Resurrection of a CFEI 100 KVA Induction Furnace

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David Jupp:
You need to work out the overall heat transfer co-efficient for the heat exchanger - I did some basic Chemical Engineering on a course years back. All the equations are probably available on-line... but not sure you'll have all the data.

It's a bit of a 'chicken and egg' situation though as the OHTC is influenced by flow rates in both circuits, any fouling, boundary layers, etc. The exchanger manufacturer should be able to provide some guidance.

First thing is do you know the heat load to be dealt with?

You're going to end up with quite a few estimated values - and of course the water flows will change when you connect things together.

awemawson:
David,

I need to remove an estimated 40 kW of heat from the primary circuit. The primary circuit is water (with a bit of glycol) that will be pump circulated at approx 25 litres / minute. I don't want the primary to go above 30 Centigrade. The secondary will be fed with water at (say) 10-15 degrees C at 14 litres / minute.

But the unknown of course is the heat exchanger and it's characteristics as I 'ain't got one yet !

I have a feeling that to err on the large size is sensible, and throttle the secondary cooling water if need be - if indeed 14 litre / min at 15 Degrees is sufficient :scratch:

I suppose if one assumed 100% H.E. efficiency and the cooling water rising from 15 to 30 degrees at the 14 litre / min rate how much energy does that take?

Now I found this 'handy rule' on the web:

Handy rule: multiply the hot water flow rate in litres/minute by the temperature rise in celsius. Divide by 100 and multiply by 7 to get the boiler DHW output in kW. Example: 12 litre/min at 40C rise: 12 x 40 = 480.

Applying that says that it takes 14.7 kW to raise a flow of 14 litre / min by 15 degrees C so I am out by quite a margin from my requirement :bang:

I need three times that amount of cooling so presumably three times the water flow rate which is a problem :(

David Jupp:
40kW = 40 kJ/sec

14 ltr / minute = (roughly) 0.25 ltr/sec. or 0.25 kg/sec assuming water s.g. = 1

A 15 C temperature rise for that = 0.25kg x 15 C x 4.2 kJ/kg/C = 15.75kJ

So yes, quite a bit out. You can assume more or less 100% efficiency of heat transfer - the heat transfer calculations relate to how big the exchanger needs to be and how fast the water has to flow in the internals.

russ57:
So, how much water can you get out of the bore... And how cold is it..

This is a bit like the penultimate sub-plot of 'Trustee from the Tool Room' where Keith discovers a discrepancy in the hydraulic cooling calculations for a saw mill and earns himself a tidy little commission...

-Russ

awemawson:
It's certainly a bit of a roller coaster isn't it !

Cheap bore hole pumps on eBay claim a figure of 2000 - 3000 litres / hour plus so are in the ball park, but whether the bore hole can sustain that sort of draw, and whether the pumps can deliver it from the depth of the hole are more variables!

. . . argh . . why is life never simple . . .

:coffee: :coffee: :coffee: :coffee: :coffee: :coffee: :coffee:

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