Heat transfer through aluminium block 1

[MichielVZ]

My problem is as follows.
There is a plate of aluminium (500mm x 480mm x 100mm) that needs to be heated. The top surface of this plate needs to be heated to 200 degrees Celsius through the use of 4 (d-16 L-500mm 2.75kW) heating elements placed a the bottom of the plate. What i would like to know is, at the moment that the top surface reaches 200 degrees, what are the temperatures throughout the rest of the height of the plate? Assume that the 100mm is the height.
So far i have already calculated the required energy as roughly 11.6 mega Joules and the rate of heat transfer as 164.6 kW.
Any assistance will be greatly appreciated.

 

[MichielVZ]

What i have seen is that if i were to plot the graph of the change in temperature against the height, then it comes out as a straight line. But thermal conductivity for aluminium increases with an increase in temperature.

 

[LittleInch]

Well aluminium has a very high thermal conductivity, x 5 compared to steel.

So at 100mm thick I would think your temperature at the bottom surface will be < 1K different.


Not sure how your're getting 164kW when you only have 11kW of heat going into the plate??

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[MichielVZ]

LittleInch

Thanx for the reply. I calculated the required energy and rate without using the information about the heating elements. So now im in the process of recalculating what is ACTUALLY going to happen. Also, some extra information, the heating elements have a max temp of 600 degrees Celsius and they reach that temperature relatively quickly so im assuming its instantaneous in my next calculations.

 

[IRstuff]

Your calculations only partly correct; I get something more like 294W through the top, but there's also loss through the bottom, unless you've insulated it somehow. At 294W flow through the aluminum, the drop is roughly 0.5C, but you calculated a steady state number. The big question, which you didn't provide, is what is the timeframe? At 294W, your 11.6MJ will take 11 hrs to get achieve.

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[MichielVZ]

IRstuff

Thanx for the reply. The process where this happens is a thermo-forming process so the time frame we are talking about is seconds...not minutes or hours.
The heating elements reach their max temp relatively instantaneously. The block is insulated at the bottom and around the sides so theoretically we say no heat is lost to the sides or bottom.

 

[LittleInch]

well even assuming no losses on the way up in temperature I get about 17 minutes heat up time from 25C. But you will have losses so maybe 25-30 mins?

All depends on what your heat loss is from the top surface. If its close to 11kW then the last 50C might take hours.

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[IRstuff]

If your four heaters are rated at 2.75 kW each, that's 11 kW total, so, 11.6 MJ at 11 kW = 1055 seconds, which is the 17 minutes that LI got. Which means that if you want seconds, say, 30 seconds, you need 387 kW Even your calculated flow of 164.6 kW, would result in 70 second warmup time.

Just as a point of reference, 387 kW = 1.3 MBTUh; a large stovetop burner can dump about 17 kBTUh, so you're looking at the equivalent of 77 large stovetop burners in roughly one-quarter square meter. Not clear to me that even an array of acetylene torches will do that.

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[3DDave]

IRstuff - thermite.

 

[IRstuff]

I did actually think of that, but I was too lazy to try and run some numbers

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[MichielVZ]

Thank you all for your input, it has been greatly appreciated. Now, having sorted out the above mentioned issues, picture the block of aluminium as two seperate blocks of aluminium with some heating element in between them. The top bit of aluminium is 100mm high and the one below the element is 20mm down. Here is the catch, at the bottom of the 20mm piece there is insulation that has a max temp of 220 degrees Celsius. Assume only losses at the top of the 100mm piece. The Energy used/power consumed during the thermo-forming process can be described graphically as a wave function of sorts but the heating element can only be fully on or fully off and is hence a step response. So my question is, how do i match these two graphs in order to have a constant temp of 200 degrees at the surface.

 

[IRstuff]

You've all but made it impossible to meet all of your requirements. Your maximum power is going to be at the start, and you've limited yourself to 104 kW, that means your best possible timeframe is 112 s.

