7. Fluids 1

DENSITY

• Fluids can flow, change shape, split into smaller portions and combine into a larger system

• One of the best ways to quantify a fluid is in terms of its density

• The density, ρ, of a material (or fluid) is defined as the mass, M, per volume, V: ρ = M/Vkg/m3

• The denser a material, the more mass it has in any given volume

• The density of a substance is the same regardless of the amount in a system

7. Fluids 2

PRESSURE

• Pressure , P, is a measure of the amount of force, F, per area, A: P = F/A N/m2

• Pressure is increased if the force applied to a given area is increased, or if a given force is applied to a smaller area

• For example pressing your finger against a balloon just causes a small indentation, whereas pushing a needle with the same force causes the balloon to burst

• The smaller surface area of the needle tip causes a large enough pressure to rupture the balloon

• Example: Find the pressure exerted on the skin of a balloon if you press with a force of 2.1N using first your finger, then a needle. Assume the area of your fingertip is 1.0×10-4m2, and the area of the needle tip is

2.5×10-7m2. Also find the minimum force necessary to pop the balloon with the needle, given that the balloon pops with a pressure of 3.0×105 N/m2.

7. Fluids 3

ATMOSPHERIC PRESSURE

• Atmospheric pressure , Pat, is a direct result of the weight of the air above us

• Pat = 1.01×105 N/m2

• A shorthand unit for N/m2 is the pascal (Pa)• 1 Pa = 1 N/m2 (Pat = 101 kPa)• In British units, pressure is measured in pounds per

square inch: Pat = 14.7 lb/in2

• A common unit for measuring atmospheric pressure in weather forecasting is the bar : 1 bar = 105Pa ≈ 1 Pat

• Example: Find the force exerted on the palm of your hand by atmospheric pressure. Assume that your palm measures 0.08m by 0.1m.

• If your hand is vertical, Pat

pushes to the right and leftequally, so your hand feelszero net force

• The forces cancel in whichever orientation your handis in

• The pressure in a fluid actsequally in all directions, andacts at rights angles to any surface

7. Fluids 4

GAUGE PRESSURE

• In many cases we are interested in the difference between a given pressure and atmospheric pressure

• For example, a flat tire does not have zero pressure in it – the pressure in the tire is atmospheric pressure

• To inflate the tire to, say, 241 kPa, the pressure inside the tire must be greater than atmospheric pressure by this amount: P = 241 kPa + Pat = 342 kPa

• Thus gauge pressure , Pg is Pg = P – Pat

• Thus it is gauge pressure that is determined by the tire gauge

• Example: Derive the formula to calculate the gauge pressure in a basketball by pushing down on it and noting the area of contact it makes with the surface on which it rests.

7. Fluids 5

STATIC EQUILIBRIUM IN FLUIDS:

PRESSURE AND DEPTH (1)

• As a submarine dives deep into the depths of the sea, its hull undergoes increasing pressure

• The increased pressure is due to the added weight of water pressing on the hull as it goes deeper

• Consider a cylindrical container filled to a height h with a fluid of density ρ

• The top surface of the fluid is exposed to the atmosphere with a pressure Pat, and the cross sectional area of the container is A

• The downward force exerted on the top surface exerted by the atmosphere is Ftop = PatA

• At the bottom of the container, the downward force is Ftop plus the weight of the fluid

• For a cylinder of height h and area A, the weight is given by W = Mg = ρVg = ρ(hA)g

• Hence Fbottom = Ftop + W = PatA + ρ(hA)g

• The pressure at the bottom of the container is found by dividing Fbottom by the area A

• Pbottom = Fbottom/A = [PatA + ρ(hA)g]/A = Pat + ρgh

• This relation holds for any depth h below the surface

• P = Pat + ρgh

7. Fluids 6

STATIC EQUILIBRIUM IN FLUIDS:

PRESSURE AND DEPTH (2)

• The relation P = Pat + ρgh can be applied to any two points in a fluid

• If the pressure at one point is P1 and the pressure P2 is at a depth h below the first point then it follows that the pressure at P2 is P2 = P1 + ρgh

7. Fluids 7

PRESSURE AND DEPTH: EXAMPLE

• A cubical box 20.0cm on one side is completely immersed in a fluid. At the top of the box, the pressure is 105kPa; at the bottom the pressure is 106.8kPa. What is the density of the fluid? What type of fluid is being used?

