Hi

I have a basement in my house. The floor is about 1.5m below
earth/ground level and it is concrete about 30cm thick

The floor is not insolated, so in order to save some money on the
heating bill I am considering insolating it with sheets of polystyrene
foam (in principle foam filled with air) with some rafters in a mesh to

lay the wooden floor on. The lastly add 20mm of wooden plates/floor

An architect has told me to break up the floor and lay a new one with
30cm of extra insolation

But, I wonder if any of you guys can help me. I am an electrical
engineer and I don't like to do this without calculating the needed
insolation instead.

My theory is that since the floor is 1.5m below ground level, the
temperature of the soil will never be very cold. Searching the net I
find something about 14degree celcius.

So if I have 60square meters of floor heated to room temperature of
20degrees, how do I calculate the heattransfer when I have the data for

the insulation and the concrete floor?

Will the earth behave as an ideal giant block that has 6degrees of
tempeature. So the gradient from the room temperature to the earth can
never be higher than 10 degrees (20-14)?

Numbers:

Concrete, k = ~1W/mK
Polystyrene, k = 0.03W/mK
Wood, k = 0.14W/mK

Power needed to keep temperature stable: P=KAT/D

Concrete using 60square meters and 30cm thick: P = 1*60*6/0.3. P =
1.2kW

Adding polystyrene: k = 0.042 , P = 0.03*60*6/0.05 = 216W

The poystyrene is in parallel with the rafters. Assuming the rafters
take up 5% of the floor instead of the polystyrene

P= 0.14*60*0.05*6/0.05 = 50W

So from these calculations it seems I need 250W to keep the room heated

(not counting the walls)

Any wrong doings in the calculations - comments?

Thanks

Klaus Kragelund

Klaus Kragelund wrote:
Hi

I have a basement in my house. The floor is about 1.5m below
earth/ground level and it is concrete about 30cm thick

The floor is not insolated, so in order to save some money on the
heating bill I am considering insolating it with sheets of polystyrene
foam (in principle foam filled with air) with some rafters in a mesh
to

lay the wooden floor on. The lastly add 20mm of wooden plates/floor

An architect has told me to break up the floor and lay a new one with
30cm of extra insolation

But, I wonder if any of you guys can help me. I am an electrical
engineer and I don't like to do this without calculating the needed
insolation instead.

My theory is that since the floor is 1.5m below ground level, the
temperature of the soil will never be very cold. Searching the net I
find something about 14degree celcius.

So if I have 60square meters of floor heated to room temperature of
20degrees, how do I calculate the heattransfer when I have the data
for

the insulation and the concrete floor?

Will the earth behave as an ideal giant block that has 6degrees of
tempeature. So the gradient from the room temperature to the earth can
never be higher than 10 degrees (20-14)?

Numbers:

Concrete, k = ~1W/mK
Polystyrene, k = 0.03W/mK
Wood, k = 0.14W/mK

Power needed to keep temperature stable: P=KAT/D

Concrete using 60square meters and 30cm thick: P = 1*60*6/0.3. P =
1.2kW

Adding polystyrene: k = 0.042 , P = 0.03*60*6/0.05 = 216W

The poystyrene is in parallel with the rafters. Assuming the rafters
take up 5% of the floor instead of the polystyrene

P= 0.14*60*0.05*6/0.05 = 50W

So from these calculations it seems I need 250W to keep the room
heated

(not counting the walls)

Any wrong doings in the calculations - comments?

Thanks

Klaus Kragelund

Well, 1.5 meters is about the break even point for your question. So I
would suggest worrying about the wall, but not the floor. However if you
have long cold winters or long hot summers, then you might want to also work
on the floor.

--
Joseph Meehan

Dia duit

Where are you, 14c is fairly warm, I am zone 5 US where the freeze zone,
0c -32f is maybe 3.5ft, I put in 2" or R10 under a new basement floor.
With that small a difference foam pad and carpet might be as good.

Klaus Kragelund wrote:

I have a basement in my house. The floor is about 1.5m below
earth/ground level...

