Density (r) Chapter 10 Fluids. Pressure 1/13/2015


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1 1/13/015 Density (r) Chapter 10 Fluids r = mass/volume Rho ( r) Greek letter for density Units  kg/m 3 Specific Gravity = Density of substance Density of water (4 o C) Unitless ratio Ex: Lead has a sp. Gravity of 11.3 (11.3 times denser than water Calculate the mass of an iron ball of radius 18 cm (V = 4/3 pr 3, r = 7800 kg/m 3 ) You have a 00 g sample of carbon tetrachloride which has a specific gravity of Water s density is 1000 kg/m 3. a) Calculate the density of carbon tetrachloride. b) Calculate the volume of the 00 g sample. (190 kg) (190 kg) Pressure Force per unit area P = F/A Unit  N/m (Pascal) The larger the area, the less the pressure Shoeshoes Elephant feet Bed of nails 1
2 1/13/015 Fluid Pressure Pressure: Example 1 A fluid exerts the same pressure in all directions at a given depth P = rgh The atmosphere is a fluid A water storage tank is 30 m above the water faucet in a house. Calculate the pressure at the faucet: We will neglect the atmospheric pressure since it is the same at the tank and at the surface DP = rgh = (1000 kg/m 3 )(9.8 m/s )(30 m) DP = 9,000 kgm /m 3 s = 9,000 kg m/s m DP = 9,000 N/m Pressure: Example The Kraken can live at a depth of 00 m. Calculate the pressure the creature can withstand: Pressure: Example DP = rgh = (1000 kg/m 3 )(g)(00 m) DP = 1.96 X 10 6 N/m Atmospheric Pressure 1 atm = X 10 5 N/m = kpa 1 bar = 1 X 10 5 N/m (used by meteorologists) Gauge pressure We usually measure gauge pressure Ex: A tire gauge reads 0 kpa, what is the absolute pressure? P = P atm + P G P = kpa + 0 kpa = 31 ka P = P atm + P G Absolute pressure atmospheric pressure Gauge pressure
3 1/13/015 Suppose a submarine is travelling 10.0 m below the surface of the ocean. a) Calculate the gauge pressure at that depth. b) Calculate the absolute pressure at that depth, Straw Example You can pick up soda in a straw using your finger. Why doesn t the soda fall out? Another Straw Example Pascal s Principle What pushes soda up a straw when you drink through it? Pressure applied to a confined fluid increases the pressure the same throughout P in = P out F in = F out A in A out Pascal s Principle A hydraulic lift can produce 00 lb of force. How heavy a car can be lifted if the area of the lift is 0 times larger that the input of the Force? Pascal s Principle A hydraulic lift has a large piston 30.0 cm in diameter and a small piston cm in diameter. a) Calculate the force required to lift a 1500 kg car. b) Calculate the pressure in the confined liquid. F in = F out A in A out F out = F in A out = (00 lb) (0) = 4000 lbs A in 1 3
4 1/13/015 Pascal s Principle An a hydraulic press, the large piston has a cross sectional area of 00 cm and the small an area of 5 cm. If a force of 50 N is applied to the small piston, calculate the force on the large piston. P = P atm + P G P = P atm + rgh Torrecilli s Work What is the highest column of water that the atmosphere can support? P = P atm + rgh 0 = X 10 5 N/m + (1000kg/m 3 )(9.8m/s )(h) h = 10.3 m No vacuum pump can pump more than ~30 feet Try the same calculation with mercury P = P atm + rgh 0 = X 10 5 N/m + (13,600kg/m 3 )(9.8m/s )(h) h = m (760 mm) 1 atm = 760 mm Hg (760 torr) Can an astronaut attach suction cups to the boots of his spacesuit to help him climb around the space shuttle while in space? 4
