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 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:

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

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?

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

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

Answer V2 = 0.075 m/s

What is the volumetric flow rate?

Volumetric flow rate = Q

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

Mass flow rate =

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.

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

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)

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

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

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

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

Example Find the velocity at point C.

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’

Determining F for Pipes and Grain

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

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

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

Total F F = Fpipe + Fexpansion + Fcontraction + Ffittings

Darcy’s Formula

Where do you use relative roughness?

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

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

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

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

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

Answer f = 0.0285

Energy Loss due to Fittings and Sudden Contractions

Energy Loss due to Sudden Enlargement

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.

Step 1:

Step 1: Calculate Re number

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

What is viscosity? What is density?

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

So Re = 154,000

f = ? Fpipe =

Ffittings = Fexpansion = Fcontraction=

Ftotal = 199.7 m

Try it yourself Find 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.

Pump Applications

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

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

Example system characteristic curve

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

Total head Power Efficiency

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

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

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

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

Pump Affinity Laws

Power out equations

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 0.005 m3/s against a head of 19.8 m? W D = 17.8 cm N = 1760 rpm 1400W 900W 9m 0.01 m3/s

What is the operating point of first pump? N1 = 1760 D1 = 17.8 cm Q1 = 0.01 m3/s Q2 = 0.005 m3/s W1 = 9m W2 = 19.8 m

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

N2 = ?

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?

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

Fan and Bin 3 2 1

static pressure velocity head total pressure

Power

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

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)

ASAE Standards graph for Ffloor

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

Shedd’s Curve (english)

Shedd’s curves (metric)

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 0.4. The area of the bin floor is 20 m^2. Find the static and total pressure when Q=4 m^3/s

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

f

Fexp

Ffloor Equ. 2.39

V = Vbin =

Of=0.1

Fgrain

1599 Pa = _________ m?

Using Shedd’s Curves V=0.2 m/s Wheat

Ftotal = 3.2 + 21.2 + 2.3 + 130 = 157 m

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%?