Conveyor Problems
Example 1:
The following parts will be painted on an overhang conveyor system. After the parts are painted, they will be transported to a drying oven which operates at 400 oF. Each part requires 10 minutes at 400 oF to dry. The hooks will be 1.5 ft apart. In a typical 8-hour shift, the conveyor -drying oven system requires an average of 50 minutes of downtime and has an efficiency of 80%. Calculate the size of the drying oven.
Figure 1
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Figure 2
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Part # |
Parts/Hook |
Quantity to be painted (per shift) |
|
A |
1 |
500 |
|
B |
4 |
300 |
|
C |
2 |
1000 |
|
D |
8 |
2000 |
|
E |
2 |
100 |
|
F |
1 |
125 |
Solution:
In an 8-hr shift (480 minutes), there will be 50 minutes of down time which will leave 430 minutes. Since we can work at 80% efficiency, the total available time for conveyor-drying oven is 80% x 430 minutes = 344 minutes.
Consider part A, which requires 1 part/hook and we have to paint 500 parts in one shift. Part A requires 500 hooks per shift. Since we have 344 minutes available in each shift, 500/344=1.453 hooks/min would be required for Part A.
Now consider part B, which requires 4 parts/hook and we have to paint 300 parts in one shift. Part B requires 300/4=75 hooks per shift. In 344 minutes, 75/344=0.218 hooks/min would be required for Part B.
If we continue with the similar calculation for parts C, D,E and F, we can generate the following results for each part:
|
Part # |
Parts/Hook |
Quantity to be painted (per shift) |
Needed Hooks/Shift |
Hooks/min |
|
A |
1 |
500 |
500 |
1.453 |
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B |
4 |
300 |
75 |
0.218 |
|
C |
2 |
1000 |
500 |
1.453 |
|
D |
8 |
2000 |
250 |
0.727 |
|
E |
2 |
100 |
50 |
0.145 |
|
F |
1 |
125 |
125 |
0.363 |
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Total hooks |
4.359 |
A total of 4.359 hooks need to pass any given point in one minute. Since we have 1.5 ft distance between two hooks, the conveyor speed should be 4.359 x 1.5 = 6.54 ft/min.
Each part requires 10 minutes of drying at 400oF. In 10 minutes at a conveyor speed of 6.54 ft/min, we need a conveyor length of 6.54 ft x 10 =65.4 ft. This means that the part of conveyor that runs through the drying oven must be 65.4 ft long. According to Figure 3, the required conveyor length of 65.4 ft includes the following:
Total conveyor length = L1 + L2 + Length of semicircle CB
Total conveyor length = 65.4 =L1 + L2 + 5p
L1 + L2 = 65.4 -5p = 49.7 ft
Since L1 = L2
Then the length of the oven must be:
L oven = 3 + 5 + (L1+L2)/2 = 8 + 49.7/2 = 32.9 ft
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Figure 3
Example 2:
In a company that is planning to manufacture charcoal grills, each grill will be shipped in a 30" x 30" x 24" box. A daily total of 2,400 grills are to be packed and transferred to the shipping center on a belt conveyor. Each box is placed on the conveyor by a push diverter. Due to the motion of the diverter, there will be a 10 inch distance between two boxes while travelling on the conveyor. Assuming a 50 minute scheduled downtime and an 80 % efficiency, calculate the minimum conveyor speed.
Solution:
In an 8-hour shift, 50 minutes will be used for scheduled downtimes. 480 -50 = 430 minutes. At 80 % efficiency, 430 x 80% = 344 minutes will be available.(344 effective minutes of work per day)
In order to accomplish 2,400 boxes a day (single shift), we have to allow 344/2,400 = 0.143 minutes / grill.
In one minute, we can transfer 1/0.143 = 7 grills/minute. (This is also the required capacity of the push diverter)
The conveyor will be :
7 grills/min x (30+10) (in/grill) = 280 in/min = 280/12 = 23.3 ft/min
Therefore, the minimum conveyor speed should be 23.3 ft/min.
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Last Update: November 13, 1999 |
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Prepared by : Serdar Z. Elgun |
