Fluid Pressure Notes. PDF

LIQUID PRESSURE Hydrostatic pressure: pressure in a fluid at rest  PRESSURE (P) Pressure is defined as the force per unit area, or 𝐹 P= 𝐴 where F is...

LIQUID PRESSURE Hydrostatic pressure: pressure in a fluid at rest  PRESSURE (P) Pressure is defined as the force per unit area, or 𝐹 P= 𝐴 where F is the force applied per unit area and A is the area over which it is applied. The definition is valid for all phases of matter: solid, liquid, or gas. The following are examples of pressure: solid: a model’s high heel exerts pressure on the floor, liquid: water near the bottom of a swimming floor exerts noticeable pressure on a swimmer’s eardrum, gas: atmospheric pressure changes with weather condition  Measurements of pressure are common: Tires must be inflated to the correct pressure Blood pressure should stay within a normal range, Too much pressure in the eye (glaucoma) can cause blindness (Source: Physics, Paul Peter Urone)  UNITS OF PRESSURE The SI unit for pressure is the pascal(Pa), or 1 Pa = 1 N/m2 Other units of pressure are 1 atm = 1.013 X 105 Pa = 760 mm Hg 1.0 mm Hg = 133 N/m2 or Pa 1.0 lb/in2 = 6.90 X 103 N/m2 or Pa 1.0 bar = 1.0 X 105 N/m2 or Pa 1 Torr = 1 mm Hg The liquid pressure (Pl) at a given depth, h, is given by Pl = ρgh where ρ = liquid density g = 9.8 m/s2 h = depth or = the height of the liquid column Example: At a depth of 20 m , what is the liquid pressure? Pl = ρgh = 1000 kg/m3(9.8 m/s2)(20 m) = 196000 N/m2 = 1.96 X 105 Pa  MICROSCOPIC ORIGIN OF PRESSURE In hydrostatics, the fluid is considered at rest as well as any solid - whether it is a vessel containing the fluid or an object submerged in the fluid – which is in contact with the fluid. The atoms and molecules which comprise these liquids and gases, however, are continually in motion. In liquids, the atoms and molecules are closely packed; in gases, the atoms and molecules are much farther apart than in liquids, so they travel greater distances between collisions. It is these collisions of the fast-moving atoms and molecules that cause fluid pressure; it is these collisions of fluid atoms and molecules with the walls of the container that constitute fluid pressure within and on the walls of the container. (Source: General Physics I, Mc Graw-Hill Publication, 3rd ed.)  TOTAL PRESSURE The total pressure (Pt), sometimes called the absolute pressure, is simply the sum of the atmospheric pressure (Patm) and the gauge pressure (Pg). Pt = Patm + Pg Gauge pressure is defined as the pressure above or below atmospheric pressure. 5 1 atm = 1.013 X 10 Pa 2 = 14.7 lb/in  ATMOSPHERIC PRESSURE On the Earth’s surface, an ocean of fluid called air exerts a force over a certain area called atmospheric pressure. This force is due to the weights of the molecules above the surface of the Earth. We are not crushed by this pressure because most of the fluids in our bodies are at approximately the same pressure as the air around us. The pressure of the fluids inside our cells matches the pressure of the surrounding fluids on the cell membrane, so the cells do not rupture. By contrast, the blood pressure in the arteries is as much as 20 kPa higher than atmospheric pressure. The strong, elastic arterial walls are stretched by the pressure of the blood inside; the walls squeeze the arterial blood to keep its higher pressure from being transmitted to other fluids in the body. Changing weather conditions cause variations of approximately 5% in the actual value of air pressure at sea level; 101.3 kPa (1 atm) is only average value. Air pressure also decreases with increasing elevation.  PASCAL’S PRINCIPLE According to Pascal’s Principle: Pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel. In equation form, 𝐹1 𝐹2 P1 = and P2 = 𝐴1 𝐴2 since P1 = P2, therefore, 𝐹1 𝐹2 = 𝐴1 𝐴2 SAMPLE PROBLEM FOR PASCAL’S PRINCIPLE A large piston supports a dentist’s chair, and the dentist wants to lift the patient by stepping on a pedal directly on top of the small piston. Calculate the force the dentist must exert if the patient plus chair have a mass of 120 kg and the small piston has a diameter of 1.0 cm, while the large piston has a diameter of 5.0 cm. Given: F2 = W = mg= (120 kg)(9.8 m/s2) F1 = ? A1 = πr2 = π(d/2)2 = π(1.0 cm/2)2 A2 = πr2 = π(d/2)2 = π(5.0 cm/2)2 From Pascal’s Law, 𝐴1 2 π(1.0 cm/2)2 F1 = F2( ) = (120 kg)(9.8 m/s )( ) = 47.0 N 𝐴2 π(5.0 cm/2)2  MORE SAMPLE PROBLEM 1. In a hydraulic lift, if the radius of the smaller piston is 2.0 cm and the radius of the larger piston is 20.0 cm, what weight can the larger piston support when a force of 250 N is applied to the smaller piston? (Ans: 25 kN) 2. A nurse applies a force of 4.40 N to the piston of the syringe. The piston has an area of 5.00 X 10-5 m2. What is the pressure increase in the fluid within the syringe?