# Thermal time scale in tubular flow reactors

MSM
So if I have a tubular reactor for nanoparticle synthesis (PTFE tubes ID:2mm). The tubes are heated in a furnace. liquid Reagents at room temperature are pumped by a syringe pump and directed toward the furnace. The reagents decompose to form nanoparticles once they reach the steady-state furnace temperature. How can I estimate how long it takes for the reagents to to reach the set temperature in the furnace once they enter the furnace?

Gold Member
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.

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MSM
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.
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).

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-1. and t~17 seconds to reach final temperature. Does this seem reasonable? because I expected few seconds or lower

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

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Gold Member
Well you got much deeper into it than I ever did! We could start a long Q&A on this but a better bet is to get the attention of those more versed in the 'non-common' situations.

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 convective heat transfer at the surface. With the temperatures you have wouldn't radiative transfer also be significant?

Cheers,
Tom

Mentor
Well you got much deeper into it than I ever did! We could start a long Q&A on this but a better bet is to get the attention of those more versed in the 'non-common' situations.

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 convective heat transfer at the surface. With the temperatures you have wouldn't radiative transfer also be significant?

Cheers,
Tom
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