Sect. 10-8: Fluids in Motion (Hydrodynamics)

2. Sect. 10-8: Fluids in Motion (Hydrodynamics) • Two types of fluid flow: 1.Laminar or Streamline:(We’ll assume!) 2. Turbulent: (We’ll not discuss!)

3. Streamline Motion

4. Mass flow rate (mass of fluid passing a point per second):ρ1A1v1 = ρ2A2v2  Equation of Continuity PHYSICS:Conservation of Mass!! • Assume incompressible fluid (ρ1 = ρ2 = ρ) Then  A1v1 = A2v2 Or: Av = constant • Where cross sectional area A is large, velocity v is small, where A is small, v is large. • Volume flow rate: (V/t) = A(/t) = Av

5. PHYSICS:Conservation of Mass!! A1v1 = A2v2 Or Av = constant • Small pipe cross section  larger v • Large pipe cross section  smaller v

6. Example 10-11: Estimate Blood Flow rcap = 4  10-4 cm, raorta = 1.2 cm v1 = 40 cm/s, v2 = 5  10-4 cm/s Number of capillaries N = ? A2 = N(rcap)2, A1 = (raorta)2 A1v1 = A2v2  N = (v1/v2)[(raorta)2/(rcap)2] N  7  109

7. Example 10-12: Heating Duct Speed in duct: v1 = 3 m/s Room volume: V2 = 300 m3 Fills room every t =15 min = 900 s A1 = ? A1v1 = Volume flow rate = (V/t) = V2/t  A1 = 0.11 m2

8. Section 10-9: Bernoulli’s Equation • Bernoulli’s Principle (qualitative): “Where the fluid velocity is high, the pressure is low, and where the velocity is low, the pressure is high.” • Higher pressure slows fluid down. Lower pressure speeds it up! • Bernoulli’s Equation (quantitative). • We will now derive it. • NOTa new law. Simply conservation of KE + PE (or the Work-Energy Principle) rewritten in fluid language!

10. Work & energy in fluid moving from Fig. a to Fig. b : a) Fluid to left of point 1 exerts pressure P1on fluid mass M = ρV, V = A11. Moves it 1. Work done: W1 = F11= P1 A11.

11. Work & energy in fluid moving from Fig. a to Fig. b : b) Fluid to right of point 2 exerts pressure P2on fluid mass M = ρV, V = A22. Moves it 2. Work done: W2 = -F22 = -P2A22.

12. Work & energy in fluid moving from Fig. a to Fig. b : a)  b) Mass M moves from height y1 to height y2. Work done against gravity: W3 = -Mg(y1 - y2)

13. Sect. 10-9: Bernoulli’s Eqtn • Total work done from a)  b): Wnet = W1 + W2 + W3  Wnet = P1A11 - P2A22 - Mg(y1-y2) (1) • Recall the Work-Energy Principle: Wnet = KE = (½)M(v2)2 – (½)M(v1)2 (2) • Combining (1) & (2): (½)M(v2)2 – (½)M(v1)2 = P1A11 - P2A22 - Mg(y1-y2) (3) • Note that M = ρV = ρA11 = ρA22 & divide (3) by V = A11 = A22

14.  (½)ρ(v2)2 - (½)ρ(v1)2 = P1 - P2- ρg(y1-y2) (4) • Rewrite (4) as: P1 + (½)ρ(v1)2 + ρgy1 = P2 + (½)ρ(v2)2 + ρgy2  Bernoulli’s Equation • Another form: P+ (½)ρ(v1)2 + ρgy1 = constant • Not a new law, just work & energy of system in fluid language. (Note:P ρg(y2-y1) since fluid is NOT at rest!) Work Done by Pressure = KE + PE

15. Sect. 10-10: Applications of Bernoulli’s Eqtn P1 + (½)ρ(v1)2 + ρgy1 = P2 + (½)ρ(v2)2 + ρgy2  Bernoulli’s Equation Or: P+ (½)ρ(v1)2 + ρgy1 = constant NOTE! The fluid is NOT at rest, so ΔP  ρgh ! • Example 10-13

