Physics 201 Chapter 13 Lecture 1 Fluid Statics Pascal s Principle Archimedes Principle (Buoyancy) Fluid Dynamics Continuity Equation Bernoulli Equation 11/30/2009 Physics 201, UW-Madison 1
Fluids Density = Mass/Volume ρ = M / V units = kg/m 3 Pressure (P) P = Force/Area [N/m 2 ] 1 N/m 2 = 1 Pascal (Pa) Pressure variation with depth P = ρ g h Atmospheric Pressure Even when there is no breeze air molecules are continuously bombarding everything around - results in pressure normal atmospheric pressure = 1.01 x 10 5 Pa (14.7 lb/in 2 ) 11/30/2009 Physics 201, UW-Madison 2
Compressiblity Density & Pressure are related by the Bulk Modulus LIQUID: incompressible (density almost constant) GAS: compressible (density depends a lot on pressure) B = Δp ( ΔV / V ) 11/30/2009 Physics 201, UW-Madison 3
Variation of pressure with depth m = ρv; V = Ah m = ρah P = F A = mg A ( ρah)g ; i.e., P = A P = hρg True for all shapes of containers 11/30/2009 Physics 201, UW-Madison 4
Pascal s Principle A change in pressure in an enclosed fluid is transmitted undiminished to all the fluid and to its container. This principle is used in hydraulic system P 1 = P 2 (F 1 / A 1 ) = (F 2 / A 2 ) F 2 11/30/2009 Physics 201, UW-Madison 5
Pascal s Principle This principle is used in hydraulic system F 1 P 1 = P 2 (F 1 / A 1 ) = (F 2 / A 2 ) Can be used to achieve a mechanical advantage F 2 = F 1 (A 2 / A 1 )» Work done is the same: height by which the surface A 2 rises is smaller than the change in the height of surface with area A 1. A 1 F 2 A 2 11/30/2009 Physics 201, UW-Madison 6
Using Fluids to Measure Pressure Use Barometer to measure Absolute Pressure Top of tube evacuated (p=0) Bottom of tube submerged into pool of mercury open to atmosphere (p=p 0 ) Pressure dependence on depth: h = p 0 ρg Barometer Use Manometer to measure Gauge Pressure Measure pressure of volume (p 1 ) relative to the atmospheric pressure ( gauge pressure ) The height difference (Δh) measures the gauge pressure: Δh = (p p ) 1 0 ρg 1 atm = 760 mm (29.9 in) Hg = 10.3 m (33.8 ft) H 2 0 p 1 Manometer p 0 Δh 11/30/2009 Physics 201, UW-Madison 7
Measurement of Pressure Manometer If both sides of an U-tube are open to atmosphere the levels of the fluid are the same on both sides If one side is connected to a pressurized side the level difference between the two sides can be used to measure pressure. 11/30/2009 Physics 201, UW-Madison 8
Measuring the tire pressure: Is this a manometer or a barometer? 11/30/2009 Physics 201, UW-Madison 9
Measuring Blood Pressure Blood pressure is quite high, 120/80 mm of Hg Use higher density fluid in a manometer: Mercury 11/30/2009 Physics 201, UW-Madison 10
Archimedes Object immersed in a fluid is subject to a buoyant force. Force on sides cancel Force on top F t = ρgh T A Force on bottom F b = ρgh B A ΔF = ρg A Δh F B = (mg) disp 11/30/2009 Physics 201, UW-Madison 11
Archimedes Object immersed in a fluid is subject to a buoyant force. Force on sides cancel Force on top F t = ρgh T A Force on bottom F b = ρgh B A ΔF = ρg A Δh F B = (mg) disp 11/30/2009 Physics 201, UW-Madison 12
Float Weight of object = ρ 0 gv Buoyant force is the weight of the displaced fluid Weight of fluid = ρ f gv Displace just enough fluid such that forces = 0! 11/30/2009 Physics 201, UW-Madison 13
Buoyant Force (B) Archimedes Principle weight of fluid displaced (P=F/A, P=ρgh)» B = ρ fluid g V displaced» W = ρ object g V object» object sinks if ρ object > ρ fluid» object floats if ρ object < ρ fluid» Eureka! If object floats.» B=W» Therefore ρ fluid g V displaced = ρ object g V object» Therefore V displaced /V object = ρ object / ρ fluid 11/30/2009 Physics 201, UW-Madison 14
Float Buoyant force is the weight of the displaced fluid Weight of object = ρ Ice V total g Weight of fluid = ρ SeaWater gv submersed Displace just enough fluid such that forces = 0! 11/30/2009 Physics 201, UW-Madison 15
The weight of a glass filled to the brim with water is W b. A cube of ice is placed in it, causing some water to spill. After the spilled water is cleaned up, the weight of the glass with ice cube is W a. How do the weights compare: 1. W b > W a. 2. W b < W a. 3. W b = W a. Archimedes Principle Archimedes Principle: The buoyant force on an object equals the weight of the fluid it displaces. Weight of water displaced = Buoyant force = Weight of ice 11/30/2009 Physics 201, UW-Madison 16
