Thermal calculator, or simple conduction formula 1

[DesignerMike]

I'm looking for a on-line thermal conduction calculator, or simple spreadsheet to do some simple calculations.

usually I'm looking at sanitary SS tubing (.065" wall) that is either jacketed with hot water, steam, or chilled water. Occasionally I am using tubing routed through a tank to maintain product temperature.

I know that product velocity, and heat transfer fluid velocity affect the efficiency greatly, but I'm not looking for an exact number, just ball-park.

A typical tank calc would go something like this:
Product temp 105 deg F, stagnant fluid in 5000 gal tank.
Heat transfer media: water/glycol @ 120 deg F max, flowing @ 30 gpm through 1" OD .065" wall tube 304SS. Can I tranfer 15,000 BTU per hour? Total surface area of tubing in contact with product is about 15 square feet.

Thanks

 

[25362]

Since you didn't state the thermophysical properties of the fluid product in the tank, apart from telling us that it is stagnant, and in which, most probably, resides the major resistance to heat transfer, not even a ballpark figure can be estimated.

It appears the tank has no agitation provisions whatsoever and the fluid may be quite viscous. If there is negligible thermal convection in the tank's fluid, the temperature difference may drop even below 15^oF. On the other hand, the tube's geometry and location inside the tank hasn't been given.

Just as an example: for a temperature difference of 15^oF and an overall heat transfer coefficient (U) ~ 10 Btu/(h.ft².^oF), the heat transferred would be UA[Δ]T = 10[×]15[×]15= 2250 Btu/h, which is much lower than the expected rate.

Whether a U value of ~10 Btu/(h.ft².^oF) is too conservative or not, is for me impossible to assess in the light of the given information.

 

[DesignerMike]

I'm at the office now and have a little more information.

Product is around 10,000 Cps, specific gravity about 1.3 no aggitation.
Tank geometry is horizontal. I intended on runing (6) tubes from one end to the other (and double elbow them together in series). Probably about 8" from the tank wall and about 12" apart along the bottom of the tank.

The tank is insulated with 2" fiberglass (customer thought it was jacketed). Product comes into tank at temp (105 F). I just need to maintain the product at 105 F, exterior temp is probably 65 F worse case.

How do you come up with a heat transfer coefficient of 10 Btu/(h.ft2.degF)??

 

[25362]

1. General: The overall heat transfer coefficient is always lower than the lowest of the transfer coefficients in the heat path. Thus the lowest coefficient, meaning the highest heat transfer resistance, determines the value of the OHTC. Consider that 1/U = 1/h₁ + 1/h₂ +..., thus U will be always lower than the lowest of the h values.

2. Natural convection: The heat transfer to the fluid in the tank by natural convection, with no agitation, depends only on buoyancy effects. The velocity of the water-glycol mixture inside the coil is of little relevance in this case.

3. OHTC: Tabulated OHTC values for medium viscosity lube oils heated by water inside coils, as given in Perry VI, table 10-13, are 8-12. In the case in hand, I assumed 10 Btu/(h.ft².^oF). Since the product is as viscous as a heavy oil the lowest limit would be even more representative.

4. Heat losses: It appears you have estimated the heat losses through the insulation at 15,000 Btu/h, therefore to keep the product at its inlet temperature, and for bettering the internal heat transfer one could apply a combination of the following alternatives:

-Relocate the coil along the walls and heads of the
horizontal (cylindrical?) tank.
-Increase [Δ]T (don't burn the product).
-Use larger coil surfaces (longer coil).
-Agitate the product in the tank to homogeinize the
product's temperature.
-Consider using finned tubes.

NB: Perry VI is the sixth edition of the Perry's Chemical Engineering Manual (McGraw-Hill)

 

[DesignerMike]

I do have quite a bit of room for more coil surface area.

Now I'm wondering if the number I got from the other engineer about expected heat loss is even remotely correct.

Can you give me a number of BTU per hour per square feet of heat loss for tank surface? Then I can re-calculate and feel more comfortable about the design

Figure .150" thick 304SS, 2" of fiberglass, covered with .075" of steel. High temp (product @ 105 F), Low temp (ambient air @ 65 F)= delta T of 40 deg F worse case.

Sorry for all the questions, but I understand the physics of the issue, but rarely use any of the formulas or calculations. I should buy a good reference book and brush up on the simple stuff.

Thanks for all the help.

 

[Montemayor]

Mike:

Pay particularly carefull and studious attention to all the important information 25362 is accurately giving you.

I would add that it appears from your opening statement where you state you're looking for a "simple, conduction calculator" that you don't have a full grasp on the mechanisms that are taking place in your system. For example, bear in mind that the predominant and controlling mechanism behind your effective heat transfer is convection - not conduction. This is the essence of much of 25362's valuable comments. You have a static, viscous batch fluid in a relatively large tank being kept heated with a relatively warm fluid (only 120 oF). This is a very tough and demanding heat transfer operation. You would be wise to heed the conservative approach that 25362 is presenting.

