Air heater algebra
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nick pine
Oct 14, 2011, 2:24:47 PM10/14/11
to sunspace, lri...@ursinus.edu, mell...@ursinus.edu
An air heater can be modeled with 2 temperatures, eg 70 F house air
and a T (F) C cfm air heater output, with an average Ta = (T+70)/2 air
temp in the heater. If a 20' wide x 24' tall 480 ft^2 patch of R2
glazing transmits 80% of 250 Btu/h-ft^2 full sun on a 14 F average
December day in Edmonton, Alberta and the solar energy that flows into
the heater equals the heat loss from the glazing to the outdoors plus
the house heat gain, 0.8x480x250 = 96K Btu/h = (Ta-14)480ft^2/R2 +
(T-70)C = 120T +5040 +(T-70)C, approximately, so 91K = 120T +(T-70)C.
With a T = 100 F outlet temp, 91K Btu/h = 120x100 + 30C, so C = 2633
cfm, with about 2633(100-70) = 79K Btu/h of heatflow and a 100x79K/
(96K/0.8) = 66% solar collection efficiency. Warm air rises, and one
empirical chimney formula says C = 16.6Asqrt(HdT) for 2 A ft^2 vents
with an H’ height difference and a dT (F) temp diff. With Ta =
(100+70)/2 = 85 F and H = 24’, C = 2633 cfm = 16.6Asqrt(24'(85-70))
makes A = 8.4 ft^2.
With A = 4 ft^2, C = 16.6x4sqrt(24(Ta-70)) = 325sqrt(Ta-70), and
(T-70)C = 230(T-70)^1.5 makes 91K = 120T +230(T-70)^1.5, ie T =
70+(396-0.522T)^(2/3). Plugging in T = 120 F on the right makes T =
117.9 on the right. Repeating makes T = 118.0, then 118.0, with Ta =
94 and C = 325sqrt(94-70) = 1592 cfm of airflow and about 1592(118-70)
= 76.4K Btu/h of heatflow and a 100x76.4K/(96K/0.8) = 64% solar
collection efficiency.
Nick
and a T (F) C cfm air heater output, with an average Ta = (T+70)/2 air
temp in the heater. If a 20' wide x 24' tall 480 ft^2 patch of R2
glazing transmits 80% of 250 Btu/h-ft^2 full sun on a 14 F average
December day in Edmonton, Alberta and the solar energy that flows into
the heater equals the heat loss from the glazing to the outdoors plus
the house heat gain, 0.8x480x250 = 96K Btu/h = (Ta-14)480ft^2/R2 +
(T-70)C = 120T +5040 +(T-70)C, approximately, so 91K = 120T +(T-70)C.
With a T = 100 F outlet temp, 91K Btu/h = 120x100 + 30C, so C = 2633
cfm, with about 2633(100-70) = 79K Btu/h of heatflow and a 100x79K/
(96K/0.8) = 66% solar collection efficiency. Warm air rises, and one
empirical chimney formula says C = 16.6Asqrt(HdT) for 2 A ft^2 vents
with an H’ height difference and a dT (F) temp diff. With Ta =
(100+70)/2 = 85 F and H = 24’, C = 2633 cfm = 16.6Asqrt(24'(85-70))
makes A = 8.4 ft^2.
With A = 4 ft^2, C = 16.6x4sqrt(24(Ta-70)) = 325sqrt(Ta-70), and
(T-70)C = 230(T-70)^1.5 makes 91K = 120T +230(T-70)^1.5, ie T =
70+(396-0.522T)^(2/3). Plugging in T = 120 F on the right makes T =
117.9 on the right. Repeating makes T = 118.0, then 118.0, with Ta =
94 and C = 325sqrt(94-70) = 1592 cfm of airflow and about 1592(118-70)
= 76.4K Btu/h of heatflow and a 100x76.4K/(96K/0.8) = 64% solar
collection efficiency.
