1. Things to grab for this session (in priority order)  Pencil  Henderson, Perry, and Young text (Principles of Process Engineering)  Calculator  Eraser  Scratch paper  Units conversion chart  Tables of fluid properties  Moody diagram  Pump affinity laws

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

3. Core Principles  Conservation of mass  Conservation of energy

4. Assumption/Conditions  Hydrodynamics (the fluid is moving)  Incompressible fluid (liquids and gases at low pressures)  Therefore changes in fluid density are not considered

5. 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:

6. For incompressible fluids – density remains constant and the equation becomes: Q is volumetric flow rate in m 3 /s A is cross-sectional area of pipe (m 2 ) and V is the velocity of the fluid in m/s

7. 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?

8. Example D 1 = 0.15 mD 2 = 0.3 m V 1 = 0.3 m/sV 2 = ? How do we find V 2 ?

9. Example

10. Answer V 2 =

11. What is the volumetric flow rate?

12. Volumetric flow rate = Q

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

15. Mass flow rate =

17. Questions?

18. Bernoulli’s Theorem (conservation of energy)  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.

19. Bernoulli’s Theorem W = work done to the fluid h = elevation of point 1 (m or ft) P 1 = pressure (Pa or psi) = specific weight of fluid v = velocity of fluid F = friction loss in the system

20. Bernoulli’s Theorem Special Conditions (situations where we can simplify the equation)

21. Special Condition 1  When system is open to the atmosphere, then P=0 if reference pressure is atmospheric (gauge pressure)  Either one P or both P’s can be zero depending on system configuration

22. Special Condition 2  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

23. Example

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

25. Preliminary Thinking  Why is total energy in units of ft? Is that a correct measurement of energy?  What are the typical units of energy?  How do we start the problem?

26. Example Total Energy A = Total Energy B Total Energy B h A =

27. Example Find the velocity at point C.

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

29. Answer

30. Bernoulli’s Equation  Adding on – how do we calculate F (instead of having it given to us or assuming it is negligible like in the previous problems)

31. Bernoulli’s Theorem h = elevation of point 1 (m or ft) P 1 = pressure (Pa or psi) = specific weight of fluid v = velocity of fluid F = friction loss in the system

32. Determining F for Piping Systems

33. Step 1  Determine Reynolds number  Dynamic viscosity units  Diameter of pipe  Velocity  Density of fluid

34. 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 m 3 /min. Calculate Re.

35.  Step 1: Calculate Re number

36.  Calculate v = ?  Calculate v 2 / 2g, because we’ll need this a lot

38.  What is viscosity? What is density?

39. Viscosity = ρ =

40. Check units – what should they be?

41. Reynolds numbers:  < 2130 Laminar  > 4000 Turbulent  Affects what?

42. Reynolds numbers:  To calculate the f in Darcy’s equation for friction loss in pipe; need Re  Laminar: f = 64 / Re  Turbulent: Colebrook equation or Moody diagram

43. Total F in piping sytem F = F pipe + F expansion + F expansion + F fittings

44. Darcy’s Formula

45.  Where do you use relative roughness?

46.  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

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

48. Solution  ε / D = 0.000046 m / 0.05 m = 0.00092  Re = 1.7 x 10 4 Re > 4000; turbulent flow – use Moody diagram

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

50. Answer  f =

51. Energy Loss due to Fittings and Sudden Contractions

52. Energy Loss due to Sudden Enlargement

53. 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 m 3 /min. Calculate F.

54.  Step 1:

55.  f = ?  F pipe =

58.  F fittings =  F expansion =  F contraction =

60.  F total =

61. Try it yourself  Find F for milk at 20.2 C flowing at 0.075 m 3 /min in sanitary tubing with a 4 cm ID through 20 m of pipe, with one type A elbow and one type A entrance. The milk flows from one reservoir into another.

62. Pump Applications

63. How is W determined for a pump?  Compute all the terms in the Bernoulli equation except W  Solve for W algebraically

64. Why is W determined for a pump?  W is used to compute the size of pump needed

65. Power  The power output of a pump is calculated by: W = work from pump (ft or m) Q = volumetric flow rate (ft 3 /s or m 3 /s) ρ = density g = gravity

66. Remember:  P o = the power delivered to the fluid (sometimes referred to as hydraulic power)  P in = P o /pump efficiency (sometimes referred to as brake horsepower)

67. To calculate Power(out) What we knowWhat we need

68. Flow rate is variable  Depends on “back pressure”  Intersection of system characteristic curve and the pump curve

69. System Characteristic Curves  A system characteristic curve is calculated by solving Bernoulli’s theorem for many different Q’s and solving for W’s  This curve tells us the power needed to be supplied to move the fluid at that Q through that system

70. Example system characteristic curve

71. Pump Performance Curves  Given by the manufacturer – plots total head against Q: volumetric discharge rate  Note: these curves are good for ONLY one speed, and one impeller diameter – to change speeds or diameters we need to use pump laws

72. Pump Performance Curve Efficiency Total head Power N = 1760 rpm D = 15 cm

73. Pump Operating Point  Pump operating point is found by the intersection of pump performance curve and system characteristic curve

74. What volumetric flow rate will this pump discharge on this system?

75.  Performance of centrifugal pumps while pumping water is used as standard for comparing pumps

76.  To compare pumps at any other speed than that at which tests were conducted or to compare performance curves for geometrically similar pumps (for ex. different impeller diameters or different speeds) – use pump affinity laws (or pump laws)

77. Pump Affinity Laws (p. 106)  Q 1 /Q 2 =(N 1 /N 2 )(D 1 /D 2 ) 3  W 1 /W 2 =(N 1 /N 2 ) 2 (D 1 /D 2 ) 2  Po 1 /Po 2 =(N 1 /N 2 ) 3( D 1 /D 2 ) 5 (ρ 1 /ρ 2 )  NOTE: For changing ONLY one property at a time

78.  Size a pump that is geometrically similar to the pump given in the performance curve below, for the same system. Find D and N to achieve Q= 0.005 m 3 /s against a head of 19.8 m? 0.01 m 3 /s Example

79. N = 1760 rpm D = 17.8 cm (Watt)

80. Procedure  Step 1: Find the operating point of the original pump on this system (so you’ll have to plot your system characteristic curve (calculated) onto your pump curve (given) or vice-versa.

81. 0.01 m 3 /s N = 1760 rpm D = 17.8 cm P(W) 882.9 W

82. What is the operating point of first pump? N 1 = 1760 D 1 = 17.8 cm Q 1 = 0.01 m 3 /s Q 2 = m 3 /s W 1 = W 2 = m

83. How do we convert P o to W?

85. Find D that gives both new W and new Q  (middle of p. 109 – can’t use p. 106 for two conditions changing)  D 2 =D 1 (Q 2 /Q 1 ) 1/2 (W 1 /W 2 ) 1/4  D 2 =

86. Find N that corresponds to new Q and D (p. 109 equ)  N 2 =N 1 (Q 2 /Q 1 )(D 1 /D 2 ) 3  N 2 =

87. Try it yourself  If the system used in the previous example was changed by removing a length of pipe and an elbow – what changes would that require you to make?  Would N 1 change? D 1 ? Q 1 ? W 1 ? P 1 ?  Which direction (greater or smaller) would “they” move if they change?