Question Suppose you float a large ice-cube in a glass of water, and that after you place the ice in the glass the level of the water is at the very brim. When the ice melts, the level of the water in the glass will: 1. Go up causing the water to spill. 2. Go down. 3. Stay the same. Archimedes Principle: The buoyant force on an object equals the weight of the fluid it displaces. Weight of water displaced = Buoyant force = Weight of ice When ice melts it will turn into water of same volume 11/30/2009 Physics 201, UW-Madison 17
Buoyancy Two cups hold water at the same level. One of the two cups has plastic balls (projecting above the water surface) floating in it. Which cup weighs more? 1) Cup I 2) Cup II 3) Both the same Cup I Cup II Archimedes principle tells us that the cups weigh the same. Each plastic ball displaces an amount of water that is exactly equal to its own weight. 11/30/2009 Physics 201, UW-Madison 18
Sunken Balls Two identical glasses are filled to the same level with water. Solid steel balls are at the bottom in one of the glasses. Which of the two glasses weighs more? 1. The glass without steel balls 2. The glass with steel balls 3. Both glasses weigh the same The steel balls sink. The buoyant force equal to the weight of the displaced water is not sufficient to counter the weight of the steel balls. Therefore, the glass with steel balls weighs more. 11/30/2009 Physics 201, UW-Madison 19
Buoyant force and depth Imagine holding two identical bricks under water. Brick A is just beneath the surface of the water, while brick B is at a greater depth. The force needed to hold brick B in place is: 1. larger 2. the same as 3. smaller than the force required to hold brick A in place. The buoyant force on each brick is equal to the weight of the water it displaces and does not depend on depth. 11/30/2009 Physics 201, UW-Madison 20
Fluid Flow Fluid flow without friction Volume flow rate: ΔV/Δt = A Δd/Δt = Av (m 3 /s) Continuity: A 1 v 1 = A 2 v 2 i.e., flow rate the same everywhere e.g., flow of river 11/30/2009 Physics 201, UW-Madison 21
Problem Two hoses, one of 20-mm diameter, the other of 15-mm diameter are connected one behind the other to a faucet. At the open end of the hose, the flow of water measures 10 liters per minute. Through which pipe does the water flow faster? 1. The 20-mm hose 2. The 15-mm hose 3. Water flows at the same speed in both cases 4. The answer depends on which of the two hoses comes first in the flow When a tube narrows, the same volume occupies a greater length. For the same volume to pass through points 1 and 2 in a given time, the velocity must be greater at point 2. The process is reversible. 11/30/2009 Physics 201, UW-Madison 22
Faucet A stream of water gets narrower as it falls from a faucet (try it & see). V 1 A 1 A 2 V 2 The velocity of the liquid increases as the water falls due to gravity. If the volume flow rate is conserved, them the cross-sectional area must decrease in order to compensate The density of the water is the same no matter where it is in space and time, so as it falls down and accelerates because of gravity,the water is in a sense stretched, so it thins out at the end. 11/30/2009 Physics 201, UW-Madison 23
Streamlines 11/30/2009 Physics 201, UW-Madison 24
Continuity equation Δm 1 = ρ 1 ΔV 1 = ρ 1 Av 1 Δt Volume Flow rate Mass flow rate I V = ΔV Δt I M 1 = Δm 1 Δt = Av = ρ 1 Av 1 In steady state Δm 1 = Δm 2 Δt Δt General case: mass may be accumulated or decreased in the volume between A1 and A2 I M 2 I M 1 = dm 2 dt dm 1 dt = dm 12 dt Continuity equation 11/30/2009 Physics 201, UW-Madison 25
Bernoulli s Equation Pressure drops in a rapidly moving fluid whether or not the fluid is confined to a tube For incompressible, frictionless fluid: P + 1 2 ρv2 + ρgh = constant 1 2 ρv2 = 1 1 2 mv2 V = KE V ρgh = mgh V = PE V Bernoulli equation states conservation of energy For Static Fluids:P 1 + ρgh 1 = P 2 + ρgh 2 Bernoulli's Principle (constant depth):p 1 + 1 2 ρv 2 1 = P 2 + 1 2 ρv 2 2 11/30/2009 Physics 201, UW-Madison 26