Maths question

Narj

Narj

Narj

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This has bugged me for a while because I was asked if I could work it out and had no idea how.

Imagine you have a "mixer" tank with two ingress pipes with water at different temperatures (eg: 50*C and 38*C), and one outlet pipe.

How do you work out the temperature of the water at the outlet? At the time I was given a diagram and various other information such as the rate of flow from both ingress being the same and maybe the size of the tank.

Really random question but it was quite some time ago and it's been nagging at me for a while now.

Thanks!
 
It would depend on a lot of things, including where in the tank the inlet/outlet pipes are, is the tank mixed properly? Do you assume it has reached equilibrium before going to outlet?

More of a chemical engineering question than a maths one.
 
Ultimately this is a fluid dynamics question and it's dependant on the internal geometry of the mixer tank. There's no simple answer to this.
 
Elixir said:
It would depend on a lot of things, including where in the tank the inlet/outlet pipes are, is the tank mixed properly? Do you assume it has reached equilibrium before going to outlet?

More of a chemical engineering question than a maths one.
Click to expand...


This.


The simplist case would be to take the ratio of the different temperature waters, and use them.
e.g. 60% 70degrees, 40% 38degrees. Resultant temperature would be 0.6*70 + 0.4*38 = 57.2degrees
 
Elixir said:
It would depend on a lot of things, including where in the tank the inlet/outlet pipes are, is the tank mixed properly? Do you assume it has reached equilibrium before going to outlet?

More of a chemical engineering question than a maths one.
Click to expand...

Aha, well that makes me feel less stupid. I was made to feel a bit stupid when I said I had no idea apart from taking the average of the two which is obviously wrong.

Assume this:

Code: 
 _| |___| |__
|            |
|____   _____|
     | |

Ingress at top, egress at bottom.

EDIT: Just saw post above... cheers. :)
 
Last edited:
Dunks said:
Probably need some sort of CFD to truly work it out.
Click to expand...

One would suspect that it is assumed that equilbrium has been reached by the time the water is leaving the tank. Also that no heat is lost to surroundings.

If this is the case it is a heat/energy balance question. Energy going in to the tank will equal the energy going out.

Delta H (Pipe 1) = CM (Pipe 1) Delta T {T (Pipe 1) - 273}

Delta H (Pipe 2) = CM (Pipe 2) Delta T {T (Pipe 2) - 273}

Delta H 1 plus Delta H 2 = Delta H 3

Delta T (Pipe 3) = Delta H 3/CM (Pipe 3)
 
As has been said, without knowing a lot more about the system, the answer will be a very rough guess based on a lot of assumptions.

Things you need to know;

Rate of flow of each pipe,
Heat capacity of the pipes and container,
Insulating properties of the container and piping,
Is the system fully sealed?,
Volume of water kept in the container at equilibrium (based on the rate of flow of each pipe).

There's probably more, but it's lunchtime and my brain is on strike for the next half hour.
 
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