Bear in mind the mechanism that you are imposing: the viscous fluid is static in the tank (you are not contributing any positive stirring or agitation on the outside surface of the heating tube ("shellside"). And this is the controlling film heat transfer coefficient. As you heat up the shellside film from 105 oF using the very weak 15 oF driving force (120 -105 oF), you start to create positive, natural, and useful convection currents but this rate of heat transfer is subject to the bad effects of high viscosity and low temperature driving force. Your overall heat transfer coefficient ("U") would be better if you were to apply mechanical agitation and a hotter heating fluid. Without your use of mechanical-induced agitation on the shell side, I would estimate that the U = 10 Btu/ft2-oF is as good an estimate as you can get. Of course, I'm assuming efficient vessel insulation to preserve the heat added and eliminate a maximum of losses to the ambient 65 oF.

I don't know what your required heating fluid flowrate would be, but I would certainly ensure that the flow inside the heating tube(s) would be in complete turbulance - as analyzed by the Reynolds number.

I hope these remarks add to the value already given by 25362.



Art Montemayor
Spring, TX

 

[25362]

DesignerMike, how does the viscosity of the product increase with a lowering of temperature ? Does it become a semi-solid ? In which case a film of undefined thickness over the tank walls would signify in a further heat transfer resistance (by conduction). Is the tank full, or only part full ?

One-dimensional heat flux is represented by

q/A = [Δ]T_overall / [Σ]R_th​

the ratio of the thermal potential difference to the sum of the thermal resistances.

After making many assumptions regarding [Σ]R_th for the case in hand, I'd say the average heat loss to the still air surroundings for a [Δ]T = 40 deg F, would be about 4 [±] 0.5 Btu/(h.ft²).

I have the feeling that this ballpark figure results in a loss significantly lower than the pre-estimated 15,000 Btu/h. Am I right ?

 

[DesignerMike]

You are very correct 25362! Using the 4 Btu/hr number I'm looking at less than 3000 Btu/hr total.

After doing some more product research, it appears I can easily increase the water temp to 160 without damaging the product (which will make a HUGE difference). I also found that the viscosity is probably going to be closer to 9000 Cps.

So, to be safe I'll assume we may loose 4500 Btu/h, and I increase the delta-T from 15 deg to 55. I SHOULD be able to introduce 8250 Btu/hr. Theoretically I should have no problems maintaining temp, and should be able to heat the product (very slowly) if we get a shipment of cold product.

Thanks for all your help!

 

[25362]

The key word is theoretically.
I don't want to dissapoint you, but look at it this way.
We know very little about the thermophysical properties of the product to enable to better assess the value of U.

If the fluid is cold and very viscous, a layer can build around the warm coil that may act as a resistance on a conductivity basis, to such an extent that the actual U could be just as low as ~5 Btu/(h.ft^2oF), or less, instead of 10, as originally assumed.

And we haven't yet brought into consideration any fouling factors on the product side. We don't yet know how the product's viscosity and density vary with temperature.

Therefore, my advice would be twofold: a. don't design the heating coil on basis of ballpark assessments;
b. if possible, carry out some simple "bain-marie" lab-bench tests on the product to see how it actually performs on heating and cooling.

 

[DesignerMike]

I understand your "disclaimer" 25362!

I did get some more information about the product.

80 deg ~ 25,000 Cps (hopefully we never see this)
100 deg ~9500 Cps
120 deg ~2900 Cps
140 deg ~900 Cps

I'm not so worried about coil fouling. They will be cleaned regularly. We are also going to pipe the tank so that I can re-circulate product with the discharge pump if we need improved heating.

I also have quite a bit of space to increase the coil size, and would probably increase to 20 sqft.

Customer initially wanted to wrap the entire (insulated) tank with heat tracing, which sounded like a major waste of energy to me.

 

[25362]

Waste of energy ? In any case both you and your customer want to compensate for the same amount of heat losses...

Or do you mean a waste of people's energy in carrying out a more difficult or troublesome job ?

One more item not related to heat transfer.
Is the fluid combustible ? Does it have a vapor pressure at the operating conditions ? If affirmative, have you considered an inert gas blanket, and tank grounding and pipe bonding to reduce any static electricity buildup to zero volts ?

Sorry for the digression.

 

[jrhols]

This is a classic example of computer/mechanical asthma.
The exact solution, and if you are employed as an engineer, you may as well be exact, is not very general and is a complex procedure. So, computers can help. A good model that will get you an easier to communicate about and think with tool can be downloaded free at

http://www.thermcoat.com

If you have excel, it is better but a very simple DOS version can be downloaded there too. It is a matter of how far you want to carry out the computations and who's text book you feel you can trust and understand and validate.