The Grain Father
Mangrove Jacks new all grain unit - its nice
Replies to This Discussion
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Permalink Reply by Dene McDonald on
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It was 37 litres which is probably a bit of an ask for a 2kw element. With 25 litres it was a lot faster but not sure exactly what as I didn't record it. It proved frustrating and hence I've upgraded to a 70 litre pot with a combined 4.8 kw of heating power.
Here's a video of the boil - possibly not classed as rolling.
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Permalink Reply by Barry on
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Yeah I would suggest it would be quite slow and potentially a bit light on the vigorous rolling boil. Be really interesting to see whether it is or not in real life.
Kelly (Ryan) and Mike (Neilsen) brewed with one the other day, sounded like it worked pretty well.
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Permalink Reply by Mark Weusten on
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Mangrove Jacks reply
Mashing temperature to boiling temperature should take no more than 15 minutes.
Also good to note that the whole process will not take longer than 4 & half hours,
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That's impressive. They must have a double skinned wall or some kind of thermal barrier.
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Permalink Reply by Peter Smith on
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Just as the urns are 38L when they say 48L you may find that the urn element is way below the rated 2kw.... Clamp the amps being used to confirm this. Given the amount of design effort in the Grainfather there is no way its not going to hold a full rolling boil. my stainless deathtrap boil kettle elements will pop a multibox circuit breaker, the urn never does this... you got me all interested so I am going to clamp the urn amps ...
Ok having said all that, anyone want to buy my 48(actually 38L) urn, great condition $80?
holds sparge temps well 8)
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Permalink Reply by Richard Jack on
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Unless I have forgotten all my physics (a possibility that I can't really understate, too many beers since 7th form) It must be a magic 2kW element, which is pretty cool.
Q = cp m dT
1g of water takes 1 calorie or 4.19 joules to raise 1°C.
So
Heat required = 4.19 * 25000 (mass of 25l of water) * 35 (delta-T between mash @65 and boil)
Heat required = 3,666,250 J
A joule is a Watt second. So you need 3,666,250 watt seconds. in a perfect system with no losses.
So:
3,666,250 (watt seconds needed) / 2000 (element output) = 1,833 seconds for a 2kW element in a perfect system.
~30 mins.
Anyone want to check my math?
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YE CANNAE CHANGE THE LAWS O' PHYSICS JIM
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Awesome. I have just emailed Mangrove Jacks with this ( my one will probably turn up with a dent in it now :-)
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Permalink Reply by mattd2 on
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Looks about right to me - The only thing I would say is perhaps they are heating the mash to 77°C for mashout (during the "mash" time) which would close the gap to boiling. And maybe they are also calculating for a "normal" batch size (lets say 20 litres) not the maximum (but also ignoring the fact you need more volume before the boil to get your final batch volume).
That would give you 16 minutes for 20 litres from 77°C - 100°C. But I would say your maths / assumptions are much more on the money :)
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