Nick
Jim
Oct 27, 2011, 6:23:42 PM10/27/11
to sunspace
Nick, In the chimney formula{C = 2633 cfm = 16.6Asqrt(24'(85-70))}
you took the difference between the average collector temp and the
interior temperature. Seems to me the temperature difference would be
measured at the separation distance of H. What am I missing. Jim
you took the difference between the average collector temp and the
interior temperature. Seems to me the temperature difference would be
measured at the separation distance of H. What am I missing. Jim
Jim
Oct 27, 2011, 6:30:21 PM10/27/11
to sunspace
Nick, Is that because the ASHRAE chimney design procedure uses mean
stack temperature? Jim
stack temperature? Jim
nick pine
Oct 28, 2011, 8:52:40 AM10/28/11
to sunspace
Jim <wmjmorri...@gmail.com> wrote:
>... Is that because the ASHRAE chimney design procedure uses mean
> stack temperature?
Yes... http://en.wikipedia.org/wiki/Stack_effect has a more precise
estimate:
Q = CAsqrt(2gh(Ti-To)/Ti)), where: Q = stack effect flow rate, ft³/
s A = area, ft² C = discharge coefficient (usually taken to be from
0.65 to 0.70) g = gravitational acceleration, 32.17 ft/s² h = height
or distance, ft Ti = average inside temperature, °R To = outside air
temperature, °R
With C = 0.675, Q = 60CAsqrt(2x32.17H(Ti-To)/(85+460)) =
13.9Asqrt(HdT) cfm.
Nick
>... Is that because the ASHRAE chimney design procedure uses mean
> stack temperature?
Yes... http://en.wikipedia.org/wiki/Stack_effect has a more precise
estimate:
Q = CAsqrt(2gh(Ti-To)/Ti)), where: Q = stack effect flow rate, ft³/
s A = area, ft² C = discharge coefficient (usually taken to be from
0.65 to 0.70) g = gravitational acceleration, 32.17 ft/s² h = height
or distance, ft Ti = average inside temperature, °R To = outside air
temperature, °R
With C = 0.675, Q = 60CAsqrt(2x32.17H(Ti-To)/(85+460)) =
13.9Asqrt(HdT) cfm.
Nick
barbara deane-gillett
Oct 28, 2011, 11:38:29 AM10/28/11
to suns...@googlegroups.com
a few thoughts
1) for a conventional chimney ( i.e. an oil or gas flue, ) the inlet and outlet temps are pretty close. so chimney effect in a house where you are looking at an actual chimney is higher (diuble) than stack effect where you are looking at house leaking. nick i think correctly uses the avg temp diff in calculating the flow for the greenhouse as an air collector
2) 1500 cfm in your example agrees well with the rule of thumb of 2-6 cfm per sq.ft of glazing for an air collector. that large cfm is fairly easy in a large opening thermosyphoning stack, but is tough to get to with a duct and fan arrangement typical of air collectors. and as you aproach 6 the parasitic electrical pumping losses get large one of the reasons why air collectors are 15 to one while water collectors are 30 to one on pumping losses relative to output. again one of the beautifies of the thermosyphoning is that the pumping is provided by solar not fans.
3) one last point is that h height comes into the equation. although not relevant here i have found it useful in increasing flow rates to put the solar storage tank above the external htx in thermosyphon dhw systems. the added h increases flow noticably on the thermosyphon side, eliminating one of the pumps
> Date: Fri, 28 Oct 2011 05:52:40 -0700
> Subject: Re: Air heater algebra
> From: ni...@early.com
> To: suns...@googlegroups.com
> Subject: Re: Air heater algebra
> From: ni...@early.com
> To: suns...@googlegroups.com
nick pine
Nov 12, 2011, 8:19:34 AM11/12/11
to sunspace
"yreysa" <gary@...> wrote:
> You want about 2.5 cfm per sqft of collector...
Why not more? In a simple model, an air heater with C cfm of airflow
and 1 ft^2 of R1 glazing with 90% solar transmission might look like
this in full 250 Btu/h-ft^2 sun on a 30 F day:
225 Btu/h
--- 1/C
|----|-->|-------www--- 70 F
--- |
|
R1 |
30 F---www---
which is equivalent to:
R1 1/C
---www-------www--- 70 F
| -->
| I
}
--- 30+225R1 = 255 F
-
|
-
with I = (255-70)/(1+1/C) = 185C/(1+C) Btu/h. More airflow makes more
heatflow, eg 93 Btu/h with C = 1 cfm, 132 with 2.5, 154 with 5, and
185 with infinite airflow.
