Ball of Wax 1

[Reid2006]

Hi all, I've got a bit of a challenge for you (I think).

We need to melt a large tote (big plastic cube) full of wax. A heat wrap blanket was purchased to accomplish this, but the results were not acceptable. The heating blanket took 5 days to melt the wax, when ideally, we'd like to have it melted in 24 hours.

Here are the vitals:

Volume of tote: 47.2" x 39.4" x 45.7" (1.39 Cubic Meters)
Weight of wax: 83 KG
Heat of Fusion (melting) for Wax: 34 Calories / Gram
Melting Temp of Wax: 27 to 32 degrees C
Specific Heat of Wax: 2.1 KJ/(KG*Kelvin) (@ 20 degrees C)
Moloar Mass: 720 to 780 g/mol

Heating Blanket:
Wattage: 1440W @ 120VAC
Max Temp: 71 degrees C
Time taken to melt the tote of wax: 5 days
The tote has a large metal cage attached to it, so the hot blanket is not in direct contact with the tote.

It is proposed to replace the heating blanket with a hot box (essentially a large oven). This oven would heat 2 totes at once. Here are it's vitals:

Wattage: 9000W
Max Temp: 135 degrees C, plus a circulating fan to imporve convection.

Here's the question:

Can you prove, using appropriate formulas, that the hot box will in fact melt the wax faster than the heating blanket (preferably in ~24 hours).

Thanks in advance,
Adam

 

[Smiley8]

Hey,
This problem is actually my problem to solve. I work with Reid2006 and he came to you guys for me. So I decided to join, which makes things a lot easier.

 

[25362]

IRstuff, the sensible heat needed to heat the molten wax is also part of the overall heating duty.

Reid2006, there is nothing strange about my estimation.

Heat required to melt and heat the wax: say, 60 kcal/kg.
Amount of wax per block: 1518 kg.
Total heat required for 2 blocks: 60[×]1518[×]2 = 182,160 kcal.
Heating elements: 6[×]1.5 kW[×]860 kcal/kW-h[×]24 hr/day = 185,760 kcal/day
Time required: 182,160[÷]185,760 = 0.98 days

 

[IRstuff]

True, but the given specific heat with a 10 ºC delta temperature is only 6 kcal/kg. That's a long way from 34 kcal/kg.

TTFN

 

[IRstuff]

Nevermind, I see what you did.

TTFN

 

[25362]

This is a unsteady state heat transfer problem with many resistances in series.
For given thermophysical properties of the wax, a higher [Δ]T and heating from all the sides would melt the blocks in a shorter time, since time is proportional to [(thickness)²[÷][Δ]T]

 

[IRstuff]

Supposedly, the heater blanket would have had that covered. ;-)

The main thing going here is that the delta temperature can be almost doubled. But the calculations haven't definitely arrived at an answer that provides margin against the unknowns in the problem.

If the time is truly proportional to thickness², the cutting the blocks into quarters with some sort of passive fin structure might provide sufficient margin against the time requirement.

TTFN

 

[Reid2006]

Thanks all, you've been a tremendous help.

I have one question:
25362, how did you arrive at 60 kcal/kg? Did you just work backwards from the known 1.44kw blanket for 5 days?

 

[insult2injury]

Splitting the blocks into smaller pieces definitely decreases the melting time for convection as there would be a significant increase in heat transfer surface area.

I2I

 

[Reid2006]

To answer my own question, working backwards using 1.44kw and 5 days, we obtain a heat of fusion of 97.9 kcal/kg.

(X Kcal/kg)*(1518 kg)= Y Kcal
(1.44KW)*(860 Kcal/KWh)*(24h)= 29721.6 Kcal/day

(Y Kcal)/(29721.6 Kcal/day) = 5 days
Y = 148608 Kcal
X = 97.9 Kcal/Kg

As originally given, the actual heat of fusion of the wax alone is 34 kcal/kg. Quite a difference - due to losses I'm assuming.

So my REAL question is, should we use 97.9 kcal/kg as the "worst case scenario" heat of fusion, as demonstrated by the use of the heat wrap?

Also, is 70 kcal/kg a better estimate for using the hot box, as there will be lower losses and a higher rate of heat transfer due to convection?

 

[IRstuff]

While your calculation is in technically error because you've not subtracted out the heat loss and the sensible heat, the number would be close to what you would use for the oven. The heat of fusion is still 34 kcal/kg. If you had preheated your wax to just before melting, prior to measuring the heat input, that's what you'd get.

If you assume a 128ºC delta temperature in the hot box, the sensible heat comes out to 65 kcal/kg, which added to the 34 kcal/kg heat of fusion, gets you 99 kcal/kg.

TTFN

 

[EdStainless]

I like the Cu or Al fin approach. I used to melt a thick substance in a process. We used an old industrial baking convection oven. We set the SP about 50F below our smoke temp.
We had a large Al "X" that would fit in the container on the diagonals. It was about 2/3 the depth of the box and it had 'wings' to keep it from sinking and keeping about 12" sticking up for heat transfer.
We would just set it on top and fire up. The X would sink into the material quickly. We got more melting from the X than from the sides of the box.
We later went ot using metal containers with a light plastic liner in them. This helped a lot.

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[25362]

Reid2006, my answer to the assumed 60 kcal/kg:

[•] latent heat of fusion: 34 kcal/kg
[•] sensible heat from 20^oC to 70^oC: 0.52[×](70-20) = 26 kcal/kg
[•] total heat required: 34+26 = 60 kcal/kg

Cp = 0.52 kcal(kg.K), is an average I assumed for both liquid and solid wax.

To estimate times for heating or cooling such as in this case, the problem can be analysed by use of computer programs (which I don't have access to) or by use of diagrams for solid bodies of various shapes appearing in specialized books. One of them is fig 10.10 in prof. Aksel L. Lydersen's Fluid flow and heat transfer (John Wiley and Sons) ISBN 0 471 99696 3.