7. Fluids 8

THE BAROMETER (1)

• A barometer makes use of the variation of pressure with depth to measure atmospheric pressure

• It works on the principle of filling a long glass tube (open at one end, and closed at the other) with a fluid of density ρ

• Next, the tube is inverted and its open end is placed below the surface of the same fluid in a bowl

• This leaves an empty space (vacuum) at the top

• Enough of the fluid remains in the tube to create a difference in level, h, between that in the bowl and in the tube

• This height difference is related to the atmospheric pressure pushing down on the fluid in the bowl

• The pressure in the vacuum at the top of the tube is zero

• Thus the pressure at a depth h below the vacuum is

0 + ρgh = ρgh

• Thus at the level of the fluid in the bowl, the pressure is known to be 1 atmosphere , and Pat = ρgh

• A change in Pat would cause a pressure difference between the fluid in the tube and that in the bowl, resulting in a net force and a flow of fluid

• Thus a measurement of h will give the atmospheric pressure

7. Fluids 9

THE BAROMETER (2)

• A fluid that is often used in barometers is mercury (Hg), with a density ρ = 1.3595×104 Pa

• The corresponding height for a column of mercury at 1 atmosphere is h = Pat/ρg = 760mm

• Atmospheric pressure is defined in terms of millimetres of mercury (mmHg)

• 1 atmosphere = Pat = 760 mmHg

7. Fluids 10

FLUIDS SEEK THEIR OWN LEVEL

• In order for fluids to seek their own level, it is necessary that the pressure at the surface of the fluid is the same everywhere over the surface

• This was not the case for the barometer, where the pressure was Pat on one portion, and zero elsewhere

• Consider a U-shaped tube containing a quantity of fluid, density ρ – the fluid rises to the same level in each arm, where it is open to the atmosphere

• The pressure at each base of the arm is the same, and is Pat + ρgh

• The fluid in the horizontal section is pushed with equal force from each side: the fluid remains at rest

• If the two arms of the U are filled to different levels, the pressure at the base of the two arms are different: the greater pressure at the base of the right arm (b)

• The fluid in the horizontal section experiences a net force to the left, causing it to move in that direction, equalising the fluid level in both arms

7. Fluids 11

FLUID LEVELS AND ENERGY

MINIMISATION

• Consider a U-tube that is initially filled to the same level in both arms, and consider moving a small element of fluid from one arm to the other to create different levels

• To extract this fluid element, it is necessary to lift it upward, thus causing its potential energy to increase

• This leads to the conclusion that when the fluid levels are different, the system’s original minimum energy when the levels were the same undergoes and increase

• If two different liquids with different densities are combined in the same U-tube, the levels in the arms are not the same

• However the pressures at the base of each are must be equal

7. Fluids 12

OIL AND WATER DO NOT MIX:

EXAMPLE

• A U-shaped tube is filled mostly with water, but a small amount of vegetable oil has been added to one side. The density of the water is 1.0×103kg/m3. The density of the oil is 9.2×102kg/m3. If the depth of the oil is 5.0cm, what is the difference in level h between the top of the oil on one side of the U and the top of the water on the other side?

7. Fluids 13

PASCAL’S PRINCIPLE

• Recall that P = Pat + ρgh, which is the pressure at a depth h below the surface when a liquid is exposed to atmospheric pressure

• If Pat is increased to Pat + ∆P, the pressure at the depth h is: P = Pat + ∆P + ρgh = (Pat + ρgh) + ∆P

• By increasing the pressure at the top of the fluid by ∆P, we have increased it by the same amount everywhere in the fluid – Pascal’s principle

• An external pressure applied to an enclosed fluid is transmitted unchanged to every point within the fluid

7. Fluids 14

PASCAL’S PRINCIPLE: EXAMPLE –

THE HYDRAULIC LIFT

• In a hydraulic lift, there are two cylinders of cross sectional areas A1 and A2, where A2 > A1

• Each cylinder is fitted with a piston, and are connected by a tube filled with a fluid

• Initially the pistons are at the same level and exposed to the atmosphere

• Suppose a force F1 pushes down on piston 1, causing an increase in pressure in that cylinder: ∆P = F1/A1

• By Pascal’s principle, the pressure in cylinder 2 increases by the same amount; the increased upward force on piston 2 is due to the fluid: F2 = (∆P)A2

• Substituting for ∆P = F1/A1, we find that:

• F2 = (F1/A1)A2 = F1(A2/A1) > F1