Then again, more airflow uses more fan power, unless it is naturally
thermosyphoning, and more air velocity lowers the thermal resistance
of the indoor glazing airfilm. Fig 1 in chapter 22 of the ASHRAE HOF
says U = 1.5+V/5 Btu/h-F-ft^2 for smooth surfaces, where V is in mph.
For V in lfm, U = 1.5+V/440. So in a fancier model with an average
3" (1/4 ft) glazing-screen gap,
R 1/C
---www-------www--- 70 F
| -->
| I
|
--- V = 30+225R
-
|
-
R = 1/3 + 1/(1.5+C/110), and I = (V-70)/(R+1/C), eg 131 Btu/h
with C = 2.5 cfm, 153 with 5 cfm, 166 with 10, and 105
with infinite airflow.
Just for grins, what's the optimal airflow?
If I = (V-70)/(R+1/C) = U/V and d(U/V) = (UdV-VdU)/V^-2 = 0, then
(V-70)d(R+1/C) = (R+1/C)d(V-70), ie
(V-70)d(R+1/C) = (R+1/C)dV, ie
(V-70)d(R+1/C) = (R+1/C)225dR, ie
(V-70)(dR-C^-2) = (R+1/C)225dR, ie
dR(V-70-225(R+1/C) = C^-2(V-70), ie
dR(30+225R-70-225(R+1/C)) = C^-2(30+225R-70), ie
dR(225R-40-225R-225/C) = C^-2(225R-40), ie
dR(-40-225/C) = C^-2(225R-40), ie
dR(225/C+40) = -C^2(225R-40), ie
dR(45/C+8) = -C^2(45R-8), ie
d(1/3+(1.5+C/110)^-1)(45/C+8) = -C^2(45R-8), ie
-(1.5+C/110)^-2/110(45/C+8) = -C^-2(45R-8), ie
(1.5+C/110)^-2/110(45/C+8) = C^-2(45R-8), ie
(1.5+C/110)^-2(45/C+8) = 110C^-2(45R-8), ie
45/C+8 = 110C^-2(45R-8)(1.5+C/110)^2, ie
45/C+8 = 110C^-2(45R-8)((165+C)/110)^2, ie
45/C+8 = C^-2(45R-8)(165+C)^2/110, ie
4950/C+880 = C^-2(45R-8)(165+C)^2, ie
4950C+880C^2 = (45R-8)(165+C)^2, ie
4950C+880C^2 = (45R-8)(27225+330C+C^2), ie
4950C+880C^2 = 1225125R+14850RC+45RC^2-217800-2640C-8C^2, ie
7590C+888C^2+217800 = 1225125R+14850RC+45RC^2, ie
7590C+888C^2+217800 = (1225125+14850C+45C^2)R, ie
7590C+888C^2+217800 = (1225125+14850C+45C^2)(1/3+1/(1.5+C/110)), ie
2640C+873C^2-190575 = (1225125+14850C+45C^2)/(1.5+C/110), ie
(2640C+873C^2-190575)(1.5+C/110) = 1225125+14850C+45C^2, ie
3960C+1309.5C^2-285862.5+(2640C+873C^2-190575)C/110 = 1225125+14850C
+45C^2, ie
3960C+1309.5C^2-285862.5+24C^2+7.936C^3-1732.5C = 1225125+14850C
+45C^2, ie
-12622.5C+1288.5C^2-1510987.5+7.936C^3 = 0, ie
7.936C^3+1288.5C^2-12622.5C-1510987.5 = 0, which produces
C = 35.31753020310051 via http://www.1728.org/cubic.htm
which looks about right in this calc:
10 FOR C=34 TO 37'airflow (cfm/ft^2)
20 RG=1/3+1/(1.5+C/110)'equivalent glazing resistance
30 VT=30+225*RG'Thevenin temp (F)
40 Q=(VT-70)/(RG+1/C)'useful heatflow (Btu/h)
45 PRINT C,Q
50 NEXT C
cfm/ft^2 Btu/h-ft^2
34 174.0801
35 174.0862
36 174.0849
37 174.0769
With a larger glazing-screen gap or double glazing, more airflow
would raise the solar collection efficiency without reducing
the indoor airfilm resistance or increasing the heat loss
to the outdoors so much.
