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Example: Exercise 5.9.4 (Pump)

Example: Exercise 5.9.4 (Pump). Pump. Flow. Oil (S=0.82). Want: Rate at which energy is delivered to oil by pump. Example: Exercise 5.9.4 (Pump). Pump. Flow. Oil (S=0.82). Want: Rate at which energy is delivered to oil by pump. Example: Exercise 5.9.4 (Pump). Pump. Flow. Oil (S=0.82).

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Example: Exercise 5.9.4 (Pump)

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  1. Example: Exercise 5.9.4 (Pump) Pump Flow Oil (S=0.82) Want: Rate at which energy is delivered to oil by pump

  2. Example: Exercise 5.9.4 (Pump) Pump Flow Oil (S=0.82) Want: Rate at which energy is delivered to oil by pump

  3. Example: Exercise 5.9.4 (Pump) Pump Flow Oil (S=0.82) Want: Rate at which energy is delivered to oil by pump Need to find hp associated with the pump:

  4. Example: Exercise 5.9.4 (Pump)

  5. Example: Exercise 5.9.4 (Pump)

  6. Example: Exercise 5.9.4 (Pump) Rate of transfer of energy =

  7. Example: Exercise 5.9.4 (Pump) • Pumps (and also turbines) are characterized by their efficiency

  8. Example: Exercise 5.9.4 (Pump) • Pumps (and also turbines) are characterized by their efficiency • Say, in exercise 5.9.4 the pump is 90% efficient and we require • 6.83 kW of output, then input = 6.83 kW / 0.9 = 7.59 kW

  9. Example: Exercise 5.9.4 (Pump) • Pumps (and also turbines) are characterized by their efficiency • Say, in exercise 5.9.4 the pump is 90% efficient and we require • 6.83 kW of output, then input = 6.83 kW / 0.9 = 7.59 kW • Pumps (and also turbines) are characterized by their efficiency. Efficiency =

  10. General Energy Equation for Steady Flow of Any Fluid First Law of Thermodynamics: For steady flow, external work done on any system plus the thermal energy transferred into or out of the system is equal to the change of energy of the system

  11. General Energy Equation for Steady Flow of Any Fluid First Law of Thermodynamics: For steady flow, external work done on any system plus the thermal energy transferred into or out of the system is equal to the change of energy of the system (I) Using the first law of thermodynamics, (II) taking into account non-uniform velocity at a cross-section of flow region, and (III) assuming flow goes from section 1 to section 2, we can derive the following:

  12. General Energy Equation for Steady Flow of Any Fluid • is a correction factor accounting for non-uniform velocity in cross-section • If velocity is uniform in cross-section, then

  13. General Energy Equation for Steady Flow of Any Fluid • is a correction factor accounting for non-uniform velocity in cross-section • If velocity is uniform in cross-section, then • This general equation also takes into account changes in density (via ) • energy changes due to machines (via ) and due to heat transfer to • or from outside the fluid (via )

  14. General Energy Equation for Steady Flow of Any Fluid • is a correction factor accounting for non-uniform velocity in cross-section • If velocity is uniform in cross-section, then • This general equation also takes into account changes in density (via ) • energy changes due to machines (via ) and due to heat transfer to • or from outside the fluid (via ) • It also accounts for the conversions of other forms of fluid energy into internal • heat ( ) internal energy per unit weight =

  15. Recall from chapter 2 that compressibility of a liquid is inversely proportional • to the bulk modulus of the liquid • From table 2.1 (page 17 of text), for a wide range of temperatures the bulk • modulus of water is very high O(100,000 psi) relative to the usual pressures • in our problems • Thus the compressibility of our most common liquid (water) is low and we • may treat it as incompressible while still undergoing • changes in temperature and pressure

  16. General Energy Equation for Steady Flow of Any Fluid • On a unit weight basis, the change in internal energy is equal to the heat • added to or removed from the fluid plus the heat generated by fluid friction:

  17. General Energy Equation for Steady Flow of Any Fluid • On a unit weight basis, the change in internal energy is equal to the heat • added to or removed from the fluid plus the heat generated by fluid friction: • The head loss due to friction is equal to the internal heat gain minus any • heat added from external sources, per unit weight of fluid

  18. General Energy Equation for Steady Flow of Any Fluid • On a unit weight basis, the change in internal energy is equal to the heat • added to or removed from the fluid plus the heat generated by fluid friction: • The head loss due to friction is equal to the internal heat gain minus any • heat added from external sources, per unit weight of fluid • Energy loss due to friction gets converted to internal energy (proportional to • temperature)

  19. Example: Exercise 5.3.5 (Friction Head Loss) S of liquid in pipe = 0.85 A Diameter at A = Diameter at B, thus by continuity B Want: Pipe friction head loss and direction of flow

  20. Example: Exercise 5.3.5 (Friction Head Loss) S of liquid in pipe = 0.85 A Diameter at A = Diameter at B, thus by continuity B Want: Pipe friction head loss and direction of flow Assume flow goes from A to B:

  21. Example: Exercise 5.3.5 (Friction Head Loss) S of liquid in pipe = 0.85 A Diameter at A = Diameter at B, thus by continuity B Thus flow goes from B to A and

  22. Example: Exercise 5.3.5 (Friction Head Loss) S of liquid in pipe = 0.85 A Diameter at A = Diameter at B, thus by continuity B Thus flow goes from B to A and Let . If flow goes from B to A, and

  23. Role of pressure difference (pressure gradient) A B Thus flow will go from B (high pressure) to A (low pressure), only if Otherwise flow will go from A (low pressure) to B (high pressure)

  24. Role of pressure difference (pressure gradient) A B Thus flow will go from B (high pressure) to A (low pressure), only if Otherwise flow will go from A (low pressure) to B (high pressure) In general, the pressure force (resulting from a pressure difference) wants to move a fluid from a high pressure region towards a low pressure region For a flow to actually go from a high pressure region towards a low pressure region, the pressure force must be higher than other forces that could be trying to move fluid in opposite direction (e.g. gravitational force in exercise 5.3.5)

  25. Energy Grade Line (EGL) and Hydraulic Grade Line (HGL) Graphical interpretations of the energy along a pipeline may be obtained through the EGL and HGL: EGL and HGL may be obtained via a pitot tube and a piezometer tube, respectively

  26. Energy Grade Line (EGL) and Hydraulic Grade Line (HGL) • head loss, say, • due to friction EGL HGL piezometer tube pitot tube Datum

  27. Energy Grade Line (EGL) and Hydraulic Grade Line (HGL) EGL Large V2/2g because smaller pipe here HGL Steeper EGL and HGL because greater hL per length of pipe Head loss at submerged discharge

  28. Energy Grade Line (EGL) and Hydraulic Grade Line (HGL) Positive Negative Positive EGL HGL If then and cavitation may be possible

  29. Energy Grade Line (EGL) and Hydraulic Grade Line (HGL) Helpful hints when drawing HGL and EGL: 1. EGL = HGL + V2/2g, EGL = HGL for V=0 2. If p=0, then HGL=z 3. A change in pipe diameter leads to a change in V (V2/2g) due to continuity and thus a change in distance between HGL and EGL 4. A change in head loss (hL) leads to a change in slope of EGL and HGL 5. If then and cavitation may be possible

  30. Helpful hints when drawing HGL and EGL (cont.): 6. A sudden head loss due to a turbine leads to a sudden drop in EGL and HGL 7. A sudden head gain due to a pump leads to a sudden rise in EGL and HGL 8. A sudden head loss due to a submerged discharge leads to a sudden drop in EGL

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