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[LittleInch]

Well how much heat energy is this process taking?

And if its seconds rather than minutes then you need to rely on the thermal mass of your block.

With AL having such a high conductivity, you can look at the block as a single mass of heat energy as any losses in the top surface will be very quickly spread around the rest of the block.

Until you start to describe this process a little more and use some consistent numbers, we can't see if this I feasible or not.

Also what range of temps can your process stand. Less than say 1 degree then you're looking for something I don't think exists. 5 to 10C then you're in business.

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[MichielVZ]

Guys, forget about the energy required and all that jazz. I've solved that problem. To give more clarity, this is a pulp molding machine. For example, the cup holders from McDonalds are made using this machine. So the current problem is this. We can get the temperature to 200 degrees with no problem. Then, the wet pulp is applied to that surface and of course absorbs some energy. As i mentioned before, the heating elements are either fully on or fully off and therefore are represented by a step response. The loss of energy because of the wet pulp being added to the mold can be represented by some sort of eave form similar to a sine wave. THE QUESTION IS THIS---> How do you combat the wave form of the lost energy with the step response of the elements?

 

[LittleInch]

Turn individual ones on and off in a smaller stepped wave form to match a sine wave?

You have four to play with.

You haven't explained what temperature drop you're looking for as acceptable or how much energy is used in the cup manufacture. Or how many cups per hour.

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[MichielVZ]

The machine produces 70kg of product per hour. The energy loss because of wet pulp being added is 3.2kWh/kg of product. We want to keep the temperature within 1 degree. So 199-201

 

[LittleInch]

sorry, somethings not right here.

3.2kWh per kg = 11.52 MJ in one hour per one kg.

you have 70 kgs in an hour = 806 MJ in one hour = 224 kW.

You only have 11 kW of power going into your block.

You are out by a factor of 10 somewhere.

Within 1 C with a varying heat output from one surface will probably require a much larger thermal mass. At 100g/cup I get that is one cup every 5 seconds or less if a cup weighs less than 100g.

And on the basis of the above a LOT more power is needed, but I think there is an error somewhere as heat density from the elements is limited.

Is this a design in practice now or are you designing it?

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[MichielVZ]

This is a working machine already.
There are 4 elements per die.
4 dies per machine.
4 x 11 = 44kW.
Right now i'm not concerned about the energy calculations and whether E in = E out because the machine works fine so the energy balances out.

What i'm working on now is how to minimize the temperature fluctuations at the surface of the dies.
The elements cannot be turned on separately, They are either all on or all off.
The machine has a built in PLC which tries to manage the temperature as best it can but we are also trying to find the optimal position for a temperature probe in order to give the PLC the best information.
Currently there is a single temperature probe on each die.
This probe is situated at the elements. So im wondering if moving the probe a bit closer to the surface will help the plc better combat the fluctuations.

 

[LittleInch]

"Right now I'm not concerned about the energy calculations and whether E in = E out ",

Well in my opinion you should be as this will inform you of the relative numbers and issues you are looking at.

So now we can gather that there are four parts to the process each with their own die / heating element?

I can only suggest you look at ways to actually re-wire the elements to give you more control.

Aluminium is a very good heat conductor so temperature probe location shouldn't make a big difference. It also has one of the highest heat capacities per kg of most common metals, though less dense which more or less equals out. My point being that you could add some steel behind the block to increase overall heat capacity of the block which could help in smoothing out your temperature dips - Do you know what sort of surface temperature dips you're getting in practice?

I think the only way you are going to smooth out the temperature dips and different wave forms is more metal in the die.

moving the probe I can't see making a big difference, but you could try. The block should remove the heat from the elements very quickly but there does need to be a temperature difference to allow the heat to flow into the block - what contact area or heat transfer mechanism is there between the elements and the block? Do you have a drawing / section / picture of the element and block? You might need to improve the heat transfer between element and block.





Remember - More details = better answers
Also: If you get a response it's polite to respond to it.