1. Conditions/Assumptions for Bernoulli Five conditions A) B) C) D) E)

2. Solution for Bernoulli Assumptions A.Steady state B.No friction C.No work D.Same streamlines E.Newtonian fluid

3. Boat Pitot Tube Problem The speed of a boat is measured by a Pitot tube, as shown below. When traveling in sea water, the tube measures a pressure of 2.5 lb f /in 2. Given the density of sea water is 64 lb m /in 3. Derive the mathematical relationship to show how the velocity of the boat relates to “L”. What is the velocity of the boat in this problem (ft/sec)? L d

4. Solution – Boat Pitot Tube

5. Velocity of fluid on a moving sheet y x u

6. Solution for Moving Sheet Problem

7. Navier Stoke Solution Method Make Good Assumptions Write Down Navier Stokes and simplify Integrate the equations Find the boundary conditions

8. Easy Boundary Conditions The fluid at the surface whether it’s a pipe or plate will move at the velocity of the surface. No Slip Condition (usually 0 because surfaces are generally stationary) The fluid in the middle of the pipe moves the fastest thus dv=0 because it’s the maximum

9. Navier Stoke Assumptions Steady State Constant Density and Viscosity Entrance effects are neglected thus only 1D motion No slip condition

10. Velocity Problem A tank containing acetone at 20°C has a drain pipe connected vertically at the bottom. The pipe is made of galvanized iron, 0.5 in NPS Schedule 40 with length L= 2m. The acetone depth in the tank is H= 5 m. Calculate the velocity and flow rate of acetone at the exit. Remember there is a pipe entrance as a “fitting” in this problem

11. Solution Find Viscosity using eq 1.17 because of a non standard temperature Assumptions P1=P2 thus deltaP=0 Z1=5+2 and z2=0 Standard energy equation simplifies to Gz1=u2^2/2+F F=2*ff*um^2*L/D+2*ff*um^2(25) 25 value comes from pipe fitting

12. Solution Cont Assume a Reynolds in order to calculate ff using charts Find ff values to find velocity Use velocity for new Reynolds. Iterate Answer: 4.53m/s

13. Momentum Balance

14. Solution Law of Continuity Q1=Q2 thus A1=A2 Sum of Forces=Momentum Y direction first to find the mass Fy-W=change in momentum W=98 so m=10 Answer: 10 kg

15. One more Momentum

16. Solutions Fx=-25.8kN Fz=-3522N Momentum Balance Fx+PinAin+PoutAout*cos(45)=- pAoutUout^2*cos45-pAinUin^2 Fz-Welbow-Wh20+PoutAout*sin45=- pAoutUout^2*sin(45)

17. Laminar Problem The distribution of velocity, u, in metres/sec with radius r in metres in a smooth bore tube of 0.025 m bore follows the law, u = 2.5 - kr 2. Where k is a constant. The flow is laminar and the velocity at the pipe surface is zero. The fluid has a coefficient of viscosity of 0.00027 kg/m s. Determine (a) the rate of flow in m 3 /s (b) the shearing force between the fluid and the pipe wall per metre length of pipe.

18. Answers [6.14x10 -4 m 3 /s, 8.49x10 -3 N] Steps Find K through boundary equation at r=.0125 velocity is 0 solve for k Dq=v*da DQ=(2.5-16000r^2)*2pi*rdr from 0 to.0125

19. Shear Force Calc Newton’s law of viscosity F=tau(2pir) Tau=viscosity*du/dr Du/dr=-32000r F=-.00027*32000*.0125*2pir F=8.48E-3 N