Physics 115A General Physics II Session 4 Fluid flow Bernoulli’s equation • R. J. Wilkes • Email: phy115a@u.washington.edu • Home page: http://courses.washington.edu/phy115a/ 4/4/14 Physics 115 1
Lecture Schedule (up to exam 1) Just joined the class? See course home page Today courses.washington.edu/phy115a/ for course info, and slides from previous sessions 4/4/14 Physics 115A 2
Example: Flotation float • A “string” of undersea oceanographic instruments is held down by an anchor, and pulled upright by a hollow steel sphere float. instruments – The float displaces 1 m 3 of seawater – It weighs 1000N when in air What upward force can it supply to hold up the string? anchor F g = m STEEL g = 1000 N Ocean floor ( ) 9.8 m / s 2 ( ) 1 m 3 ( ) = 10,025 N B = ρ SEAWATER gV FLOAT = 1023 kg / m 3 Net upward force on sphere when free: = B – F g = 9,025 N This is the max weight of instruments it F g B can ‘lift’ 4/4/14 3 Physics 115
Combine several ideas: “Cartesian diver” demo Demonstrates: Archimedes principle, Pascal’s Law, gas law (next week) • You can do this at home with a plastic soda bottle and a small packet of soy sauce or ketchup • “Diver” has some air in it, and just barely floats ( neutral buoyancy ) – Squeeze the bottle: increase P in water, compress air bubble – Compression à smaller V à higher density for diver: sinks – Release: reduced pressure à bubble re-expands à lower density: floats again (Toy said to have been invented by Descartes) Rene Descartes, 1596-1650 For how-to instructions, see http://www.stevespanglerscience.com/lab/experiments/cartesian-diver-ketchup 4/4/14 4 Physics 115
Fluids in motion: continuity equation Incompressible fluid flows in a pipe that gets narrower. What happens? Conservation of matter : mass going into pipe must = mass going out m V Av t Mass passing point 1 in time Δ t: Δ = ρ Δ = ρ Δ 1 1 1 1 1 1 Mass passing point 2 in time Δ t: m A v t Δ = ρ Δ 2 2 2 2 Fluid cannot disappear, or be compressed, so must have ρ 1 A 1 v 1 = ρ 2 A 2 v 2 If density is constant, ρ 2 = ρ 1 A 1 v 1 = A 2 v 2 → 4/4/14 5 Physics 115
Laminar flow vs turbulent flow • As always in physics, we start with the simplest case, and add complications after we solve the easy stuff • Laminar flow = smooth motion of fluid – Paths of particles of fluid (elements of mass Δ m) do not cross – “streamline flow” Streamlines: lines indicate the laminar flow path. Separation between streamlines correlates with pressure and flow speed (more on this later). 4/4/14 6 Physics 115
Turbulence – we wont go into... • Turbulent flow = complex motion of fluid mass elements – Paths of water parcels may cross • Hard to analyze! – Important topic in weather prediction, oceanography, etc – Example of chaotic behavior • parcels of water that start out next to each other have unpredictable locations farther downstream – We have equations describing motion, but they cannot be solved uniquely in most cases • Chaos theory (mathematical toolkit) helps... 4/4/14 7 Physics 115
Bernoulli’s equation: energy considerations in flow • Apply conservation of energy to fluid flow – Suppose fluid speed changes (as usual: incompressible) ρ 1 A 1 v 1 = ρ 2 A 2 v 2 – We know that – Consider a “parcel” of fluid Δ V : length Δ x , area A – If speed changes, a force must have acted: pressure x area Daniel Bernoulli (1700 – 1782) 4/4/14 8 Physics 115
