Posted by *zoe_lithoi* on December 10, 2013, 5:37 am

Greetings,

How does a family of 5 (3 adults, 2 kids) keep warm in 20degF temperature i

n their modular home when they cannot afford electricity or propane?

Let's put all 5 of them in a 10x12', R20 bedroom. Then let's put 2 layers o

f $1 R7.7 4'x8' insulation foam board on the ceiling and 4 walls. The mat

tresses would take up about the entire floor providing approix. R15 on the

floor.

The calculations after my signature indicate theoretically, that after 1 ho

ur the room temperature is 51, after 2 hours it is 63, and after 3 hours, i

t is 69degF.

Home Depot has R7.7 foam board insulation for $1 (R-Tech 2 in. x 4 ft. x 8

ft)

2 layers of this would be ~R15. To cover the walls and ceiling of a 10'*12'

room with 2 layers of this foam board insulation, you would need about 30

boards, i.e. $00. Ouch. BUT it's better than paying the 'typical' $00 per

month for the heating bill for a 6 month winter ($800).

-------------------

To heat this room to 70degF without adding any insulation would require a 5

00Watt space heater (plus the 5 people). I'm guessing it's about 20cents pe

r kilowatt, so the cost would be 10 cents per hour, or $.40 per day or $2

pre month or $32 for a 6 month winter.

Toby

The surface area of the 4 walls, floor, and ceiling would be:

A = 4*8*12 + 2*8*10 + 2*10*12 = 2^4 * [24+10+15*49 ~= 16*50 =

800sqft

There is 300 Btu/hr for a sedentary adult, so let's estimate, with kids, 25

0btu/hr and average, i.e. the amount of heat entering the room from sleepin

g people:

H = 5*250Btu/hr = 1250 Btu/hr

The Heat leaving the room, of Temperature, Trm, to the 20degF outside thru

the 800sqft of R20 walls, floor, and ceiling is:

(Trm - 20)degF*800sqft / R20 hr-sqft-degF/Btu] = (Trm - 20)* 40 Btu/hr

The amount of heat the air in the 1000cuft (8'*10'*12' = ~1000cuft) room

absorbs to go from 20degF to Trm is:

(Trm - 20)degF * 1/55 Btu/F /cuft * 1000cuft ~= (Trm - 20) * 20 Btu

I guess in the first hour this would be:

(Trm - 20) * 20 Btu/hr

The heatflow equation then, is:

1250 Btu/hr = (Trm - 20) * 20 Btu/hr + (Trm - 20)* 40 Btu/hr

1250 = (Trm - 20) * 60

Trm = 1250/60 + 20 = 41F

Since the room is now heated to 41F, the 2nd hour would be:

1250 Btu/hr = (Trm - 41) * 20 Btu/hr + (Trm - 20)* 40 Btu/hr

1250 = 60Trm - 820 -800

Trm = [2870]/60 = 48degF

Since the room is now heated to 48F, the 3rd hour would be:

1250 Btu/hr = (Trm - 48) * 20 Btu/hr + (Trm - 20)* 40 Btu/hr

1250 = 60Trm - 960 -800

Trm = [3010]/60 = 52degF

The 4th hour might be 55, the 5th hour might be 56 or 57....

Halfway thru the night, it's still cold

---------------------------------------------------------------------------

--

Suppose we added R15 foam insulation to the ceiling and walls. We'll assum

e that the mattresses covering the floor add R15 to the floor, so that now,

we have R20 + R15 = R35 everywhere

---------------------------------------------------------------------------

--

(Trm - 20)degF*800sqft / R35 hr-sqft-degF/Btu] = (Trm - 20)* 23 Btu/hr

The heatflow equation then, is:

1250 Btu/hr = (Trm - 20) * 20 Btu/hr + (Trm - 20)* 23 Btu/hr

1250 = (Trm - 20) * 43

Trm = 1250/43 + 20 = 49F

Since the room is now heated to 49F, the 2nd hour would be:

1250 Btu/hr = (Trm - 49) * 20 Btu/hr + (Trm - 20)* 23 Btu/hr

1250 = 43*Trm - 980 - 460

Trm = 2690/43 = 63F

Since the room is now heated to 63F, the 3rd hour would be:

1250 Btu/hr = (Trm - 63) * 20 Btu/hr + (Trm - 20)* 23 Btu/hr

1250 = 43*Trm - 1260 - 460

Trm = 2970/43 = 69F

After the 3rd hour, the room is comfortable (69degF).

