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Heat loss calculation questions, very long and technical

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Posted by The Bald Ass Prairie Farm on September 2, 2004, 5:22 pm
 
Hi,
another question of the Bald Ass Prairie Farm, I hope you can help me out
here, I did some searching and I'm trying to calculate the heat loss of my
house (to build) so I can adjust my Solar warm water installation and
eventually my photo-voltaic solar panels.
I did the math in a spreadsheet and calculated for the wall/floor/roof etc.:

                                                  Thick             Lambda
Ri
Concrete wall         Thick             in " Metre     W/(m.K)
concrete 7"                 7"             0.1778             1.70
0.105
insulation                      4"               0.1016         0.04
2.540
airspace                         1"                                  0.30
0.170
plasterboard                 0.5"             0.0127         1.70
0.007
aircontact I                                                           0.27
0.270
total R
3.092

wall
aircontact e
0.03         0.030
siding    0.170
insulation                     9"             0.2286                 0.04
5.715
airspace                     1"
0.17         0.170
plasterboard             0.5"             0.0127                 1.70
0.007
aircontact I
0.27         0.270
total R
6.362

ceiling
aircontact e                                                           0.03
0.030
insulation                     10"             0.254                 0.04
6.350
airspace                         1"
0.15          0.150
plasterboard                 0.5"         0.0127                 1.70
0.007
aircontact I
0.27         0.270
total R
6.807

Basement floor
aircontact I
0.27         0.270
concrete 7"                 7"             0.1778                  1.70
0.105
insulation                     1"             0.025                     0.35
0.073
wood                         3/4"         0.01905                 0.50
0.038
total R
0.485

Windows average
aircontact e
0.03         0.030
glass                         1/8"              0.00318                0.08
0.040
space/gas                 3/8"
0.30         0.300
glass                         1/8"              0.00318                0.08
0.040
aircontact I
0.27         0.270
total R
0.679

shutters
aircontact e
0.03         0.030
wood                     1/4"                 0.00635                 0.50
0.013
insulation                 3"                     0.8128
0.35         2.322
wood                     1/4"                 0.00635                 0.50
0.013
aircontact I
0.27         0.270
total R
2.648

porch floor
aircontact I
0.27         0.270
concrete 7"             7"                     0.1778                  1.70
0.105
insulation                 3"                     0.8128
0.35         2.322
total R
2.697

As I had to go from Website to Website the above is just what I found and
concluded from what I think is right, is it ? or close to what is right?
Then I did the math with the square metre thing to calculate what I will be
losing in wintertime. Also done in a spreadsheet.

                                     R                 k=1/R             M2
Ti         To         Ti-T2         Q in W         in% of total
Basement floor          0.485             2.061             93.65
15         7             8                  247.29               9 %
walls in soil                3.092             0.323             64.72
18         3            15                 87.32                 3 %
rest wall basement     3.092             0.323             38.83
18      -30            48                 279.43             10 %
walls                         6.362             0.157             271.53
20     -30             50             157.17                 6 %
windows                   0.679             1.472             65.14
20      -30             50             1471.94             52 %
roof                          6.362              0.157            71.81
20      -30             50              157.17                6 %
ceiling                       6.807              0.147             88.02
20       -30             50             146.90                 5 %
floor porch               2.697               0.371            10.06
15       -10             25             139.05                 5 %
floor breakfast nook 6.362               0.157            10.06
   20        -30            50             157.17                 6 %

