Core Ag Engineering Principles Session 1

Core Ag Engineering Principles Session 1Bernoullis EquationPump Applications1Bernoullis EquationHydrodynamics (the fluid is moving)

Incompressible fluid (liquids and gases at low pressures)Therefore changes in fluid density are not considered2Conservation of MassIf 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:

3

For incompressible fluids density remains constant and the equation becomes:

Q is volumetric flow rate in m3/sA is cross-sectional area of pipe (m2) andV is the velocity of the fluid in m/s4ExampleWater 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?5ExampleD1 = 0.15 mD2 = 0.3 mV1 = 0.3 m/sV2 = ?

How do we find V2?6ExampleD1 = 15 cm IDD2 = 30 cm IDV1 = 0.3 m/sV2 = ?

We know A1V1 = A2V2

7AnswerV2 = 0.075 m/s

8What is the volumetric flow rate?9Volumetric flow rate = Q10

11What is the mass flow rate in the larger section of pipe?12Mass flow rate =

13

14Bernoullis TheoremSince 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.15Bernoullis Theorem

h = elevation of point 1 (m or ft)P1 = pressure (Pa or psi) = specific weight of fluidv = velocity of fluid

16Bernoullis Theorem Special CasesWhen system is open to the atmosphere, then P=0 if reference pressure is atmospheric (can be one P or both Ps)17When one V refers to a storage tank and the other V refers to a pipe, then V of tank 4000; turbulent flow use Moody diagram

34Find /D , move to left until hit dark black line slide up line until intersect with Re #

35Answerf = 0.028536Energy Loss due to Fittings and Sudden Contractions

37Energy Loss due to Sudden Enlargement

38ExampleMilk 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.39Step 1:40Step 1: Calculate Re number

41Calculate v = ?

Calculate v2 / 2g, because well need this a lot42

43What is viscosity? What is density?44Viscosity = 2.13 x 10-3 Pa s = 1030 kg/m345So Re = 154,000

46f = ?

Fpipe =47

48

49Ffittings =

Fexpansion =

Fcontraction=50

51Ftotal = 199.7 m52Try it yourselfFind F for milk at 20.2 C flowing at 0.075 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.53Pump Applications54PowerThe power output of a pump is calculated by:

W = work from pump (ft or m)Q = volumetric flow rate (ft3/s or m3/s) = densityg = gravity55System Characteristic CurvesA system characteristic curve is calculated by solving Bernoullis theorem for many different Qs and solving for Ws

This curve tells us the input head required to move the fluid at that Q through that system56Example system characteristic curve

57Pump Performance CurvesGiven 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 laws58

EfficiencyTotal headPower59Pump Operating PointPump operating point is found by the intersection of pump performance curve and system characteristic curve

60What volumetric flow rate will this pump discharge on this system?

61Performance of centrifugal pumps while pumping water is used as standard for comparing pumps62To 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

63Pump Affinity Laws

64Power out equations

65A 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 0.005 m3/s against a head of 19.8 m?900W 9m1400WW0.01 m3/sD = 17.8 cmN = 1760 rpm66What is the operating point of first pump?N1 = 1760D1 = 17.8 cmQ1 = 0.01 m3/s Q2 = 0.005 m3/sW1 = 9m W2 = 19.8 m67Now we need to map to new pump on same system curve.

Substitute intoSolve for D2

68N2 = ?69

70Try it yourselfIf 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?71Bernoullis Theorem for FansPE Review Session VIB section 172Fan and Bin

12373

staticpressurevelocityheadtotal pressure74Power

75Ftotal=Fpipe+Fexpansion+Ffloor+FgrainFpipe=f (L/D) (V2/2g) for values in pipeFexpansion= (V12 V22) / 2gV1 is velocity in pipeV2 is velocity in binV1 >> V2 so equation reduces toV12/2g76FfloorEquation 2.38 p. 29 (4th edition) for no grain on floorEquation 2.39 p. 30 (4th edition) for grain on floorOf=percent floor opening expressed as decimalp=voidage fraction of material expressed as decimal (use 0.4 for grains if no better info)77ASAE Standards graph for Ffloor

78FgrainEquation 2.36 p. 29 (Cf = 1.5)A and b from standards or Table 2.5 p. 30Or use Shedds curves (Standards)X axis is pressure drop/depth of grainY axis is superficial velocity (m3/(m2s)Multiply pressure drop by 1.5 for correction factorMultiply by specific weight of air to get F in m or f79Shedds Curve (english)

80Shedds curves (metric)

81ExampleAir 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 0.4. The area of the bin floor is 20 m^2. Find the static and total pressure when Q=4 m^3/s82F=F(pipe)+F(exp)+F(floor)+F(grain)F(pipe)=83

84

85f

86

87

88

89Fexp

90

91Ffloor Equ. 2.39

92V = Vbin =

93Of=0.1

94

95Fgrain

961599 Pa = _________ m?97

98Using Shedds CurvesV=0.2 m/sWheat

99Ftotal = 3.2 + 21.2 + 2.3 + 130 = 157 m100Problem 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%?101