[Greenbuilding] energy and power terms

[Bob the Futureship]

Thank you for the scientific breakdown Mr. Pine!
I have been watching these calculations for years now
But do not know the formula's or a place for explanation
Of your calculations that I can research further.

My comments in GREEN follow:


Benjamin Pratt <benjamin.g.pratt at gmail.com> wrote:

>My house came with a small "Instant hot water dispenser" to keep 1/2 gallon

>of water at 190 degrees f...

... 1/2 gallon is 0.5/7.48 = 0.067 ft^3, eg a 0.4' cube with about 1 ft^2 of

surface.

1 cubic foot (ft^3) = 7.48051948 US gallon or 7.48 gals/ft^3.  or  .134
ft^3/gal 
your ½ gallon calculation converted to cubic feet measurement
(In the following unit factor analysis calculation 
 the gallons cancel each
other out.) 

 .5 gals x .134 ft^3/gals = .067 ft^3 for a ½ gallon container.

Just to be clear on the math not to be nit picky 


.067 ft^3  = .406 ft x .406 ft x .406 ft = .0669 ft^3   

(.0669 rounded off to .067 ft^3)  and (.406 rounded off to .4 ft)


Surface area of 0.4 cube = 6 sides x (.4 ft x .4ft)   = .96 ft^2 ( rounded
off to 1 ft^2)
   
I was able to figure the previous calculations because of my engineering
background.
But now I need your help with the formula’s you used in the building science
calculations:


>My wife wants to use it this winter and asked me how much it would cost us.

>It is 115v 6.5A 750w, and has 1/2" or so expanded foam insulation but not 
>complete on all sides.

... 1 ft^2 completely surrounded with R2 insulation with no heat loss from 
connecting pipes would have a 0.5 Btu/h-F conductance. At 190 F, it would 
lose 24h(190-70)0.5 = 1440 Btu/day (0.42 kWh/day) to 70 F room air.

Can you tell me the how you get from 1 ft^2 to 0.5 Btu/h-F conductance?
What is the formula and a suggested reference to study the calculation?

And just to be clear 
 what was your conversion factor for Btu’s to Kwh? 


>Our energy cost is about $0.09 kWh

$0.09 KWh x 0.42 KWh/day = $0.0378 per day

So it would cost about 4 cents per day to run. Or less, with more 
insulation?

Gennaro Brooks-Church - Eco Brooklyn <info at ecobrooklyn.com> wrote:

>After building a Passive House in NY I learned it only needed insulation to

>code - R18. What made it or broke it was how well the 6mil plastic air 
>barrier was sealed. Just a one inch hole in the plastic rendered the house 
>unable to heat with the tiny mini-splits.

Two of these $399 9K Btu/h COP 3.5 compressors? 
http://www.minisplitshop.com/store/product.php?productid=261&cat=0&page=1

The 99% winter design temp in NYC is 12 F, ie it's warmer 99% of the time. A

16' tall 70 F house with a 1" hole at the top and bottom would have about 
16.6/144sqrt(16'(70-12)) = 3.5 cfm of natural air leakage with about 
3.5(70-12) =  204 Btu/h (60 watts) of heat loss.

I could not follow these calculations? 
99% winter design temp ? Is this a worst case scenario and where did you get
this number?
16’ tall 70 F house?  I don’t understand the 16’ tall house
16.6 / 144sqrt(16’(70-12))  ?   Can you tell me what the formula is for this
calculation?
3.5(70-12) ?  Can you tell me what formula this is?


>Seal the hole and you can heat the house with a hair drier.


If I have not tried your patience up to this point then explanation of the
formula’s used in these
calculations would be tremendous! If possible a reference to where I can
found explanation
of formula’s. 

If the compressors supply 18K Btu/h and the house needs 18K - 204 = 17796 
Btu/h (5216 W) without the hole, you could heat it with 5216/1500 = 3.5 1500

watt hair dryers.

OTOH, NREL says 980 Btu/ft^2 falls on a south wall on an average 31.5 
January day in December. If the house with the hole has a 18K/(70-12) = 310 
Btu/h conductance, and it needs 24h(70-31.5)310 = 287K Btu/day, and 600 
kWh/mo of indoor electrical use provides 68K of that, and 1 ft^2 of R2 
twinwall polycarbonate with 80 F air inside gains 0.8x980-6h(80-31.5)/R2 = 
639 Btu/day, we can heat it on an average December day with 342 ft^2 of 
solar siding with an infinite COP. It needs about 1.1 million Btu for 5 
cloudy days in a row, which might come from 1.1M/(140-80) = 18K pounds (292 
ft^3) of water cooling from 140 to 80 F in a 4' tall x 10' diameter sheet 
metal swimming pool. R30 insulation would lower the house conductance to 186

Btu/h-F and allow using 162 ft^2 of solar siding and a 4' tall x 3' diameter

tank.

Corwyn <corwyn at midcoast.com> wrote:

>... Microwave is almost certainly the most efficient way to heat water;

I'd vote for an electric teakettle with an automatic shutoff.

Nick 

Thank you  very much,
Bob Jakaitis aka “Jake at the Lake”




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