Nick
> You want about 2.5 cfm per sqft of collector...
Why not more? In a simple model, an air heater with C cfm of airflow
and 1 ft^2 of R1 glazing with 90% solar transmission might look like
this in full 250 Btu/h-ft^2 sun on a 30 F day:
225 Btu/h
--- 1/C
|----|-->|-------www--- 70 F
--- |
|
R1 |
30 F---www---
which is equivalent to:
R1 1/C
---www-------www--- 70 F
| -->
| I
}
--- 30+225R1 = 255 F
-
|
-
with I = (255-70)/(1+1/C) = 185C/(1+C) Btu/h. More airflow makes more
heatflow, eg 93 Btu/h with C = 1 cfm, 132 with 2.5, 154 with 5, and
185 with infinite airflow.
Then again, more airflow uses more fan power, unless it is naturally
thermosyphoning, and more air velocity lowers the thermal resistance
of the indoor glazing airfilm. Fig 1 in chapter 22 of the ASHRAE HOF
says U = 1.5+V/5 Btu/h-F-ft^2 for smooth surfaces, where V is in mph.
For V in lfm, U = 1.5+V/440. So in a fancier model with an average
3" (1/4 ft) glazing-screen gap,
R 1/C
---www-------www--- 70 F
| -->
| I
|
--- V = 30+225R
-
|
-
R = 1/3 + 1/(1.5+C/110), and I = (V-70)/(R+1/C), eg 131 Btu/h
with C = 2.5 cfm, 153 with 5 cfm, 166 with 10, and 105
with infinite airflow.
Just for grins, what's the optimal airflow?
If I = (V-70)/(R+1/C) = U/V and d(U/V) = (UdV-VdU)/V^-2 = 0, then
(V-70)d(R+1/C) = (R+1/C)d(V-70), ie
(V-70)d(R+1/C) = (R+1/C)dV, ie
(V-70)d(R+1/C) = (R+1/C)225dR, ie
(V-70)(dR-C^-2) = (R+1/C)225dR, ie
dR(V-70-225(R+1/C) = C^-2(V-70), ie
dR(30+225R-70-225(R+1/C)) = C^-2(30+225R-70), ie
dR(225R-40-225R-225/C) = C^-2(225R-40), ie
dR(-40-225/C) = C^-2(225R-40), ie
dR(225/C+40) = -C^2(225R-40), ie
dR(45/C+8) = -C^2(45R-8), ie
d(1/3+(1.5+C/110)^-1)(45/C+8) = -C^2(45R-8), ie
-(1.5+C/110)^-2/110(45/C+8) = -C^-2(45R-8), ie
(1.5+C/110)^-2/110(45/C+8) = C^-2(45R-8), ie
(1.5+C/110)^-2(45/C+8) = 110C^-2(45R-8), ie
45/C+8 = 110C^-2(45R-8)(1.5+C/110)^2, ie
45/C+8 = 110C^-2(45R-8)((165+C)/110)^2, ie
45/C+8 = C^-2(45R-8)(165+C)^2/110, ie
4950/C+880 = C^-2(45R-8)(165+C)^2, ie
4950C+880C^2 = (45R-8)(165+C)^2, ie
4950C+880C^2 = (45R-8)(27225+330C+C^2), ie
4950C+880C^2 = 1225125R+14850RC+45RC^2-217800-2640C-8C^2, ie
7590C+888C^2+217800 = 1225125R+14850RC+45RC^2, ie
7590C+888C^2+217800 = (1225125+14850C+45C^2)R, ie
7590C+888C^2+217800 = (1225125+14850C+45C^2)(1/3+1/(1.5+C/110)), ie
2640C+873C^2-190575 = (1225125+14850C+45C^2)/(1.5+C/110), ie
(2640C+873C^2-190575)(1.5+C/110) = 1225125+14850C+45C^2, ie
3960C+1309.5C^2-285862.5+(2640C+873C^2-190575)C/110 = 1225125+14850C
+45C^2, ie
3960C+1309.5C^2-285862.5+24C^2+7.936C^3-1732.5C = 1225125+14850C
+45C^2, ie
-12622.5C+1288.5C^2-1510987.5+7.936C^3 = 0, ie
7.936C^3+1288.5C^2-12622.5C-1510987.5 = 0, which produces
C = 35.31753020310051 via http://www.1728.org/cubic.htm
which looks about right in this calc:
10 FOR C=34 TO 37'airflow (cfm/ft^2)
20 RG=1/3+1/(1.5+C/110)'equivalent glazing resistance
30 VT=30+225*RG'Thevenin temp (F)
40 Q=(VT-70)/(RG+1/C)'useful heatflow (Btu/h)
45 PRINT C,Q
50 NEXT C
cfm/ft^2 Btu/h-ft^2
34 174.0801
35 174.0862
36 174.0849
37 174.0769
With a larger glazing-screen gap or double glazing, more airflow
would raise the solar collection efficiency without reducing
the indoor airfilm resistance or increasing the heat loss
to the outdoors so much.
