Hey GM, I used 11/inches as length which is wrong, but the nice thing is when you are using a ratio it all works out to be the same.
Surface are of a cylinder = length*pi*2r, the volume = length *pi&r^2
for the 1/2" tube (I didn't take wall thickness into account so it is slightly off, So I was using 1/2" ID)
Sa=11"*pi*2*(.25") = 17.28 sq inches
Vol=11"*pi*(.25")^2 = 2.16 cubic inches
1 gal = 231"^3 cubic inches
2.16"^3/231"^3 = 0.0094 gal per tube
17.28"^2 / 0.0094gal = 1838 in^2/gal
I must have lost 10 square inches in my rounding somewhere
With the rods, volume of the tube - the volume of the slug = total liquid volume in pipe.
then surface are of the pipe / total liquid volume in pipe = surface area to volume ratio.
This doesn't take into account the entire volume of the kettle. just the fluid in the tubes and how much steam heat surface area they are exposed to because this is where the system is dumping the majority of the heat into the liquid. Everything else is absorbing heat from the liquid coming out of the heater.
calculating it this way is the only real way to compare different tube sizes heat transfer efficiency and keep it apples to apples. So a tube with a higher ration than another tube of a different diameter will transfer heat better.
With that said, you will also need to figure out how much water you are trying to heat at any instant. Let's continue with the following system
24x 1/2" x 11" tube
48k btu/hr of heat transfer (kladue's calc from the last run)
0.0094gal and 17.28"^2 per tube
24 * 0.0094gal = 0.2256gal ,
17.28"^2 * 24 = 415"^2 , 0.2256*1838 = 415 as well.
(48k btu/hr) / 415"^2 = 116 btu/hr per in^2
, sounds a little low to me compared to an electric heating element, that is my only point of reference tbh.
(BTW 415"^2/0.2256gal yields the same in^2/gal ratio as a single tube.)
So to get velocities up you need to put a lot of heat into a small volume as quickly as possible. So high ratio and low volume should result in higher fluid acceleration. Now that I think about it though, that will require a lot of heat to move.
Kladue, do you know what the exit velocity of the wort in one of those things? If you do we could calculate the rate of acceleration in a 12" tube is needed and some how translate that to a btu to volume. I'm thinking it is going to be a ridiculous number though.
Man my head is spinning now... I'm calling it a night