It would help if you asked specific questions. I can't tell the level of understanding you have about the process, so I dont know where to start. Is your question, where to use free or forced convection coefficients, or do you need help in understanding the process?

Minimum outside temperature and maximum wind velocity determine heat lost through the tank's surface by convection. There is also radiation loss to space. The outgoing heat is the sumation of all losses through the tank surface. Watts lost.

To maintain equilibrium, the heat transferred from heating pipe to the tank water must equal the heat lost by the tank to the outside environment.

Heat transfer of the pipe contents to the water can be calculated in terms of heat loss per foot of pipe. Watts/meter (of heating pipe). Length of the heating pipe required is Watts lost / Watts/meter.

Inside surface film transfer of the heating pipe to pipe wall is forced convection.
Outside surface film transfer from the heating pipe to tank water is free convection, as is the film coefficient for heat transfer between tank water and tank wall. Heat transfer of tank to air is free, with no wind, forced with wind. With winds, some portion of the tank wall not exposed could be free, exposed portions forced. You may be able to determine the tank heat losses as "exposed" or "sheltered", in which case heat loss may be determined more directly, as for this oil filled tank toolbox example,
https://www.engineeringtoolbox.com/heat-loss-oil-t...
You may be able to simplify the wind considerations and just use the same sheltered to exposed loss ratios of the toolbox example for the outside tank wall heat loss calculation. I'd try it, unless you have some unusually high wind or exposure considerations to account for.

So, where do we start? It would help if you ask specific questions and post the calculations you have done so far. Free convection happens when air is not being mechanically moved by some force, such as wind pressures, a fan, or a pump, or a mechanical mixer, or vortex vanes. Free convection is fluid movement basically by natural density currents alone.

The material of the tubing is nearly irrelevant.
You have a small delta T and the opportunity for very long contact times (long length).
Your outlet water will be very near 100F.
If the ambient temp is going to be below about 80F you will need to insulate the tank.
And without mixing (even just a gentle stir) you will have very nonuniform temperatures.
It is just a matter of a heat balance.
When you first start heating it will take a long time but keeping it warm is strictly related to heat loss to the environment.

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P.E. Metallurgy, consulting work welcomed

And water fill and draw rates.

Before you even start these calcs, would agree the stated feed hot water temp is too low. Tank coils such as these have low overall heat transfer coeff, since overall htc U is controlled by coil outside (free convection) htc ho, which is low. Hence the practice to make up for this poor U with high LMTD. Calcs can get quite involved, and also requires a calc to determine economically optimum insulation thickness, so you need an insulation contractor to tell you what the cost is for insulating this tank to a range of thicknesses. Adding to @1503-44's list, also include (a) radiation heat transfer from warm water surface at tank top to inside surface of tank roof when doing the calc for roof loss and (b)heat loss through tank bottom.

Also remember the feed pump should develop the dp required to get this water through the coils for all operating scenarios. Break up the coil into 2 or 3 parallel sections so as to reduce pumping dp. Add a generous fouling factor fo on the outside if this water can corrode the coils, cause scaling or is dirt laden. Avoid using externally finned pipe ( to make up for low ho) for obvious reasons.

For higher heating medium feed temp, take into account the scaling tendency of this "unmixed water" in this tank. At above a given tube OD film temp ( tube OD film temp can be calculated), the water may drop out Ca and Mg fouling scale onto the tubes, reducing U. The GPSA has guidance on the calculation of the scaling tendency ( using the Langelier Index) in the chapter on water treatment for a given TDS and pH. Check that film temp at the hot end of these coils is less the estimated scaling temp for this water. If this turns out to be a problem that is difficult to resolve economically, one option would be to inject antiscale inhibitor upstream of this tank.

Hi,
Consider the document attached about storage tank heat losses to understand the issue with your proposal .

Good Luck
Pierre

• https://files.engineering.com/getfile.aspx?folder=b8f2d250-7c96-4d02-bf91-e

Montemayor, that's a blast from the past. The link has the original printed article

Link

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Hi IRstuff,
What a contribution !
Regards

Pierre

• https://files.engineering.com/getfile.aspx?folder=d003697b-3f98-4ddf-8230-b

Thanks everyone for your input so far. I should have been more specific. I have already accounted for thermal losses to ambient and ground as well as the makeup heat needed to maintain the incoming tank flow. This set the minimum required heat input from the heating loop. What I'm really asking is for recommendations on determining the overall heat transfer coefficient for the heating loop so that I can determine the length of pipe needed.

With such a low delta T you U will be very small.
The only downside to making the loop too long is wasted pump power.
Even if you circulate the tank, the water will be nearly static, so strictly limited by conduction (with a dead boundary layer).
However you estimate the U make sure that it is conservative enough.

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P.E. Metallurgy, consulting work welcomed

Use DQ Kern " Process Heat Transfer " to compute hi(forced convection inside tubes), ho(free convection outside tubes), add in the fi and fo to get U. A more detailed account of these procedures is in Perry Chem Engg Handbook. Since carbon steel wouldnt last long in this application, suggest going for 3-4% Mo SS316L (if chloride content in this water is low enough) or else standard 22Cr duplex SS (check with your materials engineer). Beware of substandard "SS316L" that's flooded the market these days.

Hi,
You may want to use this link to get an estimate for heat transfer coefficient :
https://www.engineeringtoolbox.com/heat-transfer-c...
Note : heat loss calculator vertical tank
https://view.officeapps.live.com/op/view.aspx?src=...
Be conservative , OHT coefficient should be low .
One way to improve it could be to use submersible agitator (google Flygt submersible agitator).
Good luck
Pierre

And what George means by low Cl for good 316L is >100ppm with zero chance of scale formation.
Trying to buy 316L with Mo above the min (2.00-2.1%) requires special mill orders.
Though with Ni prices where they are you should be able to buy light gauge 2205 (sch 5) for less than good sch10 316L.
The 2205 has higher corrosion resistance, better thermal conductivity, and lower thermal expansion.
And it will have a higher pressure rating at sch5 then 316L at sch10.

= = = = = = = = = = = = = = = = = = = =
P.E. Metallurgy, consulting work welcomed