Nick
kreads
Nov 12, 2011, 11:07:22 AM11/12/11
to sunspace
lol I am swimming in calculations here! I thought I had it until I
tried again w the grins. I am trying to apply to gapping. As in, I
have double glass panes (21x67 or 27x73) and an aluminum clad wall w
perfect solar exposure. I have seen people put a break in the aluminum
siding, paint it flat black, and cut openings thru to living space
behind. So, I am calculating not only how big for openings, which
seems clear in many calculations as proportionate to glass size and
reducible by adding fans. But, I'm also working on, how deep, how much
air space created by depth of sides holding the glass doors to decide
how wide sideboards shall be. I can drop variable of transmission
since it is glass. Can you please help me plug it into these calcs?
Thanks,
Kathyann
tried again w the grins. I am trying to apply to gapping. As in, I
have double glass panes (21x67 or 27x73) and an aluminum clad wall w
perfect solar exposure. I have seen people put a break in the aluminum
siding, paint it flat black, and cut openings thru to living space
behind. So, I am calculating not only how big for openings, which
seems clear in many calculations as proportionate to glass size and
reducible by adding fans. But, I'm also working on, how deep, how much
air space created by depth of sides holding the glass doors to decide
how wide sideboards shall be. I can drop variable of transmission
since it is glass. Can you please help me plug it into these calcs?
Thanks,
Kathyann
nick pine
Nov 12, 2011, 1:00:15 PM11/12/11
to sunspace
Me too :-)
>... Fig 1 in chapter 22 of the ASHRAE HOF says U = 1.5+V/5
>So in a fancier model with an average 3" (1/4 ft) glazing-screen gap,
R 1/C
---www-------www--- 70 F
| -->
| I
|
--- V = 30+225R
-
|
-
>with R = 1/3 + 1/(1.5+C/110), and I = (V-70)/(R+1/C), eg 131 Btu/h
eg 132 Btu/h with C = 2.5 cfm, 154 with 5 cfm, 168 with 10, and
105 with infinite airflow. The optimal airflow is about 108 cfm:
10 FOR C= 107 TO 110 STEP 1'airflow (cfm/ft^2)
20 RG=1/3+1/(1.5+C/1100)'equivalent glazing resistance
50 NEXT C
107 181.5389
108 181.5391
109 181.5391
110 181.5388
> I have double glass panes (21x67 or 27x73) and an aluminum clad wall w
> perfect solar exposure.
How big?
> I have seen people put a break in the aluminum siding, paint it flat
> black, and cut openings thru to living space behind.
You might also add a layer of black aluminum window screen to increase
the sun-air heat transfer surface and lower reradiation loss.
> So, I am calculating not only how big for openings, which
> seems clear in many calculations as proportionate to glass size and
> reducible by adding fans. But, I'm also working on, how deep, how much
> air space created by depth of sides holding the glass doors to decide
> how wide sideboards shall be.
Gary made the openings about 3% of the glazing area and made sure that
the vertical airpath through his barn heater was no narrower. Larger
and
deeper are better, altho that costs more, and larger plastic flap
dampers
lose more heat at night.
Nick
>... Fig 1 in chapter 22 of the ASHRAE HOF says U = 1.5+V/5
Btu/h-F-ft^2 for smooth surfaces, where V is in mph. For V in lfm,
U = 1.5+V/440.
Oops. What happened to the 5? That shoulda been U = 1.5+V/4400...
U = 1.5+V/440.