16. Application #1:Water Storage Tank P1 + (½)ρ(v1)2 + ρgy1 = P2 + (½)ρ(v2)2 + ρgy2 (1) Fluid flowing out of spigot at bottom. Point 1  spigot Point 2  top of fluid v2  0 (v2 << v1) P2  P1 (1) becomes: (½)ρ(v1)2 + ρgy1 = ρgy2 Or, speed coming out of spigot: v1 = [2g(y2 - y1)]½“Torricelli’s Theorem”

17. Application #2:Flow on the level P1 + (½)ρ(v1)2 + ρgy1 = P2 + (½)ρ(v2)2 + ρgy2 (1) • Flow on the level  y1 = y2  (1) becomes: P1 + (½)ρ(v1)2= P2 + (½)ρ(v2)2 (2) (2) Explains many fluid phenomena & is a quantitative statement of Bernoulli’s Principle: “Where the fluid velocity is high, the pressure is low, and where the velocity is low, the pressure is high.”

18. Application #2 a)Perfume Atomizer P1 + (½)ρ(v1)2= P2 + (½)ρ(v2)2 (2) “Where v is high, P is low, where v is low, P is high.” • High speed air (v)  Low pressure (P)  Perfume is “sucked” up!

19. Application #2 b)Ball on a jet of air(Demonstration!) P1 + (½)ρ(v1)2= P2 + (½)ρ(v2)2(2) “Where v is high, P is low, where v is low, P is high.” • High pressure (P) outside air jet  Low speed (v  0). Low pressure (P) inside air jet  High speed (v)

20. Application #2 c)Lift on airplane wing P1 + (½)ρ(v1)2= P2 + (½)ρ(v2)2(2) “Where v is high, P is low, where v is low, P is high.” PTOP < PBOT  LIFT! A1  Area of wing top, A2  Area of wing bottom FTOP = PTOP A1 FBOT = PBOT A2 Plane will fly if ∑F = FBOT - FTOP - Mg > 0 !

21. Application #2 d)Sailboat sailing against the wind! P1 + (½)ρ(v1)2= P2 + (½)ρ(v2)2 (2) “Where v is high, P is low, where v is low, P is high.”

22. Application #2 e)“Venturi” tubes P1 + (½)ρ(v1)2= P2 + (½)ρ(v2)2 (2) “Where v is high, P is low, where v is low, P is high.” Auto carburetor

23. Application #2 e)“Venturi” tubes P1 + (½)ρ(v1)2= P2 + (½)ρ(v2)2 (2) “Where v is high, P is low, where v is low, P is high.” Venturi meter: A1v1 = A2v2(Continuity) With (2) this  P2 < P1

24. Application #2 f)Ventilation in “Prairie Dog Town” & in chimneys etc. P1 + (½)ρ(v1)2= P2 + (½)ρ(v2)2 (2) “Where v is high, P is low, where v is low, P is high.”  Air is forced to circulate!

25. Application #2 g)Blood flow in the body P1 + (½)ρ(v1)2= P2 + (½)ρ(v2)2(2) “Where v is high, P is low, where v is low, P is high.”  Blood flow is from right to left instead of up (to the brain)

26. Problem 46: Pumping water up Street level:y1 = 0 v1 = 0.6 m/s, P1 = 3.8 atm Diameter d1 = 5.0 cm (r1 = 2.5 cm). A1 = π(r1)2 18 m up: y2 = 18 m,d2 = 2.6 cm (r2 = 1.3 cm). A2 = π(r2)2 v2 = ? P2 = ? Continuity:A1v1 = A2v2  v2 = (A1v1)/(A2) = 2.22 m/s Bernoulli: P1+ (½)ρ(v1)2 + ρgy1= P2+ (½)ρ(v2)2 + ρgy2  P2 = 2.0 atm