Tubular cartridge heater wattage

In summary: The two steel plates are 2000mm long and the tubular heater is sandwiched between them. There is a gap of 8mm between each adjacent 2 heaters.
  • #1
AligatorAmy
28
1
Hi,
I have a machine equipped with 2000mm long tubular cartridge heater (230V, 1600W, 4mm diameter). This tubular heater is sandwiched between the two steel plates (2000mm long) i.e. the heater keeps the steel components hot. My problem is that the temperature of the steel components varies along the heater +/-15degC (nominal T=190degC).

In order to maintain more uniform temperature along the steel components, I intend to replace the exisitng heater, 2000mm long, with 6 heaters, 325mm long each (there wil be approx. 8mm gap between each nearby 2 heaters). Each 325mm long heater will be powered and controlled separately. But all 6 heaters will be controlled in the way to maintain the same temperature in the steel based on the feedback from 6 thermocouples. The total power (wattage) of all 6x325mm long heaters will be the same as in case of 2000mm heater, i.e. 1600W.
Please also refer to the attached draft.

My question is:
Question 1
Does it mean that I would need to basically divide the power of 2000mm long heater, i.e. 1600W, by 6? What gives me approx. 270W per heater.
(If yes, I would then add perhaps 20% of power to each heater for some headroom, i.e. 270W+20%= 324W).
I have some doubt in regards to this solution because 6 heaters will be spread on the length of 2000mm and I am not sure whether simple dividing 1600W by 6 is adquate here.
Question 2
Do I maintain the in-line control on a single 325mm long heater temperature basically through varying the input voltage delivered to it?
If yes, my requirement to the heater supplier would be the condition that 325mm long heater must be capable to be operational at varying input voltage, say 180V-280V (with 230V nominal).

Please help.

Regards
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  • #2
The text indicates a 2000mm long heater is currently used, but the drawing shows it as 1600mm.

Where is the present thermocouple located?
Where will the (6) planned thermocouples be located?
These steel plates that are heated - are they parts in the machine?

The idea of dividing total power into the number of heaters is valid. However, cold spots occur where heaters butt up against one another in the 'planned' drawing, partially due to the spacing between them, and partially because each heater element has a 4 to 8 mm dead spot at each end for the electrical terminations. Is it possible to stagger the heaters so they overlap?
AligatorAmy said:
Do I maintain the in-line control on a single 325mm long heater temperature basically through varying the input voltage delivered to it?
While it is possible to vary voltage by using a phase-fired power controller, the practical drawbacks of electrical noise generation and cost usually result in ceding it's use to applications that absolutely require it. More often, heater power control is time proportioned by adjusting the length of on-time proportional to total cycle time. For instance, if cycle time is 8 seconds, then the heater is turned on for 4 seconds to obtain 50% output, 2 seconds for 25% output, 8 seconds (always on) for 100%, and so on.
AligatorAmy said:
If yes, my requirement to the heater supplier would be the condition that 325mm long heater must be capable to be operational at varying input voltage, say 180V-280V (with 230V nominal).
Heaters are rated at full operating voltage (i.e., 230V), and not for a variable range (i.e., 180V-280V). If a heater is expected to withstand operation at 280V, it is rated for 280V.
 
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  • #3
@Asymptotic
Thank you for your reply.
The actual length of currently used heater is 2000mm (the length shown on the drawing is incorrect).
Presently 1 thermocouple is located close to one of the far ends of the steel plate (outside of the plate).
(The temperature variation along the steel plates was monitored with 8 thermocouples (K type) fixed at equal distances.)
Each of the planned thermocouples will be fixed to the outside wall of the steel plate ''in the centre" of each 325mm long heater.
The steel plates are the components of the machine. Unfortunately, it would be very challenging to overlap the heaters to avoid the cold spots.
These plates have hundreds of holes and grooved channels (where the polymer flows) so there is no space for other than linear array of short (325mm long) heaters.
Thank you also for your answers in regards to the temperature control.
 
  • #4
AligatorAmy said:
Thank you also for your answers in regards to the temperature control.
you might find this of help
http://www.omega.com/temperature/Z/pdf/z110-114.pdf
 
  • #5
AligatorAmy said:
@Asymptotic
The steel plates are the components of the machine. Unfortunately, it would be very challenging to overlap the heaters to avoid the cold spots.
These plates have hundreds of holes and grooved channels (where the polymer flows) so there is no space for other than linear array of short (325mm long) heaters.

I'm trying to visualize what this installation looks like in cross-section. Do the plates tangentially touch the outer heater surface, or are they machined so the heater is in good contact with them along the lines of this illustration?

If the latter, have you tried using a heat transfer compound between heater and plate? For this temperature range, very finely powered graphite and aluminum in a silicone or mineral oil base has been successfully employed, and is commercially available. If temperature variation is caused by uneven contact between heater and plates this should bring it under control.
Tubular_Heater.jpg
 
  • #6
AligatorAmy said:
My problem is that the temperature of the steel components varies along the heater +/-15degC (nominal T=190degC).

