PNEUMATIC CONVEYING SPREADSHEET-CONTENT

2 PDH Professional Development Hour course content

© Copy right John Andrew P.E., 30 July 2012

SUMMARY

Transfer System Engineering Companies

􀂃MAC
􀂃Hammertek
􀂃Fuller Bulk Handling
􀂃CoperionWaeschle

There is virtually no limit to the range of materials that can

be conveyed with dilute – phase pneumatic system.
Products commonly conveyed in dilute phase
systems include: flour, resins, specialty chemicals, ground
feeds, and granular and palletized products.
Of the various types of pneumatic
systems, a dilute phase system will generally be lowest in
capital cost. -

The dilute phase system requires relatively high conveying air

velocities depending on the material type.
This is typically in the region of 3000 fpm for a fine
powder, to 4000 fpm for a granular material, and beyond for
larger particles and higher density materials.

1. In a horizontal pipeline the velocity of the particles will

typically be about 80% of that of the air. This is usually
expressed in terms of a slip ratio, defined in terms of the
velocity of the particles divided by the velocity of the air
transporting the particles, and in this case it would be 0.8.

2. In vertically upward flow in a pipeline a typical value

of the slip ratio will be about 0.7.
At the point at which the material is fed
into the pipeline, the material will essentially have zero velocity.
In order for material to accelerate to conveying velocities, an
initial section of straight piping is necessary.
-
Good
engineering practice dictates that a straight section equal to 25
times the pipe diameter is required before the first bend.
The
conveying velocity and hence air flow rate is greatly influenced
by material characteristics. -

Particle shape, size distribution, mean particle size and particle

density; all have an effect on minimum conveying velocity,
pressure drop, air flow, etc. Properties such as moisture
content, cohesiveness and adhesiveness may cause flow
problems through vessels and valves.
-
Although both materials can be conveyed pneumatically, the
pneumatic conveying regime for cement powder is likely to be
quite different for the regime selected for wet lump coal.
-
The reason for this concerns the properties of the
bulk material and how these properties interact during the
 pressure drop, air flow, etc. Properties such as moisture
content, cohesiveness and adhesiveness may cause flow
problems through vessels and valves.
-
Although both materials can be conveyed pneumatically, the
pneumatic conveying regime for cement powder is likely to be
quite different for the regime selected for wet lump coal.
-
The reason for this concerns the properties of the
bulk material and how these properties interact during the
pneumatic conveying process.
-
For example, cement powder may
be easily fluidized and mixed with air.
When conveyed at high
velocities, it will not degrade to the detriment of the bulk
material. -
Wet, lump
coal (2" mean size), on the other hand, cannot be fluidized
without severely degrading the material to the extreme
detriment of the coal product.
-
These factors affect the choice of allowable material
velocities through the pipeline.
-
Different grades of exactly the same material can
exhibit totally totally different performances.
For practical purposes, a conservative design
approach is to keep the ratio of standard cubic feet of air to
pounds of material below a 2.25:1 proportion.
-
Successful systems have been designed using air-
material loadings of 1:1 or more when the system components
are
Ref: well-designed and eliminate sharp turns, abrupt junctions,
http://www.cedengineering.com/upload/Pneumatic%20Conveying%20Systems.pdf
or other potential points of binding, clogging, or drop-out.
different performances.
Sufficient velocities must be maintained throughout the conveying
system to avoid material settling. When settling occurs in the
horizontal plane, it is known as saltation.
-
When settling occurs in the vertical plane, it is called choking.
-
Saltation is the
process of deposition of solid particles along a horizontal pipeline.
This phenomenon occurs when the air velocity falls below the
minimum conveying value.
-
Caution – Don’t select a velocity higher than needed. The additional
velocity would be detrimental to the system by causing increased
friction, wear, and operating costs.
-
Choking in downward movement often occurs in the vertical line as a
direct result of saltation in the adjacent horizontal line. -

