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Physics 203 – College Physics I Department of Physics – The Citadel Physics 203 College Physics I Fall 2012 S. A. Yost Chapter 10 Part 2 Pascal’s Principle, Archimedes’ Principle, Buoyancy Physics 203 – College Physics I Department of Physics – The Citadel Announcements Thursday: exam on chapters 7 – 9 (only the sections covered in the homework) Today: Problem set 10A was due. (You remembered, right?) Problem set 10B is scheduled for the Tuesday after Thanksgiving, but this depends on what we discuss in today’s class. Watch the due date for possible changes. Physics 203 – College Physics I Department of Physics – The Citadel Atmospheric Pressure The weight of the air above us produces atmospheric pressure at sea level equal to 1 atm = 1.013 × 105 N/m2. Pressure is also measured in Pascals: 1 Pa = 1 N/m2. Physics 203 – College Physics I Department of Physics – The Citadel Gauge Pressure Pressure gauges are normally set to zero when only atmospheric pressure is present. Gauge pressure is then the additional pressure beyond that due to the atmosphere. The total pressure including atmospheric pressure is the absolute pressure. Physics 203 – College Physics I Department of Physics – The Citadel Suction Pump A negative gauge pressure corresponds to suction. If we produce a negative gauge pressure on a straw, water will be “sucked” up the straw. P < Patm h Why does the water go up straw? the Physics 203 – College Physics I Department of Physics – The Citadel Suction Pump What is the highest a suction pump can draw water up a tube? The absolute pressure can’t be less than zero. Assume the pressure in the tube is reduced to P = 0. P < Patm h Physics 203 – College Physics I Department of Physics – The Citadel Suction The water must be supported by atmospheric pressure. Pump P=0 rgh = 1 atm = 1.013 × 105 N/m2 h= 1.013 × 105 N/m2 (1000 kg/m3)(9.8 m/s2) = 10.3 m. h P = rgh Physics 203 – College Physics I Department of Physics – The Citadel Hydraulic Lift A hydraulic lift is a simple machine which uses the fact that any fluid pushed into a cylinder on one end must push out the same volume into a cylinder at the other end. V L1 V L2 Physics 203 – College Physics I Department of Physics – The Citadel Hydraulic Lift Since the volumes are the same, the distances the cylinders move is related to their areas: A1 L1 = V = A2 L2. → L2/L1 = A1/A2. A1 V L1 A2 V L2 Physics 203 – College Physics I Department of Physics – The Citadel Hydraulic Lift The work done by a force on one cylinder will equal the work done by the other cylinder. F2 V F1 L1 = W = F2 L2. V F1 L1 L2 Physics 203 – College Physics I Department of Physics – The Citadel Hydraulic Lift Mechanical advantage F2 F2 / F1 = L1 / L2 = A2 / A1. A1 V F1 L1 A2 V L2 Physics 203 – College Physics I Department of Physics – The Citadel Pascal’s Principle Know: F2 / F1 = A2 / A1 F2 Rewrite: A2 F2/A2 = F1/A1 → V DP2 = DP1. A1 F1 V Physics 203 – College Physics I Department of Physics – The Citadel Pascal’s Principle When an force is applied to a closed vessel, the pressure increases by the same amount throughout A2 the vessel. F2 DP A1 F1 DP = F1/A1 = F2/A2 Physics 203 – College Physics I Department of Physics – The Citadel Pascal’s Demonstration d Pascal demonstrated his principle by inserting a thin 12 m long tube of diameter 6 mm into a wine barrel of diameter 40 cm. h Filling the tube with water caused the barrel to burst! (a) What was the weight of water in the tube? (b) What was the force on the lid? D Physics 203 – College Physics I Department of Physics – The Citadel Pascal’s Demonstration d = 0.006 m What was the weight of the water in the tube? h = 12 m m = r (pr2) h r = 0.003 m = 1000 kg/m3 × (2.8 × 10-5 m2) × 12 m = 340 g. mg = 0.340 kg × 9.8 m/s2 = 3.33 N (0.75 lb) D Physics 203 – College Physics I Department of Physics – The Citadel Pascal’s Demonstration d h What was the force on the lid of the barrel? The force on the bottom of the tube is D F1 = mg = 0.75 lb on area A1 = pd2/4 (= 2.8 x 10-5 m2) Physics 203 – College Physics I Pascal’s Demonstration Department of Physics – The Citadel d h The force on the lid of the barrel is F2 = mg (A2/A1) with A2 = pD2/4 Then F2 = mg (D/d)2 Area is proportional to the length squared. D Physics 203 – College Physics I Department of Physics – The Citadel Pascal’s Demonstration d = 0.006 m h The force on the lid of the barrel is F2 = mg (D/d)2 = 0.75 lb (0.40 m / 0.006 m)2 = 0.75 lb × 4400 = 3300 lb big number! D = 0.4 m Physics 203 – College Physics I Department of Physics – The Citadel Pressure in water Balancing the net force of pressure and the weight for an imaginary box of water gives F2 – F1 = mw g where F1 = P1 A and F2 = P2 A. F1 V F2 submerged box Physics 203 – College Physics I Department