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. PHYSICS: Conservation of Mass!!
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.
NOT a new law. Simply conservation of KE + PE (or the Work-Energy Principle) rewritten in fluid language!

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

11. W2 = -F22 = -P2A22. Work & energy in fluid moving from Fig. a
to Fig. b :
b) Fluid to right of point 2
exerts pressure P2 on
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)
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.  Bernoulli’s Equation
 (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  ρ g(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 + ρgy (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) 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)
“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)
“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)
“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