Covering the ceiling with foil would give it about US R10, ie 1.76 mK/W,
with E = 0.03 and a Tc = 20 C ceiling temp and a Tf = 14 C floor temp
and a large air gap. This would reduce the radiation from the ceiling
to the floor, es((Tc+273)^4-Tf+273)^4) W/m^2, with s = 5.6697x10^-8
W/m^2-K^4. If the upper foil surface is perfectly clean, with no dust
(you are German, right? :-), this may work even better. The linearized
radiation conductance is 4esTm^3 W/m^2-K, where Tm is the approximate
mean absolute temp.

The floor is not insolated, so in order to save some money on the
heating bill I am considering insolating it with sheets of polystyrene
foam... with some rafters in a mesh to lay the wooden floor on.

The English word is "insulating." InsOlation is sunlight.

My theory is that since the floor is 1.5m below ground level, the
temperature of the soil will never be very cold. Searching the net I
find something about 14degree celcius.

The temperature of the middle of the floor might be about the same as
the yearly average air temperature. The walls and the floor near the walls
might be closer to the average daily outdoor temperature.

So if I have 60square meters of floor heated to room temperature of
20degrees, how do I calculate the heattransfer when I have the data for
the insulation and the concrete floor?

With difficulty :-) The floor surface will probably be cooler than 20 C.
That's good.

Will the earth behave as an ideal giant block that has 6degrees of
tempeature. So the gradient from the room temperature to the earth can
never be higher than 10 degrees (20-14)?

The temperature difference between the room air and the middle of the floor
might be 6 C. Then again, the room air will warm the floor, which has thermal
capacity and resistance to downwards heatflow. Some people estimate soil's
resistance to downward heatflow as US R10, ie 1.76d mK/W. Upward is less,
with evaporation from lower soil layers and condensation above. And moving
water can change this.

Concrete, k = ~1W/mK
Polystyrene, k = 0.03W/mK
Wood, k = 0.14W/mK

Air, k = 0.025 W/mK

Power needed to keep temperature stable: P=KAT/D

Concrete using 60square meters and 30cm thick: P = 1*60*6/0.3. P = 1.2kW

Adding polystyrene: k = 0.042 , P = 0.03*60*6/0.05 = 216W

Not k = 0.03, as above?

The poystyrene is in parallel with the rafters. Assuming the rafters
take up 5% of the floor instead of the polystyrene

P= 0.14*60*0.05*6/0.05 = 50W

So from these calculations it seems I need 250W to keep the room heated

You might cover the ceiling with foil and cover the walls with thin
foil-faced foamboard over spacers and carpet the floor, with no polystyrene.
Each layer of wall foil adds about US R3, plus the bulk resistance of
the foamboard. If there's no vapor barrier under the concrete, you might
put a layer of plastic film under the carpet.

Nick

Here's one way to estimate the R-value of a radiant barrier based on the air
gap and the emissivities and surface temps and the direction of heatflow from
http://www.reflectixinc.com/pdf/RIMA_Handbook.pdf

10 SCREEN 9:KEY OFFIM HC(18,6)
20 DATA 0.359,0.184,0.126,0.097,0.080,0.068
30 DATA 0.361,0.187,0.129,0.100,0.082,0.072
40 DATA 0.363,0.189,0.131,0.101,0.085,0.075
50 DATA 0.364,0.190,0.132,0.103,0.087,0.078
60 DATA 0.365,0.191,0.133,0.105,0.090,0.081
70 DATA 0.366,0.192,0.134,0.106,0.092,0.082
80 DATA 0.360,0.204,0.169,0.179,0.185,0.189
90 DATA 0.366,0.267,0.223,0.233,0.238,0.241
100 DATA 0.373,0.247,0.261,0.271,0.275,0.276
110 DATA 0.380,0.270,0.292,0.301,0.303,0.303
120 DATA 0.387,0.296,0.317,0.325,0.327,0.326
130 DATA 0.394,0.319,0.339,0.347,0.347,0.345
140 DATA 0.381,0.312,0.295,0.284,0.275,0.268
150 DATA 0.429,0.381,0.360,0.346,0.336,0.328
160 DATA 0.472,0.428,0.405,0.389,0.377,0.368
170 DATA 0.511,0.465,0.440,0.423,0.410,0.400
180 DATA 0.545,0.496,0.469,0.451,0.437,0.426
190 DATA 0.574,0.523,0.494,0.475,0.460,0.449
200 FOR I=1 TO 18'read data table
210 FOR J=1 TO 6
220 READ HC(I,J)
230 NEXT:NEXT
240 T1=105'temperature of surface 1 (F)
250 E1=.03'emissivity of surface 1
260 T2=75'temperature of surface 2 (F)
270 E2=.8'emissivity of surface 2
280 L=2'air gap (valid range: 0.5-3")
290 LI=INT(2*L+.5)'length table index
300 HF=0'heatflow 0-down,1-sideways,2-up
310 E=1/(1/E1+1/E2-1)'effective emittance
320 TM=(T1+T2)/2'mean temp (F)
330 DT=ABS(T1-T2)'temp diff (valid range: 5-30 F)
340 DTI=INT(DT/5+.5+6*HF)'temp diff table index
350 HR=.00686*((TM+459.7)/100)^3'radiant conductance
360 R=1/(E*HR+HC(DTI,LI))'US R-value (ft^2-F-h/Btu)
370 PRINT T1,E1,T2,E2
380 PRINT L,HF,R

