Need help with solar hw options

[Badfish740]

I think your system makes perfect sense me. I assume you're talking about your future long term home, given your comments on the resale value of questionable technology. 1,500 gallons seems like a good amount of storage. You could use less if it was pressurised and had a wider usable temperature range or you could use more and get away with an even lower collector temperature.

My own system is a combination of passive solar home and active solar combined with the boiler in the same system. They don't work at the same time although they could. The storage consists of the concrete, the soil underneath it and the mass of the building, sort of along the lines of annualized geo solar or AGS. It works fairly well, except for the dhw, but I'll get that done some day.

When I start from scratch, I plan to do a combination of Radiantec, doitsolar and AGS. There are lots of great systems out there on the web. Everything from evacuated tubes combined with phase change storage to pop can collectors and corrugated drain tube. To me, the main thing is how much it costs you to get the job done, it doesn't matter if you're storing lots of low temp heat or small amounts of high temp heat. Unless you're talking about butchering hogs (and I have no idea why DaveM mentioned using a hot tub for scalding, wives or otherwise) you don't need 150 degree water if you have enough 110 degree water.

Thanks for the info. I did come up with something interesting but I need some more information in order to make it useful. I found the following formula online for finding heat loss from a tank:

(Square footage of tank surface area) * (Water temp - Ambient temp) / R-value of tank walls = BTU/Hour heat loss

My calculation looks like this:

(224 SF) * (140 degrees F - 60 degrees F) / 30 = 597 BTU/Hour lost which comes to 14,328 BTU lost in a 24 hour period. What I can't seem to find is a calculation that will tell me how to use this number to determine what the temperature of my tank would be over a period of little to no sun. Any ideas?

 

[DaveM195352]

you guys are rough.

The point I was trying to offer (in my feeble way) was to point out. If your solar panels heats the water in a tank to (say) 150 degrees. You should not plan to have all 50 gallons available for use at 150 degrees. As the new water (from street/well) comes into the tank to replace the outgoing hot water, it will cool off the water in the tank.

we have a 50 gallon indirect tank that my solar panels easily heats up to 150-160 degrees. However, I can't not come close to filling our 60 gallon jetted tub and make it a comfortable 110 degrees. Thus, the furnace does come on when the tank gets down to about 120 and finishes off the requirement.

 

[Badfish740]

I may have figured something out on this-but feel free to jump in anytime folks

Ok, so I've determined that assuming tank wall r-values of 30, a basement temperature of 60 degrees F, and a water temperature of 140 degrees F, heat loss should be 14,328 BTU per day. I say should because at first glance this seemed like a large number, but when compared to the actual amount of heat energy that 1500 gallons of water heated to 140 degrees F actually holds, something smells fishy I think:

1500 gallons of water = 12,513 pounds of water, which at 140 degrees F contains a whopping 1,751,820 BTUs! So based on the amount of heat energy I'm starting with, a constant basement temperature of 60 degrees F, and the r-value of 30, after 5 days the water will have lost 71,640 BTUs, which only lowers the amount of heat energy in the tank to 1,680,180 BTUs. 1,680,180 BTUs / 12,513 pounds of water = 134.3 degrees F. I'd love to believe that this tank could stay that warm for five days but my common sense tells me it's not possible. Where am I going wrong/what am I not taking into account? I should also add that this is a completely hypothetical calculation anyway in order to establish a baseline for the tank. Obviously usage would completely throw this off. I'm trying to determine what would happen if you heated the tank to 140 degrees F, then turned the system off (closed the valve to the collector) and let it sit-no hot water usage at all. A 5.7 degree temperature drop in 5 days just doesn't seem possible.

 

[btuser]

150 degree hot water in a bathtub? I'm not making soup!

 

[benjamin]

Badfish

Your math is right on. Temperature is usually just figured relatively, as in 20 degrees change of 1,000 pound of water = about 20,000 btus, because the water still has lots of absolute heat in it even when frozen solid at 0 degrees. But who am I to judge, we're ok with that if you're ok with that.

Remember that the heat that is lost from the tank will warm up your basement, keeping it dry and also reducing the heat loss as the temp of your basement rises. Also it will take a lot of collectors and a lot of time to make up the heat lost and used over those five days.

Funny that DaveM's description of a scalding hot tub inspired me to give a boy scout speech about tempering valves or at least suggest other uses for scalding water, but the tempering valve is probably what is keeping him from being able to utilize all of the hot water in his tank, as the tempering valve dilutes the hot water with cold so he cannot use all of the available hot water in his tank. This is a purely hypothetical example, I don't want anybody to disconnect their tempering valve and risk scalding or even damaging their hot tub.

 

[WES999]

Badfish, you may find this useful
http://www.leaningpinesoftware.com/hot_water_pipes_Newtons_cooling.shtml
I put the formula in to a spread sheet, it may be a bit hard to follow but here it is.

If you want the real spreadsheet I can Email it to you.

I put in your numbers and came up with about 9 °F temp loss/24/hrs.

Electric water heater
CDS8220RT973 est hot water per day GAL 30
D98631765 est effiency % 95
0 H IN 0 0.0 FT^2 BTU to heat 1 gal water 553
0 DIA IN KWH to heat 1 gal water 0.162
KWH to heat 30 gal water 4.86
cost to heat 30 gal .17/KWA $0.80 per day
T(t) Temperature at time t cost for electric water heater $24.10 per month
Ta 60 Ambient temperature (temp of surroundings)
Th 140 Temperature of hot object at time 0 7/22/09 to 8/22/09 32.5 hr
A 224.0 Area FT^2 1.05 hr/day
m 180 gal * 8.34 lb/ga 1501.2 lb 4.725 kwh/day
c 1 1 btu/lb/F for water 0.17 cost of elect
R 30 R value 0.80 cost/day
t 24 Time 17.1

T(t) = TA + (TH - TA)e-(A/(mcR))t So the coefficient A/(mcR) = 37.5/(667*1*16) = 0.00351

Substituting into the equation:

T(t) = 60 + (120 - 60)e-0.00351 t
130.9983 0.004974
0.119371 Setting t=24 hours:
9.0 delta 2.718282
0.887478 T(24) = 115 F

 

[btuser]

Most indirects have about a degree per hour loss. R19 is about the economic break even point of a water heater/storage. After that you better start looking at heat traps and pipe insulation, or so I'm told.

The more I look into solar, the more I think its a great counter-balance to an oil boiler with indirect. I'll make the most of my hot water in the Summer, when its the least efficient to fire the boiler.

 

[mainemac]

BTUSER

I am not sure how far south you are in NH but this outfit in Portland ME comes highly recommended from a friend.

I have been saving money to get a system with them.

Good Luck

PLEASE post what you go with, savings payback etc.

Tom

(broken link removed to http://www.revisionenergy.com/solar-hot-water-solar-power.php)