of Physics – The Citadel Buoyancy Suppose I replace the imaginary box of water by an actual box. What is the net force due to pressure on the box? F2 – F1 = mw g , just as before! This is the buoyant force. Archimedes Principle: The buoyant force equals the weight of the water displaced. F1 V F2 submerged box Physics 203 – College Physics I Floating Example A plastic block with specific gravity 0.3 floats in water. What fraction of the block’s volume is under water? The weight of the block equals the buoyant force. Department of Physics – The Citadel Physics 203 – College Physics I Department of Physics – The Citadel Floating Example V If the volume of the block is V and the volume under water is Vw, we need to find Vw/V. Vw Use FB = rw Vwg = mg = rVg. Then rw Vw = rV, and Vw/V = r/rw = SG = 0.3 30% of the block is under water. Physics 203 – College Physics I Department of Physics – The Citadel Buoyancy Puzzle Two identical cups have water in them, to the same level. But one also A: has a plastic block floating in it. Which is heavier? A B C they are the same B: Physics 203 – College Physics I Department of Physics – The Citadel Buoyancy Puzzle A: V The total mass in case A is MA = rV What is the mass in case B? Don’t guess – use physics. B: v1 v2 Physics 203 – College Physics I Department of Physics – The Citadel Buoyancy Puzzle Remember Archimedes’ Principle: A: The weight of the plastic block equals the weight of the water that would B: have occupied the submerged volume of the block. Physics 203 – College Physics I Department of Physics – The Citadel Buoyancy Puzzle This means the mass of the A: whole block equals the mass of water that would occupy volume v2. Therefore the two cups have the same total mass. B: V v1 v2 Physics 203 – College Physics I Department of Physics – The Citadel Buoyancy Puzzle Another way to look at it: The force on the bottom of the cup is the weight of the contents. The force on the bottom of the cup is also the pressure times the area. Both have the same depth, so they have the same pressure. A: V B: v1 v2 Physics 203 – College Physics I Department of Physics – The Citadel Volume Rate of Flow The volume rate of flow (also called flux) is defined to be the volume of fluid that flows past a point in a pipe per unit time. Q = DV / Dt = Av where A is the area, v the flow velocity. DV A v Physics 203 – College Physics I Department of Physics – The Citadel Volume Rate of Flow The volume rate of flow is measured in m3/s. Another common unit is liters/second (L/s) 1 liter = 1000 mL = 1000 cm3 1 m3 = 1000 L. DV Physics 203 – College Physics I Department of Physics – The Citadel Volume Rate of Flow The volume rate of flow of an incompressible fluid is the same throughout a pipe. Q DV Physics 203 – College Physics I Department of Physics – The Citadel Volume Rate of Flow Q = Av is fixed: the fluid’s velocity is inversely proportional to the crosssectional area. v3 v1 v2 DV DV A DV Physics 203 – College Physics I Department of Physics – The Citadel P2 v2 Bernoulli Principle Suppose a volume V of incompressible fluid is pushed into a pipe with a pressure difference and height difference between the ends. h P1 v1 The same volume of fluid will be pushed out the other end. Physics 203 – College Physics I Department of Physics – The Citadel Bernoulli Principle Pushing a volume V of fluid into a pipe under pressure P does work on it: W = F L = F V/A = PV. F L P V A Physics 203 – College Physics I Department of Physics – The Citadel Bernoulli’s Principle is an expression of energy conservation: P2 v2 Bernoulli Principle PV + ½ m v2 + mgh = constant. h constant. P1 + potential energy = v1 Work + kinetic energy Physics 203 – College Physics I Department of Physics – The Citadel P2 v2 Bernoulli Principle Dividing by volume, P + ½ r v2 + rgh = constant. h Pressure + kinetic energy density P1 constant. v1 + potential energy density = Physics 203 – College Physics I Department of Physics – The Citadel Question 1. A fluid flows through the pipe shown. In which section is the flow velocity the greatest? Selections: A A B B C D The same C Physics 203 – College Physics I Department of Physics – The Citadel Answer The volume rate of flow Q = vA is constant for an incompressible fluid. The fluid moves fastest where the pipe is narrowest, section B. (It moves slowest in section C.) A B C Physics 203 – College Physics I Department of Physics – The Citadel Question 2. In which section is the pressure of the fluid the greatest? Selections: A A B B C D The same C Physics 203 – College Physics I Department of Physics – The Citadel Answer This is Bernoulli’s principle: The pressure in a fluid decreases when the flow velocity increases. The fluid moves most slowly at C, so the pressure is highest there (and lowest at B). A