Upward movement is often easier to control because all that is

needed is sufficient momentum (velocity) to keep the material in
suspension. -
-
All falling materials simply drop back into the airstream.
However choking in the upward flow directly above the fan
discharge will exhibit premature wear due to excessive loading.
-
To minimize the potential for saltation or choking, it is
recommended to minimize bends and elbows and also remove any
leaks because velocity will be less downstream of leaks.
-
It is good to consider some excess air in the system design
that will effectively increase velocities in the system to assist
material transportation.
Some provisions may be included to keep in the system
for bleeding excess air through adjustable vents or dampers.
discharge will exhibit premature wear due to excessive loading.
-
To minimize the potential for saltation or choking, it is
recommended to minimize bends and elbows and also remove any
leaks because velocity will be less downstream of leaks.
-
It is good to consider some excess air in the system design
that will effectively increase velocities in the system to assist
material transportation.
Some provisions may be included to keep in the system
for bleeding excess air through adjustable vents or dampers.

END OF WORKSHEET
 DISCLAIMER: The materials contained in this online course are
not intended as a representation or warranty on the part of PDH
Center or any other person/organization named herein. The
materials are for general information only. They are not a
substitute for competent professional advice. Application of this
information to a specific project should be reviewed by a
registered architect and/or professional engineer/surveyor.
Anyone making use of the information set forth herein does so at
their own risk and assumes any and all resulting liability arising
therefrom.

USEFUL LINKS
MAC
http://www.macprocessinc.com/

Hammertek
http://www.hammertek.com/index.asp

Fuller Bulk Handling

http://www.petropages.com/fuller-bulk-handling.html

CoperionWaeschle
http://www.coperion.com/en/

exair
http://www.exair.com/

SMOOT
http://www.magnumsystems.com/

http://www.khi.co.jp/english/kplant/business/energy/surround/pressure.html
http://en.bevconwayors.com/2012/03/08/coal-fuel-handling-plant/

http://www.flowmeterdirectory.com/solid_conveying_eductor.html
http://www.flowmeterdirectory.com/solid_conveying_eductor.html

Source: http://www.foxvalve.com
Fox Valve is the leading global supplier of venturi eductors
and venturi transport systems for pneumatic conveying of
powders, pellets, and bulk solids. They enable the use of low
pressure air ( below 14 psig or 1 bar) to be used to
move powders, pellets, and bulk solids with no moving parts.
Fox has sold venturi eductors continuously since our first
sale, in 1963, of a stainless eductor for use in transporting
plastic pellets with air at 4 psig.
PNEUMATIC CONVEYING SPREADSHEET-CONTENT
2 PDH Professional Development Hour course content
© Copy right John Andrew P.E., 30 July 2012

TURBULENT FLOW PNEUMATIC CONVEYOR

To scroll, roll the mouse wheel.

To zoom in, depress the Ctrl key and roll the mouse wheel away from you.
To open another, "Work Sheet Lesson" select a tab at the bottom of this spreadsheet.
To unlock the cells of this spreadsheet: Home > Format > Unprotect Sheet.
To lock the cells of this spreadsheet: Home > Format > Protect Sheet.

Dilute Conveying
Ratio of standard cubic feet of air to pounds of material below a 2.25:1 proportion.
Dilute conveying stream flow is the most common pneumatic method.
Dilute conveying stream flow; air volume / solids weight > 2.25
Dilute and pulse conveying stream flow; 2.25 < air volume / solids weight < 0.20
Pulse conveying stream flow; air volume / solids weight > 2.25
High linear air velocities (4,200 to 6,500 ft/min)

Pipe Internal Diameter

Input
Pipe outside diameter, de = 12.000 in
Pipe wall thickness, t = 0.280
Calculate
Pipe inside diameter, d = d - 2*t
= 11.440 in
Pipe inside section area, A = π*(d/12)^2 / 4
= 0.7138 ft^2

