Bernoulli’s Principle Steady, Incompressible, Non-Viscous fluid Work-Kinetic Energy Theorem ( W = KE) E = ½ mv2 + mgy Use to relate Speed, Pressure, & Elevation at different points in fluid flow: Can explain; Why when the speed of a fluid increases, the internal pressure in a fluid decreases. Wing air lift…

Bernoulli’s Principle P1 + ½v12 + gy1 = P2 + ½v22 + gy2 Or P + ½v2 + gy = a constant © Laura Fellman, PCC Rock Creek Campus

Can we apply energy concepts to fluids? 1) Potential Energy per unit volume Ug = mgh Ug / V = ρgh 2) Kinetic Energy per unit volume KE = (½)mv2 KE/V = (½)ρv2 3) Pressure = F/A P = (F.x) / V = Work / V Proof… W = ΔK + ΔU W / V = ΔK / V + ΔU / V ~ Bernoulli’s Principle!

What happens to Bernoulli’s Equation if the fluid is not moving? Let v1 = v2 (i.e. Static fluid) h P1 P1 + ρ gh1+ (½) ρ v12 = P2 + ρ gh2 + (½) ρ v22 P1 + ρ gh1 = P2 + ρ gh2 P2 = P1 + ρgΔh Static pressure Equation!

What happens to Bernoulli’s Equation if the elevation is constant? P1 + ρ gh1+ (½) ρ v12 = P2 + ρ gh2 + (½) ρ v22 P1 = P2 + (½) ρ v22 - (½) ρ v12 P1 = P2 + (½) ρ[v22 - v12] Due to the continuity principle: v2 > v1 so P1 > P2

Bernoulli Equation Example Water is flowing through as S-shaped pipe with the properties shown below. If water has a density of 1000 kg/m3, what pressure is it exerting on the pipe at position 2? P1 + ½ ρv 12 + ρgh1 = P2 + ½ ρv22 + ρgh2

Bernoulli Equation Example P1 + ½ ρv 12 + ρgh1 = P2 + ½ ρv22 + ρgh2 150,000 Pa + ½ (1,000 kg/m3)(5.0 m/s)2 + ρgh1 = P2 + ½ (1,000 kg/m3)(10 m/s)2 + (1,000 kg/m3)(9.8 m/s2)(2.0 m) P2 = 127,900 Pa

② ① P1 > P2 Hence the lift… P1 + ρ gh1+ (½) ρ v12 = P2 + ρ gh2 + (½) ρ v22

P1 + ρ gh1+ (½) ρ v12 = P2 + ρ gh2 + (½) ρ v22 Wing air lift… ② ① h2 ~ h1 v2 > v1 P1 > P2 Hence the lift… P1 + ρ gh1+ (½) ρ v12 = P2 + ρ gh2 + (½) ρ v22

① ? ② P2 > P1 P1 + ρ gh1+ (½) ρ v12 = P2 + ρ gh2 + (½) ρ v22

② ① P1 + ρ gh1+ (½) ρ v12 = P2 + ρ gh2 + (½) ρ v22

Fun demos – can you explain the effect?

Use Bernoulli’s equation … P1 + ρgh1 + ½ρv12 = P2 + ρgh2 + ½ρv22 [1] A water pipe of cross-section 0.002 m2 carries water into the basement of a house at a speed of 1.0 m/s and pressure 180 kPa. If the pipe tapers to 0.001 m2 and rises to the ground floor 4m above the basement pipe, calculate the water pressure on the ground floor. Use Bernoulli’s equation … P1 + ρgh1 + ½ρv12 = P2 + ρgh2 + ½ρv22 180000 + ½(1000)(2)2 = P2 +(1000)(9.8)(4) + ½(1000)(1)2 P2 = 142 300 pascals v2 = 2m/s

[2] Hurricane If the speed of the wind is 50 m/s and the density of the air is 1.29 kg/m3, Find the reduction in air pressure due to the wind. If the area of the roof measures 10m x 20m, what is the net upward force on the roof? [3] (a) What is the pressure drop due to Bernoulli effect as water goes into a 3 cm diameter nozzle from a 9 cm diameter fire hose while carrying a flow of 40 L/s? (b) To what maximum height above the nozzle can this water rise neglecting air resistance (hint: use projectile motion concepts).

