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Core Ag Engineering Principles – Session 1

Core Ag Engineering Principles – Session 1. Bernoulli’s Equation Pump Applications. Bernoulli’s Equation. Hydrodynamics (the fluid is moving) Incompressible fluid (liquids and gases at low pressures) Therefore changes in fluid density are not considered. Conservation of Mass.

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Core Ag Engineering Principles – Session 1

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  1. Core Ag Engineering Principles – Session 1 Bernoulli’s Equation Pump Applications

  2. Bernoulli’s Equation • Hydrodynamics (the fluid is moving) • Incompressible fluid (liquids and gases at low pressures) • Therefore changes in fluid density are not considered

  3. Conservation of Mass • If the rate of flow is constant at any point and there is no accumulation or depletion of fluid within the system, the principle of conservation of mass (where mass flow rate is in kg/s) requires:

  4. Q is volumetric flow rate in m3/s A is cross-sectional area of pipe (m2) and V is the velocity of the fluid in m/s For incompressible fluids – density remains constant and the equation becomes:

  5. Example • Water is flowing in a 15 cm ID pipe at a velocity of 0.3 m/s. The pipe enlarges to an inside diameter of 30 cm. What is the velocity in the larger section, the volumetric flow rate, and the mass flow rate?

  6. Example D1 = 0.15 m D2 = 0.3 m V1 = 0.3 m/s V2 = ? How do we find V2?

  7. Example • D1 = 15 cm ID D2 = 30 cm ID • V1 = 0.3 m/s V2 = ? • We know A1V1 = A2V2

  8. Answer V2 = 0.075 m/s

  9. What is the volumetric flow rate?

  10. Volumetric flow rate = Q

  11. What is the mass flow rate in the larger section of pipe?

  12. Mass flow rate =

  13. Bernoulli’s Theorem • Since energy is neither created nor destroyed within the fluid system, the total energy of the fluid at one point in the system must equal the total energy at any other point plus any transfers of energy into or out of the system.

  14. Bernoulli’s Theorem h = elevation of point 1 (m or ft) P1 = pressure (Pa or psi) = specific weight of fluid v = velocity of fluid

  15. Bernoulli’s Theorem Special Cases • When system is open to the atmosphere, then P=0 if reference pressure is atmospheric (can be one P or both P’s)

  16. When one V refers to a storage tank and the other V refers to a pipe, then V of tank <<<< V pipe and assumed zero • If no pump or fan is between the two points chosen, W=0

  17. Example • Find the total energy (ft) at B; assume flow is frictionless A B 125’ C 75’ 25’

  18. Example • Why is total energy in units of ft? • What are the typical units of energy? • How do we start the problem?

  19. Example Total EnergyA = Total EnergyB Total EnergyB hA = 125’ = Total EnergyB

  20. Example Find the velocity at point C. 0

  21. Try it yourself: Water is pumped at the rate of 3 cfs through piping system shown. If the pump has a discharge pressure of 150 psig, to what elevation can the tank be raised? Assume the head loss due to friction is 10 feet. 9’ 1’ x’ pump 1’

  22. Determining F for Pipes and Grain

  23. Step 1 • Determine Reynolds number • Dynamic viscosity units • Diameter of pipe • Velocity • Density of fluid

  24. Reynolds numbers: • < 2130 Laminar • > 4000 Turbulent • Affects what?

  25. Reynolds numbers: • < 2130 Laminar • > 4000 Turbulent • Affects what? • The f in Darcy’s equation for friction loss in pipe • Laminar: f = 64 / Re • Turbulent: Colebrook equation or Moody diagram

  26. Total F F = Fpipe + Fexpansion + Fcontraction+ Ffittings

  27. Darcy’s Formula

  28. Where do you use relative roughness?

  29. Relative roughness is a function of the pipe material; for turbulent flow it is a value needed to use the Moody diagram (ε/D) along with the Reynolds number

  30. Example • Find f if the relative roughness is 0.046 mm, pipe diameter is 5 cm, and the Reynolds number is 17312

  31. Example • Find f if the relative roughness is 0.046 mm, pipe diameter is 5 cm, and the Reynolds number is 17312

  32. Solution • ε / D = 0.000046 m / 0.05 m = 0.00092 • Re = 1.7 x 104 Re > 4000; turbulent flow – use Moody diagram

  33. Find ε/D , move to left until hit dark black line – slide up line until intersect with Re #

  34. Answer • f = 0.0285

  35. Energy Loss due to Fittings and Sudden Contractions

  36. Energy Loss due to Sudden Enlargement

  37. Example • Milk at 20.2C is to be lifted 3.6 m through 10 m of sanitary pipe (2 cm ID pipe) that contains two Type A elbows. Milk in the lower reservoir enters the pipe through a type A entrance at the rate of 0.3 m3/min. Calculate F.

  38. Step 1:

  39. Step 1: Calculate Re number

  40. Calculate v = ? • Calculate v2 / 2g, because we’ll need this a lot

  41. What is viscosity? What is density?

  42. Viscosity = 2.13 x 10-3 Pa · s ρ = 1030 kg/m3

  43. So Re = 154,000

  44. f = ? • Fpipe =

  45. Ffittings= • Fexpansion = • Fcontraction=

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