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CE 230-Engineering Fluid Mechanics

CE 230-Engineering Fluid Mechanics. Lecture # 25-27 ENERGY EQUATION. Bernoulli Equation did not account for 1) Energy loss due to viscous resistance 2) Energy that can be add to the system through a device 3) Energy that can be taken from the system through a device

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CE 230-Engineering Fluid Mechanics

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  1. CE 230-Engineering Fluid Mechanics Lecture # 25-27 ENERGY EQUATION

  2. Bernoulli Equation did not account for 1) Energy loss due to viscous resistance 2) Energy that can be add to the system through a device 3) Energy that can be taken from the system through a device These are some of the limitations of BE Instead we will use an equation that takes care of all these This is the Energy Equation for1-D, steady, incompressible flow: = 1 for uniform v Close to 1 turbulant Close to 2 laminar

  3. Find K.E. correction factor for flow in a circular pipe if the velocity is given by: Recall

  4. We must calculate V:

  5. Example 7.3 Compare your results with results obtained using BE)

  6. Example 7.4

  7. Example 7.5

  8. Concept of EGL & HGL The total head (energy/unit weight) can be split into three components: 1) Elevation head 2) Pressure head 3) Velocity head To see the variation of the flow energy we may draw a line called EGL

  9. EGL and HGL in a straight pipe

  10. Influence of pump on HGL and HGL

  11. Influence of size change on HGL and EGL

  12. Influence of size change on HGL and EGL

  13. What if pipe was not horizontal ?

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