PRESSURE. 7. Fluids 2


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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/V kg/m 3 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 1
2 PRESSURE Pressure, P, is a measure of the amount of force, F, per area, A: P = F/A N/m 2 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 m 2, and the area of the needle tip is m 2. Also find the minimum force necessary to pop the balloon with the needle, given that the balloon pops with a pressure of N/m Fluids 2
3 ATMOSPHERIC PRESSURE Atmospheric pressure, P at, is a direct result of the weight of the air above us P at = N/m 2 A shorthand unit for N/m 2 is the pascal (Pa) 1 Pa = 1 N/m 2 (P at = 101 kpa) In British units, pressure is measured in pounds per square inch: P at = 14.7 lb/in 2 A common unit for measuring atmospheric pressure in weather forecasting is the bar: 1 bar = 10 5 Pa 1 P at 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, P at pushes to the right and left equally, so your hand feels zero net force The forces cancel in which ever orientation your hand is in The pressure in a fluid acts equally in all directions, and acts at rights angles to any surface 7. Fluids 3
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 + P at = 342 kpa Thus gauge pressure, P g is P g = P P at 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 4
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 P at, and the cross sectional area of the container is A The downward force exerted on the top surface exerted by the atmosphere is F top = P at A At the bottom of the container, the downward force is F top 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 F bottom = F top + W = P at A + ρ(ha)g The pressure at the bottom of the container is found by dividing F bottom by the area A P bottom = F bottom /A = [P at A + ρ(ha)g]/a = P at + ρgh This relation holds for any depth h below the surface P = P at + ρgh 7. Fluids 5
6 STATIC EQUILIBRIUM IN FLUIDS: PRESSURE AND DEPTH (2) The relation P = P at + ρgh can be applied to any two points in a fluid If the pressure at one point is P 1 and the pressure P 2 is at a depth h below the first point then it follows that the pressure at P 2 is P 2 = P 1 + ρgh 7. Fluids 6
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 7
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 P at = ρgh A change in P at 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 8
9 THE BAROMETER (2) A fluid that is often used in barometers is mercury (Hg), with a density ρ = Pa The corresponding height for a column of mercury at 1 atmosphere is h = P at /ρg = 760mm Atmospheric pressure is defined in terms of millimetres of mercury (mmhg) 1 atmosphere = P at = 760 mmhg 7. Fluids 9
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 P at on one portion, and zero elsewhere Consider a Ushaped 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 P at + 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 10
11 FLUID LEVELS AND ENERGY MINIMISATION Consider a Utube 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 Utube, the levels in the arms are not the same However the pressures at the base of each are must be equal 7. Fluids 11
12 OIL AND WATER DO NOT MIX: EXAMPLE A Ushaped 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 kg/m 3. The density of the oil is kg/m 3. 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 12
13 PASCAL S PRINCIPLE Recall that P = P at + ρgh, which is the pressure at a depth h below the surface when a liquid is exposed to atmospheric pressure If P at is increased to P at + P, the pressure at the depth h is: P = P at + P + ρgh = (P at + 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 13
14 PASCAL S PRINCIPLE: EXAMPLE THE HYDRAULIC LIFT In a hydraulic lift, there are two cylinders of cross sectional areas A 1 and A 2, where A 2 > A 1 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 F 1 pushes down on piston 1, causing an increase in pressure in that cylinder: P = F 1 /A 1 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: F 2 = ( P)A 2 Substituting for P = F 1 /A 1, we find that: F 2 = (F 1 /A 1 )A 2 = F 1 (A 2 /A 1 ) > F 1 7. Fluids 14
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