T1 (F) E1 T2 (F) E2

105 .03 75 .8

gap heatflow US R-value

2" 0 (down) 7.146456

With more than one space in series (eg double-foil foamboard spaced away
from a basement wall), we can't just add R-values. We only know the overall
temp diff, so we have to iterate to find a solution. It's no surprise that
the FTC prohibits makers from advertising R-values for radiant barriers
to avoid confusing the public.

Nick

Yea nick well if you were correct foil faced foamboard both sides would
not have the R rating it has, which is verified, it would be R6 more.
Foil is a Radiant barrier only, it has no R value to speak of, or gee,
wouldn`t the big manufacturers like to be as smart like you and
capitalise on extra performance.

Nick you should go into business, mortage everything, buy 1"
Polyisocyanurate foilfaced foamoard and add R6 to the rating to tell
everyone its R13.2 and sell it, and see what Gov agencys come knocking
to make you prove it.

m Ransley errs again:

... if you were correct foil faced foamboard both sides would
not have the R rating it has.

Nope. FTC rules mostly prohibit advertising installed R-values
to avoid confusing people smarter than you :-)

Nick

wrote:

Here's one way to estimate the R-value of a radiant barrier based on the air
gap and the emissivities and surface temps and the direction of heatflow from
http://www.reflectixinc.com/pdf/RIMA_Handbook.pdf

Nick,

How does this work out for the double bubble?

URL: http://www.blueridgecompany.com/radi...ic/189#pricing

It seems to me there are two ways to go for the underfloor insulation
for staple up radiant.

One is foil backed fiberglass insulation with an airspace. That is hard
to find! The other would be double bubble stapled on the joists. It
seems to me that would minimize heat loss through the joists themselves
as they would be uninsulated elsewise. Any thoughts?

Jeff

10 SCREEN 9:KEY OFFIM HC(18,6)
20 DATA 0.359,0.184,0.126,0.097,0.080,0.068
30 DATA 0.361,0.187,0.129,0.100,0.082,0.072
40 DATA 0.363,0.189,0.131,0.101,0.085,0.075
50 DATA 0.364,0.190,0.132,0.103,0.087,0.078
60 DATA 0.365,0.191,0.133,0.105,0.090,0.081
70 DATA 0.366,0.192,0.134,0.106,0.092,0.082
80 DATA 0.360,0.204,0.169,0.179,0.185,0.189
90 DATA 0.366,0.267,0.223,0.233,0.238,0.241
100 DATA 0.373,0.247,0.261,0.271,0.275,0.276
110 DATA 0.380,0.270,0.292,0.301,0.303,0.303
120 DATA 0.387,0.296,0.317,0.325,0.327,0.326
130 DATA 0.394,0.319,0.339,0.347,0.347,0.345
140 DATA 0.381,0.312,0.295,0.284,0.275,0.268
150 DATA 0.429,0.381,0.360,0.346,0.336,0.328
160 DATA 0.472,0.428,0.405,0.389,0.377,0.368
170 DATA 0.511,0.465,0.440,0.423,0.410,0.400
180 DATA 0.545,0.496,0.469,0.451,0.437,0.426
190 DATA 0.574,0.523,0.494,0.475,0.460,0.449
200 FOR I=1 TO 18'read data table
210 FOR J=1 TO 6
220 READ HC(I,J)
230 NEXT:NEXT
240 T1=105'temperature of surface 1 (F)
250 E1=.03'emissivity of surface 1
260 T2=75'temperature of surface 2 (F)
270 E2=.8'emissivity of surface 2
280 L=2'air gap (valid range: 0.5-3")
290 LI=INT(2*L+.5)'length table index
300 HF=0'heatflow 0-down,1-sideways,2-up
310 E=1/(1/E1+1/E2-1)'effective emittance
320 TM=(T1+T2)/2'mean temp (F)
330 DT=ABS(T1-T2)'temp diff (valid range: 5-30 F)
340 DTI=INT(DT/5+.5+6*HF)'temp diff table index
350 HR=.00686*((TM+459.7)/100)^3'radiant conductance
360 R=1/(E*HR+HC(DTI,LI))'US R-value (ft^2-F-h/Btu)
370 PRINT T1,E1,T2,E2
380 PRINT L,HF,R

