conservation and continuity Fluid Flow conservation and continuity § 11.7–11.8

Volume Flow Rate Volume per time through an imaginary surface perpendicular to the path DV/Dt units: m3/s

Volume Flow Rate DV/Dt = v·A if v is constant over A

Mass Flow Continuity Constant mass flow for a closed system Dm = Dt 1 2 r1A1v1 = r2A2v2

Flow Continuity For an incompressible fluid: constant r Dm = Dt 1 2 r1A1v1 = r2A2v2 Because r1 = r2, A1v1 = A2v2 DV Dt = 1 2

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

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

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

Bernoulli’s Equation Energy in fluid flow § 11.9–11.10

Incompressible Fluid Recall: Continuity condition: constant volume flow rate DV1 = DV2 v1A1 = v2A2

Poll Question Where is the kinetic energy of a parcel greatest in this stream of incompressible fluid? Up here. Down here. Same for both. Can’t tell.

Changing Cross-Section Fluid speed varies Faster where narrow, slower where wide Kinetic energy changes Work is done (somehow).

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

Poll Question Where would the pressure be greatest if the fluid were stationary? Up here. Down here. Same for both. Can’t tell.

Conservation of Energy K1 + Ug1 + Wnon-g = K2 + Ug2 Wnon-g = K2 + Ug2 – K1 – Ug1 Wnon-g = K2 – K1+ Ug2 – Ug1 Wnon-g = DK + DUg What is this “Wnon-g”? 15

Work done by Pressure W = F·Ds Work done on fluid at bottom: W1 = p1A1·Ds1 Work done on fluid at top: W2 = –p2A2·Ds2 Non-g work done on fluid : Wnon-g = p1A1·Ds1–p2A2·Ds2 = (p1 – p2)DV

Kinetic Energy Change Steady between “end caps” Lower cap: K1 = ½ mv12 Upper cap: K2 = ½ mv22 m = rDV DK = 1/2 rDV (v22–v12)

Potential Energy Change Steady between “end caps” Lower cap: U1 = mgy1 Upper cap: U2 = mgy2 m = rDV DU = rgDV (y2–y1)

Put It All Together Wnon-g = DK + DU (p1 – p2)DV = 1/2 rDV (v22–v12) + rgDV (y2–y1) (p1 – p2) = 1/2 r (v22–v12) + r g(y2–y1) p1 + 1/2 rv12 + rgy1 = p2 + 1/2 rv22 + rgy2 This is a conservation equation Strictly valid only for incompressible, inviscid fluid

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

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?

Torricelli’s Law p1 + 1/2 rv12 + rgy1 = p2 + 1/2 rv22 + rgy2 v22 = 2gh 1/2 rv22 = rg(y2–y1) + (p2–p1) – 1/2 rv12 1/2 rv22 = rgh v22 = 2gh v2 =    2gh look familiar?

Exercise P. 333 Question 65 A Venturi meter is a device for measuring the speed of a fluid within a pipe. A gas with density 1.30 kg/m3 flows at speed v through a pipe with cross section A2 = 0.0700 m2. The meter has cross section A1 = 0.0500 m2 and is substituted for a section of the pipe. The pressure difference is 120 Pa. What is the speed v? What is the volume flow rate?