Why curveballs curve Bernoulli's equation can be used to explain why curveballs curve. Let's say the ball is thrown so it spins. As air flows over the ball, the seams of the ball cause the air to slow down a little on one side and speed up a little on the other. The side where the air speed is higher has lower pressure, so the ball is deflected toward that side. To throw a curveball, the rotation of the ball should be around a vertical axis. It's a little more complicated than that, actually. Although the picture here shows nice streamline flow as the air moves left relative to the ball, in reality there is some turbulence. The air does exert a force down on the ball in the figure above, so the ball must exert an upward force on the air. This causes air that travels below the ball in the picture to move up and fill the space left by the ball as it moves by, which reduces drag on the ball.

[4] Water is pumped from a pipeline 2 m above the ground to a water tower 15 m above the ground. If the pipeline velocity is 8 m/s, its pressure is 3.103 x 105 Pa, and water enters the tank at a pressure of one atmosphere, with what velocity does the water enter the tank? Ans: 15.09 m/s [5] A gas is flowing through a pipe whose cross-sectional area is 0.07 m2. The gas has a density of 1.30 kg/m3. A Venturi meter is used to measure the speed of the gas. It has a cross-sectional area of 0.05 m2. The pressure difference between the pipe and the Venturi meter is found to be 120 Pa. Find a) the speed of the gas in the pipe b) the volume rate of flow of the gas. Ans: 13.87 m/s; 0.97 m3/s

[6] Using the value of atmospheric pressure at sea level, 1 x 105 Pa, estimate the total mass of the earth's atmosphere above a 5 m2 area. A. 5 x 104 kg B. 9 x 102 kg C. 2 x 10–4 kg D. 4 x 10–2 kg [7]  To drink from a waterhole, a giraffe lowers its head 4.8 m from the upright position. What is the change in blood pressure at the brain when the giraffe raises its head back to the upright position? A. 980 kPa B. 48.9 kPa C. 5.0 x 104 Pa D. 101.3 Pa [8] A balloon inflated with helium gas (density = 0.2 kg/m3) has a volume of 6 x 10–3 m3. If the density of air is 1.3 kg/m3, what is the buoyant force exerted on the balloon? A. 0.01 N B. 0.08 N C. 0.8 N D. 1.3 N [9] A 2 kg block displaces 10 kg of water when it is held fully immersed. The object is then tied down as shown in the figure; and it displaces 5 kg of water. What is the tension in the string? A. 10 N B. 20 N C. 30 N D. 70 N

Variation in Pressure with Depth

P1 + ρ gh1+ (½) ρ v12 = P2 + ρ gh2 + (½) ρ v22

Common cases: Find the velocity of a liquid flowing out of a spigot: The pressure is the same P1 = P2 The velocity v1 = 0 Bernoulli’s equation is: This is called Torricelli’s Theorem.

= 9.9 m/s R = v2 A2 = 9.9 (π (1.5x10-2)2) = 7x10-3 m3/s (60 s/min) What volume of water will escape per minute from an open top tank through an opening 3.0 cm in diameter that is 5.0 m below the water level in the tank? r = 1.5x10-2 m h1 = 5 m v1 = 0 m/s h2 = 0 = 9.9 m/s R = v2 A2 = 9.9 (π (1.5x10-2)2) = 7x10-3 m3/s (60 s/min) = 4.2x10-1 m3/min

② ① Bernoulli's equation + Torricelli's law: does the speed of the fluid change if we change the area of the hole but not the height?

The velocity would stay the same no matter whether the area is slightly increased or decreased provided the hole is at the same height and is quantified by V= (2gh)0.5 And at the same time continuity principle isn't violated. Where you've gone wrong is comparing the same thing. It should be like .. Case 1 : Area increased Velocity stays the same, but the flow rate increases. Area at exit point (A2) increased and velocity (V2) stays the same. Similarly cross-sectional area of the tank (A1) is the same but the velocity with which the water moves down (V1) increases emptying the tank quickly as the flow rate is high. So (A1V1)=(A2V2) ------ since if A2 raises then V1 raises. Case 2 : Area decreased Velocity stays the same, but the flow rate decreases. Area at exit point (A2) decreased and velocity (V2) stays the same. Similarly cross-sectional area of the tank (A1) is the same but the velocity with which the water moves down (V1) decreases emptying the tank comparatively slowly as the flow rate is low. So (A1V1)=(A2V2) ------ since if A2 decrease then V1 too decreases. https://physics.stackexchange.com/questions/285059/bernoullis-equation-torricellis-law-does-the-speed-of-the-fluid-change-if-w