Why is that ? Does cold process fluid enter at one end ? Might a little bit of recirculation preheat incoming fluid ?

AligatorAmy said:
Presently 1 thermocouple is located close to one of the far ends of the steel plate (outside of the plate).
(The temperature variation along the steel plates was monitored with 8 thermocouples (K type) fixed at equal distances.)
Each of the planned thermocouples will be fixed to the outside wall of the steel plate ''in the centre" of each 325mm long heater.
It's time lag that makes temperature control difficult. Thermal capacity and transport time from heater to sensor cause overshoot and hunting. The closer you can place your sensors to the heaters the better.
 
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  • #7
@jim hardy
Thank you for the link to the controllers brochure. It is very useful.
The molten polymer (180degC) which folws through the steel plates has its inlet at one of the plate ends. The polymer flow channels are not 'symmetrical' along the steel plates length (2000mm) and it is supposedly one of the thermal issues here.
The two steel plates are part of a larger assembly and more uniform temperature is required also in other nearby parts of the assembly. Ideally the 6 thermocouples would be embeded inside the heater. However, there are various factors which affect 'locally' uniformity of the temperature and placing the thermocouples 'too close' to the heater may mask the thermal problems residing in the assembly. Although indeed, possible decay in the heater 'actual response' to the thermocouple feedback may not be optimum either.

@Asymptotic
Thank you for your question.
Unfortunately, the channel which accommodates the heater is of square shape and is too deep, please refer to the attached draft on left side, therefore is not ideal for the round heater.
The steel parts are of old design but for a moment I need to go ahead with them because they are too expensive.
I considered two actions in order to temporarily fix the problem of square channel and round heater. Please refer to the draft on right side.
I think about making the channel slightly deeper with a round mill (4mm dia), so at least half of the channel (bottom part) would be in a good contact with the heater and then I would add a mild steel strip over the heater. The compound would be added there. However, the problem is that the two steel plates needs to be firmly tighted (there is 1mm aluminium gasket between them) two avoid any polymer or hot compressed air (which also flows through hundreds of channels in these plates) leakage. Thus the strip inside the channel would need to be only in a slight touch with the heater.
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  • #8
Do you have issues with limited heater lifespan, too? 1600W in a 2000mm x 4mm tube is about 41 watts/square inch of watt density. This yields higher than desirable sheath temperatures wherever heater-to-heated object contact is poor.

I'd try rounding the bottom of the channel with a 4mm ball mill as you've described. Rather than a flat strip on top, however, make it appropriately deeper, and round it with the same 4mm ball mill so it also is in good thermal contact. Do this before experimenting with multiple elements and heat zones - basic geometry improvements will go a long way to even out thermal distribution within the plates, and perhaps enough that multiple zones will no longer be necessary.
 
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  • #9
AligatorAmy said:
Ideally the 6 thermocouples would be embeded inside the heater.

Indeed a temperature measurement as you describe that's ahead of the process's thermal inertia is de-facto feedforward.
I see you have understanding of your process..

I guess i don't quite understand why the structure is steel rather than aluminum for best heat transfer. Do you need the strength ?

old jim
 
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  • #10
jim hardy said:
... I guess i don't quite understand why the structure is steel rather than aluminum for best heat transfer. Do you need the strength ?
When @AligatorAmy said "These plates have hundreds of holes and grooved channels (where the polymer flows)" and "The molten polymer (180degC) which flows through the steel plates has its inlet at one of the plate ends. The polymer flow channels are not 'symmetrical' along the steel plates length (2000mm) and it is supposedly one of the thermal issues here." it put me in mind of injection molding.

Injection molding is a completely different critter than sheet thermoforming (which I've had experience with) but from what I've picked up by talking to folks in that field, and in an article I've just read, strength factors into it, as do coefficient of expansion, and wear rate.
 
  • #11
AligatorAmy said:
... the channel which accommodates the heater is of square shape and is too deep, please refer to the attached draft on left side, therefore is not ideal for the round heater.

Is it possible the tool builder originally designed it using a square or rectangular heater? I've never used them (in fact, I didn't know they existed until now), but several firms manufacture such heaters. Here are links to two of them.
http://www.fastheatuk.com/cartridge.html
http://www.duratherm.com/square-cartridge-heaters/

4.2mm x 5.3mm isn't a standard size, and 3/16" (4.76mm) is the smallest I've seen so far (there may be issues fabricating cartridge elements under this size), but it's worth discussing the possibility of manufacturing a custom rectangular heater with your supplier.
 
  • #12
Or how about filling/packing that void around the heater with ceramic powder as a thermal transfer medium? Possibly without even having to disassemble it.

P.S. Rather than paying for ceramic powder, you could even use sand as used for sand blasting.
 
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  • #13
Tom.G said:
Or how about filling/packing that void around the heater with ceramic powder as a thermal transfer medium? Possibly without even having to disassemble it.

P.S. Rather than paying for ceramic powder, you could even use sand as used for sand blasting.
Not a bad idea, so long as the powder is very fine (like flour) and tightly packed. Usually magnesium oxide is used due to high thermal conductivity for an electrical non-conductor (45 to 60 W/mK).

