The document discusses Bernoulli's equation and its applications in fluid dynamics, including calculations for pressure, velocity, and force in various scenarios. It provides examples involving water and air flow, as well as real-life applications such as airplane wings and fountain pressure. The document also includes several calculations demonstrating how to derive velocities and pressures in different setups.

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

2. 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
2
2
2
1
2
2
2
1
2
1
1
1 v
gh
P
v
gh
P 


 




Derive

3. Bernoulli’s Equation:
(The Data Packet)

4. Example 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?
25.8 m/s
ρgh = ½ρv2
gh = ½v2
v = √(2gh)
Torricelli's Law!!! (demo – bucket with holes)

5. 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 1.00x105
Pa. What is the
a. velocity and b. pressure
when the pipe narrows to
12.0 cm?
32.0 m/s, 0.993E5Pa
(π(0.242
))(2) = (π(0.062
))(v), v = 32.0 m/s
P + ½ρv2
= P + ½ρv2
1.00E5 + .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

6. myth – Bernoulli’s equation
explains all of how airplane wings
generate lift.
http://popphysics.com/chapter-4-fl
uid-mechanics/bernoullis-principl
e-and-airplanes/

7. Bernoulli
1-5

8. 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
1.013x105
Pa.
B. What is the net upward force on the roof?
P = P + ½ρv2
so the difference is just ½ρv2
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)

9. 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 P1 is 2000 PSI (convert), v1 is zero?, P2 is 1.013E5
Pa, solve for v2. 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-xmaPSZ6GM
https://www.youtube.com/watch?v=ejEJGNLTo84
P = P + ½ρv2
(2000/14.7)*1.013E5 = 1.013E5 + .5(1.17)v2
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)

10. 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
, P2 = 1.013E5 Pa
P = P + ρgh
P = 1.013E5 + (1000)(9.81)(23.2) = 328,892 Pa ≈ 3.29E5 Pa
328,892 Pa – 101,300 Pa = 2.28x105
Pa
3.29x105
Pa, 2.28x105
Pa

11. 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
, P2 = 1atm = 1.013E5 Pa. 1 atm = 14.7 PSI
P = P + ρgh + ½ρv2
(21.0+14.7)/14.7*1.013E5 = 1.013E5 + (1000)(9.81)(4.80) + .5(1000)v2
v = 13.973… m/s ≈ 14.0 m/s
14.0 m/s

12. Water flows at 2.00 m/s at an elevation of 1.50 m with a
pressure of 1.15x105
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 + ½ρv2
= P + ρgh + ½ρv2
1.15E5 + (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 ≈ 6.72E4 Pa
6.72x104
Pa