>So in a fancier model with an average 3" (1/4 ft) glazing-screen gap,
R 1/C
---www-------www--- 70 F
| -->
| I
|
--- V = 30+225R
-
|
-
with C = 2.5 cfm, 153 with 5 cfm, 166 with 10, and 105
with infinite airflow.
That shoulda been R = 1/3 + 1/(1.5+C/1100), and I = (V-70)/(R+1/C),
with infinite airflow.
eg 132 Btu/h with C = 2.5 cfm, 154 with 5 cfm, 168 with 10, and
105 with infinite airflow. The optimal airflow is about 108 cfm:
10 FOR C= 107 TO 110 STEP 1'airflow (cfm/ft^2)
20 RG=1/3+1/(1.5+C/1100)'equivalent glazing resistance
30 VT=30+225*RG'Thevenin temp (F)
40 Q=(VT-70)/(RG+1/C)'useful heatflow (Btu/h)
45 PRINT 100+C,Q
40 Q=(VT-70)/(RG+1/C)'useful heatflow (Btu/h)
50 NEXT C
107 181.5389
108 181.5391
109 181.5391
110 181.5388
> I have double glass panes (21x67 or 27x73) and an aluminum clad wall w
> perfect solar exposure.
> I have seen people put a break in the aluminum siding, paint it flat
> black, and cut openings thru to living space behind.
the sun-air heat transfer surface and lower reradiation loss.
> So, I am calculating not only how big for openings, which
> seems clear in many calculations as proportionate to glass size and
> reducible by adding fans. But, I'm also working on, how deep, how much
> air space created by depth of sides holding the glass doors to decide
> how wide sideboards shall be.
the vertical airpath through his barn heater was no narrower. Larger
and
deeper are better, altho that costs more, and larger plastic flap
dampers
lose more heat at night.
Nick
barbara deane-gillett
Nov 13, 2011, 8:55:10 AM11/13/11
to suns...@googlegroups.com
now nick, something wrong if infinite less than optimal. also don't forget fan horsepower electrical cost and how that increases with cube of velocity. i'll stick with rule of thumb for powered systems of 2-5 cfm. 10 at max. as we dicussed some time ago that reults in a cop ( energy from system over energy into fan) of about 30 for air systems. 100 for liquid.
also thermosyphoning is not free, it adds to temp in collector and therefore losses.
also thermosyphoning is not free, it adds to temp in collector and therefore losses.
> Date: Sat, 12 Nov 2011 10:00:15 -0800
nick pine
Nov 14, 2011, 8:07:16 AM11/14/11
to sunspace
drew gillett <dea...@hotmail.com> wrote:
>... something wrong if infinite less than optimal.
More airflow made more heatflow in the first simple model with "R1
glazing" with an inside airfilm resistance that didn't depend on the
indoor air velocity. The second model had an R1/3 outdoor film
resistance and a 1.5+V/5 Btu/h-F-ft^2 inside airfilm conductance,
where V is the air velocity in mph.
>... thermosyphoning is not free, it adds to temp in collector and therefore losses.
It seems close to free with R2 twinwall polycarbonate glazing, which
now costs less than Dynaglas single glazing where I live.
Nick
>... something wrong if infinite less than optimal.
More airflow made more heatflow in the first simple model with "R1
glazing" with an inside airfilm resistance that didn't depend on the
indoor air velocity. The second model had an R1/3 outdoor film
resistance and a 1.5+V/5 Btu/h-F-ft^2 inside airfilm conductance,
where V is the air velocity in mph.
>... thermosyphoning is not free, it adds to temp in collector and therefore losses.
It seems close to free with R2 twinwall polycarbonate glazing, which
now costs less than Dynaglas single glazing where I live.
Nick
barbara deane-gillett
Nov 14, 2011, 9:30:07 AM11/14/11
to suns...@googlegroups.com
still - more airfolow equals higher conductance equals more heat transfer. still something wrong if infinite flow is loess than any integer flow. perhaps try a near infinite number in equation. this kind of problem often occurs with math models near extremes witness the confusion over solar radiation data when sun near the horizon
still - thermosyphoning not free as it does add to losses to outside ( even thru twinwall) because of higher air temp. my wag is 5% of enrgy transfered.
basic physics as you know. i count on you to sort it out.
still - thermosyphoning not free as it does add to losses to outside ( even thru twinwall) because of higher air temp. my wag is 5% of enrgy transfered.
basic physics as you know. i count on you to sort it out.
