9 replies to this topic

[populas]

Hello all,

Following the reading of this very interesting article http://www.cheresour...el_design.shtml

I share you some questions regarding the explanation given and hope somebody can enlighten me.
I have to calculate the heat transfer coefficient of an external half pipe coil. In the explanation, we have to calculate the external diameter De with the Dci, but I fail to understand this equation :
De = PI / (4.Dci)
I even fail to understand its meaning, why the PI, what about the units, why calculate the Ax too,etc... I am sure I am missing something and I hope someone here can help me.

Thank you very much.

[Profe]

Hi Populas


If you review the article, D_ciis the central angle.

Half-pipe coil jackets are generally manufactured with either 180° or 120° central angles (D_ci)

or consult with the autor of that article.

By: Santosh Singh, Guest Author

Please direct inquiries to: process.santos@googlemail.com

[populas]

thank you Profe, I will contact him right after this message.

Actually one of the problem I have is with the interpretation of the "central angle" Dci value. That's the first time I see this concept. Is it an angle or the value of the "diameter" (if 180°) in meter? What about the PI symbol. Is it really the number PI or is there another meaning?

again, thanks for all

[Profe]

Hi Populas

I think that D_ci is the central angle in radians. But what about the advice from the author Santosh Singh?

Good luck.

Edited by Profe, 20 April 2010 - 08:53 AM.

[populas]

Hi Populas

I think that D_ci is the central angle in radians. But what about the advice from the author Santosh Singh?

Good luck.

mmm, I don't think so Profe. I was first thinking like you but then, I was wondering... this formula is done for the 180° case, why then putting again this value in it... which will give the same value of De for all the 180° cases.

I sent an email to process.santosh@googlemail.com (corrected with a "h" at the end of his name) but for the moment, I haven't received any answer.

Imagine, you are doing a big calculation with so many parameters, hypothesis,etc. and then you block on this value I really hope he will answer me.

Thanks for your support

[populas]

no answer from Mr. Santosh Singh...

did anybody make this kind of calculation?

Edited by populas, 22 April 2010 - 10:18 AM.

[Art Montemayor]

Populas:

I've said it before many times, and I'll state it here again in case it can help you out.

No one - to my knowledge - has had any reasonable justification for investing time and effort to design the required area or heat transfer characteristics of a heating/cooling jacket on a chemical reactor or process vessel. The everyday, normal scheme of things is to FIRST DESIGN THE SIZE AND TYPE OF REACTOR VESSEL, and subsequently fit the selected vessel size with as much jacket area as one can economically and practically justify. In other words, one doesn't design the shoe size and then fits the foot to the shoe; the process is the other way around: The existing foot is fitted with the shoe that fits it.

The heating and cooling requirements of a reactor for a specific set of chemical processes never conveniently results in a size that fits the size of jacket that can fit the required vessel. Therefore, what one does in the real world is that once the required reactor or vessel size is selected, a jacket type is selected and applied. Whatever heating and cooling capacities result from such a jacket are accepted as "what you see is what you get" - and one goes about calculating how to obtain additional heat and cooling inputs if one is not satisfied with the time it takes to effect a complete heat transfer. That is - more often than not - the usual, normal case.

Of course, some people take it upon themselves to spend time designing a jacket on a vessel. They do this for various reasons - most of them are for selling jackets. Other reasons are academic or heat transfer exercises. You haven't stated WHY you want to design a half-pipe jacket on a vessel. I suspect you want to do it because you believe you can obtain sufficient heat transfer for what you want to do. Unless your vessel is an extremely large one, I would simply apply as much jacket to the external surface and be done with it. When you make the calculations you will find that the overall "U" will be approximately 70 btu/h-ft2-oF. The pressure drop for the required cooling water will be extremely high and you will not achieve the cooling rate that you would prefer.

I hope this experience helps you out.

[populas]

Thank you Montemayor, I really appreciate your experience's share.

You are right, I haven't stated yet why I want to design this external coil. I will try to explain it as clearly as possible.

