/V)=5490(((QuantityUnit))/(((QuantityUnit))^3))

. - Pressure in a Fluid

Pascal’s Principle

Pressure P is defined as force per unit area,

P=(F/A)=(Force/Area)

In the SI system, we measure pressure in (QuantityUnit)=pascal.

Units:  The SI unit for pressure is: (QuantityUnit)=(QuantityUnit)/(((QuantityUnit))^2)

Suppose you have a vertical cylindrical column of some solid. Pushing downward at the top of the column will increase the force per area on a horizontal surface inside the column, but it will not affect the force per area on any vertical surfaces. A downward force on the column will not produce any horizontal outward forces.

In the case of a fluid, a liquid or a gas, pushing downward on the top of a vertical cylindrical piston, the pressure (force/area) is the same for all orientations of a surface; pushing downward on a piston will also increase the outward force of the fluid in all directions.

This piston example illustrates Pascal’s principle:

Pressure pushes in all directions equally. The force on a flat surface of area A inside a surface is has the same magnitude of F=P A , independent of the orientation of the surface.

Atmospheric Pressure and Gauge Pressure

Atmospheric pressure varies. The current pressure a topic in discussion of the current weather; typically high pressure corresponds to good weather and low pressure to bad weather. When a standard value of atmospheric pressure is given, it is an average value; we will call this 1(QuantityUnit).

1(QuantityUnit)=1.013×(10^5)(QuantityUnit)=14.7 lb/(((QuantityUnit))^2)

The listed value in pounds per square–inch or psi=lb/(((QuantityUnit))^2) is familiar to most. Forces due to atmospheric pressure are huge. Why do we not notice them? A pressure of 1(QuantityUnit) acting on the large area of a wall gives a huge force but there is the same huge force acting on the other side of the wall to cancel it. If all the air in an adjacent room were evacuated making it a vacuum, then essentially any wall would collapse under those huge forces. It is pressure differences that cause forces.

Suppose a car’s tire pressure is badly depleted and a pressure gauge reads 13psi=13 lb/(((QuantityUnit))^2). Does that mean the pressure in the tire is less than outside? Clearly this is not the case; air will flow out of the tire. What the gauge reads is called the gauge pressure, which is the difference between the absolute pressure and atmospheric pressure. So in the tire example the pressure in the tire is larger than 1 (QuantityUnit) by 13 psi. An absolute pressure of zero is a perfect vacuum. A gauge pressure of zero means the pressure is 1 (QuantityUnit).

gauge pressure=P-1(QuantityUnit)

Variation of Pressure with Depth

It is a well-known fact that as one goes deeper under water, the pressure increases. This is a general feature of fluids and it is caused by the added weight of the fluid above it.

Consider a fluid and isolate a rectangular block of the same fluid, shown in red above. The rectangular block has an area A at the top and the bottom and a height h. Take the pressures at the top and bottom of the block to be (Subscript[P, top]) and (Subscript[P, bottom]), respectively, and the forces at the top and bottom are (Subscript[P, top])A and (Subscript[P, bottom])A. The net force on a rectangular block of fluid sitting in the same fluid must be zero. Pressure exerts forces on the sides of the rectangular block, but those forces will cancel. Since this is a case of static equilibrium, the net force must vanish. Equating the total force upward (Subscript[P, bottom])A with the total downward force (Subscript[P, top])A+W gives:

(Subscript[P, bottom])A=(Subscript[P, top])A+W

where W is the weight of fluid in the block. The mass in the block is m=ρ V and the volume is V=A h.

W=m g=ρ V g=ρ A h g

Combining these two expressions and canceling the area A gives the expression for variation of pressure with depth.

(Subscript[P, bottom])=(Subscript[P, top])+ρ g h

For water open to the air, the pressure at the top is 1(QuantityUnit). At what depth (Subscript[h, 0]) does the pressure increase by another 1(QuantityUnit).

1(QuantityUnit)=(Subscript[ρ, water]) g (Subscript[h, 0])

Using the density of water we can find (Subscript[h, 0]).

1(QuantityUnit)=1.013×(10^5)(QuantityUnit), (Subscript[ρ, water])=1000(((QuantityUnit))/(((QuantityUnit))^3))   and  g=9.8(((QuantityUnit))/(((QuantityUnit))^2)) ⟹  (Subscript[h, 0,water])=((1(QuantityUnit))/((Subscript[ρ, water])g))=10.3(QuantityUnit)

With water then, every additional (Subscript[h, 0])=10.3(QuantityUnit) of depth gives another atmosphere of pressure. So at a depth of (Subscript[h, 0]) the pressure is 2(QuantityUnit), at 2(Subscript[h, 0]) it is 3(QuantityUnit), etc.

Example . - Water Pressure at the Titanic Wreck

Solution

ρ=1027(((QuantityUnit))/(((QuantityUnit))^3)) , h=3800(QuantityUnit)   and   (Subscript[P, top])=1(QuantityUnit)=1.013×(10^5)(QuantityUnit)

P=(Subscript[P, bottom])=(Subscript[P, top])+ρ g h=3.83×(10^7)(QuantityUnit)=379(QuantityUnit)

Torricelli Barometer

Imagine a sink full of water and a glass in the sink. With the glass full of water invert the cup and lift it with the lip still below the surface. The pressure at the surface is 1(QuantityUnit), so the pressure at the top of the inverted glass is less by ρ g h, where h is the height of the glass.

