material handling

1. Pneumatic Conveyors
•Pneumatic conveying is a method of transporting bulk
materials in the form of powder, short fiber and
granules over a pipeline as a mixture with air or due to
pressure of air. There are three basic system used.
a) Suction or Vacuum System utilizes a vacuum created
in the pipeline to draw the material with the surrounding
air.

2. • These systems are particularly
suited to moving material from
• multiple pickup points to a
the reason
single location,
being that the bulk
expense
system's
terminal
receiver,
end
rotary
where
valves,
of the
is in the
the
and
vacu u m source are located.

3. Pneumatic Conveyor - Vacuum System

4. b) Pressure-type System is ideally suited for
conveying from one pickup Location to many discharge
locations. Generally,this type of system is more
economical when going from one point to several.
A pressure system of this type generally conveys with a
product-to-air ratio of about 20kg of material per kg of air,
or approximately 24kg of material/m3 of air (or 20 m3 of
air/m3 of product).

5. Cont’d
Pneumatic Conveyor -Pressure System

6. C)Combination System (Push-Pull system)
 This is a system in which a sanction (permit) system is
used to convey material from a number of loading points
and pressure system is employed to deliver it to a
number of unloading points.
Such installations are utilized when conveying over a long
distance is required.
Applications and Limitations:
Pneumatic conveyors have many advantages:
•delivery of materials over a path capable of changing its
direction in any plane,

7.  processing of the material simultaneously with its
conveying,
 an almost limitless number of loading and unloading points
served by a single system,
 air and gas tightness eliminating dust nuisance (pains) and
dust hazards
 an almost totally automated conveying with considerable
reduction of losses of material,
 improved labor conditions and minimum of human
attendance.

8. The limitations of the system are
high power requirements (15kWh/t, 10 to 15 times higher
than mechanically conveyors),
rapid wear of equipment, the problem of dust recovery from
the exhaust air
inability to convey wet, caking (block) and sticky loads.

9. Pneumatic Conveyor Components
Intake Units
One of the most delicate problems in pneumatic
conveyors is the introduction of material to the flow of
air at a regulated rate
a. Nozzle Injector
Nozzle Injector

10. b. Rotating Valve
Rotating Valve
Stationary Screw Feeder

11. Cont’d
Suction Nozzle
Conveying Pipe and Changeover Valves

12. Cont’d
Separator

13. Design Considerations
In pneumatic conveyor calculations given are properties of
the material, required capacity Q tons per hour and the
configuration of the conveying pipe.
Required are:
1. Calculated (reduced) conveying length, Lred [m]

14. Values of Equivalent Lengths for Elbows
Material
4 6 10 20
Powdered 4-8 5-10 6-10 8-10
Granular Homogenous - 8-10 12-16 16-20
Small Lumped Irregular - - 28-35 38-45
Large Lumped Irregular - - 60-80 70-90
The commonly used value for a two-way changeover valve
L=
8m.
eqv

15. 2. Conveying air stream velocity, = [m/s]
air 1
v    BL2
red
Values of Factor for the Size of Load Particles

Material
Particle Size

Powdered 1-1000(micron) 10-16
Granular Homogenous 1-10mm 17-20
Small Lumped Homogenous 10-20mm 17-22
Medium Lumped Homogenous 40-80mm 22-25
Where,  = factor for the size of load particles
γ1 = specific weight of the load particles[tons/m3 ]
B = factor assumed as equal to (2 to 5) 10-5 , the lower values being taken
for dry powder materials.

16. 3. Weight concentration of the mixture, :

Graph Showing the Dependence of the Weight Concentration of
the Mixture  on the Reduced Conveying Length
.
Lred

17. Note:Graph (1):
4
air
2
p
air
ol
3.6 
Q v
d
V 
where Q = capacity of installation [tons/hour]

18. 5. Conveying pipe inner diameterd
, p
[m]
air
p
v 
4Vol
d 
6. The required air pressure in the pipe,
[kg/cm2]
104

Hair 
cs
P
Where,
air= specific weight of the air
(average for a given vertical section).

19. For pressure conveying system;
cs
p
red air
i P
d
2
P  1
L v
red air
d p
2
 L v
Where,
= a factor; for pressure conveying systems,
depends on the value of s 
= 1.510 7
and for suction and for suction conveying system:

And for suction conveying system
cs
p
red air
f P
d
2
L v
P  1

20. • The plus sign before in equation is
taken for upward, the minus sign for
downward movement.
Graph Showing the Dependence of Factor
the Value of s
on

21. 7. The required air pressure of the
compressor or air blower, [kg/cm2]
Pb Pw  Ploss
w
Where,
P
= working pressure
= for pressure conveying system
,
= suction conveying syste,m
Pi
Po  Pf

loss
P
Ploss
0
= 1.15 to 1.25 factor for losses in the
intake
unit,
= pressure loss in the supplying air for
compressors,
P = 0.3 kg/cm2,
= atmospheric pressure = 1atm.

22. 8. The required capacity of the
compressor
or blower, [m3/min]
4
2
p air
v '
o ol
d
V V '
where ,
=' factor for losses due to leaks = 1.1.

23. 9. The required motor power, [kW]
1 m3 drawn in during isothermal
compression [kgm/m3].
= total efficiency of compressors =0.55 to 0.75.
Where, Lb = theoretical work of the blower reduced to

.
0
0
P
Pb
b
L 23,030 P log
60102
LbVo
b
N 