The tank with its external coil already exists. I cannot change the design anymore. However, the goal of my calculation is to define the flowrate of the thermal oil needed to do the heating in a defined time.
I found a way to calculate this flow rate for this specific case, but to complete it, I still need to estimate the overall U. For info, it is an agitated tank. That's why I found the explanation of Mr. Singh quite interesting as it gives a good way to calculate both most influencal heat transfert coefficients : h(jacket) and h(inside vessel).

I found an overall U of 30 btu/h-ft2-oF. Of course, it depends on many parameters but if I change this overall U and put 70btu/h-ft2-oF (as you proposed), it doesn't change as much the flow rate of the thermal oil (about 5%). Is it normal according to you?

[Art Montemayor]

Populas:

Yes, the effect is slight. That is a problem with spiral, half-pipe heating/cooling jackets on cylindrical vessels. What you are experiencing is normal. The half-pipe, in order to be effectice, turns out to be so long and reduced in cross-sectional area that you obtain a very high pressure drop through it. The result is less fluid flow through the pipe - with a resultant lower heat transferred quantity.

Additionally - on the practical, mechanical fabrication side of the application - you will find that you have coil fitting and welding issues on applying the idea to a vessel. The inherent thermal and fluid flow inefficiency of the coil makes you want to reduce the pitch between adjacent pipes in order to add more effective heat transfer area to the external side of the vessel. This however, complicates the already difficult (and taxing) job of welding the pipe to the vessel. You will find that the amount and RATE of welding heat applied to the external vessel wall sometimes affects the parent material to the point where it weakens and warps the vessel. And the quantity of welding to be done is a very labor-intensive chore - and expensive. But all these agravations and problems with half-pipe coils on vessels are to be taken as what you inherit when you try to apply a process where it is ill prepared to be applied. Again, as I stated in the prior post, you should be satisfied in getting only the heat transfer that the geometry and configuration inherently offers - which is very little. You cannot squeeze blood out of a turnip.

The real, and effective contributor to good heat transfer through the walls of a vertical, cylindrical vessel is the internal mixing device that you can have installed. It is the turbulance and mixing (forced convection) that this device can add that really determines the internal heat transfer film coefficient and, thus, sets the effective heat transferred. Your time and effort are best spent in concentrating on the vessel mixer than on the vessel jacket.

If you are after real, effective, and predictable engineered heat transfer in a vertical cylindrical vessel, the answer lies in circulating a heated portion of the contained liquid through an external heater or heat exchanger. This allows you to control the rate of mixing and heating at the same time and is the quickest and most efficient way to heat up a batch system.

I hope this experience helps you out.

[populas]

Thank you again for you answer Montemayor.

I would like to share with you the exchange of emails with the author of the article (Mr. Santos Singh). Maybe it can help you. There are some clues (P is the pitch between coils) but I still have some doubts and difficulties with the Dci. Why is the Dci a parameter (angle in radian) in an equation which can only work with angle of 180°? The units are also weird...

1- First email with questions

(...)
I have to calculate the heat transfer coefficient of an external half pipe coil. In the explanation, we have to calculate the external diameter De with the Dci, but I fail to understand this equation : De = PI / (4.Dci)
One of the problem I have is with the interpretation of the "central angle" Dci value. That's the first time I see this concept. Is it an angle in degree/radian, or the value of the "diameter" (if 180°) in meter? What about the PI symbol. Is it really the number PI or is there another meaning? I even fail to understand the meaning of the equation; why the PI, what about the units, why calculate the Ax too, why is it completely different for the 120° case,etc... I am sure I am missing something very big here.
Thank you very much for your help.

2- the first answer of Mr. Singh

Sorry for answering late, Values of angle are in degree only, Please refer http://www.cheresour...l_design2.shtml and regarding PI this is not PI but De = P / (4 Dci ), where P is the pitch of half pipe coil, other values in the equations are in FPS unit.
Hope this clarify your question.

3- my answer

If the pitch of half pipe coil is 150mm (0.15m) or 0.5 feet, and if the central angle is 180°, then we would have :
De = 0.5 / (4*180) --> De = 0.0007 ft
I find this value very strange and yet, as you can notice, I have used normal values. Reynolds number will always remain very low.
Regarding the units, is there any impact if I use metric units (SI) instead of FPS unit in the entire equation?

4- last answer of Mr.Singh

In the equation you have to put the value in radian for angle.