What would happen if the glass were taller than (Subscript[h, 0])=10.3(QuantityUnit)? The absolute pressure cannot be negative so the maximum height of the column of water is (Subscript[h, 0]); above that would be a vacuum. This could be made into a barometer, a device to measure atmospheric pressure. If a column of water is taller than (Subscript[h, 0])=10.3(QuantityUnit), then watching the small changes in the height is measuring the atmospheric pressure. Because it is so tall, it would not be a practical device. Using a more dense fluid would improve this.

A much more dense fluid is mercury, with a density of (Subscript[ρ, mercury])=13600(QuantityUnit)/(((QuantityUnit))^3). This is the basis of the barometer invented by Torricelli.

(Subscript[ρ, mercury])=13600(((QuantityUnit))/(((QuantityUnit))^3))  ⟹  (Subscript[h, 0,mercury])=((1(QuantityUnit))/((Subscript[ρ, mercury])g))=0.760(QuantityUnit)=760(QuantityUnit)

Pressure is sometimes measured in mmHg or millimeters of mercury with a standard value of 760 mmHg for atmospheric pressure.

Example . - Blaise Pascal’s Very French Barometer

Solution

1(QuantityUnit)=1.013×(10^5)(QuantityUnit)=(Subscript[ρ, wine]) g (Subscript[h, 0,wine])  ⟹  (Subscript[ρ, wine])=984(((QuantityUnit))/(((QuantityUnit))^3))  ⟹  (Subscript[h, 0,wine])=((1(QuantityUnit))/((Subscript[ρ, wine])g))=10.5(QuantityUnit)

. - Buoyancy and Archimedes’ Principle

Archimedes (287 - 212 BC) was a remarkable polymath and genius. Much of his work was lost to history, but some somewhat recently discovered documents illustrate how advanced he was for the time; he had anticipated and used much of what later became calculus almost two millenia before its invention. The common story of the discovery of Archimedes’ principle is perhaps apocryphal but is too good to not repeat. While taking a bath, he discovered the principle of buoyancy. Thrilled by this discovery, he ran into the street naked screaming “Eureka”.

The buoyant force B is the net upward force on a floating or submerged object caused by the variation of pressure with depth. To see there is a net upward force consider the submerged block in the diagram below.

Since the forces on the sides of the block must cancel, the net upward force, the buoyant force B, becomes B=(Subscript[P, bottom])A-(Subscript[P, top])A. Using the variation of pressure with depth we get (Subscript[P, bottom])-(Subscript[P, top])=(Subscript[ρ, fluid])g h. Since the volume of the block is V=A h. Combining these results gives an expression for the buoyant force (Subscript[F, b]).

B=(Subscript[ρ, fluid])V g

This is Archimedes principle: the buoyant force is the weight of the displaced fluid.

The derivation above illustrates that the variation of pressure with depth, but this mathematical result could not motivate anyone to run down the street naked screaming “Eureka”. A more general derivation involves replacing the submerged object with the fluid. The force of water pressure pushing up on the submerged object is the buoyant force. Now replace the submerged object with an imaginary surface, shown dashed in the diagram below to the right, that contains the same fluid. The force on the fluid sitting in the same fluid bust be zero, so the buoyant force must equal the weight of fluid that occupies that volume. Thus, the buoyant force is the weight of the displaced fluid.

Example . - Just the Tip of the Iceberg

Solution

(Subscript[ρ, ice])=917(QuantityUnit)/(((QuantityUnit))^3)  and   (Subscript[ρ, sw])=1027(QuantityUnit)/(((QuantityUnit))^3).

(Subscript[W, ice])=(Subscript[ρ, ice])(Subscript[V, tot])g  and  (Subscript[W, disp])=(Subscript[ρ, sw])(Subscript[V, disp])g  ⟹  (Subscript[W, ice])=(Subscript[W, disp])  ⟹  (Subscript[ρ, ice])(Subscript[V, tot])g=(Subscript[ρ, sw])(Subscript[V, disp])g

(({{fraction}, {below}}))=((Subscript[V, disp])/(Subscript[V, tot]))=((Subscript[ρ, ice])/(Subscript[ρ, sw]))=0.893  ⟹  (({{fraction}, {above}}))=(((Subscript[V, tot])-(Subscript[V, disp]))/(Subscript[V, tot]))=1-((Subscript[V, disp])/(Subscript[V, tot]))=1-((Subscript[ρ, ice])/(Subscript[ρ, sw]))=0.107

Example . - Lighter than Air

Solution

V=0.012(((QuantityUnit))^3), (Subscript[m, rubber])=1.4(QuantityUnit)=1.4×(10^(-3))(QuantityUnit), (Subscript[ρ, helium])=0.123(QuantityUnit)/(((QuantityUnit))^3), 
(Subscript[ρ, air])=1.20(QuantityUnit)/(((QuantityUnit))^3)  and  g=9.8(QuantityUnit)/(((QuantityUnit))^2)

(Subscript[m, tot])=(Subscript[m, rubber])+(Subscript[ρ, helium])V=0.003956(QuantityUnit)  ⟹  W=(Subscript[m, tot])g=0.03877(QuantityUnit)

B=(Subscript[ρ, air])g V=0.14112(QuantityUnit)

T=B-W=0.102(QuantityUnit)