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More Heat Loss/delivery calculations!

Kestrel Member Posts: 102
Hello again - I'm hoping all the brain power here can come to my aid again!

I'm working on the calculation for my first floor.  House is 105 year old in Seattle.  Basement presently torn open for remodel, and joist space accessible.  Plan is to ditch forced air and switch to gas-fired hydronic  - tubing in floor joists for first floor, radiators on second floor and soon-to-be finished basement.

Based on heat loss calc at design temp of 20'F, total for the first floor is about 18500 BTUH (see below).  The layout is very open, with entry/living/dining and kitchen pretty contiguous; a single small bathroom is the only really separate space.  Significant floor area taken up by kitchen and bathroom cabinetry, and spaces for up and down stairs (two different locations)  I also calculated for heat loss at the much more frequent (relentless?) ambient temp around here of 40'F:

Room..........BTUH20........BTUH40........Total area........Available area.....BTUH/avail

Living            8250              4950               300                     280                   29.5

Dining            3655              2190               140                     140                  26.1

Kitchen          2775              1665               125                      45                   61.7

Back Entry     3045              1825               100                      57                   53.4

Bath                785                 470                 71                      57                   13.8

total BTUH20/total area = 25

total BTUH20/avail area = 32 = q(up)

(I think I'm going to have to use a kick plate n the kitchen at least)

From Modern Hydronic Heating I adapted calc for in slab tubing (I[m using the first edition):


q(down) = q(u)(Rff+Rair)+T(above)-T(below) / R(down) = 3.8

I ignored R(slab), and for Rff used R for 1/2 inch subfloor (fir, measured) and 3/4 inch jatoba, and got 1.34.


It's above a heated basement, so T(above)=67 and T(below)=67, and they cancel.

Thus, q(down)=3.8

T(ave) = T(air) + q(up) x (R(ff) + R(air)) = 67 + 32 x (1.95) = 129.4

If deltaT is 10, then T(in) must be about 135.

a = [ 1 / (Rff+Ra) ] x (1+ [q(down) / q (up)]) = 0.57

Looking at the tubing layout (may need work), Im using 4 lengths on average 240 feet each (vary 200-260, 0.5 inch)

Total BTUH = GPM x deltaT x 500, and design delta T is 10,

total GPM=3.7, split among 4 circuits, ave=0.925 GPM each

therefore, b = a x L/500 x f = 0.57 x 240 / 500 x 0.925 = .293

So at this point, Solving for T(out) = T(room) + (T(in)-T(room)) x e^-b

and T(in)=135, and T(room)=67

I get T(out) = 117

This gives me a delta T of 18 and seems very wide of the mark of delta T=10 that I was aiming for.

Can anyone help spot the mistake(s) I'm making??
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