Fluid Mechanics with teacher's notes

More documents

Recommendations

Info

$Holt Math Minnesota Test Prep Workbook for Grade 9$

$Math Skills: More Practice$

The expression for the conservation of energy in fluids is called Bernoulli’s equation, and it can be expressed as follows: BERNOULLI’S EQUATION P + ⎯ 1 2 ⎯rv 2 + rgh = constant pressure + kinetic energy per unit volume + gravitational potential energy per unit volume = constant along a given streamline Note that Bernoulli’s equation differs slightly from the law of conservation of energy given in Chapter 5. First of all, two of the terms on the left side of the equation look like the terms for kinetic energy and gravitational potential energy, but they contain density, r, instead of mass, m. That is because the conserved quantity in Bernoulli’s equation is energy per unit volume—not just energy—and density is equivalent to mass per unit volume. This statement of the conservation of energy in fluids also includes an additional term: pressure, P. Note that the units of pressure are equivalent to the units for energy per unit volume. If you wish to compare the energy in a given volume of fluid at two different points, Bernoulli’s equation takes the following equivalent form: P 1 + ⎯ 1 2 ⎯rv 1 2 + rgh1 = P 2 + ⎯ 1 2 ⎯rv 2 2 + rgh2 Bernoulli’s principle is a special case of Bernoulli’s equation Two special cases of Bernoulli’s equation are worth mentioning here. First, if the fluid is not moving, then both speeds are zero. This case is a static situation, such as a column of water in a cylinder. If the height at the top of the column, h 1, is defined as zero, and h 2 is the depth, then Bernoulli’s equation reduces to the equation for pressure as a function of depth, introduced in Section 9-2: P 1 = P 2 + rgh 2 (static fluid) Second, imagine again a fluid flowing through a horizontal pipe with a constriction. Because the height of the fluid is constant, the gravitational potential energy does not change. Bernoulli’s equation then reduces to the following: P 1 + ⎯ 1 2 ⎯rv 1 2 = P2 + ⎯ 1 2 ⎯rv 2 2 (horizontal pipe) This equation suggests that if v 1 is greater than v 2 at two different points in the flow, then P 1 must be less than P 2. In other words, the pressure decreases as speed increases. This is Bernoulli’s principle again, which we now can see as a special case of Bernoulli’s equation. The conditions required for this case tell us that Bernoulli’s principle is strictly true only when elevation is constant. Copyright © by Holt, Rinehart and Winston. All rights reserved. P1A1 y1 ∆x 1 v1 Bernoulli’s Principle MATERIALS LIST ✔ single sheet of paper You can see the effects of Bernoulli’s principle on a sheet of paper by holding the edge of the sheet horizontally and blowing across the top surface. The sheet should rise as a result of the lower air pressure above the sheet. NSTA TOPIC: Bernoulli’s principle GO TO: www.scilinks.org sciLINKS CODE: HF2094 y2 ∆x2 v2 P 2A 2 Figure 9-15 As a fluid flows through this pipe, it may change velocity, pressure, and elevation. SECTION 9-3 Teaching Tip The equivalence of units in Bernoulli’s equation can be seen more clearly if you multiply the units of pressure, N/m 2 , by a factor equivalent to unity, m/m: N ⎯ m 2⎯ × ⎯ m m ⎯ = = ⎯ N •m J ⎯m 3 m 3⎯ TEACHER’S NOTES For this lab to be effective, it is best to hold a piece of very light paper or soft plastic with both hands and blow over it through a straw. Demonstration 9 Atomizer Purpose Demonstrate an application of Bernoulli’s principle. Materials glass jar, water, plastic straw, bicycle-horn bulb attached to thin hollow tube Procedure Place the straw in the water. Use the bulb and tube to blow air across the top of the straw. Water will come up the straw and spray out. The increased air flow above the straw results in a pressure decrease inside it. Fluid Mechanics 335 335
