Things to grab for this session (in priority order)

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

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Core Ag Engineering Principles Session 1

• Bernoullis Equation
• Pump Applications

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Core Principles

• Conservation of mass
• Conservation of energy

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Assumption/Conditions

• Hydrodynamics (the fluid is moving)
• Incompressible fluid (liquids and gases at low
  pressures)
• Therefore changes in fluid density are not
  considered

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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

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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

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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?

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Example

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

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Example

• D1 0.15 m D2 0.3 m
• V1 0.3 m/s V2 ?
• How do we find V2?
• We know A1V1 A2V2

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Answer

• V2 0.075 m/s

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What is the volumetric flow rate?
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Volumetric flow rate Q
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What is the mass flow rate in the larger section
of pipe?
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Mass flow rate
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Questions?
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Bernoullis 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.

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Bernoullis Theorem
W work done to the fluid h elevation of
point 1 (m or ft) P1 pressure (Pa or psi)
specific weight of fluid v velocity of fluid F
friction loss in the system
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Bernoullis Theorem Special Conditions

• (situations where we can simplify the equation)

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Special Condition 1

• When system is open to the atmosphere, then P0
  if reference pressure is atmospheric (gauge
  pressure)
• Either one P or both Ps can be zero depending on
  system configuration

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Special Condition 2

• When one V refers to a storage tank and the other
  V refers to a pipe, then V of tank ltltltlt V pipe
  and assumed zero

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Special Condition 3

• If no pump or fan is between the two points
  chosen, W0

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Example
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• Find the total energy (ft) at B assume flow is
  frictionless

A
B
125
C
75
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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?

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Preliminary Thinking

• Feet is a measure of pressure it can be
  converted to more traditional pressure units.
• Typical units of energy energy work Nm or
  ft-lb. If we multiply feet or meter by the
  specific weight of the fluid, we obtain units of
  pressure.
• How do we start the problem?

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Example

• Total EnergyA Total EnergyB

Total EnergyB
hA
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Example

• Total EnergyA Total EnergyB

Total EnergyB
hA 125 Total EnergyB
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Example

• Find the velocity at point C.

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Example

• Find the velocity at point C.

0
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Try it yourself
Water is pumped at the rate of 3 cfs through
piping system shown. If the energy of the water
leaving the pump 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.
9
1
x
pump
1
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Answer

• 150 psig 62.4 lb/ft3 144 in2 / ft2 346 ft
• 1ft 346 ft 0 0 10 11 x
• X 326

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Bernoullis 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)

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Bernoullis Theorem
h elevation of point 1 (m or ft) P1
pressure (Pa or psi) specific weight of
fluid v velocity of fluid F friction loss in
the system
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Determining F for Piping Systems
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Step 1

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

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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 Re.

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• Step 1 Calculate Re number

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• Calculate v ?
• Calculate v2 / 2g, because well need this a lot

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• What is viscosity? What is density?

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• Viscosity 2.13 x 10-3 Pa s
• ? 1030 kg/m3

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• So Re 154,000

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Reynolds numbers

• lt 2130 Laminar
• gt 4000 Turbulent
• Affects what?

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Reynolds numbers

• To calculate the f in Darcys equation for
  friction loss in pipe need Re
• Laminar f 64 / Re
• Turbulent Colebrook equation or Moody diagram

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Total F in piping sytem

• F Fpipe Fexpansion
• Fexpansion Ffittings

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Darcys Formula
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• Where do you use relative roughness?

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• Relative roughness is a function of the pipe
  material for turbulent flow it is a value needed
  to use the Moody diagram (e/D) along with the
  Reynolds number

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Example

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

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Solution

• e / D 0.000046 m / 0.05 m 0.00092
• Re 1.7 x 104
• Re gt 4000 turbulent flow use Moody diagram

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

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Answer

• f 0.0285

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Energy Loss due to Fittings and Sudden
Contractions
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Energy Loss due to Sudden Enlargement
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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.

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• Step 1

59

• Step 1
• Re 154,000

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• f ?
• Fpipe

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• Ffittings
• Fexpansion
• Fcontraction

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Ftotal 199.7 m
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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.

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Pump Applications
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How is W determined for a pump?

• Compute all the terms in the Bernoulli equation
  except W
• Solve for W algebraically

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Why is W determined for a pump?

• W is used to compute the size of pump needed

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Power

• 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) ? density g gravity
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Remember

• Po the power delivered to the fluid (sometimes
  referred to as hydraulic power)
• Pin Po/pump efficiency
• (sometimes referred to as brake horsepower)

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To calculate Power(out)

• What we know

• What we need

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To calculate Power(out)

• What we know

• What we need

• W
• ?
• g

• Q

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Flow rate is variable

• Depends on back pressure
• Intersection of system characteristic curve and
  the pump curve

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System Characteristic Curves

• A system characteristic curve is calculated by
  solving Bernoullis theorem for many different
  Qs and solving for Ws
• This curve tells us the power needed to be
  supplied to move the fluid at that Q through that
  system

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Example system characteristic curve
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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

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Pump Performance Curve
Total head
Power
Efficiency
N 1760 rpm D 15 cm
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Pump Operating Point

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

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What volumetric flow rate will this pump
discharge on this system?
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• Performance of centrifugal pumps while pumping
  water is used as standard for comparing pumps

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• 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)

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Pump Affinity Laws (p. 106)

• Q1/Q2(N1/N2)(D1/D2)3
• W1/W2(N1/N2)2(D1/D2)2
• Po1/Po2(N1/N2)3(D1/D2)5(?1/?2)
• NOTE For changing ONLY one property at a time

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Example

• 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 m3/s against a head of 19.8 m?

0.01 m3/s
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(Watt)
N 1760 rpm D 17.8 cm
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Procedure

• Step 1 Find the operating point of the original
  pump on this system (so youll have to plot your
  system characteristic curve (calculated) onto
  your pump curve (given) or vice-versa.

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P(W)
N 1760 rpm D 17.8 cm
91
P(W)
882.9 W
0.01 m3/s
N 1760 rpm D 17.8 cm
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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 W2 19.8 m

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How do we convert Po to W?
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How do we convert Po to W?

• W Po/Q?g
• W(882.9 Nm/s)/(0.01 m3/s 1000 kg/m3 9.81
  m/s2)
• W 9 m

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Find D that gives both new W and new Q

• (middle of p. 109 cant use p. 106 for two
  conditions changing)
• D2D1(Q2/Q1)1/2(W1/W2)1/4
• D2
• D2

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Find D that gives both new W and new Q

• (middle of p. 109)
• D2D1(Q2/Q1)1/2(W1/W2)1/4
• D20.178m(.005 m3/s/0.01m3/s)1/2(9 m/19.8 m)1/4
• D20.141 m

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Find N that corresponds to new Q and D (p. 109
equ)

• N2N1(Q2/Q1)(D1/D2)3
• N2

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Find N that corresponds to new Q and D

• N2N1(Q2/Q1)(D1/D2)3
• N2 1770 rpm

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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?

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Answers

• Removing pipe elbow would reduce F and
  therefore reduce W (increase Q). System curve
  would move to the right.
• N would likely change. D- no. Q- yes. P depends
  on the shape of the power curve but likely it
  will change.
• N would increase (N2N1(Q2/Q1)(D1/D2)3)
• Q increase P depends.