1. Chapter 15B - Fluids in Motion
A PowerPoint Presentation by
Paul E. Tippens, Professor of Physics
Southern Polytechnic State University
© 2007

2. The lower falls at Yellowstone National Park: the water at the top of the falls passes through a narrow slot, causing the velocity to increase at that point. In this chapter, we will study the physics of fluids in motion.
Fluid Motion
Paul E. Tippens

3. Objectives: After completing this module, you should be able to:
Define the rate of flow for a fluid and solve problems using velocity and cross- section.
Write and apply Bernoulli’s equation for the general case and apply for (a) a fluid at rest, (b) a fluid at constant pressure, and (c) flow through a horizontal pipe.

4. All fluids are assumed in this treatment to exhibit streamline flow.
Fluids in Motion
All fluids are assumed in this treatment to exhibit streamline flow.
Streamline flow is the motion of a fluid in which every particle in the fluid follows the same path past a particular point as that followed by previous particles.

5. Assumptions for Fluid Flow:
All fluids move with streamline flow.
The fluids are incompressible.
There is no internal friction.
Streamline flow
Turbulent flow

6. Rate of flow = velocity x area
The rate of flow R is defined as the volume V of a fluid that passes a certain cross-section A per unit of time t.
The volume V of fluid is given by the product of area A and vt: vt
Volume = A(vt) A
Rate of flow = velocity x area

7. Constant Rate of Flow
For an incompressible, frictionless fluid, the velocity increases when the cross-section decreases: A1 A2
R = A1v1 = A2v2 v1 v2

8. The area is proportional to the square of diameter, so:
Example 1: Water flows through a rubber hose 2 cm in diameter at a velocity of 4 m/s. What must be the diameter of the nozzle in order that the water emerge at 16 m/s?
The area is proportional to the square of diameter, so:
d2 = cm

9. Example 1 (Cont.): Water flows through a rubber hose 2 cm in diameter at a velocity of 4 m/s. What is the rate of flow in m3/min?
R1 = m3/s
R1 = m3/min

10. Problem Strategy for Rate of Flow:
Read, draw, and label given information.
The rate of flow R is volume per unit time.
When cross-section changes, R is constant.
Be sure to use consistent units for area and velocity.

11. Problem Strategy (Continued):
Since the area A of a pipe is proportional to its diameter d, a more useful equation is:
The units of area, velocity, or diameter chosen for one section of pipe must be consistent with those used for any other section of pipe.

12. The Venturi Meter h A B C
The higher velocity in the constriction B causes a difference of pressure between points A and B.
PA - PB = rgh

13. Demonstrations of the Venturi Principle
Examples of the Venturi Effect
The increase in air velocity produces a difference of pressure that exerts the forces shown.

14. Work in Moving a Volume of FluidP2 A2
Note differences in pressure DP and area DA P1 A1
Volume V A2 P2
, F2 F1 h P1 A1
Fluid is raised to a height h.

15. Net Work = P1V - P2V = (P1 - P2) V
Work on a Fluid (Cont.)
F1 = P1A1
F2 = P2A2 v1 v2 A1 A2 h2 h1 s1 s2
Net work done on fluid is sum of work done by input force Fi less the work done by resisting force F2, as shown in figure.
Net Work = P1V - P2V = (P1 - P2) V

16. Conservation of Energy
F1 = P1A1
F2 = P2A2 v1 v2 A1 A2 h2 h1 s1 s2
Kinetic Energy K:
Potential Energy U:
Net Work = DK + DU
also
Net Work = (P1 - P2)V

17. Conservation of Energy
Divide by V, recall that density r = m/V, then simplify: v1 v2 h1 h2
Bernoulli’s Theorem:

18. Bernoulli’s Theorem (Horizontal Pipe):
Horizontal Pipe (h1 = h2)
h1 = h2 r v1 v2 h
Now, since the difference in pressure DP = rgh,
Horizontal Pipe

19. Bernoulli’s Equation (h1 = h2)
Example 3: Water flowing at 4 m/s passes through a Venturi tube as shown. If h = 12 cm, what is the velocity of the water in the constriction? r
v1 = 4 m/s v2 h
h = 6 cm
Bernoulli’s Equation (h1 = h2)
Cancel r, then clear fractions:
2gh = v22 - v12
v2 = 4.28 m/s
Note that density is not a factor.