T1 (F) E1 T2 (F) E2

105 .03 75 .8

gap heatflow US R-value

2" 0 (down) 7.146456

With more than one space in series (eg double-foil foamboard spaced away
from a basement wall), we can't just add R-values. We only know the overall
temp diff, so we have to iterate to find a solution. It's no surprise that
the FTC prohibits makers from advertising R-values for radiant barriers
to avoid confusing the public.

Nick

Jeff wrote:

... one way to estimate the R-value of a radiant barrier based on the air
gap and the emissivities and surface temps and the direction of heatflow
from http://www.reflectixinc.com/pdf/RIMA_Handbook.pdf

How does this work out for the double bubble?

Haven't tried that. You might work it out, if you know the gap width, etc.

http://www.blueridgecompany.com/radi...ic/189#pricing

It seems to me there are two ways to go for the underfloor insulation
for staple up radiant.

One is foil backed fiberglass insulation with an airspace. That is hard
to find! The other would be double bubble stapled on the joists. It
seems to me that would minimize heat loss through the joists themselves
as they would be uninsulated elsewise. Any thoughts?

I would staple on foil or thin double-foil foamboard.

Nick

wrote in message
...

Here's one way to estimate the R-value
of a radiant barrier based on the air
gap and the emissivities and surface
temps and the direction of heatflow from

It is?

The British Advertising Standards Authority got Actis, a French company,
claiming their reflective foil insulation is 'Equivalent to 200mm of
traditional Rockwoool insulation'. A complaint has been upheld after ASA
went to independent technical experts.

The judgement can be seen at:
http://tinyurl.com/s6c2p

Think hard before you buy.

wrote in message

Covering the ceiling with foil would give it about US R10, ie 1.76 mK/W,
with E = 0.03 and a Tc = 20 C ceiling temp and a Tf = 14 C floor temp
and a large air gap. This would reduce the radiation from the ceiling
to the floor, es((Tc+273)^4-Tf+273)^4) W/m^2, with s = 5.6697x10^-8
W/m^2-K^4. If the upper foil surface is perfectly clean, with no dust
(you are German, right? :-), this may work even better. The linearized
radiation conductance is 4esTm^3 W/m^2-K, where Tm is the approximate
mean absolute temp.

Su, what the **** does all that gibberish mean? Oh, I know, it means Nick
can show off he knows a couple of equations and has a calculator. So,
rather than impress us with all of your learning, why not take the time to
explain if this is good or bad and translates to dollars (or Euros) saved.

Yes, that closed cell EPS will be a good insulator.

Gibberish is right, "Covering the ceiling with foil gives you R10"
Covering nick with foil and putting him in " Nicks Lightbulb Sauna"
[easybake oven] of previous fame, for a month would do it, or maybe he
just got out of his Easy Bake Oven [sauna].

Edwin Pawlowski wrote:

wrote:

Covering the ceiling with foil would give it about US R10, ie 1.76 mK/W,
with E = 0.03 and a Tc = 20 C ceiling temp and a Tf = 14 C floor temp
and a large air gap. This would reduce the radiation from the ceiling
to the floor, es((Tc+273)^4-Tf+273)^4) W/m^2, with s = 5.6697x10^-8
W/m^2-K^4. If the upper foil surface is perfectly clean, with no dust
(you are German, right? :-), this may work even better. The linearized
radiation conductance is 4esTm^3 W/m^2-K, where Tm is the approximate
mean absolute temp.

Su, what the **** does all that gibberish mean?

