1. Continuity of Fluid Flow & Bernoulli’s Principle

2. The Flow of a Fluid The flow of a fluid is very complicated. Many aspects of fluid flow are not well understood yet. We will limit our study to these three assumptions: 1)Incompressible fluids – this means the fluid does not change its density. 2)Steady (laminar) flow – this means the fluid flows in layers (streamlines) rather than in a chaotic fashion as in turbulent flow. 3)Non-viscous fluids – this means there is no friction in the fluid. A non- viscous fluid would be like water as opposed to pancake batter.

3. Mass Flow Rate (kg/s) Volume Flow Rate (m 3 /s) A ≡ Cross-sectional Area (m 2 ) m ≡ mass (kg) t ≡ time (s) v≡ velocity (m/s) V ≡ volume (m 3 ) ρ ≡ density (kg/m 3 ) Continuity of Fluid Flow As a fluid flows through a pipe which changes its cross-sectional area, its mass flow rate must remain constant. If the fluid is incompressible, then the volume flow rate must remain constant.

4. Relates the pressure (Pa), the height of the fluid (m) and flow speed of the fluid (m/s) at one point in a laminar flow to another point in a laminar flow. A special case of Bernoulli’s Principle occurs when the flow speed of a fluid is zero at one point in the flow and the pressure at two points is the same. In this case Toricelli’s Equation gives the speed of the fluid at the other point.

5. Momentum & Energy Example 29: Fluid Velocity & Pressure in a Pipe Water is flowing through a 19-cm diameter pipe at 75-m/s. The pipe’s diameter decreases to 10-cm. The pipe is level. a)What is the speed of the air in the 10-cm diameter pipe? I’m baaaaaaaaaaack!!!!!!! First, draw a picture of the pipe and label it. Don’t forget to convert the diameters to meters. D 1 = 0.19-m D 2 = 0.10-m v 1 = 75-m/sv2v2 Now apply continuity of fluid flow and solve for v 2, just like all good AP Physics students do!!! Calculate the areas, A 1 & A 2. Remember to use the radius of each rather than the diameter. Finally, substitute into the equation for v 2 and simplify. Change this

6. Momentum & Energy Example 29: Fluid Velocity & Pressure in a Pipe Water is flowing through a 19-cm diameter pipe at 75-m/s. The pipe’s diameter decreases to 10-cm. The pipe is level. b)If you wanted the airflow to reach the speed of sound (345-m/s), what would the diameter of the smaller pipe need to be? Draw and label the picture!! D 1 = 0.19-m D2D2 v 1 = 75-m/sv 2 =345-m/s Apply continuity of fluid flow and solve for the area at point 2, A 2. Calculate A 1, and substitute into the equation to find A 2. Now find the radius and then the diameter.

7. Momentum & Energy Example 29: Fluid Velocity & Pressure in a Pipe Water is flowing through a 19-cm diameter pipe at 75-m/s. The pipe’s diameter decreases to 10-cm. The pipe is level. c)The pressure in the large diameter pipe is 1.2-atm. If the 10-cm pipe is placed 1.5-m above the 19-cm diameter pipe, what will the pressure in the 10-cm pipe be? Draw and label the diagram. Continuity of fluid flow still gives you a speed of 270-m/s at the 10-cm end. h =1.5-m v 1 = 75-m/s v 2 = 270-m/s Convert the pressure to Pa. P 1 = 1.2 X 10 5 -Pa Write out Bernoulli’s Equation. We will assume the density of the water remains constant. (1000-kg/m 3 ) Since h 1 = 0, we can drop this term as we solve for P 2. Substitute in and simplify.

8. Momentum & Energy Example 30:Water Flow from a Tank A water tank with a valve at the bottom is shown below. Assume the cross-sectional area at A is very large compared with that at B, and the acceleration of gravity is 9.81-m/s 2. a)What is the speed the water leaves the spigot at B after the valve is opened? Apply Bernoulli’s principle. At A & B, the pressure is the same so we can eliminate these terms. We need to find h B using a bit of trigonometry. We can also assume the speed at A is zero. v A = 0-m/s Solve the equation for v B. This is Torricelli’s Equation. Substitute and find v B.

9. Momentum & Energy Example 30:Water Flow from a Tank A water tank with a valve at the bottom is shown below. Assume the cross-sectional area at A is very large compared with that at B, and the acceleration of gravity is 9.81-m/s 2. b)What is the maximum height above the opening of the spigot (∆y max ) attained by the water stream coming out of the spigot at B after the valve is opened? This is just a projectile motion problem. So we can use our equations of motion. At maximum height v y = 0, so solving for ∆y max we get this equation. We need to use trigonometry, to find v oy. then substitute to find the solution. 10-m/s v oy = 10sin(49) = 7.5-m/s)