Posted by Ed on August 19, 2006, 4:45 am
Many years ago on Okinawa I was faced with a requirement to deliver hot
water on demand to a technical application. To do so I installed a
series of flash heaters in line and measured the output temperature as
the water exited each flash heater. My memory is that I had to deliver
170 degrees to the application. Input water temperatures varied
somewhat, but not a lot. My recollection is that the rainey season
water was a few degrees cooler than the hot season water, but not by
much. Generally speaking, Okinawa is consistently warm. Overall nice
weather. Low thermal mass solar might have worked during the daylight
hours most of the time, but often there would be several days with rain
and overcast skies.
Installing a 200 gallon tank in my small furnace room would take up
quite a bit of room. Presently have the furnace, an 80 gallon electric
HW tank, a large well pressure tank, a limestone iron filter tank, plus
the normal assortment of shelfs for paint, stain, emergency water, etc.
In order to put in a 200 gallon tank I'd have to do some additional
excavation and pour some additional concrete flooring, and shore up the
walls with some cinder blocks. It might cost me more to add the extra
basement space for a 200 gallon tank than it would to dig a 6' X 6' X
10' pit for an insulated concrete tank.
Digging a tank pit, putting some crushed rock in the bottom, tamping
and leveling it, sealing the outside of the tank and affixing the
insulation is all straight forward and fairly simply work. Installing
the coiled poly lines through the manhole access and mounting them
before the tank is lowered into the pit seems simple enough.
Predrilling holes through the solid top part of the tank, grouting and
insulating the lines, and adding the connectors doesn't appear to be a
very difficult task. And yes, adding heavy insulation in the trenches
is important and is planned.
You got it right. Hot water and supplemental heat. Here are a couple
of questions that you might be able to help me with. If I install an
insulated concrete tank underground that measures 4' X 4' X 8' and
heated the water using coiled closed loop glycol lines that were heated
by three 4' X 8' rack mounted solar panels, what do you think the
average daily temperature of the water inside the concrete tank would
be at Latitude 37.74 N and Longitude 78.98W in January, April, July,
and October? And, if I wanted to heat 80 gallons of domestic hot water
for a family of four, and also heat 1,800 sf (later 3,000 sf) of living
space, what average daily inside air temperatures should I expect to be
generated by that tank and those solar panels for the same months?
Posted by Jeff on August 19, 2006, 5:52 am
Asuming inside dimensions that is 128 cubic feet of water. At 62.5
pounds per cubic foot you have 8000 pounds.
It takes one BTU to raise 1 pound of water one degree.
Lets add a therm, or 100,000 BTU's, you'd have a temperature rise of
12.5F. You could store several therms. Remember you'll be losing BTUs
through the insulation, and of course using them. The useable BTU's you
can store are determined by the difference between lowest temp that you
can use and the highest temp you can get the tank to.
36 lat, single glazing, .2 ground reflectance, clear day.
Date IDN vertical
Jan 21 2340 1688
Apr 21 3088 838
Jul 21 3038 582
Oct 21 2584 1550
That's BTU/square foot, you'll want to multiply that by the efficiency,
perhaps 45%... I'm not exactly sure what the IDN is, but I think it is
the optimum collector tilt. Anyone know?
So for a vertical collector on Jan 21 with 45% efficiency you'd need
about 130 square feet of collector to get one therm.
Figures from the "Passive Solar Energy Book" by Edward Mazria.
And, if I wanted to heat 80 gallons of domestic hot water
Calculate your heat loss through all the surfaces and add in the loss
through air changes and you'll have a rough idea how many BTU you need.
My guess is that you'll need a therm or so on an average cold day, but I
really really don't know.
Note also that you'll have to figure out how to use those BTU's you
get from the sun. Some methods require higher temps than others. Higher
temps mean that your tank losses will be higher and the amount of usable
heat you can store will be less because you can't run the tank down to
as low a temperature. I would think that tank you have in mind would
work well, it's bigger than what I have in mind here.
Posted by nicksanspam on August 19, 2006, 11:19 am
That depends the hot water usage and tank insulation.
I'm afraid that's irrelevant here.