20. Bernoulli’s Theorem for Fluids at Rest.
For many situations, the fluid remains at rest so that v1 and v2 are zero. In such cases we have:
P1 - P2 = rgh2 - rgh1
DP = rg(h2 - h1)
This is the same relation seen earlier for finding the pressure P at a given depth h = (h2 - h1) in a fluid. h
r = 1000 kg/m3

21. Torricelli’s Theorem When there is no change of pressure, P1 = P2.
Consider right figure. If surface v2  0 and P1= P2 and v1 = v we have: h1 h2 h
v2  0
Torricelli’s theorem:

22. Interesting Example of Torricelli’s Theorem:v
Torricelli’s theorem:
Discharge velocity increases with depth.
Maximum range is in the middle.
Holes equidistant above and below midpoint will have same horizontal range.

23. Torricelli’s theorem:
Example 4: A dam springs a leak at a point 20 m below the surface. What is the emergent velocity?
Torricelli’s theorem: h
Given: h = 20 m g = 9.8 m/s2
v = 19.8 m/s2

24. Strategies for Bernoulli’s Equation:
Read, draw, and label a rough sketch with givens.
The height h of a fluid is from a common reference point to the center of mass of the fluid.
In Bernoulli’s equation, the density r is mass density and the appropriate units are kg/m3.
Write Bernoulli’s equation for the problem and simplify by eliminating those factors that do not change.

25. Strategies (Continued)
For a stationary fluid, v1 = v2 and we have: h
r = 1000 kg/m3
DP = rg(h2 - h1)
For a horizontal pipe, h1 = h2 and we obtain:

26. Strategies (Continued)
For no change in pressure, P1 = P2 and we have:
Torricelli’s Theorem

27. General Example: Water flows through the pipe at the rate of 30 L/s
General Example: Water flows through the pipe at the rate of 30 L/s. The absolute pressure at point A is 200 kPa, and the point B is 8 m higher than point A. The lower section of pipe has a diameter of 16 cm and the upper section narrows to a diameter of 10 cm. Find the velocities of the stream at points A and B.
8 m A B
R=30 L/s
R = 30 L/s = m3/s
AA = (0.08 m)2 = m3
AB = (0.05 m)2 = m3
vA = 1.49 m/s
vB = 3.82 m/s

28. General Example (Cont.): Next find the absolute pressure at Point B.
R=30 L/s
Given: vA = 1.49 m/s vB = 3.82 m/s PA = 200 kPa hB - hA = 8 m
Consider the height hA = 0 for reference purposes.
PA + rghA +½rvA2 = PB + rghB + ½rvB2
PB = PA + ½rvA2 - rghB - ½rvB2
PB = 200,000 Pa Pa –78,400 Pa – 7296 Pa
PB = 200,000 Pa + ½(1000 kg/m3)(1.49 m/s)2
– (1000 kg/m3)(9.8 m/s2)(8 m) - ½(1000 kg/m3)(3.82 m/s)2
PB = 115 kPa

29. Summary Streamline Fluid Flow in Pipe: Fluid at Rest:
PA - PB = rgh
Horizontal Pipe (h1 = h2)
Fluid at Rest:
Bernoulli’s Theorem:
Torricelli’s theorem:

30. Summary: Bernoulli’s Theorem
Read, draw, and label a rough sketch with givens.
The height h of a fluid is from a common reference point to the center of mass of the fluid.
In Bernoulli’s equation, the density r is mass density and the appropriate units are kg/m3.
Write Bernoulli’s equation for the problem and simplify by eliminating those factors that do not change.

31. CONCLUSION: Chapter 15B Fluids in Motion