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MSM

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MSM

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Tom.G

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To *Estimate*:

Find the thermal capacity of the fluid.

Multiply by the thermal resistance of the tube.

This will give you the Time Constant for the heating process. (the time it takes to reach 63% of final temperature)

Multiply by 5.

Cheers,

Tom

p.s. If you are familiar with electrical circuits, the equivalent is a series RC being charged by a voltage source.

p.p.s. You can use this same approach in your previous thread regarding a continuously flowing fluid. It won't be exact (the calculated time will be too short) but is at least a starting point for low flow rates.

Find the thermal capacity of the fluid.

Multiply by the thermal resistance of the tube.

This will give you the Time Constant for the heating process. (the time it takes to reach 63% of final temperature)

Multiply by 5.

Cheers,

Tom

p.s. If you are familiar with electrical circuits, the equivalent is a series RC being charged by a voltage source.

p.p.s. You can use this same approach in your previous thread regarding a continuously flowing fluid. It won't be exact (the calculated time will be too short) but is at least a starting point for low flow rates.

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- #3

MSM

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I tried the lumped system analysis, but my understanding is that for this method to be valid, Biot Number (Bi) has to be less than 0.1 (Bi<0.1).ToEstimate:

Find the thermal capacity of the fluid.

Multiply by the thermal resistance of the tube.

This will give you the Time Constant for the heating process. (the time it takes to reach 63% of final temperature)

Multiply by 5.

Cheers,

Tom

p.s. If you are familiar with electrical circuits, the equivalent is a series RC being charged by a voltage source.

p.p.s. You can use this same approach in your previous thread regarding a continuously flowing fluid. It won't be exact (the calculated time will be too short) but is at least a starting point for low flow rates.

Now consider a fluid with thermal conductivity k=0.14 W/m K, and heat capacity Cp= 2500 J/kg K, and density of 780 kg/m3.

Now for long pipes, Nu=3.68, and with a tube internal diameter of 2 mm, I get a heat transfer coefficient of 257 W/m2 K. This gives us a Bi=0.9. Now with this information , I get a thermal time constant of ~3 s

In this case, can I still use the method you suggested?

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Tom.G

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Let's see if paging @Chestermiller gets better input for you.

If I understand the numbers you supplied, the thermal conductivity of the liquid is extremely low with a moderately high thermal capacity. If that is the case, how about modeling the liquid as a solid thermal barrier with an equivalent 'perfect' (zero size, infinite conductivity) thermal mass at the center?

As I understand the Biot number, it considers

Cheers,

Tom

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Chestermiller

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See my response to the other thread. https://www.physicsforums.com/threads/heat-transfer-in-thin-tubes.1003289/#post-6495102

Let's see if paging @Chestermiller gets better input for you.

If I understand the numbers you supplied, the thermal conductivity of the liquid is extremely low with a moderately high thermal capacity. If that is the case, how about modeling the liquid as a solid thermal barrier with an equivalent 'perfect' (zero size, infinite conductivity) thermal mass at the center?

As I understand the Biot number, it considersconvectiveheat transfer at the surface. With the temperatures you have wouldn't radiative transfer also be significant?

Cheers,

Tom

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