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CE 230-Engineering Fluid Mechanics Lecture # 18 CONTINUITY EQUATION Section 5.3 (p.154) in text.

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Presentation on theme: "CE 230-Engineering Fluid Mechanics Lecture # 18 CONTINUITY EQUATION Section 5.3 (p.154) in text."— Presentation transcript:

1 CE 230-Engineering Fluid Mechanics Lecture # 18 CONTINUITY EQUATION Section 5.3 (p.154) in text

2 Time rate of change of a system extensive property = time rate of change of that property in CV + Net outflow rate of that property across CS Property B can be selected as mass, momentum, energy etc

3 Selecting the extensive property, B, to me the system mass M And therefore the intensive property, b, to be 1 If we substitute in the control volume equation we get the continuity equation Rate of mass accumulation in CV + net outflow rate of mass across CS=0 Assuming constant velocity over flow area Deriving mass balance equation

4 Example 5.9 A 120-cm pipe in series with a 60 cm pipe. The flow rate of water is 2 m 3 /s. What is the velocity in each pipe?

5 Example 5.5 what is the rate of water accumulation in the tank?

6 Example 5.6 what is the rate of water rise in reservoir?

7

8 Example 5.7 A 10-cm water jet leaves the 1-m diameter tank as shown. If the jet velocity to be (2gh) 0.5. How long will it take to drop the water surface from 2m to 0.5 m?


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