Bernoulli’s equation: changing speed • F 1 = force due to pressure on parcel of fluid Δ V when at location 1, from fluid behind it F 1 = P 1 A Δ V 1 = Δ x 1 A 1 1 • F 2 = force due to pressure on same parcel of fluid Δ V when at location 2, from fluid ahead of it F 2 = P 2 A 2 Δ V 2 = Δ x 2 A 2 • Work done by forces: Δ W 1 = F 1 Δ x 1 = P 1 A 1 Δ x 1 = P 1 Δ V 1 Δ W 2 = − F 2 Δ x 2 = − P 2 Δ V 2 (F 2 points backward) • Incompressible: Δ V 1 = Δ V 2 = Δ V • Net work is thus ( ) Δ V Δ W = Δ W 1 + Δ W 2 = P 1 Δ V − P 2 Δ V = P 1 − P 2 • Net work = change in KE of parcel 2 = 1 where K i = 1 2 Δ W = K 2 − K 1 , 2 Δ mv i 2 ρ Δ V v i 2 − 1 ( ) Δ V ( 1 2 Δ W = Δ K → P 1 − P ) Δ V = 2 ρ v 2 2 ρ v 1 2 2 = P 2 ρ v 2 = constant or 2 P + 1 2 + 1 1 + 1 P 2 ρ v 2 2 ρ v 1 So The Bernoulli Equation (part of it...) 4/4/14 9 Physics 115
Pressure/speed of flow • The part of Bernoulli’s eq’n we did so far tells us that pressure must drop (P 2 < P 1 ) when speed increases – Makes sense: P 1 pushes fluid forward, P 2 pushes backward! • Example: Hose nozzle: circular cross-section, diameter reduced ½ P 2 = 110 kPa (water) Air d 1 = 2 d 2 à A 1 = 4A 2 If v 2 = 25 m/s, Hose what is P 1 and v 1 ? ( ) 25 m / s 1 v 1 = ρ A 2 v 2 → v 1 = A 2 ρ A v 2 = = 12.5 m / s Continuity eqn says: A 4 1 2 = P 2 → P 2 − v 1 ( ) 2 + 1 1 + 1 2 + 1 2 P 2 ρ v 2 2 ρ v 1 1 = P 2 ρ v 2 Bernoulli: 2 − 6.25 m / s ( ) 2 ( ) 25 m / s ( ) + 1 2 1000 kg / m 3 P 1 = 110 kPa ! # ! # " $ " $ ) (586 m 2 / s 2 ) = 403 kPa = 110 kPa + 500 kg / m 3 ( 4/4/14 10 Physics 115
Another part of Bernoulli’s eqn: changing height • Suppose the pipe changes height but not diameter: – The gravitational PE of the fluid changes – Raising the fluid: pressure forces have to do work on the parcel of fluid • Or: gravity does negative work on it (falling) • Work done raising a mass distance y = -mgy Δ W GRAV = − mg ( y 2 − y 1 ) = − ρ Δ Vg ( y 2 − y 1 ) • Total work done on parcel = W done by gravity + W by forces (pressure difference) If pipe area A does not change, v = constant, so no change in KE Δ W TOTAL = Δ W GRAV + Δ W FORCES = Δ K = 0 ( ) Δ V − ρ Δ Vg ( y 2 − y 1 ) = 0 P 1 − P 2 P 2 + ρ gy 2 = P 1 + ρ gy 1 or P + ρ gy = constant 4/4/14 11 Physics 115
Doing The Full Bernoulli The Bernoulli Equation • We can combine both pieces: 2 = P – Pressure vs speed 2 + ρ gh 2 + 1 1 + 1 2 P 2 ρ v 2 1 + ρ gh 2 ρ v 1 – Pressure vs height 1 2 P gh v constant or + ρ + ρ = 2 • This turns out to be conservation of total energy – Multiply both sides by Δ V = A Δ x y 2 y 1 2 = P 1 + 1 2 A 2 Δ x 2 + ρ A 2 Δ x 2 gh 2 + 1 2 P 1 A 1 Δ x 1 + ρ A 1 Δ x 1 gh 2 ρ A 1 Δ x 1 v 1 2 ρ A 2 Δ x 2 v 2 2 = F 2 Δ x 2 + mgh 2 + 1 1 + 1 2 F 1 Δ x 1 + mgh 2 mv 1 2 mv 2 Work on m by PE KE = Work by m PE KE upstream mass on downstream mass 4/4/14 12 Physics 115
Example: raised and narrowed pipe • Garden hose connected to raised narrower hose A 1 = 2 A 2 (so what is ratio of diameters?) Δ y = h = 20 cm, v 1 = 1.2 m/s, P 1 = 143kPa , fluid is fresh water Find P and v at location 2 1 v 1 = A 2 v 2 → v 2 = A 1 v 1 – Use continuity eqn to get v 2 : A = 2 v 1 = 2.4 m / s A 2 – Use Bernoulli to relate P’s : 2 = P 2 → P 2 − v 2 1 + 1 2 + ρ gh 2 + 1 ) + 1 ( 2 ) ( P 1 + ρ gh 2 ρ v 1 2 ρ v 2 2 = P 1 + ρ g h 1 − h 2 2 ρ v 1 2 − 2.4 m / s " % ( 2 ) ) 9.8 m / s 2 − 0.2 m 2 = 143 kPa + 1000 kg / m 3 ( ( ) + 1 P { 1.2 m / s } { } $ ' 2 # & = 143 kPa + − 1,960 Pa − 2,160 Pa ( ) = 138.9 kPa 4/4/14 13 Physics 115
Torricelli’s Law A large tank of water, open at the top, has a small hole through its side a distance h below the surface of the water. Find the speed of the water as it flows out the hole. v = 0 P P P = = a b at a Evangelista Torricelli (1608 - 1647) (P at surface, and outside the hole) 1 2 P gh 0 P gh v + ρ + = + ρ + ρ a a b b b 2 1 2 g h v v 2 g h ρ Δ = ρ = Δ b 2 b Torricelli’s Law: Speed of water jet at depth h is same as speed of object dropped from height h 4/4/14 14 Physics 115
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