help regarding drawing ladder diagram and modification

Looks a lot like homework to me.....Do you have anything so far? If so, post what you have and you will get help here. No one will just do it for you though.

Dave

answer please check whether it is correct

01-part in sensing region
02-parts good
03-parts bad
10-robot picks up part
20-part goes in buffer a
30-part goes in buffer b
40-alarm,conveyor stops
c100-downcounter for a
c200-downcounter for b
t100-timer for process

Answer please: Check whether it is correct.

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No, it is not correct. There are a few things missing or not done correctly. See the attached PDF file. Because you did not state what kind of PLC, I had to choose one, and I picked an Omron ZEN. I hope you can figure out how it works!

Because you did not use any rung comments on your version, then I did not use any either. I hope you can figure out what I did! I hope I can remember what I did if you ask me how it works!

What modification or alternate process and hardware would you suggest to the existing system to improve the process or logic?

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Because you asked, the following underlined rquirement must be a mistake in the specifications:

If during the cycle, buffer B has more that 2 parts, the conveyor is stopped, an alarm is sounded and the clock is reinitialized. If during the same cycle if Buffer A has less than 100 parts, the same steps are taken.

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If you look at my Rung 11, you can see that I tried to follow those directions even though it does not make much sense or produce a logical process. Because Buffer A contains the number of good parts, then why would anyone want to stop the conveyor, sound the alarm, and reset the clock at any time when C200 < 100? It seems that it would make more sense to run the process UNTIL Buffer A conatains 100 parts (or whatever the desired number of good parts). There should be no reason to shut down just because the required number of good parts has not yet been produced! Perhaps the spec was meant to say "...if Buffer A has MORE than 100 parts, the same steps are taken."

PS: It appeared to me like you need to use Counters that count up, not down. The ZEN has both, so I used the Count-UP type. It seems more logical to start with 0 good parts and count up to 100 good parts, and with 0 bad parts and count up to 2 bad parts before setting off the alarm and stopping the process.

16091988 said:

...A timer keeps track of the process for 60 minutes after which it is reinitialized. If during the cycle, buffer B has more that 2 parts, the conveyor is stopped, an alarm is sounded and the clock is reinitialized. If during the same cycle if Buffer A has less than 100 parts, the same steps are taken. The cycle continues otherwise...

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Lancie1 said:

Because you asked, the following underlined rquirement must be a mistake in the specifications:

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Lancie1 said:

Because Buffer A contains the number of good parts, then why would anyone want to stop the conveyor, sound the alarm, and reset the clock at any time when C200 < 100? It seems that it would make more sense to run the process UNTIL Buffer A conatains 100 parts (or whatever the desired number of good parts). There should be no reason to shut down just because the required number of good parts has not yet been produced! Perhaps the spec was meant to say "...if Buffer A has MORE than 100 parts, the same steps are taken."​

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If less than 100 good parts have been counted within the 1 hour cycle, or more than 2 bad parts, then you stop the process, alarm for inspection, and reset the clock for the start of the next cycle.

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G, the key phrase words are the first part of the requirement "If during the same cycle....".

If during the same cycle if Buffer A has less than 100 parts, the same steps are taken. The cycle continues otherwise...

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As the problem reads, there are only three conditions that cause the cycle to end: the timer reaches 60 minutes, or there are more than 2 bad parts before 60 minutes, or there are less than 100 parts before 60 minutes. The last condition is not logical. In a real assembly line, the setpoint would be "100 parts", not "60 minutes". Now if there is a production goal of 100 parts per hour, there might be a flag or message that gets sent to the plant supervisor and maybe the maintenance department, but I doubt if the line would be shut down due to slow production. That would kind of defeat the goal to get the parts produced and would only delay the required quota even more.

If it had said "after the same cycle", or even "during the next cycle when the last cycle did not have 100 good parts", then your theory would be valid. However it does not say either of those, but the action must be done BEFORE the cycle ends (during the cycle), not after it is finished or after 60 minutes is completed. Also it is not clear in the second part of the sentence as to what the subject modifier "same" means. It could mean that the cycle restarts, or it could mean that the alarm sounds, and the conveyor stops. But it makes no logical sense to stop before 100 good parts are finished. This entire problem is poorly written by someone who is not familar with good English usage, or does not really know what he is doing.

Another failure in the problem is the obvious lack of Start and Stop methods. We have to assume that the hidden Conveyor magically starts and carries the part down to the Robot, without any signal. Later we are told to stop the conveyor under certain conditions, even though we never got it started in the first place. I suggest the addition of Start and Stop pushbuttons to control the System Run Relay.

I agree that it is an assigment, and that is all the more reason that the instructions are written clearly. When I received this sort of specifications for a real PLC programming job, I found out the hard way that it does not work to simpy make assumptions based on what I thought it meant. What I interpreted was often wrong because usually there were hidden issues that were not mentioned. I responded with questions to clear up the misused and confusing language, just as a student should do with this type of garbled poorly-written problem.

Ok Lancie, now don't give out to me! My intentions were/are good. I was sort of turning you into the student for a minute to make you think about why it was written that way.

You're thinking that they got the functional design specification (FDS) wrong. I agree that as it reads, it makes no sense. But remember what was asked at the end...

16091988 said:

What modification or alternate process and hardware would you suggest to the existing system to improve the process or logic?

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What you've identified is an intentional error in the functional design to make the student think, "hey that doesn't look right!?", "that can't work?".

That's why I posted the highlighted quotes in the order I did. First to recap on the FDS...

16091988 said:

...A timer keeps track of the process for 60 minutes after which it is reinitialized. If during the cycle, buffer B has more that 2 parts, the conveyor is stopped, an alarm is sounded and the clock is reinitialized. If during the same cycle if Buffer A has less than 100 parts, the same steps are taken. The cycle continues otherwise...

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Then your questioning of the design...

Lancie1 said:

...must be a mistake...why would anyone want to stop the conveyor, sound the alarm, and reset the clock at any time when C200 < 100?...

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Not only is it a mistake, it cannot work.

Remember I wrote...

Geospark said:

If it's assignment work, the why is sort of irrelevant. The how is what's important. It's a theoretical scenario to get you to think about how you would achieve this, not so much why.

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How can you stop for having less than 100 good parts during the 60mins? Any number, during the 60mins, from 0 to 99 good parts, will stop the cycle. So from the second the cycle is started it's stopped. It's a catch 22.

I then outlined what I think it should be to correct the FDS...






In their defense, the people that set these questions are actually quite clever, that's why they're the ones that are doing it.

What you've identified is an intentional error in the functional design to make the student think, "hey that doesn't look right!?", "that can't work?".

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Yes, I agree, and that is exactly what I was doing. I also pointed out another design failure, the lack of any way to easily stop the system.

There are other design failures. The biggest one is the expectation that beginning PLC programmers will know enough to find this type of subtle error. At the beginning level, they barely know the difference between input and output. Any instructor who expects that this type of trick question will do any good is just fooling himself. I know, I have been there and done that.