---------------------------------------------------------------------------

--

Suppose we added R20 foam insulation to the ceiling and walls. We'll assum

e that the mattresses covering the floor add R20 to the floor, so that now,

we have R40 everywhere

---------------------------------------------------------------------------

--

(Trm - 20)degF*800sqft / R40 hr-sqft-degF/Btu] = (Trm - 20)* 20 Btu/hr

The heatflow equation then, is:

1250 Btu/hr = (Trm - 20) * 20 Btu/hr + (Trm - 20)* 20 Btu/hr

1250 = (Trm - 20) * 40

Trm = 1250/40 + 20 = 51F

Since the room is now heated to 51F, the 2nd hour would be:

1250 Btu/hr = (Trm - 51) * 20 Btu/hr + (Trm - 20)* 20 Btu/hr

1250 = 40*Trm - 1020 - 400

Trm = 2670/40 = 67F

Since the room is now heated to 67F, the 3rd hour would be:

1250 Btu/hr = (Trm - 67) * 20 Btu/hr + (Trm - 20)* 20 Btu/hr

1250 = 40*Trm - 1340 - 400

Trm = 2990/40 = 75F

After the 2nd hour, the room is comfortable (67degF). After the 3rd hour, t

he room is getting toasty (75degF).

-----------------------

2 layers of the R7.7 4'x8' 2inch thick, foam board on the 4 walls and ceili

ng works out to about 30 boards.

2 layers * [2*8*10 + 2*8*12 + 10*12]/(4*8)

= 2 layers * 8*[20 + 24 + 15]/(4*8)

= [20 + 24 + 15]/2)

= 29.5 boards

-------------------------------------

How big of a space heater would you need if you didn't insulate the room at

all?

Let's say we wanted the room to be 70degF. and the heat from the space heat

er is Hsh

The heatflow equation given above becomes:

Hsh + 1250 Btu/hr = (70 - 20) * 20 Btu/hr + (70 - 20)* 40 Btu/hr

Hsh + 1250 = 50 * (40+20)

Hsh + 1250 = 50 * (60)

Hsh + 1250 = 3000

Hsh = 3000 - 1250 = 1750 Btu/hr = 1750 Btu/hr * 0.293 Watts/BTU/hr

= 512Watts

Posted by *Morris Dovey* on December 10, 2013, 6:37 am

On 12/9/13 11:37 PM, zoe_lithoi wrote:

*> How does a family of 5 (3 adults, 2 kids) keep warm in 20degF *

*> temperature in their modular home when they cannot afford electricity *

*> or propane? *

Depending on how the home is sited, you might be able to try something

like this: http://www.iedu.com/Solar/Panels/index.html

--

Morris Dovey

http://www.iedu.com/Solar/

Posted by *zoe_lithoi* on January 10, 2014, 5:56 pm

Greetings,

It appears there are many online articles claiming they can, along a turned

on computer, for example, heat a room with candles... and include making a

radiant heater out of 4 'tea candles' costing 1 british pound ($.65) for

100 of them, the candles burn for about 4 hours. Is this fact or fiction? O

ne of the complaints was that no thermal calculations were done to support

said claim, taking into account the size and insulation of the room, nor re

lating it's performance to how cold it is outside.

My calc's below show that the claim is reasonable for a well-insulated room

(R20) heated by 5 people and 4 British Tea candles.... 58degF for the 1st

hour, and 70degF for the next.

One video is at:

http://www.whydontyoutrythis.com/2013/11/how-to-easily-heat-your-home-using

-flower-pots-and-tea-lights.html

The 'Characterization of Candle Flames' paper can be read at:

http://fire.nist.gov/bfrlpubs/fire05/PDF/f05141.pdf

The candle was calculated as 77±9Watt which can be converted to be 263 BT

U/h

So, 4 candles would yield about 1000 Btu/hr for 4 hours. Our goals is to ma

ke it thru an 8 hour nite, so we would need 8 candles per nite which would

cost 8 *$.64/100 = 0.13 (13 cents), so over 1 month, it would cost about

50cents.