Conveyor Pipe Friction Head Loss Quiz 6

Input-1 Input-1
Material conveyed is; Sawdust Sawdust
Bulk mMaterial flow rate is, W = 1800 lb/hr 1800
Bulk material density is, ρm = 11.0 lb/ft^3 11.0
Air volume flow rate, q = 785.0 cfm 785.0
Pipe internal diameter, d = 6.065 in 6.065
Roughness in inches, ei = 0.0018 in 0.0018
Suction + Discharge length of straight pipe, L = 100.0 ft 100.0
Air viscosity, µ = 0.0000003750 ft^2/sec 0.0000003750
Air density, ρ = 0.075 lbs/ft^3 0.075
Blower efficiency, EffP = 60% % 60%
Drive efficiency - Gear or Belt, EffD = 80% % 80%
Motor efficiency, EffM = 90% % 90%
 Output-1
Solids volume flow rate, Qs = W / 60
= 30.000 lbs/min
Ratio: air cfm volume / lb/min solids weight , R = Qs / q
= 26.2 R > 2.5 OK, dilute air flow
Gravitational constant, g = 32.2 ft/sec^2
Round pipe hydraulic dia = internal diameter, D = d/12
= 0.5054 ft
Pipe internal area, A= 3.1416*D^2/4
= 0.2006 ft^2
Air flow volume, Q = q / 60
= 13.083 ft^3/sec
Air flow velocity, V = Q / A
= 65.21 ft/sec
= 3913 ft/min
Air mass density, ρM = ρ / 32.2
= 0.002329 slugs/ft^3
Reynolds number, Re = D * V * ρM / µ
= 204,716 OK, turbulent air flow
Note that the velocity of the flow in the Reynolds calculation
is based on the actual cross section area of the duct or pipe.
Reynolds number Re

The flow is:

laminar when Re < 2300
transient when 2300 < Re <
4000
turbulent when Re > 4000
Click GOAL SEEK tab below for friction factor f calculation method.
Darcy-Weisbach friction factor f calculated from the Colebrook-White equation.
http://www.engineeringtoolbox.com/darcy-weisbach-equation-d_646.html
Input-2
Guess friction factor, f = 0.01616 ft 0.01616
Output-2
Roughness, e = ei / 12
= 0.00015 ft
Relative roughness, e / D = 0.00030 << See Moody diagram >>
1 / f^0.5 = -2*LOG10((e / 3.7*D) + (2.51 / (Re*f^0.5)))
Both sides of equation are divided by (f^0.5), 1 = f^0.5* (-2*LOG10((e / 3.7*D) + (2.51 / (Re*f^0.5))))
SOLVED IF: 0.9990 < f < 1.0002, 1 = 0.9997
Output-3
2
Air flow head loss, ΔP = f*(l / dh) (ρ v / 2)
ΔP = f*(L/d)*(ρ*V^2 /2)
ΔP = 42.49 in H2O/100 ft
Minimum blower motor power, PM = q * (ΔP /12) / (3960 * EffP * EffD * EffM)
 = 1.62 hp
Use a standard blower motor = 2.0 hp

Ref: http://www.engineeringtoolbox.com/darcy-weisbach-equation-d_646.html
Pressure Loss
The pressure loss (or major loss) in a pipe, tube or duct can be
expressed with the Darcy-Weisbach equation
Δp = f (l / dh) (ρ v2 / 2)
where
Δp = pressure loss (Pa, N/m2)
f = Darcy-Weisbach friction coefficient
l = length of duct or pipe (m)
dh = hydraulic diameter (m)
ρ = density (kg/m3)

The friction coefficients used to calculate pressure loss (or

major loss) in ducts, tubes and pipes can be calculated with the
Colebrook equation
1 / f1/2 = -2 log [ 2.51 / (Re f1/2) + (k / dh) / 3.72 ] (1)
where
f = Darcy-Weisbach friction coefficient
Re = Reynolds Number
k = roughness of duct, pipe or tube surface (m, ft)
dh = hydraulic diameter (m, ft)
The Colebrook equation is only valid at turbulent flow
conditions.

