Heat transfer of a hot plastic thread in a cold water bath

In summary, the thread is cooled by water bath using conduction and convection. The water bath is kept at a temperature lower than the temperature of the thread. The effect of convection is negligible due to the high speed of the cooling.
  • #1
Rotzooi
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Hi,

I'll describe my problem more specific.
I have a continuous flow of this polymer thread (polystyrene) at 465 °K (190°C sorry I'm metric) that needs cooling down. It passes trough a water bath which is kept at 310°K (35°C). Now i try to calculate the heat exchange into the water, to determine the speed of the cooling, or to calculate how hot the thread still is after a time.

approach:
I was thinking using conduction from the thread's core to the exterior and then convection into the water. (water is not moving btw).

Now another look at this approach:
The core of the polymer will remain hot when the exterior is cold. Because there is a very low thermal conductivity (0,187 W/m.K). So maybe i ought to use another formula. Because my first approach doesn't cover this. But this is where my knowledge stops. And research doesn't seem to deliver results. (When a metal bar is put in water u can take a temperature that is uniform over the bar because the heat exchange in metal goes so fast, here it is not the case.)

Any ideas or remarks would be of great use!
 
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  • #2
I'll leave better answers to others here, but it's probably also worth pointing out that the matter will be confounded by the convection effects of the water. The water will heat and thus rise up, causing turbulences and pulling cooler water from below. Calculating that on paper, not sure it can be done.
 
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  • #3
This is a problem in man-made fiber spinning into a water bath. What is the diameter of the capillary at the spinneret, and what is the quenched diameter of the filament? I'm guessing that most of the filament deformation takes place within the air gap, although most of the cooling takes place below the water surface. As a first approximation, you can assume that the water is dragged along with the filament velocity, so that, at each cross section of filament, you are dealing with transient 1D conductive heat transfer into an infinite medium.

I'm also guessing that you have multiple filaments in an array, and not just one filament. If that is the case, the thermal boundary layers from the filaments can interfere with one another, and inhibit the heat transfer. The drag flow from outer rows of filaments can also prevent cooler water from reaching the inner rows. This then becomes a pretty complicated problem. Because of the high speed involved, natural convection is going to be negligible, but the main flow effect will be filaments dragging fluid downward and inward towards and through the array.

You need to hit the literature on fiber spinning.

Chet
 
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  • #4
rumborak said:
I'll leave better answers to others here, but it's probably also worth pointing out that the matter will be confounded by the convection effects of the water. The water will heat and thus rise up, causing turbulences and pulling cooler water from below. Calculating that on paper, not sure it can be done.

Hi, thanks for the response.
Ok, i get what u want to say but this is thinking a little too far. I will never be able to clalculate this indeed.
What is your opinion on the effect that i describe in approach two?
 
  • #5
Chestermiller said:
This is a problem in man-made fiber spinning into a water bath. What is the diameter of the capillary at the spinneret, and what is the quenched diameter of the filament? I'm guessing that most of the filament deformation takes place within the air gap, although most of the cooling takes place below the water surface. As a first approximation, you can assume that the water is dragged along with the filament velocity, so that, at each cross section of filament, you are dealing with transient 1D conductive heat transfer into an infinite medium.

I'm also guessing that you have multiple filaments in an array, and not just one filament. If that is the case, the thermal boundary layers from the filaments can interfere with one another, and inhibit the heat transfer. The drag flow from outer rows of filaments can also prevent cooler water from reaching the inner rows. This then becomes a pretty complicated problem. Because of the high speed involved, natural convection is going to be negligible, but the main flow effect will be filaments dragging fluid downward and inward towards and through the array.

You need to hit the literature on fiber spinning.

Chet
Hi, thanks, fiber spinning is indeed a bit similar! Will search the term for more!

My problem however has nothing to do with fiber spinning, its just for cooling so cutting it to a granulate form goes easier.
The diameter of the thread is 3milimeter. (no quenched diameter of filament available) And there is only one filament at a time in the water bath.
 
  • #6
Rotzooi said:
Hi, thanks, fiber spinning is indeed a bit similar! Will search the term for more!

My problem however has nothing to do with fiber spinning, its just for cooling so cutting it to a granulate form goes easier.
The diameter of the thread is 3milimeter. (no quenched diameter of filament available) And there is only one filament at a time in the water bath.
OK. You're making pellets. That's a pretty big diameter thread. If there's any kind of circulation in the water bath, that's going to enhance the heat transfer.

If it were I working on this, I would solve the problem using the 1D transient heat transfer approximation I mentioned in post #3. This would probably require numerical analysis. You probably would have to take into account the heat of fusion, although, if you want to do it crudely, that could be lumped in with the heat capacity, particularly since the phase change is going to occur over a range of temperatures (in the case of a polymer).

So basically, you have a cylinder with one set of thermal properties embedded within semi-infinite medium with a different set of thermal properties, where, at time zero the cylinder is uniformly hot and the medium is uniformly cold. You then solve this transient heat conduction problem.

Chet
 
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  • #7
It is indeed an interesting approach. However, difficult to calculate.
Anyway you pushed me in the right direction. So thank u Chet!

This summed it up pretty good! Thanks alot!
Chestermiller said:
So basically, you have a cylinder with one set of thermal properties embedded within semi-infinite medium with a different set of thermal properties, where, at time zero the cylinder is uniformly hot and the medium is uniformly cold. You then solve this transient heat conduction problem.
Still a little unclear what u meant by this
Chestermiller said:
I'm guessing that most of the filament deformation takes place within the air gap
 
  • #8
Rotzooi said:
It is indeed an interesting approach. However, difficult to calculate.
Anyway you pushed me in the right direction. So thank u Chet!

This summed it up pretty good! Thanks alot!
Still a little unclear what u meant by this
That's when I still thought we were dealing with spinning.

Check out Carslaw and Jaeger.

Chet
 

1. What is heat transfer?

Heat transfer is the process of thermal energy being exchanged between two objects or systems at different temperatures. This can occur through conduction, convection, or radiation.

2. How does heat transfer occur in a hot plastic thread and cold water bath?

In this scenario, heat transfer occurs through conduction. The hot plastic thread, which has a higher temperature, comes into direct contact with the cold water bath, which has a lower temperature. The thermal energy from the hot plastic thread is transferred to the cold water bath, causing the thread to cool down and the water to heat up.

3. What factors affect the rate of heat transfer in this scenario?

The rate of heat transfer in this scenario can be affected by several factors, such as the temperature difference between the two objects, the surface area of contact, and the thermal conductivity of the materials involved.

4. Why does the hot plastic thread cool down and the cold water bath heat up?

This is due to the second law of thermodynamics, which states that heat will naturally flow from a higher temperature to a lower temperature until thermal equilibrium is reached. In this scenario, the hot plastic thread has a higher temperature and the cold water bath has a lower temperature, so heat will transfer from the thread to the water until both reach the same temperature.

5. How can heat transfer be controlled in this scenario?

Heat transfer can be controlled by adjusting the factors that affect its rate. For example, the temperature difference can be reduced by using a less extreme temperature for the water bath, or the surface area of contact can be increased by using a larger container for the water. The materials used can also affect the rate of heat transfer, so choosing materials with different thermal conductivities can alter the rate of heat transfer in this scenario.

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