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1 Pneumatic Conveying System Design Session 101
Transcript
  • 1

    Pneumatic Conveying System DesignSession 101

  • The design procedure is taken from the book “Fluidization and Fluid Particle Systems” by

    Zenz and Othmer

    2

  • 3

    The Effective Forces to Address

    1. Friction of the gas against the pipe wall

    2. Force required for moving the solids mass through the conveying line

    3. In vertical pipes, the force required to support the weight of the solids

    4. The forces required to support and accelerate the gas

    5. The force required to accelerate the solids

    6. The friction between the pipe and the solids

  • GLOSSARY OF TERMSC = Capacity of conveying system - pounds/minuteDpl = Diameter of largest particle - inchesDpm = Diameter of mean particle - inchesDt = Inside diameter of conveying line - inchese = Surface roughness - use 0.0018f = Fanning friction factorfP = Coefficient of friction, solid to pipe wallG = Gravity - 32.2 feet/second/secondL = Length of conveying line - feetPx = Pressure at point "x" - psigRe = Reynolds numberSCFM = Air volume - standard cubic feet / minuteTx = Temperature at point "x" - degree Fahrenheitu = Viscosity - pounds/second footVg = Velocity of the gas - feet/secondVp = Velocity of the material - feet/secondW = Material loading - pounds/second/square foot∆ p = Change in pressure - pounds/square inchρ g = Density of Gas - pounds/cubic footρ p = True density of material - pounds/cubic foot

    4

  • Formula (11.1) is used to calculate the pressure drop in Stream Flow Horizontal conveying

    systems

    5

    Acceleration of the Gas Acceleration

    of the Solids

    Friction of the Gas and the Pipe Wall

    Flow of solids through the pipe

  • Formula (11.2) is used to calculate the pressure drop in vertical conveying systems, either

    stream flow or two phase flow.

    6

    This is the additional term to considerfor vertical conveying

  • 7

    ∆P = L x 379 W x [ Dp ] 0.25

    144 (Vg)0.55 [ Dt ]

    Formula (11.10) is used to calculate the pressure drop in horizontal conveying systems below the saltation velocity

  • 8

    HORIZONTAL PHASE DIAGRAM

  • Horizontal Conveying Stream Flow

    Fluid Acceleration

    • The first term of equation 11.1 is for thecalculation of the Kinetic Energy of the gasdischarging from the conveying system.

    • Since the gas is picked up from a stationary room,and discharges from the conveying line at avelocity, this requires energy, which we willcalculate in the form of a pressure drop.

    • It is only calculated once for a given system, and iscalculated for the conditions at the end of theconveying system

    9

  • Horizontal Conveying Stream Flow

    • The first term is :

    • (11.1.1)

    10

  • Horizontal Conveying Stream Flow

    • The velocity of the gas can be calculated bydividing the air volume by the cross sectional areaof the conveying line, and correcting for pressureand temperature.

    • The calculation for gas velocity would be:• (11.1.1.A)

    11

  • Horizontal Conveying Stream Flow • The density of the gas can be obtained by correcting the

    standard density of the gas for pressure and temperature.

    • Since the density of air is 0.075 @ standard pressure and70 degrees F.

    • For air the density can be expressed by the followingformula.

    • (11.1.1.B)

    12

  • Horizontal Conveying Stream Flow

    • Substituting equations (11.1.1.a) and (11.1.1.b) in equation (11.1.1) we can combine terms and obtain:

    • (11.1.1)’

    13

  • PARTICLE ACCELERATION• The second term of equation 11.1 is for the

    calculation of the Kinetic Energy of the solidsdischarging from the conveying system.

    • Since the solids are picked up from a stationarycondition, and discharge from the conveying lineat a velocity, this requires energy, which we willcalculate in the form of a pressure drop.

    • (11.1.2)

    14

  • PARTICLE ACCELERATION

    • In the above term, "W" is material loading in pounds per second per square foot.

    • This can be calculated by:• (11.1.2.A)

    15

  • PARTICLE ACCELERATION

    • Substituting equation (11.1.2.a) in equation (11.1.2) we can combine terms and obtain:

    • (11.1.2)’

    16

  • PARTICLE ACCELERATION

    • In equation (11.1.2) we need the velocity of theparticle at the terminal end of the conveying line.

    • It is obvious that the velocity of the particle willbe less than the air velocity at this point.

    • If the material is fine and light weight, it mayclosely approach the gas velocity.

    • The larger the particle and the higher the density,the greater the difference between the gas andsolids velocity.

    17

  • 18

    • We have inserted the calculation for the gasvelocity and included a term Vg-p, whichallows a correction for the differencebetween solids and gas velocity.

    • A number can be inserted here for thedifference between the two velocities if thisvalue is known.

    • The formula can easily be modified to use amultiplication factor, such as 75%, etc.

    • If this information is not known, use the gasvelocity as the particle velocity, and this willresult in a slightly conservative solution.

    PARTICLE ACCELERATION

  • Fluid To Pipe – Friction Loss • This pressure calculation is a function of the density of the

    gas, which is affected by system pressure.

