# An Alberta bungalow

Posted by nicksanspam on September 6, 2007, 1:02 pm

A 1024 ft^2 house with a 640 ft^2 loft might look like this, in a fixed font:

.
.   fan .
.               .
.               . 12'
8'.               .
.               .
. . . . . .     .
.               .
8'.               . 8'
.               .
.               .
..........................
32'

. . . . . . . . .
.         .     .
.         .     .
.         .     .
.         .     .
.         . fan . 32'
.         .     .
.         .     .
.         .     .
.         .     .
. . . . . . . . .

If made entirely of R32 8" Structural Insulated Panels (18 SIPs?)
with 1024 ft^2 of ceiling and 2304 ft^2 of walls and no air leaks,
its thermal conductance would be 3328ft^2/R32 = 104 Btu/h-F.

The 1981 NRC Solarium Workbook says a south wall in Edmonton gets 2369
Wh/m^2 (751 Btu/ft^2) of sun on an average -10 C (14 F) December day.

At an average 65 F indoors, we need 24h(65-14)104 = 127K Btu of heat.
A frugal 300 kWh/mo of indoor electrical use could provide 34K of that,
leaving 93K Btu heat from 480 ft^2 of \$/ft^2 Thermaglas Plus U0.58
twinwall polycarbonate "solar siding" with 80% solar transmission over
a 1' air gap over a dark south wall.

If the siding supplies 93K Btu/day in 6 hours at an average temp T
and loses the rest of the solar heat to the outdoors and 0.8x480x751
= 288K Btu = 93K + 6h(T-14)480x0.58, T = 131 F, hot enough for showers.
Adding more twinwall would raise this temp.

If the house is 70 F for 12 hours per day and 60 for the other 12 and
sunspace air keeps it 70 F for 6 hours with 6h(70-14)104 = 35K Btu,
we need to store 93K - 35K = 58K Btu of "overnight heat."

If the house has, say, 4000 Btu/F of thermal mass, including furnishings,
(3328 ft^2 of 1/2" drywall would have 1664), that would store 4K(70-60)
= 40K Btu of overnight heat, leaving 58K-40K = 18K Btu. (If it were
70 - 58K/4K = 56 vs 60 F for 12 hours, the mass could store 58K Btu,
but with the lower night temp, we wouldn't need that much heat :-)

We might warm 12x32 = 384 ft^2 of shiny cathedral ceiling mass with sunspace
air to store 18K Btu. At 60 F, the house needs (60-14)104 = 4784 Btu/h. With
a 1.5 Btu/h-F-ft^2 slow-moving airfilm conductance, 60+4784/(384x1.5) = 68 F
mass could keep the house 60 F. A slow ceiling fan and a setback thermostat
could keep the room exactly 70 F for 12 hours and exactly 60 F for the other
12 hours on an average December day.

How much ceiling mass? Here's an equivalent circuit for the sunspace:

288K/6 = 48K Btu/h    (35K+40K)/6h = 12.5K Btu/h
---                    ---
|---|-->|------------------|-->|--- 70 F
---             |      ---
|
1/480x0.58 = 1/278   |
|   1/(384x1.5) = 1/576
14 F ---www--------------------www---------------
|  C Btu/F
---
---
|
-

If the sunspace provides 35K Btu to keep the house warm and 40K Btu
to recharge its mass from 60 to 70 F, that removes 12.5K Btu/h from
the sunspace current source, so we have:

35.5K Btu/h
---
|---|-->|------------
---             |
|
1/278            |     1/576
14 F ---www--------------------www---------------
|  C Btu/F
---
---
|
-
which is equivalent to:

1/278                  1/576
---www--------------------www---------------
|                                            |
| 14+ 35.5K/278 = 142 F                      |  C Btu/F
---                                          ---
-                                           ---
|                                            |
-                                            -

which is equivalent to:

1/188
-----www----- T
|           |
| 142 F     |  C Btu/F
---         ---
-          ---
|           |
-           -

If C warms from 68 to T and stores 18K Btu in 6 hours, T = 68+18K/C
= 142+(68-142)e^-(6/RC), so 18K/C = 74(1-e^-(6/RC)). RC = C/188, so
C = 243/(1-e^-(1128/C)). Plugging in C = 250 Btu/F on the right makes
C = 246 on the left, then 245 Btu/F, with T = 140 F. We might have 384 ft^2
of 1/2" SIP OSB with 0.31x384 = 119 Btu/F plus (246-119)/0.5 = 127 ft^2
of 1/2" drywall or another 384 ft^2 layer of roof OSB with foil beneath.

