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Pulsar2003
I enjoy reading other topics, some evenings my eyes are hurting because I spend allot of time on forums like this one.

Here is a problem that I have and I was just wondering if it could be resolved easily. There are many experienced engineers and programmers that offer their help and knowledge. I have many topics here an I appreciate the help and suggestions.

I live in a Canadian city about 45 minutes north from the Vermont-New Hampshire border. This will help to understand the climate. I work in the plastic industry. During the summer it can be really warm while in the winter it is cold. But worse it is the spring and fall the problem and not summer or winter. In the spring and fall the days are rather warm while at night it can get really cool.

The problem with this is that the molds come out of the oven at 300C (572F) so the cooking time never really varies, it's the cooling time. I'm currently working to improve those automated ovens. The PLC has now a 1746-INT4 input module with only one input used.

I was wondering if the cooling time could be set at a minimum. Once the time is nearing the end I could pick up the nearby room temperature and have an extra step where the preset value of a timer would be set according to this temperature.

Even better... their could be a minimum time for cooling but as the buggy gets out of the oven and his nearing it's destination the PLC would monitor the room temperature and determine the correct cooling time.

I know it would be hard at first to set the parameters. Lots of Maths instructions or Compute.

Is it better to use a thermocouple to pick up the room temperature?

The bits 4 and 5 should be set to what? 0,0 for engineering units or 1,1 for proportional counts?

Suggestions???
BobLfoot
I'd approach this with simple Algebra myself and once that works then I'd get fancy.

Let's define 4 cases to start.

Case #1 - Room Ambient 40 F.

Case #2 - Room Ambient 60 F.

Case #3 - Room Ambient 80 F.

Case #4 - Room Ambient 100 F.

This will most likely cover the winter and summer cases and everything in between. If you know your cooling time for each case then the time for any temp inbetween will be an algebra solution.

Hope I am making sense.
Leitmotif




you did not specify clearly what the heated temperature is and what the desired temp is when they come out
I will pick em 550 is heated and 100 is cooled

You also do not say if you are cooling the oven with fans or just letting heat go to ambient

Heat transfer is proportinal to the temp difference ie heat (BTU) = m (BTU per pound of cooled) x Flow rate (if you have fans) x Th (550) - Tc (100)
Now you have the totall heat removed from object (if cooling the oven you would have to calculate that also and add)
If you are cooling you have to consider what the heat trasnfer rate thru the oven wall is so you need to know what the K is (heat transfer value)
So after you get all these calcs done you still have an estimate.
It may be best because this is what you are going to do after you do all these calcs to run a few temperature measurements including ambient and find out empirically how long it takes to cool at ambient and then use these values to derive your algorythm.

Dan Bentler
paulengr
QUOTE (Leitmotif @ Oct 26 2009, 01:29 PM) *
Heat transfer is proportinal to the temp difference ie heat (BTU) = m (BTU per pound of cooled) x Flow rate (if you have fans) x Th (550) - Tc (100)
Now you have the totall heat removed from object (if cooling the oven you would have to calculate that also and add)
If you are cooling you have to consider what the heat trasnfer rate thru the oven wall is so you need to know what the K is (heat transfer value)
So after you get all these calcs done you still have an estimate.
It may be best because this is what you are going to do after you do all these calcs to run a few temperature measurements including ambient and find out empirically how long it takes to cool at ambient and then use these values to derive your algorythm.

Dan Bentler


Heat transfer at the SURFACE is relatively constant but controlled by the temperature delta...simple version:
BTU's removed per unit of time = K * (temperature delta)

This by itself gives you an exponential...as the temperature difference decreases, the rate of cooling decreases linearly, but this translates into an exponentially longer time.

BUT, it's also dependent on the surface area if you are just doing ambient air cooling, so the shape and configuration of the mold will change things a bit.

AND, it depends on the density of air. Moist air transfers heat better than cold air. Depending on your particular temperature ranges and the swings in humidity, this may have a huge impact (it does on the coast line) or very little.

AND, it depends on conduction...how quickly the internal heat can migrate up through the mold to the surface. Thick section parts cool much slower than thin section parts, especially with plastics which aren't very thermally conductive to start with in most cases.

So those are the basics without getting into the math. The best thing to do is to get two thermocouples. Put one in the room where it can measure the ambient air temperature. Put the other one some place where it can measure something close to the mold wall. It doesn't have to be in the exact perfect spot but just somewhere nearby. You might also want to consider an infrared pyrometer since they are relatively cheap these days and "shoot" the part if you can. Run some experiments and chart the results.

You will probably find that as you start to collect data that you can pretty easily figure out what "close enough" means without going nuts over heat transfer equations.

Pulsar2003
[quote name='Leitmotif' date='Oct 26 2009, 01:29 PM' post='88429']
you did not specify clearly what the heated temperature is and what the desired temp is when they come out
I will pick em 550 is heated and 100 is cooled

You also do not say if you are cooling the oven with fans or just letting heat go to ambient

When the buggy gets out of the oven and reaches its destination then a set of fans blow air onto them. These molds also turn on two axis to cover all angles.

