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Cooling 2 kettles - will this work?

by Octavius
Tags: cooling, kettles, work
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Octavius
#1
Feb2-14, 11:52 AM
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The two kettles contain a boiling aqueous solution. They must be cooled to room temp as fast as possible (otherwise adverse chemical reactions will take place). Only one pump and one cooling coil are available.

It would seem the second kettle would fill up. However, would the increased "head" then allow more of the liquid from this kettle to flow to the tee, at the expense of the liquid from the first kettle? Thus equalizing the height of liquids in the two kettles.
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mfb
#2
Feb4-14, 09:57 AM
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They must be cooled to room temp as fast as possible (otherwise adverse chemical reactions will take place).
How to weight cooling stuff in kettle 1 versus cooling stuff in kettle 2?

Only one pump and one cooling coil are available.
Does that mean you can mix the liquids without any issues?
If not, what does the pump do?

Is the cooling coil connected to anything? Can you move it between the kettles?

It would seem the second kettle would fill up. However, would the increased "head" then allow more of the liquid from this kettle to flow to the tee, at the expense of the liquid from the first kettle?
Where is the point in the exit of the second kettle? Can you close it?

Thus equalizing the height of liquids in the two kettles.
If the setup is completely symmetric... could happen.
Octavius
#3
Feb4-14, 10:58 AM
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Sorry for the confusion.
The liquids are the same.
The cooling coil has water in and out (just didn't show it in the diagram).
Cooling coil is too heavy to move.
No exit in second kettle.
Setup is symmetrical.

I'm trying to cool both pots at the same time - I was hoping there would be sufficient exchange between the pots to accomplish this.

I thought it would be interesting (from a theoretical position) - will the second pot fill up or will the levels remain the same? If they remain the same will there be sufficient movement from pot to pot?

Thanks!

mfb
#4
Feb4-14, 11:19 AM
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Cooling 2 kettles - will this work?

Can you move the pump exit? Otherwise I don't see how you want to cool kettle 1. Where would an exchange come from?
I thought it would be interesting (from a theoretical position) - will the second pot fill up or will the levels remain the same? If they remain the same will there be sufficient movement from pot to pot?
In your setup, if both valves are open, I think you pump some (probably small) amount of liquid from kettle 1 to kettle 2 until the system is in equilibrium (the pressure drop in the pipe cancels the pressure difference between the kettles), afterwards you are just circulating liquid in kettle 2.
Octavius
#5
Feb4-14, 11:23 AM
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Hmm, OK.
Back to the drawing board.
Thanks for the input!
Wellesley
#6
Feb4-14, 09:59 PM
P: 276
Quote Quote by Octavius View Post
Sorry for the confusion.
The liquids are the same.
The cooling coil has water in and out (just didn't show it in the diagram).
Cooling coil is too heavy to move.
No exit in second kettle.
Setup is symmetrical.

I'm trying to cool both pots at the same time - I was hoping there would be sufficient exchange between the pots to accomplish this.

I thought it would be interesting (from a theoretical position) - will the second pot fill up or will the levels remain the same? If they remain the same will there be sufficient movement from pot to pot?

Thanks!
Levels would be the same. Think of Bernoulli


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