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. For incompressible fluids – density remains constant and the equation becomes:
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

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 = m/s

9. What is the volumetric flow rate?

10. Volumetric flow rate = Q

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

13. Mass flow rate =

15. 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.

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

17. 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)

18. 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

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

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

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

22. Example
Find the velocity at point C.

23. 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’

24. Determining F for Pipes and Grain

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

26. Reynolds numbers:
< Laminar
> 4000 Turbulent
Affects what?

27. 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

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

29. Darcy’s Formula

30. Where do you use relative roughness?

31. 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

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

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

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

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

36. Answer
f =

37. Energy Loss due to Fittings and Sudden Contractions

38. Energy Loss due to Sudden Enlargement

39. 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.

40. Step 1:

41. Step 1: Calculate Re number

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

44. What is viscosity? What is density?

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

46. So Re = 154,000

47. f = ?
Fpipe =

50. Ffittings =
Fexpansion =
Fcontraction=

52. Ftotal = m

53. Try it yourself
Find F for milk at 20.2 C flowing at m3/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.

54. Pump Applications

55. Power W = work from pump (ft or m)
The power output of a pump is calculated by:
W = work from pump (ft or m)
Q = volumetric flow rate (ft3/s or m3/s)
ρ = density
g = gravity

56. 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 input head required to move the fluid at that Q through that system

57. Example system characteristic curve

58. Pump Performance Curves
Given by the manufacturer – plots total head against 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

59. Total head
Power
Efficiency

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

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

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

63. To compare pumps at any other speed than that at which tests were conducted or to compare performance curves for geometrically similar pumps – use affinity laws

64. Pump Affinity Laws

65. Power out equations

66. A pump is to be selected that is geometrically similar to the pump given in the performance curve below, and the same system. What D and N would give m3/s against a head of m? W
D = 17.8 cm
N = 1760 rpm
1400W
900W 9m
0.01 m3/s

67. What is the operating point of first pump?
N1 = 1760 D1 = 17.8 cm Q1 = 0.01 m3/s Q2 = m3/s W1 = 9m W2 = 19.8 m

68. Now we need to “map” to new pump on same system curve.
Substitute into
Solve for D2

69. N2 = ?

71. 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 N1 change? D1? Q1? W1? P1?
Which direction (greater or smaller) would “they” move if they change?

72. Bernoulli’s Theorem for Fans
PE Review Session VIB – section 1

73. Fan and Bin 3 2 1

74. static
pressure
velocity
head
total
pressure

75. Power

76. Ftotal=Fpipe+Fexpansion+Ffloor+Fgrain
Fpipe=f (L/D) (V2/2g) for values in pipe
Fexpansion= (V12 – V22) / 2g
V1 is velocity in pipe
V2 is velocity in bin
V1 >> V2 so equation reduces to
V12/2g

77. Ffloor Equation 2.38 p. 29 (4th edition) for no grain on floor
Equation 2.39 p. 30 (4th edition) for grain on floor
Of=percent floor opening expressed as decimal
εp=voidage fraction of material expressed as decimal (use 0.4 for grains if no better info)

78. ASAE Standards graph for Ffloor

79. Fgrain Equation 2.36 p. 29 (Cf = 1.5)
A and b from standards or Table 2.5 p. 30
Or use Shedd’s curves (Standards)
X axis is pressure drop/depth of grain
Y axis is superficial velocity (m3/(m2s)
Multiply pressure drop by 1.5 for correction factor
Multiply by specific weight of air to get F in m or f

80. Shedd’s Curve (english)

81. Shedd’s curves (metric)

82. Example
Air is to be forced through a grain drying bin similar to that shown before. The air flows through 5 m of 0.5 m diameter galvanized iron conduit, exhausts into a plenum below the grain, passes through a perforated metal floor (10% openings) and is finally forced through a 1 m depth of wheat having a void fraction of The area of the bin floor is 20 m^2. Find the static and total pressure when Q=4 m^3/s

83. F=F(pipe)+F(exp)+F(floor)+F(grain)

86. f

90. Fexp

92. Ffloor Equ. 2.39

93. V = Vbin =

94. Of=0.1

96. Fgrain

97. 1599 Pa = _________ m?

99. Using Shedd’s Curves
V=0.2 m/s
Wheat

100. Ftotal =
= 157 m

101. Problem 2.4 (page 45)
Air (21C) at the rate of 0.1 m^3/(m^2 s) is to be moved vertically through a crib of shelled corn 1.6 m deep. The area of the floor is 12 m^2 with an opening percentage of 10% and the connecting galvanized iron pipe is 0.3 m in diameter and 12 m long. What is the power requirement, assuming the fan efficiency to be 70%?