I'm afraid clear day numbers are also irrelevant. NREL says 750 Btu/ft^2
of sun falls on the ground and 1180 falls on a south wall on an average
34.2 F day with a 43.6 daily max and an average daytime temp of 38.9 F
in Lynchburg, VA (37.33 N, 79.20 W), so a collector with an atn(750/1180)
= 64 degree tilt would get sqrt(750^2+1180^2) = 1398 Btu/ft^2, or more, with
more ground reflectance. The average yearly (deep ground) temp is 55.9 F.
With a single layer of R1 glazing with 90% solar transmission and perfect
collector back insulation and a large water flow and a perfect tank heat
exchanger and 50K Btu/day of hot water use and a tank with 160 ft^2 of R30
insulation surrounded by 56 F R10 soil and 3x4x8x0.9x1398 = 120.8K Btu/day
= 6h(T-38.9)3x4x8/R1 + 50K + 24h(T-56)160ft^2/R40, ie 70.8K = 576(T-38.9)
+ 96(T-56), T = 146.7 F, if I did that right.
... 80x8.33(105-56) = 32.7K Btu. A greywater heat exchanger could help.
Low. The tank might be fine for a few cloudy days in a row, but the house
needs lots of heat on an average day.
It's more interesting to calculate how much fuel would be needed for comfort.
Ignoring air changes and electrical energy use, an 1800 ft^2 x 8' tall square
house with 8% of the floorspace (144 ft^2) as R2 windows and 1214 ft^2 of R30
walls and an R60 ceiling would have a conductance of 144ft^2/R2 = 72 for the
windows + 1214/30 = 40 for walls + 1800/60 = 30 for ceiling = 142 Btu/h-F, so
it would need about 24h(65-34.2)142 = 105K Btu on an average January day.
A square foot of R1 Orangerie glazing with 90% solar transmission might
collect 0.9x1180 = 1062 Btu and lose 6h(70-38.9)1ft^2/R1 = 187 on an average
January day, for a net gain of 875, so we might heat the house with 105K/875
= 120 ft^2 of glazing with a dark mesh curtain to keep 70 F house vs solar-
warmed air near the glazing and make the space more useful for people. If
air enters the house at 130 F and ceiling mass cools from 120 to 60 on an
average day, we might store 18h/24hx105K = 79K Btu of overnight heat in
79K/(120-60) = 1317 pounds of water under the living space ceiling.
Posted by Ed on August 20, 2006, 6:18 am
Very interesting calculations. Thanks. I think you are fairly close in
your assumptions regarding the interior dimensions, except that the
ceiling heights downstairs (1,800 sf) are 10' and the upstairs rooms
(1,200 sf) are 8' ceilings. Seven of the downstairs windows are
Jeffersonian style and have approximately 18sf of surface glass each,
while the remaining 6 windows are about 9sf each and 4 doors average
4sf of glass for each door.
You mentioned, "...a perfect tank heat exchanger..." and that got me to
thinking about the coiled loops in the tank. My idea is to use 1"
polyethelene coiled loops that are each about 4' in diameter. One loop
would be hooked to the solar panels, the other loop would be for the
house. My thoughts are to have one coiled loop at each end of the 8'
insulated concrete tank. Do you believe this is an adequate
arrangement for the configuration of the system or should the coils
perhaps be something like 8' L X 4' W X 2' H oval shaped, with one
coiled loop on the bottom of the tank being heated by the solar panels,
while the other loop of the same size would be affixed toward the top
of the tank and supplying the house? Should the poly pipe size be 1"
or some other diameter?
You also mentioned, "...a large water flow..." and I wondered what you
meant by "large"? Volume? Flow rate?
Am I correct in interpreting your January calculations that I would
have 79K Btu stored in 1,317 pounds of water? If so, is it correct to
roughly extrapolate those calculations to 8,000 pounds of water for a
474K stored BTUs in the tank?
I gather from your calculations, "...so we might heat the house with
105K/875 = 120 ft^2 of glazing ..." that I need 120sf of glazing, which
means I should install 4 ea 4' X 8' collectors instead of three as
presently planned. Is that correct?
Posted by nicksanspam on August 20, 2006, 10:49 am
I'd use a draindown system and try to avoid antifreeze and heat exchangers.
A large enough flow rate to make the temp gain through the collector
less than (say) 10 F. One gpm (8.33x60 = 500 lb/h) can collect 5K Btu/h
with a 10 F temp gain.
Yes, under the ceiling. It might be in 24 10'x4" thinwall PVC pipes.
That would be a seperate system for domestic hot water and cloudy days.
No. I was talking about your new orangerie glazing.