total house
2843.44 Watt/sec
The windows are a big leak, I'll go for shutters.
Now the question(s)
This seems like an awful lot 2.8 KW per second, but as my gas bill is in BTU
it doesn't tell me anything useful. I haven been able to find the missing
link to go from Watt/second to BTU or a way to check weather the calculation
principle is correct. Any smart people on line today?.
As I want to heat (in floor heating by water) the house by (solar) warm
water with an additional heater on natural gas what would be the right
amount of storage water?.
Searching the net I found this.
Heat accumulation W== . c = in Joule, where is the mass of the material
an c is a coefficient specific for that material in Joule per cubic metre
per degree Kelvin. For water that would be 4.18*10  Joule / kg Kelvin. Or
it takes 4.18 Joule to heat up one litre (=1/1000 of a cubic metre) one
degree. There should be a link between Joule and Watt, but I haven't found
it yet, any ideas?
I thought when I can calculate the heat loss of the whole house over a 1
week period and I have a storage of that amount of heat in my water tank of
X gallons then only in case of sudden temperature changes my auxiliary heat
source will kick in. Right? And how do I relate Joule to BTU? I love the
struggle between Imperial and Metric values, but it gets a bit confusing and
frustrating when you think you can and then you can't calculate.
Thanks for reading anyway and if you can throw some light, be my lighthouse.

Richard




--
The Bald Ass Prairie Farm
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Posted by The Bald Ass Prairie Farm on September 2, 2004, 5:26 pm
 
SNIP    If I had know that it would look like this I would have done it
different Sorry

The Bald Ass Prairie Farm
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Posted by The Bald Ass Prairie Farm on September 2, 2004, 7:16 pm
 Found finally part of the answer at http://www.simetric.co.uk/sibtu.htm
"SNIP"
The rest is still open for improvement :)

Thanks Richard



Posted by Gunnar on September 2, 2004, 10:27 pm
 Some conversion factors: 1 GJ = 277.8 kWh = 947817 BTU



lighthouse.


Posted by Gary on September 3, 2004, 2:43 am
 The Bald Ass Prairie Farm wrote:

 To         Ti-T2         Q in W         in% of total

 7             8          247.29             9 %

 3            15          87.32              3 %

-30            48         279.43             10 %

 -30          50         157.17              6 %

-30           50         1471.94             52 %

-30             50       157.17               6 %

-30            50         146.90               5 %

-10           25         139.05               5 %

-30           50         157.17               6 %

Its just 2843 watts -- watts are already a RATE of heat loss, Watt/sec
does not make much sense.

The watt is a rate of using energy, and the BTU is a quantity of
energy -- they are not directly comparable.  It would be like trying
to compare gallons/minute to liters.

A 2843 watt heat loss rate means that you would use 2842 watt-hours
or, 2.843 kilowatt hours of energy in each hour to heat the house.
There are 3412 BTU per KWhr, so 2.84 KWhr = 9700 BTU/hr.
I'm not that familiar with the metric R units in your calc above, but
this works out to about 110 BTU/hr-F, which does not seem too
unreasonable for a house of about 1000 ft^2 (88 m^2).

I don't know where you live, but the -30C (or -22F) looks more like a
design temperature than like an average daily temperature.  If this is
your design temperature, than it would represent the maximum heat loss
rate that your house would ever experience -- it would be appropriate
to use this to size your furnace, but not to estimate average heat loss.

It does not look like you have accounted heat loss due to
infiltration, which can be a significant part of your total heat loss,
depending on how tight the house is.



A Joule is a quantity of energy and the watt is a rate of using
energy, so they are not comparable, but you can compare KWhr or BTU to
the Joule:

1,000,000 Joules = 1 megaJoule = 0.278 KWhr = 947 BTU

This site has unit conversions from just about anything to just about
any (compatible) thing:
http://www.onlineconversion.com/energy.htm




One simple way to look at it is that you need to have a balance
between the solar energy you collect, and the heat loss from your
house on an "average" day -- this will involve some form of storage
for the night time period, since the solar gain comes all in about 6
hours, and the loss is over all 24 hours.  In addition (if you want a
high fraction of solar heating), you need to have some additional
storage to tide you over some number of cloudy days, and there has to
be solar collection associated with charging this storage.
You might start to get a feel for this by looking up the weather and
solar radiation data for your area for the mid-winter months.  If you
live in the US, the NREL site has this data for 240 or so cities.

Gary

Right? And how do I relate Joule to BTU? I love the


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