2. RECIRCULATING CONVEYORS
These are Continuous loop conveyors can be used
for storage, if loaded carries are allowed to flow
back on the return leg raising the following
possibilities:
Empty carriers are not available when needed for
loading
Full carriers are not immediately available for
unloading
3. CONTINUOUS LOOP CONVEYORS
Consider a continuous closed-loop conveyor, such as
An overhead trolley system with one load
one unload station.
Assume that all carriers are emptied at the unload station.
Return leg
Le
Delivery leg Ld
Empty carrier
Load
station
Unload
station
np
Full
carrier
4. CONTD.
nc = number of carriers in the
system
Ld = length of the delivery leg (m)
Le = length of the return leg (m)
sc = carriers spacing (m/carrier)
Td = deliverytime(min/carrier)
Ld = conveying distance between
load and unload stations (m)
Np = total number of parts in the
system
np = number of parts in each
carrier
Rf = part feed rate (parts/min)
vc = conveyor speed (m/min)
TL= loading time(min/carrier)
TU= unloading time(min/carrier)
c
ed
c
s
LL
n
ed
dcp
c
dp
p
LL
Lnn
s
Ln
N
L
p
c
cp
f
T
n
s
vn
R
c
d
d
v
L
T
L
p
c
cp
Lf
T
n
s
vn
RR
TU < TL
5. CONVEYOR ANALYSIS (KWO’S PRINCIPLE)
Kwo’s recirculating conveyor design requirements with ONE
load station and ONE unload station. Three rules
Speed rule: Operating speed within a certain limit determined
by #carriers/min (vc/ sc ). The lower limit should be greater
than or equal the required loading or unloading rate
whichever is the greater.
c
cp
LU
s
vn
, RMax R parts/min
6. The upper limit should be less or equal to the capabilities of
the material handlers.
c
c
s
v
<Min 1
TL TU
1
,
min/carrier
• Capacity constraint: The flow rate capacity of the conveyor
must be at least equal to the flow rate requirement to
accommodate reserve stock and allow for the time elapsed
between loading and unloading due to delivery distance.
c
cp
fs
vn
R
•Uniformity Principle: Empty and full carriers should be
uniformly distributed along the line to avoid excessive waiting
for carriers.
7. CONVEYOR ANALYSIS: EXAMPLE
A recirculation conveyor has a total length of 300m. Its
speed is 60m/min, and the spacing of part carriers along
its length is 12m. Each carrier can hold 2 parts. The time
required to load 2 parts into each carrier is 0.20min and
the unload time is the same. The required loading and
unloading rates are both defined by the specified flow
rate, which is 4parts/min. Evaluate the conveyor system
design with respect to Kwo’s three principle.
Specified flow rate , Rf = 4 parts/min
Conveyor speed, vc = 60m/min
Spacing of carriers, Sc =12m
Number of parts per carrier, np = 2 parts
Loading TL = Unloading TU = 0.2 min/carrier
8. CONTD.
Solution: 1) Speed Rule The lower limit on speed is set by the
required loading and unloading rates, which is 4 part/min.
Checking this by
c
cp
LU
s
vn
, RMax R
c
c
s
v
<Min 1
TL TU
1
,
12 m/carrier
(2 part/carrier)
=10 parts/min >4 parts/min
(60 m/min)
Checking the lower limit
9. CONTD.
12 m/carrier
60 m/min 1
0.2 0.2
1
,
= 5 carrier/min <Min
=Min {5,5} = 5
• 2.Capacity Constraint: Conveyer flow rate capacity =
10parts/min as computed above since this is substantially
greater than the required delivery rate of 4 parts/min, the
capacity constraint is satisfied. KWO provides guidelines for
determining the flow rate requirement that should be
compare to be conveyer capacity.
10. CONTD.
• 3.Uniformity Principle : The conveyor is assumed to be
uniformly loaded throughout its length, since the
loading and unloading rates equal and the flow rate
capacity is substantially greater than the load/unload
rate.