Let's add candle light heat to the calculations I made earlier in this thre

ad about heating a a 10x12', R20 bedroom with the heat from 5 people, total

ing about 1250Btu/hr with an outside temperature of 20degF.... resulting in

a 41F room temperature. So the 4-candle heater 'claim' doesn't seem like i

t would do much good on a 20degF night.... even if combined with heat from

an approix. 200 Watt (680btu/hr) computer.

Calculations are below my signature.

Toby

From before, the surface area of the 4 walls, floor, and ceiling would be:

A = 4*8*12 + 2*8*10 + 2*10*12 = 2^4 * [24+10+15*49 ~= 16*50 =

800sqft

There is 300 Btu/hr for a sedentary adult, so let's estimate, with kids, 25

0btu/hr and average, i.e. the amount of heat entering the room from sleepin

g people:

H (people) = 5*250Btu/hr = 1250 Btu/hr

H (4 candles) = 1000 Btu/hr

H (total) = 2250 Btu/hr

The Heat leaving the room, of Temperature, Trm, to the 20degF outside thru

the 800sqft of R20 walls, floor, and ceiling is:

(Trm - 20)degF*800sqft / R20 hr-sqft-degF/Btu] = (Trm - 20)* 40 Btu/hr

The amount of heat the air in the 1000cuft (8'*10'*12' = ~1000cuft) room

absorbs to go from 20degF to Trm is:

(Trm - 20)degF * 1/55 Btu/F /cuft * 1000cuft ~= (Trm - 20) * 20 Btu

I guess in the first hour this would be:

(Trm - 20) * 20 Btu/hr

The heatflow equation then, is:

2250 Btu/hr = (Trm - 20) * 20 Btu/hr + (Trm - 20)* 40 Btu/hr

1250 = (Trm - 20) * 60

Trm = 2250/60 + 20 = 57.5F

Since the room is now heated to 57.5F, the 2nd hour would be:

2250 Btu/hr = (Trm - 57.5) * 20 Btu/hr + (Trm - 20)* 40 Btu/hr

2250 = 60Trm - 1150 -800

Trm = [4200]/60 = 70degF

Toby

On Monday, December 9, 2013 9:37:00 PM UTC-8, zoe_lithoi wrote:

*> Greetings, *

*> *

*> *

*> *

*> How does a family of 5 (3 adults, 2 kids) keep warm in 20degF temperature *

in their modular home when they cannot afford electricity or propane?

*> *

*> *

*> *

*> Let's put all 5 of them in a 10x12', R20 bedroom. Then let's put 2 layers *

of $1 R7.7 4'x8' insulation foam board on the ceiling and 4 walls. The m

attresses would take up about the entire floor providing approix. R15 on th

e floor.

*> *

*> *

*> *

*> The calculations after my signature indicate theoretically, that after 1 *

hour the room temperature is 51, after 2 hours it is 63, and after 3 hours,

it is 69degF.

*> *

*> *

*> *

*> Home Depot has R7.7 foam board insulation for $1 (R-Tech 2 in. x 4 ft. x *

8 ft)

*> *

*> *

*> *

*> 2 layers of this would be ~R15. To cover the walls and ceiling of a 10'*1 *

2' room with 2 layers of this foam board insulation, you would need about 3

0 boards, i.e. $00. Ouch. BUT it's better than paying the 'typical' $00 p

er month for the heating bill for a 6 month winter ($800).

*> *

*> *

*> *

*> ------------------- *

*> *

*> *

*> *

*> To heat this room to 70degF without adding any insulation would require a *

500Watt space heater (plus the 5 people). I'm guessing it's about 20cents

per kilowatt, so the cost would be 10 cents per hour, or $.40 per day or $

72 pre month or $32 for a 6 month winter.

*> *

*> *

*> *

*> Toby *

*> *

*> *

*> *

*> *

*> *

*> The surface area of the 4 walls, floor, and ceiling would be: *

*> *

*> *

*> *

*> A = 4*8*12 + 2*8*10 + 2*10*12 = 2^4 * [24+10+15*49 ~= 16*50 *

= 800sqft

*> *

*> *

*> *

*> There is 300 Btu/hr for a sedentary adult, so let's estimate, with kids, *

250btu/hr and average, i.e. the amount of heat entering the room from sleep

ing people:

*> *

*> *

*> *

*> H = 5*250Btu/hr = 1250 Btu/hr *

*> *

*> *

*> *

*> The Heat leaving the room, of Temperature, Trm, to the 20degF outside thr *

u the 800sqft of R20 walls, floor, and ceiling is:

*> *

*> *

*> *

*> (Trm - 20)degF*800sqft / R20 hr-sqft-degF/Btu] = (Trm - 20)* 40 Btu/hr *

*> *

*> *

*> *

*> The amount of heat the air in the 1000cuft (8'*10'*12' = ~1000cuft) roo *

m absorbs to go from 20degF to Trm is:

*> *

*> *

*> *

*> (Trm - 20)degF * 1/55 Btu/F /cuft * 1000cuft ~= (Trm - 20) * 20 Btu *

*> *

*> I guess in the first hour this would be: *

*> *

*> *

*> *

*> (Trm - 20) * 20 Btu/hr *

*> *

*> *

*> *

*> The heatflow equation then, is: *

*> *

*> *

*> *

*> 1250 Btu/hr = (Trm - 20) * 20 Btu/hr + (Trm - 20)* 40 Btu/hr *

*> *

*> 1250 = (Trm - 20) * 60 *

*> *

*> Trm = 1250/60 + 20 = 41F *

*> *

*> *

*> *

*> Since the room is now heated to 41F, the 2nd hour would be: *

*> *

*> *

*> *

*> 1250 Btu/hr = (Trm - 41) * 20 Btu/hr + (Trm - 20)* 40 Btu/hr *

*> *

*> 1250 = 60Trm - 820 -800 *

*> *

*> Trm = [2870]/60 = 48degF *

*> *

*> *

*> *

*> Since the room is now heated to 48F, the 3rd hour would be: *

*> *

*> *

*> *

*> 1250 Btu/hr = (Trm - 48) * 20 Btu/hr + (Trm - 20)* 40 Btu/hr *

*> *

*> 1250 = 60Trm - 960 -800 *

*> *

*> Trm = [3010]/60 = 52degF *

*> *

*> *

*> *

*> The 4th hour might be 55, the 5th hour might be 56 or 57.... *

*> *

*> Halfway thru the night, it's still cold *

*> *

*> *

*> *

*> ------------------------------------------------------------------------- *

----

*> *

*> Suppose we added R15 foam insulation to the ceiling and walls. We'll ass *

ume that the mattresses covering the floor add R15 to the floor, so that no

w, we have R20 + R15 = R35 everywhere

*> *

*> ------------------------------------------------------------------------- *

----

*> *

*> *

*> *

*> (Trm - 20)degF*800sqft / R35 hr-sqft-degF/Btu] = (Trm - 20)* 23 Btu/hr *

*> *

*> *

*> *

*> The heatflow equation then, is: *

*> *

*> 1250 Btu/hr = (Trm - 20) * 20 Btu/hr + (Trm - 20)* 23 Btu/hr *

*> *

*> 1250 = (Trm - 20) * 43 *

*> *

*> Trm = 1250/43 + 20 = 49F *

*> *

*> *

*> *

*> Since the room is now heated to 49F, the 2nd hour would be: *

*> *

*> *

*> *

*> 1250 Btu/hr = (Trm - 49) * 20 Btu/hr + (Trm - 20)* 23 Btu/hr *

*> *

*> 1250 = 43*Trm - 980 - 460 *

*> *

*> Trm = 2690/43 = 63F *

*> *

*> *

*> *

*> Since the room is now heated to 63F, the 3rd hour would be: *

*> *

*> *

*> *

*> 1250 Btu/hr = (Trm - 63) * 20 Btu/hr + (Trm - 20)* 23 Btu/hr *

*> *

*> 1250 = 43*Trm - 1260 - 460 *

*> *

*> Trm = 2970/43 = 69F *

*> *

*> *

*> *

*> After the 3rd hour, the room is comfortable (69degF). *

*> *

*> *

*> *

*> ------------------------------------------------------------------------- *

----

*> *

*> Suppose we added R20 foam insulation to the ceiling and walls. We'll ass *

ume that the mattresses covering the floor add R20 to the floor, so that no

w, we have R40 everywhere

*> *

*> ------------------------------------------------------------------------- *

----

*> *

*> *

*> *

*> (Trm - 20)degF*800sqft / R40 hr-sqft-degF/Btu] = (Trm - 20)* 20 Btu/hr *

*> *

*> *

*> *

*> The heatflow equation then, is: *

*> *

*> 1250 Btu/hr = (Trm - 20) * 20 Btu/hr + (Trm - 20)* 20 Btu/hr *

*> *

*> 1250 = (Trm - 20) * 40 *

*> *

*> Trm = 1250/40 + 20 = 51F *

*> *

*> *

*> *

*> Since the room is now heated to 51F, the 2nd hour would be: *

*> *

*> *

*> *

*> 1250 Btu/hr = (Trm - 51) * 20 Btu/hr + (Trm - 20)* 20 Btu/hr *

*> *