That's called "elementary engineering" :-) The OP Klaus is an engineer.

Nick

wrote:
Klaus Kragelund wrote:

I have a basement in my house. The floor is about 1.5m below
earth/ground level...

Covering the ceiling with foil would give it about US R10, ie 1.76 mK/W,
with E = 0.03 and a Tc = 20 C ceiling temp and a Tf = 14 C floor temp
and a large air gap. This would reduce the radiation from the ceiling
to the floor, es((Tc+273)^4-Tf+273)^4) W/m^2, with s = 5.6697x10^-8
W/m^2-K^4. If the upper foil surface is perfectly clean, with no dust
(you are German, right? :-), this may work even better. The linearized
radiation conductance is 4esTm^3 W/m^2-K, where Tm is the approximate
mean absolute temp.

The floor is not insolated, so in order to save some money on the
heating bill I am considering insolating it with sheets of polystyrene
foam... with some rafters in a mesh to lay the wooden floor on.

The English word is "insulating." InsOlation is sunlight.

My theory is that since the floor is 1.5m below ground level, the
temperature of the soil will never be very cold. Searching the net I
find something about 14degree celcius.

The temperature of the middle of the floor might be about the same as
the yearly average air temperature. The walls and the floor near the walls
might be closer to the average daily outdoor temperature.

So if I have 60square meters of floor heated to room temperature of
20degrees, how do I calculate the heattransfer when I have the data for
the insulation and the concrete floor?

With difficulty :-) The floor surface will probably be cooler than 20 C.
That's good.

Will the earth behave as an ideal giant block that has 6degrees of
tempeature. So the gradient from the room temperature to the earth can
never be higher than 10 degrees (20-14)?

The temperature difference between the room air and the middle of the floor
might be 6 C. Then again, the room air will warm the floor, which has thermal
capacity and resistance to downwards heatflow. Some people estimate soil's
resistance to downward heatflow as US R10, ie 1.76d mK/W. Upward is less,
with evaporation from lower soil layers and condensation above. And moving
water can change this.

Concrete, k = ~1W/mK
Polystyrene, k = 0.03W/mK
Wood, k = 0.14W/mK

Air, k = 0.025 W/mK

Power needed to keep temperature stable: P=KAT/D

Concrete using 60square meters and 30cm thick: P = 1*60*6/0.3. P = 1.2kW

Adding polystyrene: k = 0.042 , P = 0.03*60*6/0.05 = 216W

Not k = 0.03, as above?

The poystyrene is in parallel with the rafters. Assuming the rafters
take up 5% of the floor instead of the polystyrene

P= 0.14*60*0.05*6/0.05 = 50W

So from these calculations it seems I need 250W to keep the room heated

You might cover the ceiling with foil and cover the walls with thin
foil-faced foamboard over spacers and carpet the floor, with no polystyrene.
Each layer of wall foil adds about US R3, plus the bulk resistance of
the foamboard. If there's no vapor barrier under the concrete, you might
put a layer of plastic film under the carpet.

Thankyou for the very good points

The basement is going to be used for office, so it will be heated all
year round.

I should have said I'm situated in Denmark (just north of Germany :-)

Perhaps then my number of 6 degrees gradient is overly optimistic due
to fringe effects near the outer walls. I want as little insulation as
possible since adding too much to the floor will limit the height of
the doors.

Currently no mostiture film is used. I want to add a wooden floor with
a carpet on top of this. The basement is dry, but my intention was to
add ventilation holes in the wooden floor and use no platic film to
allow the construction to"breathe". But adding plastic film might be a
good idea since this also limits heat flow caused by travelling
moisture.

Thanks

Klaus

Klaus Kragelund wrote:

Thankyou for the very good points

You are welcome.

I should have said I'm situated in Denmark (just north of Germany :-)

Danish people might permit a tiny amount of dust on the upper foil.
Then again, they build airtight houses, so excess moisture could be
a problem in summertime.

Currently no mostiture film is used. I want to add a wooden floor with
a carpet on top of this.

You may not need the wood.

The basement is dry, but my intention was to add ventilation holes in
the wooden floor and use no platic film to allow the construction to
"breathe". But adding plastic film might be a good idea since this also
limits heat flow caused by travelling moisture.

Upwards heatflow could be nice in wintertime.

Nick