An automobile radiator in a vertical duct that returns air to the sunspace
without mixing with room air could make solar hot water and store it in
an unpressurized tank. On cloudy days, we could pump tank water up through
the radiator to warm the house. A pipe coil heat exchanger in the tank could
make hot water for showers, with the help of a graywater heat exchanger.

Nick

Posted by bealiba on September 7, 2007, 7:50 am

It really is a sad joke. Nick could really be right, BUT, only if the
house is not used for anything.

three children (aged 7, 10, 17)

and a medium sized dog and a standard sized cat.

Two sixty watt lamps in each room for an average of 3 hours/day.

Oven in kitchen 2Hrs/day

Two TVs, 5 hours each

Play station 3 hrs/day

What the hell are you going to do with all this heat?

Posted by nicksanspam on September 8, 2007, 8:31 am

... 500Wx12h/day = 6 kWh/day

+ 124Wx18h = 8.2 kWh for an ASHRAE-standard 50 lb dog and 6.61 lb cat

+ 2 14W CFLs x 3h = 8.3 kWh

+ 1500 W toaster oven x 5 min = 8.4 kWh

+ 2 flat screens x 20W x 3h = 8.5 kWh

+ 2 books x 0W = 0 Wh/day = 8.5 kWh, totaling 255 kWh/mo.

That's less than the 300 kWh/mo assumed below :-)

And 255 would only provide 29K, leaving 98K Btu of heat from 480 ft^2
of \$/ft^2 Thermaglas Plus U0.58 twinwall polycarbonate "solar siding"
with 80% solar transmission...

If the siding supplies 98K Btu/day in 6 hours at an average temp T
and loses the rest of the heat to the outdoors and 0.8x480x751
= 288K Btu = 98K + 6h(T-14)480x0.58, T = 128 F.

If the house is 70 F for 12 hours per day and 60 for the other 12 and
sunspace air keeps it 70 F for 6 hours with 6h(70-14)104 = 35K Btu,
we need to store 98K - 35K = 63K Btu of "overnight heat."

...4000 Btu/F of inherent thermal mass could store 4K(70-60) = 40K Btu
of overnight heat, leaving 63K-40K = 23K Btu.

Let's warm 2 shiny 4'x8' platforms with 128 ft^2 of 2-sided surface hanging
below the ceiling with sunspace air to store 23K Btu. At 60 F, the house
needs 4784 Btu/h. With a 1.5 Btu/h-F-ft^2 slow-airfilm conductance,
60+4784/(128x1.5) = 85 F mass could keep the house 60 F.

Here's an equivalent circuit for the sunspace and platforms:

288K/6 = 48K Btu/h    (35K+40K)/6h = 12.5K Btu/h
---                    ---
|---|-->|------------------|-->|--- 70 F
---             |      ---
|
1/480x0.58 = 1/278   |   1/(128x1.5) = 1/192      C Btu/F...
|                         ||
14 F ---www--------------------www----------------||---|
||
ie         1/114
-----www----- T
|           |
| 142 F     |  C Btu/F
---         ---
-          ---
|           |
-           -

How could a meticulous solar consultant like George miss that mistake? :-)

If the platforms warm from 85 to T and store 23K Btu in 6h, T = 85+23K/C
= 142+(85-142)e^-(6/RC), so 23K/C = 57(1-e^-(6/RC)). RC = C/114, so
C = 404/(1-e^-(684/C)). Plugging in C = 500 Btu/F on the right makes
C = 542 on the left, then 564, 575, 581, and 584 Btu/F, with T = 124 F
and 292 pounds of water 2" deep in a poly film duct on each platform.

An automobile radiator in a vertical duct that returns air to the sunspace
without mixing with room air could make solar hot water and store it in
an unpressurized tank. On cloudy days, pump water up through the radiator
to warm the house. Make hot water for showers with a \$0 1"x300' PE pipe
coil heat exchanger in the tank and a graywater heat exchanger.

At dawn with no internal heat gain, the house needs about (70-14)104 = 5824
Btu/h. An 800 Btu/h-F radiator could do this with 70+5824/800 = 77 F tank
water. We need 5x98K = 490K Btu for 5 cloudy days in a row. If the tank is
140 F on an average day, with the help of some Big Fins behind the twinwall,
we need 490K/(140-77) = 7.8K pounds of water, eg a 4'x8'x4'-tall plywood
tank lined with a single folded 12'x16' piece of EPDM rubber.

