Help with friction loss formula

Jamie_5

I think the problem is the flow rate. It looks too high for 5/8" piping. Checking your calculations against Siegenthaler's pipe sizer, I came up with head loss of over 23 ft. per 100 feet of tubing. That is equal to 70 feet of head for 300 feet of piping. If you need that flow rate, I think you're going to have to use a larger tube.

John McArthur_2

Tried to do some friction loss calcs with this formula and found the results to be strange. What am I doing wrong?

H=k*l*f1.75(exp)

Did a calculation for a 300' loop of 5/8" pex with a k factor of 0.014 and at 5 gpm. Got a result of 70' of head loss for water and 94' for 50% glycol. I re-checked numbers several times. This does not make sense. I have checked my reference for the k factors and it is correct.

Can anyone explain what I am doing wrong? Also, the references indicate the k factor is per foot.

John McArthur_2

Thanks, I thought I was just doing something wrong. I got the formula and k factors from Siggy.

I get a different number when compared to the results from my Radiant Works software. Have not figured it out yet.

Snowmelt applications are the biggest issue. I have gone to several jobs where I found 300' loops of 5/8" pex installed for S/M. With a 1,200 sqft area at (conservatively) 120 Btu/hr/ft, thats 144,000 Btu. Even at 30dT it's still about 10 gpm. With 50% glycol that's about 315' head by the formula. That does not seem correct.

It seems as though I'm totally screwing something up.

Mike T., Swampeast MO

Simple Little Rad Pad

Shows 300' of 5/8" pex at 5 gpm WAY off the scale--both in flow and in length.

At 2.4 gpm head loss through 300' of 5/8" pex is a much more reasonable 16'

(for water only)

Mike T., Swampeast MO

Only one 300' loop of 5/8" pex for 1,200 square feet of snowmelt? If so, I doubt it would melt a flurry by tomorrow in 32° air...

John McArthur_2

Sorry, I was unclear. The S/M job I was refering to actually has 4-300' loops of 5/8" pex.

Still, the formula indicates the flow rate for the entire panel through the loop length (300'). I realize that the gpm flow rate is for the entire load and not just one loop. In this case should I have divided the load by 4 and used that portion of the load to base gpm?

Radiant Works tells me that I need 21gpm at 75' of head for the job with 50% glycol. I am in the Colorado Rockies with a snowfall of 18" a day as an average.

I am curious about proper application of the formula for field calculations. I have several software porograms for use at home.

Just hoping to find the light of understanding to proper application for the formula. I appreciate the responses.

Terry

2.5 gpm

10 gpm divided by 4 loops is 2.5 gpm.

do the headloss calc on 300 ft of pipe @ 2.5 gpm...

unless all 4 loops are piped in series,. then it won't work anyway...

EIN

John McArthur_2

Thank you, exactly what I needed to know.