Ok I'll do it *scratches his arm nervously*
There is all kinds of snow and how it is packed effects not only the mass but also the heat transfer... *happy sigh*
Here is an
ASHRAE paper on the subject
QUOTE qo = qs + qm + Ar (qe + qh)
where
qs = sensible heat transferred to the snow (Btu/h.ft2)
qm = heat of fusion (Btu/h.ft2)
Ar = ratio of snow-free area to total area (dimensionless)
qe = heat of evaporation (Btu/h.ft2)
qh = heat transfer by convection and radiation (Btu/h.ft2)[/quote]
QUOTE During the summer period approximately 20% of the incident solar radiation on the activated road surface can be collected, corresponding to 150,000 kWh (512 million Btu). Losses amount to approximately 35% of this quantity, the remaining energy being available to keep the bridge surface free of ice during the winter period.[/quote]
150,000 - 150,000*.35 = 97,500 kw of heat energy from a system of pipes (they used mechanical heat transfer (a higher efficiency btw rather than solar panels) but no matter this is cards stacked in favor of the new system.... ok so we have a grand total of 100,000 lw of heat
You can now figure out how many inches of snow (based on some bassic assumptions of course) you can melt on the road surface that is 14,000 ft^2.
We need to get the values for heat flux to those in the paper Btu/h*ft^2
You have the conversion factor for Kw to Btu/hr and the total kw -> 341214200 Btu/hr
Now divide by the area for the
heat flux -> 24372 BTU/hr*ft^2
So let's get calculating!
QUOTE qs = s cp ? (32 - ta) / c1
where
qs = sensible heat transferred to the snow (Btu/h.ft2)
s = rate of snowfall (inches of water equivalent per hour)
cp = specific heat of snow (0.5 Btu/lb.oF)
? = density of water equivalent of snow (62.4 lbs/ft3)
ta = air temperature (oF)
c1 = conversion factor (12 in./ft).[/quote]
An important point of the discussion is how much is the ambient air effecting the melt. Let's make it 32F to simply things and please note that temperatures colder than 32F mean some of the total heat availible goes to just warming the ambient air in contact with the snow
qs = s * cp * (32-32)/ c1 -> this equals 0 because 32-32 = 0
s = unknown (total inches we could melt of the road)
qs = 0
QUOTE qm = s hf ? / c1
where
hf = enthalpy of fusion for water (143.5 Btu/lb).
For hot water (hydronic) systems, the above reduces
to:
qm = 746 s[/quote]
nothing more to do here
QUOTE Ar = ratio of snow-free area to total area (dimensionless)[/quote]
Once all the snow has melted how much is still on the road? The new road system is not zero but lets say it is
QUOTE Ar (qe + qh)[/quote]
so if we assune 100% of the water drains off the road... obviously horse poop but I'm lazy as heck
nothing more to do here either
therefore qo = qs + qm + Ar (qe + qh)
wittles down to (if we neglact everything but just what is needed to melt the snow):
24372 BTU/hr*ft^2 = 746 BTU/hr*ft^2 * s inch/hr
32.6 in of snow could be melted in total
IF
the ambient temperature was always 32F when it snowed, the water always completely drained off (including any film) the road, snow density was light, the efficiency of the solar cells approached the mechanical system, 100% of the energy was used to melt snow, and there where no other loses.
So Fuzz my man the answer is a resounding no for your old stomping grounds
