1. Fluid Flow conservation and continuity § 12.4

2. Volume Flow Rate Volume per time through an imaginary surface perpendicular to the velocity dV/dtunits: m 3 /s

3. Volume Flow Rate dV/dt = v·A if v is constant over A = ∫v·dA if it is not

4. Flow Continuity Constant mass flow for a closed system 1A1v1 = 2A2v21A1v1 = 2A2v2 dm dt dm dt = 12

5. Flow Continuity For an incompressible fluid: constant   1 A 1 v 1 =  2 A 2 v 2 dm dt dm dt = 12 dV dt dV dt = 12 A 1 v 1 = A 2 v 2

6. Poll Question Where is the velocity greatest in this stream of incompressible fluid? A.Here. B.Here. C.Same for both. D.Can’t tell.

7. Quick Question Where is the density greatest in this stream of incompressible fluid? A.Here. B.Here. C.Same for both. D.Can’t tell.

8. Poll Question Where is the volume flow rate greatest in this stream of incompressible fluid? A.Here. B.Here. C.Same for both. D.Can’t tell.

9. Bernoulli’s Equation Energy in fluid flow § 12.5

10. Incompressible Fluid Continuity condition: constant volume flow rate dV 1 = dV 2 v 1 A 1 = v 2 A 2

11. Poll Question Where is the kinetic energy of a parcel greatest in this stream of incompressible fluid? A.Here. B.Here. C.Same for both. D.Can’t tell.

12. Changing Cross-Section Fluid speed varies –Faster where narrow, slower where wide Kinetic energy changes Work is done!

13. Ideal Fluid No internal friction (viscosity) No non-conservative work!

14. Poll Question Where would the pressure be greatest if the fluid were stationary? A.Here. B.Here. C.Same for both. D.Can’t tell.

15. Work-Energy Theorem W net =  K K 1 + W net = K 2 K 1 + U 1 + W other = K 2 + U 2 W other =  K +  U dW = dK + dU

16. Work done by Pressure dW = F·ds Work done on fluid at bottom: dW 1 = p 1 A 1 ·ds 1 Work done on fluid at top: dW 2 = –p 2 A 2 ·ds 2 = (p 1 – p 2 )dV Total work done on fluid : dW = p 1 A 1 ·ds 1 –p 2 A 2 ·ds 2

17. Kinetic Energy Change Steady between “end caps” Lower cap: dK 1 = 1/2 dmv 1 2 Upper cap: dK 2 = 1/2 dmv 2 2 dm =  dV dK = 1/2  dV (v 2 2 –v 1 2 )

18. Potential Energy Change Steady between “end caps” Lower cap: dU 1 = dmgy 1 Upper cap: dU 2 = dmgy 2 dm =  dV dU =  gdV (y 2 –y 1 )

19. Put It All Together dW = dK + dU (p 1 – p 2 )dV = 1/2  dV (v 2 2 –v 1 2 ) +  gdV (y 2 –y 1 ) (p 1 – p 2 ) = 1/2  (v 2 2 –v 1 2 ) +  g(y 2 –y 1 ) p 1 + 1/2  v 1 2 +  gy 1 = p 2 + 1/2  v 2 2 +  gy 2 – This is a conservation equation – Strictly valid only for incompressible, inviscid fluid

20. What Does It Mean? Faster flow  lower pressure Maximum pressure when static pV is energy

21. Example problem A bullet punctures an open water tank, creating a hole that is a distance h below the water level. How fast does water emerge from the hole?

22. Torricelli’s Theorem p 1 + 1/2  v 1 2 +  gy 1 = p 2 + 1/2  v 2 2 +  gy 2 1/2  v 2 2 =  g(y 2 –y 1 ) + (p 2 –p 1 ) – 1/2  v 1 2 1/2  v 2 2 =  gh v 2 2 = 2gh h v v 2 = 2gh look familiar? 1 2

23. Example problem Water emerges from a downward-facing tap with a diameter of 2.0 cm at a flow rate of 40 L/min. As the water falls, it accelerates downward and the stream becomes thinner. What is the diameter of the stream after it has fallen a distance y from the tap?