> Date: Mon, 14 Nov 2011 05:07:16 -0800
nick pine
Nov 14, 2011, 1:33:57 PM11/14/11
to sunspace
drew gillett <dea...@hotmail.com> wrote:
> still - more airfolow equals higher conductance equals more heat transfer.
Nonono. More airfolow means more air velocity near the glazing, with
> still - more airfolow equals higher conductance equals more heat transfer.
raises the glazing's indoor airfilm conductance and lowers the series
glazing resistance and raises the heat loss to the outdoors.
>... this kind of problem often occurs with math models near extremes witness the confusion over solar radiation data when sun near the horizon
That's different.
> still - thermosyphoning not free as it does add to losses to outside ( even thru twinwall) because of higher air temp. my wag is 5% of enrgy transfered.
and deep enough air heaters can move as much air as any fan, but we
might talk about Btu/$, including higher losses at night through
larger vent openings and more daytime heat loss through deeper
sidewalls.
Nick
nick pine
Dec 2, 2011, 5:48:26 PM12/2/11
to sunspace
"Bob Waldrop" <bwal...@cox.net> wrote:
>I recently came across this interesting take on the traditional solar window heater.
http://blog.imehrle.com/2011/02/10/solar-heating-made-easy----very-easy.aspx
>In discussing this elsewhere, however, someone mentioned that since it hangs on the inside of a window, it blocks sunlight from the window.
Views too.
>That person thought that the heat resulting from the flow of sunlight into a room without any obstruction in the window would be greater than the heat resulting from the operation of this device in a window, which would have the incidental effect of blocking some sunlight from the room.
Is that correct?
Yes, but it also increases the heat loss from the window when the sun
is shining.
http://rredc.nrel.gov/solar//old_data/nsrdb/1961-1990/bluebook/data/14821.SBF
says 780 Btu/ft^2 falls on a south wall on an average January day in
Columbus, OH with a 26 F outdoor temp. With no solar heating panel, a
1 ft^2 R2 window with 80% solar transmission in a 70 F house would
gain 0.8x780 = 624 Btu and lose 24h(70-26)1ft^2/R2 = 528, for a net
gain of 96 Btu/day. With a 143 F panel next to the glass for 6 hours,
it would lose 6h(143-26)1ft^2/R2 + 18h(70-26)1ft^2/R2 = 747 Btu, for a
net loss of 123 Btu/day, or more, if the panel reflects some
sunlight back out of the window.
And how can Mike circulate solar warm air through the rest of the
house with the furnace blower if he leaves the return vent open but
closes the supply vents to the solar room? Airflow requires supply and
return paths. Where's the supply path to the room?
If Mike knew more about airflow or heatflow, he might suggest hanging
huge ugly boxes and cutting holes in house walls or turning a south
window into an air heater with a foamboard insert that collects less
solar heat than the window but greatly reduces the heat loss at
night...
If a 2' wide x 3' tall window gains 6ft^2x624/6h = 624 Btu/h by day
and a 1" R10 foil-foamboard panel with 1.5" slots at the top and
bottom with dark high-temp BBQ paint facing the window and foil facing
the room loses about 18h(70-26)6ft^2/R12 = 396 Btu at night and
16.6(36in^2/144in^2/ft^2)sqrt(3?(70+dT-70)) = 7.2dT^0.5 cfm of 70+2dT
(F) air flows out of the top slot and provides (70+2dT)-70)7.2dT^0.5 =
14.4dT^1.5 Btu/h to the room and 624 = (70+dT-26)6ft^2/R2+14.4dT^1.5 =
132+3dT+14.4dT^1.5, ie dT = ((492-3dT)/14.4)^(2/3). Plugging in dT =
10 F on the right makes dT = 10.07 on the right, then 10.07, with 23
cfm of 80 F airflow and 230 Btu/h of heatflow, for a net gain of
6hx230-396 = 984 Btu/day, vs 96x6ft^2 = 576 for the window alone, or a
net loss of 738 with Mike's panel.
Nick
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