END OF WORKSHEET
DENSITY CONVERTER
http://www.flowmeterdirectory.com/densitycalculator.php
PNEUMATIC CONVEYING SPREADSHEET-CONTENT
2 PDH Professional Development Hour course content
© Copy right John Andrew P.E., 30 July 2012

HYDRAULIC DIAMETER
http://www.engineeringtoolbox.com/hydraulic-equivalent-diameter-d_458.html
The hydraulic diameter - dh - is used to calculate the dimensionless Reynolds Number
to determine if a flow is turbulent or laminar.
A flow is:
laminar if Re < 2300
transient for 2300 < Re < 4000
turbulent if Re > 4000

Note that the velocity of the flow in the Reynolds calculation is based on the
actual cross section area of the duct or pipe.
The hydraulic diameter is also used to calculate the pressure loss in a ducts or pipe.

The hydraulic diameter is not the same as the geometrical diameter in a non-circular
duct or pipe and can be calculated with the generic equation.

Hydraulic Diameter of a Circular Tube or Duct = the diameter

Input
Duct internal diameter, d = 10.136 in
Calculate
Duct internal section area, A = π*(d/12)^2 / 4
= 0.5604 ft^2
Wetted perimeter of the duct , p = π*d / 12
2.653605 ft
Hydraulic diameter , dh = 4*A / p
= 0.8447 ft
= 10.136 in

Hydraulic Diameter of a rectangular duct.

Input
Duct internal section width, w = 10 in
Duct internal section height, h = 5 in
Calculate
Duct internal section area, A = w*h/144
= 0.3472 ft^2
Wetted perimeter of the duct , p = 2*(w + h)
= 2.500 ft
Hydraulic diameter , dh = 4*A / p
= 0.5556 ft
= 6.667 in
0.1
END OF WORKSHEET
Pipe Internal Diameter
Input
Pipe external diameter, de = 10.75 in
Pipe wall thickness, t = 0.25
Calculate
Pipe internal diameter, d = d - 2*t
= 10.250 in
Duct internal section area, A = π*(d/12)^2 / 4
= 0.5730 ft^2
Standard Pipe Fittings Dimensions
http://www.steelindiaco.com/pdf/buttweld%20fittings.pdf
http://www.hackneyladish.com/DimensionData-pr1.aspx

Standard Pipe Dimensions

http://www.hackneyladish.com/DimensionData-pr5.aspx
PNEUMATIC CONVEYING SPREADSHEET-CONTENT
2 PDH Professional Development Hour course content
© Copy right John Andrew P.E., 30 July 2012

GOAL SEEK

Some advantages of spreadsheet calculations over

hand written include:
1. easier to read.
2. better recall from archives.
3. greater accuracy.
4. faster with repeat use.
5. graphs are created automatically.
6. numerous useful formulas.
7. "Goal Seek" enables optimization.
8. solve any equation with, "Solver".
9. solve sets of linear and non-linear equations.

To scroll, roll the mouse wheel.

To zoom in, depress the Ctrl key and roll the mouse wheel away from you.
To open another, "Work Sheet Lesson" select a tab at the bottom of this spreadsheet.
To unlock the cells of this spreadsheet: Home > Format > Unprotect Sheet.
To lock the cells of this spreadsheet: Home > Format > Protect Sheet.