SECTION 9-3 Classroom Practice The following may be used as teamwork exercises or for demonstration at the chalkboard or on an overhead projector. PROBLEM Bernoulli’s equation A camper creates a shower by attaching a tube to the bottom of a hanging bucket that is open to the atmosphere on top. If the water level in the tank is 3.15 m above the end of the tube (the shower head), then what is the speed of the water exiting the tube? Answer 7.86 m/s A pipe narrows from a cross section of 2.0 m 2 to 0.30 m 2 .If the speed of the water flowing through the wider area of the pipe is 8.0 m/s, what is the speed of the water flowing through the narrow part? Answer 53 m/s Assuming incompressible flow, what is the change in pressure as the pipe narrows? Answer 1.4 × 10 6 Pa 336 336 SAMPLE PROBLEM 9D PROBLEM SOLUTION 1. DEFINE 2. PLAN 3. CALCULATE 4. EVALUATE Chapter 9 Bernoulli’s equation A water tank has a spigot near its bottom. If the top of the tank is open to the atmosphere, determine the speed at which the water leaves the spigot when the water level is 0.500 m above the spigot. Given: h 2 − h 1 = 0.500 m Unknown: v 1 = ? Diagram: 2 h2 P2 h1 Choose an equation(s) or situation: Because this problem involves fluid flow and differences in height, it requires the application of Bernoulli’s equation. P 1 + ⎯ 1 2 ⎯rv 1 2 + rgh1 = P 2 + ⎯ 1 2 ⎯rv 2 2 + rgh2 Point 1 is at the hole, and point 2 is at the top of the tank. If we assume that the hole is small, then the water level drops very slowly, so we can assume that v 2 is approximately zero. Also, note that P 1 = P 0 and P 2 = P 0 because both the top of the tank and the spigot are open to the atmosphere. P 0 + ⎯ 1 2 ⎯rv 1 2 + rgh1 = P 0 + rgh 2 Rearrange the equation(s) to isolate the unknown(s): 1 ⎯ 2 ⎯rv 2 1 = rgh2 − rgh1 v 1 2 = 2g(h2 − h 1) v1 = 2g (h 2− h 1) Substitute the values into the equation(s) and solve: v1 = 2( 9. 81 m/s 2 (0.500 ) v 1 = 3.13 m/s A quick estimate gives the following: 1 v1 ≈ 2( 10 )(0. 5) ≈ 3 A2 A1 P1 v1 m) Copyright © by Holt, Rinehart and Winston. All rights reserved.
Page 1 and 2: Chapter 9 Overview Section 9-1 defi
Page 3 and 4: Section 9-1 Demonstration 1 Volume
Page 5 and 6: SECTION 9-1 The Language of Physics
Page 7 and 8: SECTION 9-1 Demonstration 4 Lifting
Page 9 and 10: SECTION 9-1 PRACTICE GUIDE 9A Solvi
Page 11 and 12: SECTION 9-2 STOP 326 Misconception
Page 13 and 14: SECTION 9-2 Visual Strategy Figure
Page 15 and 16: SECTION 9-2 Classroom Practice The
Page 17 and 18: Section 9-3 Demonstration 7 Fluid f
Page 19: SECTION 9-3 Misconception STOP Aler
Page 23 and 24: Section 9-4 Demonstration 10 Temper
Page 25 and 26: SECTION 9-4 Classroom Practice The
Page 27 and 28: Chapter 9 Summary Teaching Tip Ask
Page 29 and 30: 9 REVIEW & ASSESS 12. because the p
Page 31 and 32: 9 REVIEW & ASSESS 42. 7.1 m 43. 17
Page 33 and 34: 9 REVIEW & ASSESS 63. 60.9 m/s 2 64
Page 35 and 36: Chapter 9 Laboratory Exercise NOTE
Page 37 and 38: CHAPTER 9 LAB Boyle’s Law Apparat

Fluid Mechanics with teacher's notes

You also want an ePaper? Increase the reach of your titles

Delete template?

Save as template?