(Ans: 88.0 kPa)  MEASURING PRESSURE Blood pressure, the difference between the pressure in the blood and atmospheric pressure, is measured in millimeters of mercury(mm Hg), also called the torr. Millimeters of mercury may seem like a strange units for pressure: how can a force per unit area be equal to a distance (so many mm Hg)? The Manometer A mercury manometer consists of a vertical U-shaped tube, containing some mercury, with one side typically open to the atmosphere and the other connected to a vessel containing a gas whose pressure we want to measure.(Source: General Physics I, pp. 348 – 349)  The pressure at B is PB = PC + ρgh where ρ is the density of mercury. The difference in pressure on the two sides of the manometer is ΔP = PB – PC = ρgh Thus, the difference in mercury levels h is a measure of the pressure difference – commonly reported in millimeters of mercury. The pressure measured when one side of the manometer is open is the difference between atmospheric pressure and the gas pressure rather than the absolute pressure. This difference is the gauge pressure which represented by the formula, Pgauge = Pabs – Patm If h = 1.0 mm, the equivalent pressure in pascals is 1.0 mm Hg = ρgh = (13,600 kg/m3)(9.8 m/s2)(0.0010 m) = 133 Pa  SAMPLE PROBLEM A manometer is attached to a container of gas to determine its pressure. After the container is attached, the mercury (ρ = 13,600 kg/m3) on the side attached to the gas container rises 12 cm above its previous level. (a) What is the gauge pressure of the gas in Pa? (b) What is the absolute pressure of the gas in Pa? (a) Pg = -32,000 Pa = - 32 kPa, (b) 69.3 kPa (a) Pg = ρgh = (13,600 kg/m3)(9.8 m/s2)(-0.24 m) = - 31,987.2 Pa ≈ -32 kPa (b) Pabs = Patm + Pg = 1.013 X 105 Pa + (- 3.2 X 105 Pa) = 69,300 Pa = 69.3 kPa  ANOTHER SAMPLE PROBLEM A mercury manometer is connected to a container of gas. (a) The height of the mercury column on the side connected to the gas is 22.0 cm (measured from the bottom of the manometer). What is the height of the mercury column on the open side if the gauge pressure is measured to be 13.3 kPa? (b) If the gauge pressure of the gas doubles, what are the new heights of the two columns? Ans: (a) 32.0 cm, (b) 17 cm and 37 cm.  ARCHIMEDES’S PRINCIPLE When a body is completely or partially immersed in a fluid, the fluid exerts an upward force on the body equal to the weight of the fluid displaced by the body.  The upward force exerted by a fluid on the object immersed on it is called the buoyant force (FB). FB = the weight of the fluid displaced = mg = ρgV from m = ρV where ρ = density of the fluid V = volume of the fluid displaced The buoyant force is also determined by displacement method. FB = weight of the object in air – weight of the object immersed in the fluid If the weight of the object in air is 500 N and its weight when it is submerged in water is 300 N. The buoyant force is 200 N (from 500 N – 300 N). This is the reason why you find an object less heavier when it is submerged in water. This is due to the buoyant force on the object.  FRACTION OF THE OBJECT SUBMERGED When an object is floating on the fluid, a fraction of it is submerged in the fluid. What fraction of the object is submerged and what fraction is not? When the object floats the weight of the object (Wo) is equal to the weight of the fluid displaced (Wf). Wo = Wf mog = mfg ρoVo = ρfVf ρo/ρf = Vf/Vo ρo/ρf = fraction of object submerged The volume of the object submerged is equal to the volume of the fluid (Vf) displaced.  SAMPLE PROBLEM What fraction of an iceberg is submerged in seawater? The density of ice is 917 kg/m3 and the density of seawater is 1030 kg/m3. ρo/ρf = 917 kg/m3/1030 kg/m3 ρo/ρf = 0.89 Therefore, 89% of iceberg is submerged and only 11% is from the surface of seawater up to the tip of the iceberg.  