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Lecture 04 bernouilli's principle

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- 1. Lecture 4 Bernouilli’s equation.
- 2. ACT: Aluminum and lead Two blocks of aluminum and lead with identical sizes are suspended from the ceiling with strings of different lengths and placed inside a bucket of water as shown. In which case is the buoyant force greater? A. Al ceiling B. Pb C. It’s the same for both The displaced volume (= volume of the block) is the same in both cases. Al Depth or object density do not play any role. Pb The different weight is compensated with a different tension in the strings.
- 3. Work by pressure As an element of fluid moves during a short interval dt, the ends move distances ds1 and ds2. If the fluid is incompressible, the volume should remain constant: dV dV = Ads1 = A2ds2 1 dV v2 A2 ρ2 A1 v1 ρ1 ds1 ds2 Work by pressure during his motion: dW = p1Ads1 − p2A2ds2 = ( p1 − p2 ) dV 1
- 4. Kinetic and gravitational potential energy Change in kinetic energy: dK = Change in potential energy: 1 1 ρdV v22 − ρdV v12 2 2 ( ) ( ) dU = g ρdV ( y2 − y1 ) y: height of each element relative to some initial level (eg: floor) v2 ρ2 A1 v1 ρ1 A2 ds1 ds2
- 5. Bernouilli’s equation Putting everything together: d ( K + U ) = dWother 1 1 2 ρdV v2 − ρdV v12 + g ρdV y 2 − y1 = p1 − p2 dV 2 2 ( ) ( ) ( 1 1 p1 + ρv12 + g ρ y1 = p2 + ρv22 + g ρ y2 2 2 ) ( p+ ) 1 2 ρv + g ρ y = constant 2 NB: Bernouilli’s equation is only valid for incompressible, nonviscous fluids with a steady laminar flow!
- 6. Static vs flowing fluid Cylindrical container full of water. Pressure at point A (hA below surface): pA = patm + ρ ghA hA xA gauge = pA − patm = ρ ghA Or gauge pressure: pA Now we drill a small hole of radius at depth hA. Point A is now open to the atmosphere! pA = patm hA Ax
- 7. Container with hole Assume the radius of the container is R = 15 cm, the radius of the hole is r = 1 cm and hA = 10 cm. How fast does water come out of the hole? Bernouilli at points A and B (on the surface): 1 2 1 2 pA + ρv A + g ρ yA = pB + ρvB + g ρ yB 2 2 where pA = pB = patm and yB − y A = hA 2 2 vA − vB = 2 ghA (Eqn 1) Continuity at points A and B: AAv A = AvB B r 2vA = R 2vB R = 15 cm hA = 10 cm Ax yB yA (Eqn 2) Bx
- 8. 2 2 v A − vB = 2 ghA 2 r vB = ÷ v A R r v A = R vB 2 2 4 4 r 1 cm = ÷ ÷ = 0.000020 R 15 cm For once, let’s plug in some numbers before the end: Therefore, r 4 2 1 − ÷ v A = 2 ghA R 4 r 1− ÷ ~1 R ⇒ 2 v A = 2 ghA This is equivalent to taking vB ~ 0 (the container surface moves very slowly because the hole is small ―compared to the container’s base) ( vA = 2 ghA = 2 9.8 m/s 2 ) ( 0.10 m) = 1.4 m/s DEMO: Container with holes
- 9. Measuring fluid speed: the Venturi meter A horizontal pipe of radius RA carrying water has narrow throat of radius RB. Two small vertical pipes at points A and B show a difference in level of h. What is the speed of water in the pipe? h Statics: pA − pB = ρ gh ●A flow ● B Continuity: AAv A = AvB B 2 → RAv A = RB2vB 2 equations for vA, vB Bernouilli: pA + 1 2 1 2 ρv A = pB + ρvB 2 2 (y A = yB ) Venturi effect: High speed, low pressure Low speed, high pressure
- 10. 1 2 1 2 ρv A = pB + ρvB 2 2 2 RAvA = RB2vB pA + 1 2 2 ρ vB − v A 2 ( 1 RA ρ 2 RB ) RA vB = R B 2 ÷ vA ÷ = pA − pB 4 2 ÷ − 1v A ÷ = pA − pB vA = and pA − pB = ρ gh 2 gh 4 RA ÷ −1 R ÷ B DEMO: Tube with changing diameter
- 11. Partially illegal Bernouilli Gases are NOT incompressible Bernouilli’s equation cannot be used But… It can be used if the speed of the gas is not too large (compared to the speed of sound in that gas). Ie, if the changes in density are small along the streamline
- 12. Example: Why do planes fly? High speed, low pressure DEMO: Paper and spool. Low speed, high pressure Net force up (“Lift”) ptop + Bernouilli: ( 1 2 1 2 ρv top = pbottom + ρv bottom 2 2 ) Lift = pbottom − ptop ( area of wing ) = ( ρ g ∆h is negligible ) ρ 2 2 v top − v bottom 2 ( ) ( area of wing )
- 13. Aerodynamic grip Race cars use the same effect in opposite direction to increase their grip to the road (important to increase maximum static friction to be able to take curves fast) Tight space under the car ↔ fast moving air ↔ low pressure Higher pressure Lower pressure Net force down
- 14. ACT: Blowing across a U-tube A U-tube is partially filled with water. A person blows across the top of one arm. The water in that arm: A. Rises slightly B. Drops slightly C. It depends on how hard is the blowing. The air pressure at A is lower where the air is moving fast. This is how atomizers work!
- 15. Tornadoes and hurricanes Strong winds ↔ Low pressures vout = 250 mph (112 m/s) vin = 0 2 1 2 1 3 pout − pin = ρv out = 1.2 kg/m ( 112 m/s ) = 7500 Pa 2 2 2 F = ( 7500 Pa ) ( 10 m ) = 7.5 × 105 N Upward force on a 10 m x 10 m roof: ( ) Weight of a 10 m x 10 m roof (0.1 m thick and using density of water –wood is lighter than water but all metal parts are denser): ( )( ) mg = 10 4 kg 10 m/s2 = 105 N The roof is pushed off by the air inside !
- 16. The suicide door Air pressure decreases due to air moving along a surface. The high speed wind will also push objects when the wind hits a surface perpendicularly! Modern car doors are never hinged on the rear side anymore. If you open this door while the car is moving fast, the pressure difference between the inside and the outside will push the door wide open in a violent movement. In modern cars, the air hits the open door and closes it again.
- 17. Curveballs Lower speed (relative to ball) Higher speed (relative to ball) Boomerangs are based on the same principle.
- 18. Beyond Bernouilli In general, when there is no friction (no viscosity), the Ventury effect applies (it’s just a consequence of the work-kinetic energy theorem). In the presence of viscosity, lower speed does not necessarily mean higher pressure. Example: Punctured hose (with steady flow): Ideal fluid (no viscosity) Real fluid (with viscosity) Here friction accounts for the decrease in pressure. (Speed must remain constant due to continuity equation).

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