Bernoullis Equation Contents How to calculate Whiteboards Bernoullis

Bernoulli’s Equation Contents: • How to calculate • Whiteboards

Bernoulli’s Equation: P = Pressure in Pa ρ = Density of fluid in kg m-3 g = 9. 81 N kg-1 h = Height in m v = velocity in m/s Derive

Bernoulli’s Equation: (The Data Packet)

Example ρgh = ½ρv 2 gh = ½v 2 v = √(2 gh) Torricelli's Law!!! 25. 8 m/s Water with a density of 1000. kg m-3 pours from a very large tank of water from a pipe that is 34. 0 m below the surface of the water. What is its velocity? (demo – bucket with holes)

Example Air with a density of 1. 29 kg m-3 flows at 2. 00 m/s where a duct is 48. 0 cm in diameter under a pressure of 1. 00 x 105 Pa. What is the a. velocity and b. pressure when the pipe narrows to 12. 0 cm? (π(0. 242))(2) = (π(0. 062))(v), v = 32. 0 m/s P + ½ρv 2 = P + ½ρv 2 1. 00 E 5 +. 5(1. 29)(22) = P +. 5(1. 29)(322), P = 99, 342. 1 Pa How is it less? – It has to be. . why does it speed up/slow down? What is the molecule spacing? Standing waves in rivers Demo – 2 papers, blowing over one paper, blower and water tube, blower and ball, flapping flags, waves (Kelvin-Helmholz instability ) airplane wings? ? ? – maybe not 32. 0 m/s, 0. 993 E 5 Pa

myth – Bernoulli’s equation explains all of how airplane wings generate lift. http: //popphysics. com/chapter-4 fluid-mechanics/bernoullisprinciple-and-airplanes/

Bernoulli 1 -5

The wind is moving horizontally at 12. 0 m/s over a level rectangular roof that measures 4. 50 m by 8. 00 m. A. What is the pressure difference between the bottom (still air) and the top (moving air) of the roof surface? Use 1. 29 kg m-3 for the density of the air, neglect the change in height, and assume (if you need to) that the pressure underneath is 1. 013 x 105 Pa. B. What is the net upward force on the roof? P = P + ½ρv 2 so the difference is just ½ρv 2 so the difference is. 5(1. 29)(122) = 92. 88 Pa multiply this by the area of the roof to get the force. 92. 9 Pa, 3340 N (751 lbs)

A very large Nitrogen tank is at 2000. PSI. If nitrogen at STP has a density of 1. 17 kg m-3, how fast is the gas going if the valve breaks off when the tank is horizontal (assume P 1 is 2000 PSI (convert), v 1 is zero? , P 2 is 1. 013 E 5 Pa, solve for v 2. Ignore change in height. ) 1 atm = 14. 7 PSI EC – if the opening is 1. 2 cm in diameter, what thrust does the tank develop? (? ? ? ) https: //www. youtube. com/watch? v=f-xma. PSZ 6 GM https: //www. youtube. com/watch? v=ej. EJGNLTo 84 P = P + ½ρv 2 (2000/14. 7)*1. 013 E 5 = 1. 013 E 5 +. 5(1. 17)v 2 v = 4835. 9 m/s ≈ 4840 m/s F = Δp/t = m Δv/t = (Avρ)v = 3094. 6 N ≈ 3090 N 4835. 9 or roughly 4840 m/s, 3090 N (695 lbs)

What pressure is needed in a fountain if it is spraying water straight up to a height of 23. 2 m? What is the gauge pressure? ρ = 1000. kg m-3, P 2 = 1. 013 E 5 Pa P = P + ρgh P = 1. 013 E 5 + (1000)(9. 81)(23. 2) = 328, 892 Pa ≈ 3. 29 E 5 Pa 328, 892 Pa – 101, 300 Pa = 2. 28 x 105 Pa 3. 29 x 105 Pa, 2. 28 x 105 Pa

A water faucet breaks in the Physics room, spraying water upwards. If the gauge pressure in the water mains is 21. 0 PSI, (at v = 0) with what speed does the water hit the ceiling 4. 80 m above the faucet? How much time does it take a custodian to come down and fix the leak? ρ = 1000. kg m-3, P 2 = 1 atm = 1. 013 E 5 Pa. 1 atm = 14. 7 PSI P = P + ρgh + ½ρv 2 (21. 0+14. 7)/14. 7*1. 013 E 5 = 1. 013 E 5 + (1000)(9. 81)(4. 80) +. 5(1000)v 2 v = 13. 973… m/s ≈ 14. 0 m/s

Water flows at 2. 00 m/s at an elevation of 1. 50 m with a pressure of 1. 15 x 105 Pa through a 10. 0 cm diameter pipe. What is the pressure if it is at an elevation of 5. 00 m going through a 6. 00 cm diameter pipe? (Find the second speed first. ρ = 1000. kg m-3) v = (10/6)2(2. 00 m/s) = 5. 55555 m/s P + ρgh + ½ρv 2 = P + ρgh + ½ρv 2 1. 15 E 5 + (1000)9. 81(1. 50) +. 5(1000)(2. 00)2 = P + (1000)9. 81(5. 00) +. 5(1000)(5. 5555)2 P = 67, 232. 9 Pa ≈ 6. 72 E 4 Pa 6. 72 x 104 Pa