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Fluid Flow

Fluid Flow. conservation and continuity. § 12.4. Volume Flow Rate. Volume per time through an imaginary surface perpendicular to the velocity. dV / dt. units: m 3 /s. Volume Flow Rate. dV / dt = v · A if v is constant over A. = ∫ v · d A if it is not. dm. dm. =. dt. dt. 1.

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Fluid Flow

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  1. Fluid Flow conservation and continuity § 12.4

  2. Volume Flow Rate Volume per time through an imaginary surface perpendicular to the velocity • dV/dt • units: m3/s

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

  4. dm dm = dt dt 1 2 Flow Continuity • Constant mass flow for a closed system r1A1v1=r2A2v2

  5. dm dm dV dV = = dt dt dt dt 1 1 2 2 Flow Continuity • For an incompressible fluid: constantr r1A1v1 = r2A2v2 A1v1 = A2v2

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

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

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

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

  10. Incompressible Fluid • Continuity condition: constant volume flow rate • dV1 = dV2 • v1A1 = v2A2

  11. Poll Question Where is the kinetic energy of a parcel greatest in this stream of incompressible fluid? Here. Here. Same for both. 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 begreatest if the fluid were stationary? Here. Here. Same for both. Can’t tell.

  15. Work-Energy Theorem Wnet = DK K1 + Wnet = K2 K1 + U1 + Wother = K2 + U2 Wother = DK + DU dW = dK + dU

  16. Work done on fluid at bottom: dW1 = p1A1·ds1 • Work done on fluid at top: dW2 = –p2A2·ds2 • Total work done on fluid : dW = p1A1·ds1–p2A2·ds2 Work done by Pressure • dW = F·ds = (p1 – p2)dV

  17. Kinetic Energy Change • Steady between “end caps” • Lower cap: dK1 = 1/2dmv12 • Upper cap: dK2 = 1/2dmv22 • dm = rdV • dK = 1/2rdV (v22–v12)

  18. Potential Energy Change • Steady between “end caps” • Lower cap: dU1 = dmgy1 • Upper cap: dU2 = dmgy2 • dm = rdV • dU = rgdV (y2–y1)

  19. Put It All Together • dW = dK + dU • (p1 – p2)dV = 1/2 rdV (v22–v12) +rgdV (y2–y1) • (p1 – p2) = 1/2r(v22–v12) +r g(y2–y1) • p1 + 1/2rv12 + rgy1= p2+ 1/2rv22 +rgy2 • 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. 1 2 h v v2 =    2gh Torricelli’s Theorem p1 + 1/2rv12 + rgy1 = p2 + 1/2rv22 + rgy2 1/2rv22 = rg(y2–y1) + (p2–p1) – 1/2rv12 1/2rv22 = rgh v22 = 2gh look familiar?

  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?

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