*> 1250 = 40*Trm - 1020 - 400 *

*> *

*> Trm = 2670/40 = 67F *

*> *

*> *

*> *

*> Since the room is now heated to 67F, the 3rd hour would be: *

*> *

*> *

*> *

*> 1250 Btu/hr = (Trm - 67) * 20 Btu/hr + (Trm - 20)* 20 Btu/hr *

*> *

*> 1250 = 40*Trm - 1340 - 400 *

*> *

*> Trm = 2990/40 = 75F *

*> *

*> *

*> *

*> After the 2nd hour, the room is comfortable (67degF). After the 3rd hour, *

the room is getting toasty (75degF).

*> *

*> *

*> *

*> ----------------------- *

*> *

*> 2 layers of the R7.7 4'x8' 2inch thick, foam board on the 4 walls and cei *

ling works out to about 30 boards.

*> *

*> *

*> *

*> 2 layers * [2*8*10 + 2*8*12 + 10*12]/(4*8) *

*> *

*> = 2 layers * 8*[20 + 24 + 15]/(4*8) *

*> *

*> = [20 + 24 + 15]/2) *

*> *

*> = 29.5 boards *

*> *

*> *

*> *

*> ------------------------------------- *

*> *

*> *

*> *

*> How big of a space heater would you need if you didn't insulate the room *

at all?

*> *

*> *

*> *

*> Let's say we wanted the room to be 70degF. and the heat from the space he *

ater is Hsh

*> *

*> *

*> *

*> The heatflow equation given above becomes: *

*> *

*> *

*> *

*> Hsh + 1250 Btu/hr = (70 - 20) * 20 Btu/hr + (70 - 20)* 40 Btu/hr *

*> *

*> Hsh + 1250 = 50 * (40+20) *

*> *

*> Hsh + 1250 = 50 * (60) *

*> *

*> Hsh + 1250 = 3000 *

*> *

*> Hsh = 3000 - 1250 = 1750 Btu/hr = 1750 Btu/hr * 0.293 Watts/BTU/hr *

= 512Watts

Posted by *songbird* on January 11, 2014, 1:48 pm

Bob F wrote:

...

*> You'd really want to breath the sooty htdrocarbon crud from many candles all *

*> night every night? Not me! *

*> *

*> Not to mention the fire hazard. *

i was thinking suffocation hazard in any room that

is tight enough to keep that much heat in.

heat losses from highly insulated rooms are via

the doors/windows and without those you are asking

for suffocation.

air exchangers will be very important if you want

this to fly, and then you will need a proper design

to make sure the heat is captured and not leaking.

agreed on fire hazard too. too many people in a

small room full of mattresses and blankets, not a

good combination with candles.

i like the overall idea, but we keep the thermstat

set much lower so the heat needed to keep it warm is

much less. our whole house costs about 1500/yr to

keep warm per heating season. not bad, but it could

be better. i just don't want to sacrifice the views

out the patio doors and windows. in the middle of

the winter any light from outside is psychologically

important.

when it was just me here most of the time i would

drop the thermostat down to 50-55F for the rest of

the house and keep this room a little warmer. worked

fine for me, but it was too cold for visitors and that

wasted a lot of energy cycling the heat up and down.

songbird

Posted by *Morris Dovey* on January 11, 2014, 7:44 pm

On 1/11/14 11:50 AM, Bob F wrote:

*> It won't waste nearly as much as keeping the thermostat up all the time. Wearing *

*> insulation is certainly way cheaper than heating. *

That depends on the cost of the heating.

Doing whole-structure solar heating allows (on really sunny winter days)

opening a window for comfort and fresh air. :-)

--

Morris Dovey

http://www.iedu.com/Solar/

> How does a family of 5 (3 adults, 2 kids) keep warm in 20degF> temperature in their modular home when they cannot afford electricity> or propane?