GOAL SEEK OPTIMIZATION

Use these live cells for Goal Seek
Conveyor Pipe Air Flow Velocity
Input
Material conveyed is; Sawdust
Air volume flow rate, q = 785.0 cfm
Round pipe internal diameter, d = 7.500 in
Air density, ρ = 0.075 lbs/ft^3
Calculate
Round pipe hydraulic dia = internal diameter, D = d/12
Insert "Goal Seek"
= 0.6250 ft dialog box here.
Pipe internal area, A= 3.1416*D^2/4
= 0.3068 ft^2
Air flow velocity, V = Q / A
= 2559 ft/min or fpm

Step-1 Select cell containing a formula: G38 (Yellow)

Step-2 Pick drop-down menu: Data > What-If Analysis > Goal Seek
Step-3 Pick the "To value" cell and type 1
Step-4 Pick the "By changing cell" next pick cell that needs to change "G30" > OK
GOAL SEEK EQUATION SOLVING
Use these live cells for Goal Seek
Friction Factor f Calculation
Input-2
Guess friction factor, f = 0.02 ft Insert "Goal Seek"
Round pipe hydraulic dia = internal diameter, d = 10.000 in dialog box here.
Roughness, e = 0.000150 ft
Reynolds number, Re = 165,000 -
Colebrook-White Equation Output-2
Round pipe hydraulic dia = internal diameter, D = d/12
= 0.8333 ft
1 / f^0.5 = -2*LOG10((e / 3.7*D) + (2.51 / (Re*f^0.5)))
Both sides of equation are divided by (f^0.5), 1 = f^0.5* (-2*LOG10((e / 3.7*D) + (2.51 / (Re*f^0.5))))
SOLVED IF: 0.9990 < f < 1.0002, 1 = 1.0889

GOAL SEEK METHOD

Step-1 Select cell containing a formula: G60 (yellow)
Step-2 Pick drop-down menu: Data > What-If Analysis > Goal Seek
Step-3 Pick the "To value" cell and type 1
Step-4 Pick the "By changing cell" next pick cell that needs to change "G52" > OK > OK
Darcy-Weisbach friction factor f calculated from the Colebrook-White equation.
http://www.engineeringtoolbox.com/darcy-weisbach-equation-d_646.html

END OF WORKSHEET
GOAL SEEK OPTIMIZATION
Example only cells are locked
Conveyor Pipe Air Flow Velocity
Input
Material conveyed is; Sawdust
Air volume flow rate, q = 785.0 cfm
Round pipe internal diameter, d = 7.500 in
Air density, ρ = 0.075 lbs/ft^3
Calculate
Round pipe hydraulic dia = internal diA, D = d/12
= 0.6250 ft
Pipe internal area, A= 3.1416*D^2/4
= 0.3068 ft^2
Air flow velocity, V = Q / A
= 2559 ft/min or fpm

Step-1 Select cell containing a formula: G38 (Yellow)

Step-2 Pick drop-down menu: Data > What-If Analysis > Goal Seek
Step-3 Pick the "To value" cell and type 1
Step-4 Pick the "By changing cell" next pick cell that needs to change "G30" > OK
 GOAL SEEK EQUATION SOLVING
Example only cells are locked
Friction Factor f Calculation
Input-2
Guess friction factor, f = 0.01694 ft
Round pipe hydraulic dia = internal dia, d = 8.000 in
Roughness, e = 0.000150 ft
Reynolds number, Re = 165,000 -
Output-2
Round pipe hydraulic dia = internal diameter, D = d/12
= 0.6667 ft
1 / f^0.5 = -2*LOG10((e / 3.7*D) + (2.51 / (Re*f^0.5)))
Re*f^0.5)))) Both sides of equation are divided by (f^0.5), 1 = f^0.5* (-2*LOG10((e / 3.7*D) + (2.51 / (Re*f^0.5))))
SOLVED IF: 0.9990 < f < 1.0002, 1 = 1.0001

GOAL SEEK METHOD

Step-1 Select cell containing a formula: G62 (yellow)
Step-2 Pick drop-down menu: Data > What-If Analysis > Goal Seek
Step-3 Pick the "To value" cell and type 1
Step-4 Pick the "By changing cell" next pick cell that needs to change "G52" > OK > OK
/ (Re*f^0.5))))