5 1/13/015 Buoyancy Buoyancy The lift provided by water Objects weight less in water than out Caused by pressure differential between top and bottom of an object. F bouyant = rgv Derivation of the Buoyancy Formula F b = F F 1 P = F/A F = PA F = rgha F b = rgh A rgh 1 A F b = rga(h  h 1 ) F b = rgv Archimedes Principle The bouyant force on an object equals the weight of fluid displaced by the object w = weight of an object in water (or any liquid) w = mg  F b Buoyancy: Example 1 A 7000kg ancient statue lies at the bottom of the sea. Its volume is 3.0 m 3. How much force is needed to lift it? F b = rgv F b = (1000 kg/m 3 )(9.8 m/s )(3.0m 3 ) F b =.94 X 10 4 kgm/s F b =.94 X 10 4 N F b mg w = mg  F b w = (7000 kg)(9.8m/s ) .94 X 10 4 N w = 3.9 X 10 4 N A block of wood massing 7.6 kg is tied to a string and immersed in water. The wood has a density of 750 kg/m 3. Say, isn t w just the sum of the forces? Yep. F b a) Calculate the volume of the object b) Calculate the buoyant force on the wood. c) Calculate the tension in the string. SF = w mg 5
6 1/13/015 Archimedes tested a crown for the king. Out of water, it masses 14.7 kg. In water, it massed 13.4 kg. Was the crown gold? w = m cr g F b w = m cr g rgv cr (13.4 kg)(g) = (14.7 kg)(g) (1000 kg/m 3 )(g)(v cr ) 131 N = 144 N (9800 kg/ms )(V cr ) V cr = m 3 Now we can calculate the density of the crown: r = m/v = 14.7 kg/ m 3 r = 11,053 kg/m 3 Gold s density is about 19,000kg/m 3. This is much closer to lead. A metal ball weighs N in air, and N in water. Calculate the density of the metal. Floating Objects that are less dense than water will float Part of the object will be above the water line A case of static equilibrium SF = 0 F b (3840 kg/m 3 ) mg Floating A 100 kg log is floating in water. What volume of the log is under water? SF = 0 V log = m (Hey, the g s cancel!) r V log = 100 kg = 1. m kg/m 3 SF = 0 = mg F b F b mg = F b mg = rgv log V log = mg rg mg 6
7 1/13/015 Floating A wooden raft has a density of 600 kg/m 3, an area of 5.7 m, and a volume of 0.60 m 3. How much of the raft is below water in a freshwater lake? Let s first calculate the mass of the raft: r = m/v m = rv = (600 kg/m 3 )(0.60 m 3 ) = 360 kg Now we can worry about the raft. SF = 0 SF = 0 = mg F b mg = F b mg = rgv submerged mg = rgh submerged A mg = rgh submerged A m = rh submerged A (Hey, the g s cancelled!) h submerged = m/ra h submerged = 360 kg = m (1000 kg/m 3 )(5.7 m ) Floating: Example 3 Suppose a continent is floating on the mantle rock. Estimate the height of the continent above the mantle (assume the continent is 35 km thick). SF = 0 = mg F b 0 = m c g r man gv c(submerged) m c g = r man gv c(submerged) m c = r man V c(submerged) We don t know the mass of the continent r c = m c /V c(total) m c = r c V c(total) m c = r man V c(submerged) m c = r man V c(submerged) m c = r c V c(total) r man V c(submerged) = r c V c(total) V c(submerged) = r c = (800 kg/m 3 ) = 0.85 V c(total) r man (3300 kg/m 3 ) This means that 85% of the continent is submerged, and only 15% is above: (0.15)(35 km) = 5.5 km 7
8 1/13/015 Fluid Flow Laminar Flow Smooth, streamline flow (laminar means in layers ) Turbulent Flow erratic flow with eddies Viscosity Internal friction of a liquid High viscosity = slow flow Viscosity is NOT the same as density Equation of Continuity Equation of Continuity A 1 v 1 = A v A 1 v 1 = A v A = Area of a pipe v = velocity of the liquid Fluid will flow faster through a smaller opening Placing your finger over a hose opening. The term Av is the volume rate of flow Eqn. Of Continuity: Example 1 A = m v = m/s Av = m 3 /s A garden hose has a radius of 1.00 cm and the water flows at a speed of 0.80 m/s. What will be the velocity if you place your finger over the hose and narrow the radius to 0.10 cm? A 1 = pr = (3.14)(0.01 m) = 3.14 X 104 m A = pr = (3.14)(0.001 m) = 3.14 X 106 m 8