FLUID FLOW An ideal fluid is a fluid that is incompressible (that is, its density cannot change) and has no internal friction (called viscosity). Liquids are approximately incompressible in most situations, and we may also treat a gas as incompressible if the pressure differences from one region to another are not too great. What is the difference between a moving fluid and a static fluid? One difference is that a moving fluid can exert a force parallel to any surface over or past which it flows; a static fluid cannot. This parallel force is called viscous force which opposes the flow of the fluid; it is equivalent to the kinetic frictional force between solids. We consider only nonviscous fluids about which viscous force are negligibly small. We also ignore surface tension, a phenomenon which arises because the molecules of the liquid exert attractive forces on each other.  TYPES OF FLUID FLOW Fluid flow can be either steady or turbulent. In a steady flow the velocity of the fluid at any point is constant in time. The velocity is not necessarily the same everywhere, but at any particular point, the velocity of the fluid passing that point is constant in time. The density and pressure at any point in a steadily flowing fluid are also constant in time. Steady flow is also described as laminar (A lamina is a thin sheet). In a laminar flow, adjacent layers of fluid slides smoothly past each other and the flow is steady. The path that the fluid follows, starting from any point is called a streamline.  When the fluid velocity at a given point changes, the flow is unsteady or turbulent. In turbulent flow, swirling vortices – whirlpool of fluids- appear. The vortices are not stationary; they move with the fluid. The flow velocity at any point changes erratically; predictions of the direction and speed of fluid under turbulent conditions is difficult. The flow of any fluid is governed by two principles: the continuity equation and the Bernoulli’s equation. The continuity equation is an expression of conservation of mass for an incompressible fluid: since no fluid is created or destroyed, the total mass of the fluid must be constant. Bernoulli’s equation is a form of the energy conservation law applied to fluid flow. Together, these two equations enable us to predict the flow of an ideal fluid.  CONTINUITY EQUATION The continuity equation relates the speed of flow to the cross- sectional area of the fluid. For a steady flow, the mass passing through one section of pipe in time interval Δt must pass through any other section of the pipe in the same time interval. Therefore, Δm1 = Δm2 ρA1v1Δt = ρA2v2Δt A1v1 = A2v2 → the continuity equation The continuity equation for an incompressible fluid equates the volume flow rate past two different points. The same volume of fluid that enters the pipe in a given time interval exits the pipe in the same time interval. Where the radius is large, the speed of the fluid is small: where the radius is small, the fluid speed is large.  SAMPLE PROBLEM FOR CONTINUITY EQUATION The heart pumps blood into the aorta, which has an inner radius of 1.0 cm. The aorta feeds 32 major arteries. If blood in the aorta travels at a speed of 28 cm/s, at approximately what speed does it travel in the arteries? Assume that blood can be treated as an ideal fluid and that the arteries each have an inner radius of 0.21 cm. A1 = πr2aorta A2 = πr2artery A1v1 = A2v2 A1 v2 = v1 A2 π 0.010 2 V2 = 0.28 m/s ( 2) = 0.20 m/s 32π 0.0021  BERNOULLI’S EQUATION Using energy ideas, it can be shown that the pressure of the fluid in one constriction cannot be the same as the pressure before and after the constriction. For horizontal flow the speed is higher where the pressure is lower. This principle is often called the Bernoulli effect. For an incompressible fluid, V = A1Δx1 = A2Δx2 Fluid behind pushes forward, doing positive work, while fluid ahead pushes backward, doing negative work. The total work done by the neighboring fluid is W = P1 A1Δx1 – P2A2Δx2 = (P1 – P2)V  BERNOULLI’S EQUATION: A DERIVATION The energy change is ΔE = ΔK + ΔU 1 = m(v22 – v12) + mg(y2 –y1) 2 Substituting m = ρV and equating the work done on the fluid to the change in its energy yields 1 (P1 – P2)V = ρV (v22 – v12) + ρVg(y2 –y1) 2 Therefore, the Bernoulli’s equation (for ideal fluid flow) is 1 1 P1 + ρgy1 + ρv12 = P2 + ρgy2 + ρv22 2 2 1 (or P + ρgy + ρv2 = constant) 2  BERNOULLI’S EQUATION: AN ANALYSIS Bernoulli’s equation relates the pressure, flow speed, and height at two points in an ideal fluid. Although we derived Bernoulli’s equation in a relatively simple situation, it applies to the flow of any ideal fluid as long as two points 1 and 2 are on the same streamline. Each term in Bernoulli’s equation has units of pressure; each term represents work per unit volume or energy per unit volume. The pressure is the work done by the fluid on the fluid ahead of it per unit 1 volume of flow. The kinetic energy per unit volume is ρv2 and the 2 gravitational potential energy per unit volume is ρgy.  