Nick

Posted by bealiba on September 8, 2007, 12:53 pm
On Sep 8, 6:31 pm, nicksans...@ece.villanova.edu wrote:

No Nick, the question is "what about the heat".

Heat - noun: A form of energy that is transferred by a difference in
temperature.

Doors opening, movement of people and animals in and out of the house,
appliances being used, you know, all of the heat that you have not
included in your calculations. All the heat that is associated with an
occupied house.

An empty house is one thing, a house being lived in is another.

Show us the calculations for the energy use of an occupied house.

Posted by nicksanspam on September 9, 2007, 11:15 am
At an average 65 F indoors, we need 24h(65-14)104 = 127K Btu of heat.
If a frugal 300 kWh/mo of indoor electrical use provides 34K Btu and
a 4'x8'x4'-tall R30 140 F plywood heat storage tank in the house loses
24h(140-65)128ft^2/R30 = 8K, we need another 85K Btu of solar heat.

If the house is 70 F for 12 hours per day and 60 at "night" and sunspace
air keeps it 70 F for 6 hours with 6h(70-14)104 = 35K Btu, we might store
85K-35K = 50K Btu of "overnight heat" in 4000 Btu/F of inherent mass with
4K(70-60) = 40K Btu, and 50K-40K = 10K Btu in 384 ft^2 of shiny ceiling
mass C that cools from 80 to 60 F, with C = 10K/(80-60) = 500 Btu/F.

If sun arrives at a constant 48K Btu/h and we subtract 35K/6h = 5.8K
for sunspace air that warms the house, we can model the sunspace
itself like this, with air temp T, viewed in a fixed font:

42.2K Btu/h
---                                              1/278
|---|-->|---------------- T     -----------------------www--- T
---             |         | 14+42.2K/278 = 166F
|    -   ---
1/480x0.58 = 1/278   |    -    -
|    -    |
14 F ---www-------------          -

If RC = 4500/278 = 16.2 hours with lots of airflow, we can charge
the house and ceiling from 60 to 70 F in -16.2ln((70-166)/(60-166))
= 1.6 hours:

1/278
---------www----------------------- T 60->70 F
| 166 F             | 4000    | 500
---                 ---       ---
-                  ---       ---
|                   |         |
-                   -         -

With RC = 500/278 = 1.8 hours, we can charge the ceiling alone from
70 to 80 in another -1.8ln((80-166)/(70-166)) = 0.2 hours, leaving
4.2 hours like this

1/278  T = 147   1/800
-----------www--------------www-----
| 166 F           I -->              | 140
---      (166-140)/(1/278+1/800)     ---
-            = 5364 Btu/h            -
|                                    |
-                                    -

with an 800 Btu/h-F auto radiator and fan at the top of a duct that returns
air to the sunspace without mixing with room air and stores 4.2I = 22.5K
Btu/day of 140 F water in the tank. Subtracting the 8K tank loss leaves
14.5K. With 60% greywater heat recovery, a pressurized \$0 1"x300' PE pipe
coil in the tank could make 14.5K/(1-0.6) = 36K Btu/day of hot water.

We could do this more efficiently and simultaneously with Big Fins
in the sunspace, but that would cost more money.

Nick

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 Re: An Alberta bungalow nicksanspam 09-08-2007
 Re: An Alberta bungalow nicksanspam 09-09-2007
 Re: An Alberta bungalow wmbjkREMOVE 09-09-2007
 Re: An Alberta bungalow pbrannen 09-09-2007
 Re: An Alberta bungalow beemerwacker 09-11-2007
 Re: An Alberta bungalow wmbjkREMOVE 09-11-2007
 Re: An Alberta bungalow wmbjkREMOVE 09-12-2007
 Re: An Alberta bungalow Anthony Matonak 09-13-2007
 Re: An Alberta bungalow Anthony Matonak 09-13-2007
 Re: An Alberta bungalow wmbjkREMOVE 09-14-2007
 Re: An Alberta bungalow wmbjkREMOVE 09-13-2007
 Re: An Alberta bungalow nicksanspam 09-13-2007
 Re: An Alberta bungalow nicksanspam 09-15-2007
 Re: An Alberta bungalow nicksanspam 09-12-2007