Heat transfer is proportinal to the temp difference ie heat (BTU) = m (BTU per pound of cooled) x Flow rate (if you have fans) x Th (550) - Tc (100)
Now you have the totall heat removed from object (if cooling the oven you would have to calculate that also and add)
If you are cooling you have to consider what the heat trasnfer rate thru the oven wall is so you need to know what the K is (heat transfer value)
So after you get all these calcs done you still have an estimate.
It may be best because this is what you are going to do after you do all these calcs to run a few temperature measurements including ambient and find out empirically how long it takes to cool at ambient and then use these values to derive your algorythm.

I will take some data to build a chart. The air flow is quite powerfull.

Thank you.

Pulsar2003
QUOTE (BobLfoot @ Oct 26 2009, 12:39 PM) *
I'd approach this with simple Algebra myself and once that works then I'd get fancy.

Let's define 4 cases to start.

Case #1 - Room Ambient 40 F.

Case #2 - Room Ambient 60 F.

Case #3 - Room Ambient 80 F.

Case #4 - Room Ambient 100 F.

This will most likely cover the winter and summer cases and everything in between. If you know your cooling time for each case then the time for any temp inbetween will be an algebra solution.

Hope I am making sense.


Any suggestions on the formula?

Thanks by the way.
Mickey
QUOTE (Pulsar2003 @ Oct 27 2009, 05:49 AM) *
QUOTE (BobLfoot @ Oct 26 2009, 12:39 PM) *
I'd approach this with simple Algebra myself and once that works then I'd get fancy.

Let's define 4 cases to start.

Case #1 - Room Ambient 40 F.

Case #2 - Room Ambient 60 F.

Case #3 - Room Ambient 80 F.

Case #4 - Room Ambient 100 F.

This will most likely cover the winter and summer cases and everything in between. If you know your cooling time for each case then the time for any temp inbetween will be an algebra solution.

Hope I am making sense.


Any suggestions on the formula?

Thanks by the way.



Why not just take the ambient range and rescale it to your timer preset range. Use a SCP instruction to do this.

What SLC do you have? What are the cool down times for 40 deg and 100 deg?
Pulsar2003
I have a SLC5/04 that would be a 1747-L542.

I know about the SCP but never used it.

For cooling time it varies and I never made statistics about it: Temperature of the mold has it comes out of the oven, once the cooling over and the plastic part.

I did take some down a little while back and if I recall well it's near 100C. I will need to verify the mold's temperature just out which would be near 300C.

There are also infrared thermocouples. Only as the molds turns it may hit times where there are no metal. I would need to make sure that the value remains stable for a while before ending the cooling. But that is only optional.

There are now other methods where you have an isolated metal box with an electronic device inside and a compartment to put ice, communicates to the PLC via radio waves. I don't know how they are called. There are thermocouples fixed to the mold. Once the proper temperature reached the cooking is stopped and same goes as the desired cooled temperature is reached the fans and axis stops.

Thank you Mikey.
paulengr
QUOTE (Pulsar2003 @ Oct 27 2009, 02:07 PM) *
There are also infrared thermocouples. Only as the molds turns it may hit times where there are no metal. I would need to make sure that the value remains stable for a while before ending the cooling. But that is only optional.


That's called a pyrometer, not a thermocouple. At the temperatures in question, actually RTD's are more accurate and simpler. Pyrometers depend on the material that they are looking at, and strangely enough, the temperature. Ideally, all materials are either black body emitters or white body emitters (which emit nothing). In reality, everything is a grey body emitter. The emissivity is a fudge factor correcting for the difference between the object's actual infrared energy emission, and that of a perfect black body emitter. Trouble is that some objects (notably aluminum, especially molten aluminum) are very close to white body emitters. And some materials, especially iron/steel, tend to have temperature-dependent emissivity. Worse yet, emissivity is a wavelength-dependent property. Do not buy a pyrometer without adjustable emissivity. The equation for it is 4th order so it is difficult to "fix" one that doesn't handle things correctly in PLC code.

By far the best (though not cheap) infrared pyrometers I've used come from Williams. FLIR and LAND are pretty good too but their specialty is handhelds. At one time Ircon was a very big name at least in the iron/steel business, but they've fallen way behind technologically and I would not recommend them these days even if you already have a model from them.

The Williams units are fast enough that I've managed to (through some creative PLC programmed filtering) capture the temperature of the inside surface of a pipe from inside an annealing oven as it slowly rolled through it. The pyrometer was looking through a 25 mm hole mounted on roughly a 45 degree angle looking at a target approximately 2 meters away (with a 25 mm spot size) where the temperature was in the range of about 1500-1800 Fahrenheit. I could pick out the inside temperature because as the pipe rolled by, the outside was hotter than the inside. When the temperature suddenly dropped, an edge detection algorithm picked out the "low spot". When the pipe was in the cooling section of the oven, the center was hot and the edges were cold so this was much easier.
JeffKiper
I second Paul's comments on IRCON's
SloooooowRCON we say on avg. 1.5 sec for the signal to stabilize on the newest lastest and greatest at the time in 2005. The analog loop out took so long we went with the RS 485 and got lighting fast 0.75 sec. only thing slooooooower than these Pyros. where temp sticks.

Our Land Pyros where quick we didn't have to lead them. We read bar stock on an indexing conveyor.
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