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

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 m 3 /s A is cross-sectional area of pipe (m 2 ) 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 D 1 = 0.15 mD 2 = 0.3 m V 1 = 0.3 m/sV 2 = ? How do we find V 2 ?

Example D 1 = 15 cm IDD 2 = 30 cm ID V 1 = 0.3 m/sV 2 = ? We know A 1 V 1 = A 2 V 2

Answer V 2 = 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) P 1 = 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

 Find the total energy (ft) at B; assume flow is frictionless A B C 125’ 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 Energy A = Total Energy B Total Energy B h A = 125’ = Total Energy B

Example Find the velocity at point C. 0

Try it yourself: pump 9’ 1’ x’ 1’ 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.

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 = F pipe + F expansion + F expansion + F fittings

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 mm, pipe diameter is 5 cm, and the Reynolds number is 17312

Solution  ε / D = m / 0.05 m =  Re = 1.7 x 10 4 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 =

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 m 3 /min. Calculate F.

 Step 1:

 Step 1: Calculate Re number

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

 What is viscosity? What is density?

Viscosity = 2.13 x Pa · s ρ = 1030 kg/m 3

 So Re = 154,000

 f = ?  F pipe =

 F fittings =  F expansion =  F contraction =

 F total = m

Try it yourself  Find F for milk at 20.2 C flowing at 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.

Pump Applications

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

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

Efficiency Total head Power

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  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 )

 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 m 3 /s against a head of 19.8 m? 900W 9m 1400W W 0.01 m 3 /s D = 17.8 cm N = 1760 rpm

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 = 9m W 2 = 19.8 m

Now we need to “map” to new pump on same system curve. Substitute into Solve for D 2

 Q 1 /Q 2 =(N 1 /N 2 )(D 1 /D 2 ) 3  N 2 = N 1 (Q 2 /Q 1 )(D 1 /D 2 ) 3 N 2 = ?

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?