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update on shop floor heat help
jtpfarm
Member Posts: 35
First off, thanks for all the information and reply on my previous thread.
Second, I think I have found a part of my issue. My pump on my primary loop (1-1/4) is 17 gpm with 8 ft of max head.
My pump on my secondary loop is 33 gpm also with 8 ft head max. The loop is 1 inch copper into six 250 ft runs of 3/4 pex.
Am I correct in thinking that the pump on the secondary loop is too small?
Second, I think I have found a part of my issue. My pump on my primary loop (1-1/4) is 17 gpm with 8 ft of max head.
My pump on my secondary loop is 33 gpm also with 8 ft head max. The loop is 1 inch copper into six 250 ft runs of 3/4 pex.
Am I correct in thinking that the pump on the secondary loop is too small?
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Comments
I believe your just listing it's max flow rate, and head. Max flow rate only happens with the least amount of head in the pump curve.
You can choose to do what you want, but you cannot choose the consequences.
If you ran a 10* delta T, then you would need 14 gpm. At this flow rate, you would only have 3.25 ft. of head. Your secondary pump is way over-sized.
You can choose to do what you want, but you cannot choose the consequences.
7 GPM seems low for 1500' of tubing.
Am I correct? What size pump would you recommend?
We are reverse engineering the process because initial design did not take place.
Divide that 7 gpm between the 6 loops 1.17 gpm per loop providing it's all balanced for flow.14 gpm would be 2.33 gpm per loop.
To calculate head It's the longest loop, plus piping from the pump to the loop, and back. So if your piping is 50' to the loop plus 300' for the loop plus 50' back equals 400' we don't know what size piping feeds the loops (remote manifold?) so you would have to calculate the pressure drop for that pipe size, and developed length of fittings.
Now you select a pump with a curve that matches the design criteria. Ideally you want to fall in the middle third of the curve for efficient.
The 100 circ as ironman said is way over sized if you look at the curve chart.
As you said in,your initial thread your delta t is 11*"was this temp measured at the supply return where the supply /return t's are off the primary loop, or at each loop? Have you shot the floor with an infrared thermometer to see if all loop fields are evenly distributing the heat that is generated?
"Note, I did not pay any attention to how many loops you have or their spacing or length. Also, I did not account for a high water table, within 6' of the slab."
The rest of the info is on the reports.
Your downward heatloss is significant. Your current boiler is not big enough to meet the load.
Your radiant to room load is 70k. The other 35k is being sucked away into the ground.
There are different ways you could approach this. You have to decide what is important to you... warm floors, achieving set point with alternate means or using a hybrid approach.
Either way, efficiency in the slab as a radiant heat source, is out to lunch.
Harvey
ramermechanical.com
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Before I saw Harvey's heat loss calc of 105k btu's, I simply took the output of the boiler that you gave: 70k btu's. I'll still use that for the moment to demonstrate how I arrived at the numbers I gave you.
The heat loss of the structure is the foundation of designing. Your questions about flow rates and pump curves, etc. are like jumpiming into the middle of the design process and then hopscotching around. I realize you're not doing it intentionally, but it ends up with the same result: confusion. It has to be done step by step in order.
Starting with the heat loss: il still use the 70k btu number because I want you to see how it works out, even though we now know that number is wrong.
The universal hydronics formula: btu = (delta T x 500) x gpm.
In order to get 70k btu's with a 20* delta T, you need 7 gpm. You can do the math.
As I stated, most shop/garages can get by using a 20* delta T. There may be some degree of "striping", but if it's an open area, a 20* delt T will work. Therefore, you would need 7gpm. As Gordy said, that would be devided between 6 loops, which = 1.166 gpm, per loop. You don't figure as though the entire 7 gpm is going through one loop, which appears you are doing. A 3/4" pex tube that is 300' long has about 2.78 ft of head @ 1.167 gpm. Add the manifold and supply/return lines and you're about 3.25 ft of head @ 1.67 gpm. You only take the head of the longest loop, not the sum of them all. You use the sum of all of them for gpm.
Now, the problem with using a 20* with YOUR tubing layout is that it would cause the velocity to be too slow to get turbulent flow and proper air elimination. With 1.67 gpm a 3/4" pex line is moving water at about 1ft per second. This is too slow. We want a minimum of 2 ft per sec and a maximum of 4 ft per sec.
If we designed with a 10* delta T, you would need twice the gpm of 20* = 14 gpm. This divided by 6 loops gives 2.33 gpm per loop and a velocity of 2 ft per sec. The head would 10.4 ft @ a total gpm of 14.
Therefore, you would need to design closer to a 10* delta T to get enough velocity.
I skipped over several formulas and just used sizing tables for some of the data in order to keep it as simple as possible an explanation.
Now that we know what the heat loss is, Harvey's recommendation of a fan coil looks like you most viable option. Your boiler will still be under-sized during the colder days, but it will do better with the fan coil.
You can choose to do what you want, but you cannot choose the consequences.
I just have one question. How come the block wall looses that many BTU even having R12 on the inside of it?
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