    • This calculation is therefore changing as the pressurechanges throughout the conveying system.

    • We will derive this formula as a function of pressure andapply it to short lengths of the conveying line.

    • (11.1.3)

    19

  • Fluid To Pipe – Friction Loss• The friction factor (f) is a function of the roughness of

    the pipe and the Reynolds number.

    • This relationship is expressed by "e/Dt" .

    – Both of these terms must be in similar units, such asinches.

    • It is suggested with pneumatic conveying, the inside surfaceof the conveying line becomes coated with a fine dusting ofthe material being conveyed.

    – Thus the rough surfaces appear smoother, since thevalleys become filled with fine material.

    – Experience has shown that the most accurate results canbe obtained by using e = 0.0018".

    20

  • Fluid To Pipe – Friction Loss

    An approximation of the friction factor (f) can be obtained from the following formula:

    (11.1.3.A)

    21

    • The above equation can be solved or determined by graphic solution from an air flow chart

  • Fluid To Pipe – Friction Loss

    22

  • Fluid To Pipe – Friction Loss

    • In the above equation the Reynolds Number, Re, can be calculated by:

    • (11.1.3.B)

    23

  • Fluid To Pipe – Friction Loss

    • In the above equation the viscosity for aircan be obtained by the following formula:

    • (11.1.3.C)

    24

  • Fluid To Pipe – Friction Loss

    • Substituting (11.1.3.c) in (11.1.3.b) we obtain the modified equation for the Reynolds Number as:

    • (11.1.3.B)’

    25

  • Fluid To Pipe – Friction Loss

    • Substituting (11.1.3.b)' in equation (11.1.3.a) we obtain the modified equation for friction factor as:

    • (11.1.3.A)’

    26

  • Fluid To Pipe – Friction Loss

    • Substituting in equation (11.1.3) we get the modified equation:

    • (11.1.3)’

    27

  • Fluid To Pipe – Friction Loss

    • The above equation is often presented in airflow vs. pipe diameter vs. pressure drop plot,for air flows at atmospheric pressure.

    • Thus the above equation can be solved bycomputer with the above equation or bygraphic solution from an air flow chart.

    28

  • Ratio of Particle Friction Loss to Fluid Friction Loss

    (11.1.4)

    • According to the theory for the derivation of the aboveequation, the first term, the ratio of the friction factorstimes the ratio of the velocities can not exceed unity.

    • In pneumatic conveying the velocity of the gas and particleare similar, but the friction of the solid is much greaterthan the friction of the gas.

    • The solution for this first term would result in a numbergreater than unity, therefore, this fraction can be ignored.

    29

  • Ratio of Particle Friction Loss to Fluid Friction Loss

    • The remaining fraction in the above equation isoften expressed as a representation of the loadingof the system. By substitution the equationreduces to:

    • (11.1.4)’

    30

  • Ratio of Particle friction loss to Fluid Friction Loss

    • Since standard air has a density of 0.075 #/cu. ft., the above formula can also be represented as:

    • (11.1.4)’’

    • It is obvious from the above that the ratio is reduced to POUNDS OF MATERIAL PER POUND OF AIR.

    31

  • VERTICAL CONVEYING Dilute and Dense Phase

    • Equation 11.2 is the same as equation 11.1,except for the addition of the fourth termwhich accounts for the energy used inelevating the material.

    • This term is also a function of absolutepressure in the conveying system, and will besolved as a function of the line pressure.

    32

  • Static Pressure Due to Inventory in the Vertical Pipe

    (11.2.4)

    • Again we need the velocity of the particlefor the solution of this term.

    • Since the conveying is vertical, the particlevelocity can be defined as the vertical gasvelocity minus the free fall terminal velocityof the particle. 33

  • Static Pressure Due to Inventory in the Vertical Pipe

    • To calculate the free fall terminal velocity, Vg-p, we can use the following equation:

    • (11.2.4A)

    34

  • Static Pressure Due to Inventory in the Vertical Pipe

    • The term Cd can be calculated in the following equation:

    • (11.2.4.B)

    • Since the Reynolds numbers for pneumaticconveying are large, the second term of the aboveequation will be small.

    • We can simplify the calculations by substituting

    – Cd = 0.40 35

  • Static Pressure Due to Inventory in the Vertical Pipe

    • By substitution, equation (11.2.4.a) becomes:

    • (11.2.4.a)’

    36

  • Static Pressure Due to Inventory in the Vertical Pipe

    and equation (11.2.4) becomes:

    (11.2.4)’

    37

  • HORIZONTAL CONVEYING Below the Saltation Velocity

    (11.10)

    38

  • HORIZONTAL CONVEYING Below the Saltation Velocity

    • The above equation can be simplified by substitutions to the following:

    • (11.10)’

    39

  • We have modified Formula (11.10) as follows:

    (1) .25 power has been changed to .125 power

    (2) .55 power has been changed to .44 power

    (3) 3.79 has been changed to 2.96

    40

  • Questions

    ?

    Jack D. Hilbert P.E.610-657-5286

    [email protected]

    41


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