SAMPLE PROBLEMS FOR BERNOULLI’S EQUATION A barrel of rainwater has a spigot near the bottom, at a depth of 0.80 m beneath the water surface. (a) When the spigot is directed horizontally and is opened, how fast does the water come out? (b) If the opening points upward, how high does the resulting “fountain” go? (a) The water at the surface is at atmospheric pressure. The water emerging from the spigot is also at atmospheric pressure since it is in contact with the air. If the pressure of the emerging water were different than that of the air, the stream would expand or contract until the pressure were equal. Since P1 = P2 and y1 – y2 = 0.8 m, Bernoulli’s equation is 1 1 ρgy1 + ρv12 = ρgy2 + ρv22 2 2  The speed of the emerging water is v2. What is v1, the speed of the water at the surface? The water at the surface is moving very slow, since the barrel is draining. The continuity equation requires that A1v1 = A2v2. Since the cross-sectional of the spigot A2 is much smaller than the area of the barrel A1, the speed of the water at the surface is negligibly small compared with v2. Setting v1 = 0, Bernoulli’s equation reduces to 1 ρgy1 = ρgy2 + ρv22 2 After dividing through by ρ, we solve for v2: 1 2 g(y1 – y2) = v 2 2 v2 = 2𝑔 𝑦1 − 𝑦2 = 4.0 m/s (b) We take point 2 to be at the top of the fountain. Then v2 = 0 and Bernoulli’s equation reduces to ρgy1 = ρgy2 where y1 = y2 = 0.8 m.  SAMPLE PROBLEM 1: AN ANALYSIS We find that the ‘fountain” goes right back to the top of the water in the barrel. The result of part (b) is called Torricelli’s Theorem. In reality, the fountain does not reach as high as the original water level; some energy is dissipated due to viscosity and air resistance. We can verify that the speed of found in part (a) is the same as if the water just fell 0.8 m straight down. This shouldn’t be surprising since Bernoulli’s equation is an expression of energy conservation. Use v22 = v12 – 2g(y2 – y1).  NO. 2 SAMPLE PROBLEM A constriction(of cross-sectional area A2) is put in a pipe of normal cross-sectional area A1. Two vertical tubes, open to the atmosphere, rise from two points, one of which is in the constriction. The vertical tubes function like manometers, enabling the pressure to be determined. Suppose that the pipe in question carries water, A1 = 2.0A2, and the fluid heights in the vertical tubes are h1 = 1.20 m and h2 = 0.80 m. (a) Find the ratio of the flow speeds v2/v1. (b) Find the gauge pressure P1 and P2. (c) Find the flow speed v1 in the pipe. (a) From the continuity equation, 𝑣1 𝐴1 = = 2.0 𝑣2 𝐴2 Therefore, the water flows twice as fast as in the rest of the pipe. (b) The gauge pressures are: P1 = ρgh1 = (1000 kg/m3)(9.8 m/s2)(1.20 m) = 11.8 kPa P2 = ρgh2 = (1000 kg/m3)(9.8 m/s2)(0.80 m) = 7.8 kPa (c) Since the tube is horizontal, y1 = y2, and we can ignore the small change in gravitational potential energy density ρgy. Hence, 1 1 P1 + ρv12 = P2 + ρv22 2 2 By substituting v2 = 2.0v1, we get 1 1 P1 + ρv12 = P2 + ρ(2.0v1)2 2 2 Simplifying, P1 –P2 = 1.5ρv12 and finally, 11800 𝑃𝑎 −7800 𝑃𝑎 v1 = 𝑘𝑔 = 1.6 m/s 1.5 (1000𝑚3)  SAMPLE PROB. 2: AN ANALYSIS The assumption that y1 ≈ y2 is fine as long as pipe radius is small compared to the difference between the static water heights (40 cm). Otherwise, we could have to account for the different y values in Bernoulli’s equation. One subtle point: recall that we assumed that the fluid pressure at the bottom of the vertical tubes was the same as the pressure of the moving fluid just beneath. Does that contradict Bernoulli’s equation? Since there is an abrupt change in fluid speed, shouldn’t there be a significant difference in the pressures? No, because these points are not on the same streamline. Practice Problem: Water flows horizontally through a garden hose of radius 1.0 cm at a speed of 1.4 m/s. The water shoots horizontally out of the nozzle of radius 0.25 cm. What is the gauge pressure of the water inside the hose? (Ans: 250 kPa)