B C Physics 203 – College Physics I Department of Physics – The Citadel Water Tower and Fountain A water tower feeds a fountain, which shoots water straight up in the air. h How fast does the water leave the fountain? Assume the top of the water is a height h = 55 m above the fountain. v Physics 203 – College Physics I Department of Physics – The Citadel Water Tower and Fountain 1 We’ll assume the tank is big, so the top of the water stays fixed: h h1 = 55 m, v v1 = 0, P1 = 0 gauge pressure Physics 203 – College Physics I Department of Physics – The Citadel Water Tower and Fountain 1 At the fountain, h2 = 0 v2 = v unknown h What is P2? v 2 Physics 203 – College Physics I Department of Physics – The Citadel Water Tower and Fountain 1 Careful! This is not hydrostatics. If the fountain were turned off, the pressure would be P2 = rgh = 1000 kg/m3 x 9.8 m/s2 x 55 m = 5.4 x 105 N/m2. h no flow! 2 Physics 203 – College Physics I Department of Physics – The Citadel Water Tower and Fountain 1 When the fountain is flowing, this changes! The pressure just outside the pipe is P2 = 0, normal atmospheric pressure. h P2 v 2 Physics 203 – College Physics I Department of Physics – The Citadel Water Tower and Fountain 1 The velocity is given by Bernoulli’s equation with P1 = P2 = 0 h1 = h, h2 = 0 v2 = 0, v1 = v h P2 v 2 Physics 203 – College Physics I Department of Physics – The Citadel Water Tower and Fountain 1 The only terms remaining are ½ r v2 = rgh h The result is the same as if the water had fallen from the top of the tower: v = √ 2gh = 33 m/s. v 2 Physics 203 – College Physics I Department of Physics – The Citadel Water Tower and Fountain 1 What is the volume rate of flow if the pipe has diameter 1 cm? h Q = Av A = p (0.5 cm)2 = 0.785 cm2 v 2 Physics 203 – College Physics I Department of Physics – The Citadel Water Tower and Fountain 1 Q = Av A = 0.785 cm2 h v = 33 m/s = 3300 cm/s Q = 2600 cm3 /s = 2600 mL /s = 2.6 L/s v 2 Physics 203 – College Physics I Department of Physics – The Citadel Water Tower and Fountain How high does the water rise from the fountain? 1 2 Bernoulli’s equation between points 1 and 2: P1 = P2 = 0, v1 = v2 = 0 implies rgh1 = rgh2. h v Physics 203 – College Physics I Department of Physics – The Citadel Water Tower and Fountain The water rises to the height of the tower. 1 2 This assumes energy conservation: No friction (viscosity or air resistance) or turbulence is considered. h v Physics 203 – College Physics I Department of Physics – The Citadel Physics 203 – College Physics I Department of Physics – The Citadel Water Tower and Fountain What is the pressure inside the pipe feeding the fountain? If there is no nozzle constricting the flow, then v is the same in both places, so the pressure must be zero both inside and outside the pipe. v 2 1 v Physics 203 – College Physics I Department of Physics – The Citadel Water Tower and Fountain Putting a nozzle on the hose would change this. Then v is bigger outside the hose, and the pressure is higher inside the hose. If the height is about the same, then v2 2 P1 + ½ rv12 = P2 + ½ rv22. 1 v1 Physics 203 – College Physics I Department of Physics – The Citadel Water Tower and Fountain The gauge pressure in the hose is P1 - P2 = ½ r(v22 – v12). v2 Continuity: A1 v1 = A2 v2. P1 – P2 = ½ rv2 2 (1 - 2 A2 A22/A12) A1 1 v1 Physics 203 – College Physics I Department of Physics – The Citadel Water Tower and Fountain What is the gauge pressure in the pipe, assuming it’s diameter is still 1 cm at point 1, but the nozzle has diameter 1/8 cm at point 2. v2 2 P1 – P2 = ½ rv22 (1 - A22/A12) A2/A1 = (1/8)2 = 1/64 A2 A1 1 v1 Physics 203 – College Physics I Department of Physics – The Citadel Water Tower and Fountain P1 – P2 = ½ rv22 (1 - A22/A12) = 0.5 (1000 kg/m3)(33 m/s)2 (1 – 1/4096) = 5.4 x 105 Pa = 0.9998 Note that P1 – P2 ≈ rgh = 5.4 x 105 Pa The pressure inside the pipe is nearly the same as when the pipe is shut off entirely! v2 2 1 v1 Physics 203 – College Physics I Department of Physics – The Citadel Water Tower and Fountain If v2 is the same with or without the nozzle, the water shoots as high either way. What good is the nozzle? h v2 Physics 203 – College Physics I Department of Physics – The Citadel Water Tower and Fountain This was an idealized problem: we assumed the water loses no energy in the pipe, not matter how fast it goes. In reality, there is viscosity, a fluid friction that increases with speed. And 33 m/s is really fast! Bernoulli’s equation would be a poor approximation without the nozzle. h v2 v1 Physics 203 – College Physics I Department of Physics – The Citadel Water Tower and Fountain The nozzle allows the water to move more slowly through the pipe: h v1 = A2v2/A1 = v2/64 = 0.52 m/s just inside the nozzle, decreasing the energy lost to viscosity. v2 v1