9 1/13/015 A 1 v 1 = A v v = A 1 v 1 A v = (3.14 X 104 m )(0.80 m/s) = 80 m/s (3.14 X 106 m ) Eqn. Of Continuity: Example A water hose 1.00 cm in radius fills a 0.0liter bucket in one minute. What is the speed of water in the hose? A 1 = pr = (3.14)(1 cm) = 3.14 cm Remember that Av is volume rate of flow. A v =0.0 L 1 min 1000 cm 3 = 333 cm 3 /s 1 min 60 s 1 L A 1 v 1 = A v v 1 = A v /A 1 v 1 = 333 cm 3 /s = 3.14 cm 160 cm/s or 1.60 m/s 10 m 3 /h of water flows through a pipe with 100 mm inside diameter. The pipe is reduced to an inside dimension of 80 mm. a) Convert the flow rate to m 3 /s. b) Calculate the initial velocity (Av = flow rate(m 3 /s) c) Calculate the velocity in the narrow section of the pipe. (A 1 v 1 = A v ) Eqn. Of Continuity: Example 3 A sink has an area of about 0.5 m. The drain has a diameter of 5 cm. If the sink drains at 0.03 m/s, how fast is water flowing down the drain? A d = pr = (p)(0.05 m) = 1.96 X 103 m 3 A d v d = A s v s v d = A s v s /A d =[(0.5 m )(0.03 m/s)]/(1.96 X 103 m 3 ) v d = 3.8 m/s Eqn. Of Continuity: Example 4 The radius of the aorta is about 1.0 cm and blood passes through it at a speed of 30 cm/s. A typical capillary has a radius of about 4 X 104 cm and blood flows through it at a rate of 5 X 104 m/s. Estimate how many capillaries there are in the human body. 9
10 1/13/015 A a v a = NA c v c (N is the number of capillaries) Eqn. Of Continuity: Example 5 A a = pr = (3.14)(0.01 m) = 3.14 X 104 m A c = pr = (3.14)(4 X 106 cm) = 5.0 X cm How large must a heating duct be to replenish the air in a room 300 m 3 every 15 minutes? Assume air moves through the vent at 3.0 m/s. N = A a v a / A c v c N = (3.14 X 104 m )(0.30 m/s) = ~ 4 billion (5.0 X cm )(5 X 104 m/s) A d v d = A r v r A r v r = volume rate of flow: A r v r = 300 m 3 1 min = m 3 /s 15 min 60 s Bernoulli s Principle The velocity and pressure of a fluid are inversely related. A d = A r v r /v d A d = m 3 /s = 0.11 m 3.0 m/s Applications of Bernoulli s Principle Why does a shower curtain sometimes attack a person taking a shower? 1. Airplane wing What will happen to closed windows during a tornado? Will they blow in or out? 10
11 1/13/015 Applications of Bernoulli s Principle. Prairie Dog Burrows 1. Air moves faster (lower pressure) at the top. Draws air through the burrow 3. The exact same thing happens with our chimneys Applications of Bernoulli s Principle 3. Spray Paint Flow of air (low pressure) Applications of Bernoulli s Principle 4. Dime in a cup Bernoulli s Equation P t + ½rv t + rgy t = P b + ½rv b + rgy b (note: you often have to use the Eqn. Of Continuity in these situations:) A 1 v 1 = A v Often useful when you have both a change in height and area: If there is no change in altitude, the equation simplifies: Pipe from a water reservoir to a house Pipe from a house into a sewer pipe P t + ½rv t + rgy t = P b + ½rv b + rgy b P t + ½rv t = P b + ½rv b 11
12 1/13/015 Bernoulli s Equation: Example 1 A water heater in the basement of a house pumps water through a 4.0 cm pipe at 0.50 m/s and 3.0 atm. What will be the flow speed and pressure through a.6 cm spigot on the second floor, 5.0 m above? 3.0 atm X 10 5 N/m = 3.0 X 10 5 N/m 1 atm Flow speed: A t v t = A b v b v t = A b v b/ A t (Remember A = pr ) v t = pr bv b (Hey, the p s cancel!) pr t v t = r bv b = (0.0 m) (0.50m/s) = 1. m/s r t (0.013 m) Now the pressure: P t + ½rv t + rgy t = P b + ½rv b + rgy b P t + ½(1000)(1.) + (1000)(9.8)(5) = 3.0X ½(1000)(0.50) + (1000)(9.8)(0) P t =.5 X 10 5 N/m Bernoulli s Equation: Example A drunken redneck shoots a hole in the bottom of an aboveground swimming pool. The hole is 1.5 m from the top of the tank. Calculate the speed of the water as it comes out of the hole. y t = 1.5 m y b = 0 m The top of the pool is a much larger area than the hole. We will assume that the v t = 0. P t + ½rv t + rgy t = P b + ½rv b + rgy b P t + rgy t = P b + ½rv b + rgy b Also, both the top and the hole are open to the atmosphere, so P t = P b Set the bottom of the pool as y b = 0. rgy t = ½rv b + rgy b rgy t = ½rv b v b = rgy t /r v b = gy t v b = ()(9.8m/s )(1.5 m) v b = 5.4 m/s P t + rgy t = P b + ½rv b + rgy b rgy t = ½rv b + rgy b 1
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