If the blasting 'sand' is aluminum oxide and has been pulverized and spent (not fresh - the grain size is too large pack effectively) it'll work nearly as well (~35 W/mK thermal conductivity), but good old-fashioned silica sand won't (~2 W/mK thermal conductivity).
 
  • #14
@Asymptotic
The lifetime of the heaters does not seem to be short. Although I would need to have round finish of the heater channel first and introduce it into the production, for comparison.
I will definitely try the option with re-mill of the channel bottom (4mm dia ball mill) along with application of the rounded top strip sitting on the heater.

@Asymptotic, jim hardy
The steel parts are the components of the melt blowing equipment. The plates and other components design is outdated and 'not perfect' and the unsymmetry of the polymer flow channels along the plates is only one example. For this moment I need to go ahead with current design because re-design and manufacturing of new components would automatically enforce change of others...

@Asymptotic, Tom G
Application of the compound is defintely worth consideration. Thank you for the examples of the materials that you have suugested.
 
  • #15
Hello,
I have decided to go ahead with the solution involving split of 2000mm long heater into the 6 separate sections (6 heaters, 320mm long each).
I am looking for the PID temperature controllers for these heaters. Preferably 1 controller which could accommodate at least 6 inputs (from TCKs). I have came across Eurotherm company and multiple analog input T controller (with 8 in/out):

http://www.eurotherm.com/products/temperature-controllers/multi-loop/2704

Do you recommend other providers/makes?

P.S. The input from all 6 TCKs would need to be simultaneously displayed on a screen of such a controller to allow the operator easy access to cureent temperatures status.
 
  • #16
A Eurotherm 2704 handles up to three control loops each, so you'd need two of them.
It would probably be cheaper to use six single loop controllers. Among the least expensive are 1/32 DIN sized model SL4824 for $90/ea marketed by AutomationDirect.
 
  • #17
I would also look into www.watlow.com controllers. Asymptotic is probably right, 6 individual controllers may be cheaper and easier to setup. If you want to go the fancy route, Watlow also makes an F4T touch screen controller.
 
  • #18
AligatorAmy said:
My problem is that the temperature of the steel components varies along the heater +/-15degC (nominal T=190degC).

Ok so it's 15 degrees hotter at one end compared to the other. That works out at 7.5C/m. If you replace it with 6 heaters each 0.3m (and each temperature controlled) I would expect the gradient along each strip to be reduced to around 7.5*0.3 = 2.25C. That's a crude calculation. Could be less as the hot end of one section should raise the cold end of the adjacent section.

Re post #7: Copper strip instead of mild steel?
 
  • #19
CWatters said:
Ok so it's 15 degrees hotter at one end compared to the other. That works out at 7.5C/m. If you replace it with 6 heaters each 0.3m (and each temperature controlled) I would expect the gradient along each strip to be reduced to around 7.5*0.3 = 2.25C. That's a crude calculation. Could be less as the hot end of one section should raise the cold end of the adjacent section.

Re post #7: Copper strip instead of mild steel?

He likely means the sheath temperature varies 15C at any given point along the length of the element, not necessarily a relatively uniform gradient. The coiled wire inside a tubular element is vibrated while filling with MgO, and then roll reduced, so the coil is rarely centered on the heater and not always uniformly stretched (due to gravity and vibration during the fill process).

6 smaller heaters, set to the same setpoint temperature, may not fix this issue. There will be finer control over individual zones, however, which should still reduce the variability that OP is looking for. Down to 2.25C? Highly doubtful, and likely unmeasureable.
 

Related to Tubular cartridge heater wattage

1. What is a tubular cartridge heater wattage?

A tubular cartridge heater wattage refers to the amount of power or energy required to heat a specific tubular cartridge heater. It is typically measured in watts (W) and is an important factor in determining the heating capacity and efficiency of the heater.

2. How is tubular cartridge heater wattage calculated?

Tubular cartridge heater wattage is calculated by multiplying the voltage (V) and current (I) of the heater. This can be represented as W = V x I. However, it is important to note that other factors such as the resistance of the heater and the ambient temperature may also affect the wattage.

3. Why is tubular cartridge heater wattage important?

Tubular cartridge heater wattage is important because it determines the heating capacity of the heater. It is essential to choose the right wattage for a specific application to ensure efficient and effective heating. Using a heater with too low wattage may result in insufficient heating, while using a heater with too high wattage can lead to overheating and potential damage.

4. How do I select the right tubular cartridge heater wattage for my application?

The right tubular cartridge heater wattage for your application depends on several factors such as the size and material of the heater, the desired temperature, and the surrounding environment. It is best to consult with a heating expert or refer to the manufacturer's specifications to determine the appropriate wattage for your specific needs.

5. Can the wattage of a tubular cartridge heater be adjusted?

Yes, in some cases, the wattage of a tubular cartridge heater can be adjusted by changing the voltage or current. However, it is important to note that altering the wattage may affect the performance and lifespan of the heater. It is recommended to consult with a professional before making any changes to the wattage of a heater.

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