Snowmelt Design

Paul Rohrs

I am working on a snowmelt design and am experiencing some glitches with the math on the secondary loop.

I have a secondary snowmelt loop of 16GPM @ 135F. I have a 30 Delta T with returning water of 105F. I have 180F boiler water for injection supply. My math tells me:

*16GPM x 135F = 2160 (load of secondary system)

so.....

9.6GPM x 105F = 1008 (load of return water)

6.4GPM x 180F = 1152 (load of injection needed)

*16GPM 135F = 2160 (secondary load)

(6.4GPM needed on injection pump)

If I use the formula out of the Wirsbo Snow & Ice Melting Design manual I get the following. (page 110)

F1 = 16 (System Flow GPM)

T1 = 180 (Boiler Temp)

T2 = 135 (Secondary Supply temp)

TR = 105 (Secondary Return temp)

TD = 30 (Delta T)

So.....

FV =(16 x 30) (180 - 135)

FV = 480 45

FV = 10.666 GPM needed at injection pump



What am I missing? Architects and engineers are saying my 34" injection loop will not sustain secondary loop temps and want me to increase to 1". I don't think it's necessary. My math shows me I'm right,Wirsbo design verifies their answer. This is a Government job and the Engineers are dotting all I's and crossing all T's.

Regards,

PR

Paul Rohrs_2

Problem Solved

My oversight was on the Wirsbo side when I took 180F - 135F and it should have been 180F - 105F (return). Now my numbers jive at 6.4 GPM and I feel much better.

One thing I will say about architects and engineers, they reaffirm my belief to always do the math.

Thanks,

PR