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Old April 6th, 2021, 08:58 PM   #16
Aintbad
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i did some changes to the program and its working just fine, as it supposed to according to the exercise, is there anything wrong where you can help me to improve it?
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Old April 6th, 2021, 09:20 PM   #17
drbitboy
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Quote:
Originally Posted by Aintbad View Post
i did some changes to the program and its working just fine, as it supposed to according to the exercise, is there anything wrong where you can help me to improve it?



Well done!


I would do the flip-flop (rung 005) a little differently, but that's just a style issue. You might want to ask The Google about [ladder logic flip-flop] and see if there are any arrangements that are more clear. I'm partial to the top two rungs of the image below.


Does LogixPro allow for comments?




You can get rid of Rungs 001 and 002, and also intermediate bits B3:0/2 and B3:0/4 by
  1. Putting O:0/2 at the start of rungs 003 and 004,
  2. Replacing the two [XIC B3:0/2] instructions with [XIO I:1/2],
  3. Replacing the two [XIC B3:0/4] instructions with [XIO I:1/3].
That would not change the behavior, but I think it will be a bit more clear; that said, the inverted logic on the inputs might be confusing, so maybe change the names of I:1/2 and I:1/3 to something like [0 for low pressure, one motor one] and [0 for low low pressure, both motors on], respectively.



Last edited by drbitboy; April 6th, 2021 at 09:26 PM.
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Old April 6th, 2021, 09:48 PM   #18
Aintbad
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Originally Posted by drbitboy View Post
Well done!


I would do the flip-flop (rung 005) a little differently, but that's just a style issue. You might want to ask The Google about [ladder logic flip-flop] and see if there are any arrangements that are more clear. I'm partial to the top two rungs of the image below.


Does LogixPro allow for comments?




You can get rid of Rungs 001 and 002, and also intermediate bits B3:0/2 and B3:0/4 by
  1. Putting O:0/2 at the start of rungs 003 and 004,
  2. Replacing the two [XIC B3:0/2] instructions with [XIO I:1/2],
  3. Replacing the two [XIC B3:0/4] instructions with [XIO I:1/3].
That would not change the behavior, but I think it will be a bit more clear; that said, the inverted logic on the inputs might be confusing, so maybe change the names of I:1/2 and I:1/3 to something like [0 for low pressure, one motor one] and [0 for low low pressure, both motors on], respectively.


oh okay! thank you and yes you can add comments in every rung!
ill start the Part 4 and see how it goes...
ill post again if i have any more questions...
Thanks again.
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Old April 7th, 2021, 07:04 AM   #19
drbitboy
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I thought of one more thing: a few posts ago you showed PE2 (I:1/3) parameters of high limit = 118 and span = 20. So I:1/3 will become 1 when the pressure is above 118; I:1/3 will become 0 when the pressure is below 98 (=118-20); I:1/3 will not change otherwise.


With the current logic, if the load is such that two compressors are required (as determined by the pressure continuing to drop to 98 after one compressor starts running at 100), then that PE2 high limit of 118 means the second compressor will turn off when I:1/3 becomes 1, which will happen before the system pressure rises to 120. Assuming the load is still high, the remaining compressor will not be able to get the system to 120. In that case, as long as the load stays high, I:1/2 will never become 1, and the same compressor will always come on first.
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Old April 7th, 2021, 03:31 PM   #20
Aintbad
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Originally Posted by drbitboy View Post
I thought of one more thing: a few posts ago you showed PE2 (I:1/3) parameters of high limit = 118 and span = 20. So I:1/3 will become 1 when the pressure is above 118; I:1/3 will become 0 when the pressure is below 98 (=118-20); I:1/3 will not change otherwise.


With the current logic, if the load is such that two compressors are required (as determined by the pressure continuing to drop to 98 after one compressor starts running at 100), then that PE2 high limit of 118 means the second compressor will turn off when I:1/3 becomes 1, which will happen before the system pressure rises to 120. Assuming the load is still high, the remaining compressor will not be able to get the system to 120. In that case, as long as the load stays high, I:1/2 will never become 1, and the same compressor will always come on first.

Yes you're right, that was my bad.
i was playing with the values to test it and for some reason when i took the screenshot it was set like that but in my assignment the values i got to use are 120 / 10 for the first motor and 120 11 for the second one.
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Old April 7th, 2021, 05:29 PM   #21
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aintbad, in the download section under AB, is the dual compressor ex#4, done by the master of logixpro 10,000 poster Lance1, you wont get a better example of how to do things in programming than by Lancie1.
But of course you need to try and do the exercises without any help, that is how you learn. Keep trying/thinking about what should/shouldn't happen then trial it in code. LogixPro is a fantastic learning/training tool, but be aware it isn't 100% exactly how a AB processor works
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