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Straw-infilled-Pallet Winter Greenhouse.
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zoe_lithoi
Nov 30, 2014, 12:28:27 PM11/30/14
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Straw-infilled-Pallet Winter Greenhouse.
Greetings,
My son and I are building a 80" x 80" x 8'(tall) 'winter' Greenhouse made of pallets screwed together and stuffed with straw for insulation (R10?). It has plastic stapled to the inside which should prevent heat loss by convection. The Roof has masonite siding on it in case of rain. We will have 200 of grow lights which will also serve as the primary heat source. The ground will be either a heat source or a heat sink (not sure yet.....). The goal is to keep the temperature above freezing so the seedlings don't die.
The surface area of the 4 walls, and ceiling would be:
A = 4*7*8 + 7*7 = 273 sqft
The heat from the growlights
Hgl = 200W * 3.41 Btu/1Watt-hr = 682 Btu/hr
Hgnd = ground heat will be called:
For now, we will assume that heat from the growlites will enter the ground.
Hrm = heat leaving the room of Temperature, Trm, thru the 273sqft of R10 walls and ceiling to the 20degF outside is:
Hrm = (Trm - 20)degF*273sqft / R10 hr-sqft-degF/Btu] = (Trm - 20)* 27 Btu/hr
If the room gets new air each hour equivalent to it's volume, then the air-exchange heat loss for the amount of heat the air in the 400cuft (8'*7'*7') room absorbs to go from 20degF to Trm is:
Hair = (Trm - 20)degF * 1/55 Btu/F /cuft * 400cuft ~= (Trm - 20) * 5 Btu/hr
The heatflow equation then, is:
Hgl = Hgnd + Hrm + Hair
682 = Hgnd + (Trm - 20)* 27 + (Trm - 20) * 5
For now, let's assume the heat flow into or out of the ground is 0, i.e.:
Hgnd = 0
682 = (Trm - 20)* 27 + (Trm - 20) * 5
682 = 33*Trm - 20* (27 + 5)
682 = 33*Trm - 660
1342 = 33*Trm
Trm = 1342/33 = 41degF
If the outside temperature was 0degF, then
Trm = 682/33 = 21degF
--- daid seedlings
So we need to look at the ground temperature.
In the above calc's, Trm is between 20 and 40degF. The Ground temperature, for the southwest (NEvada, Utah, Arizona), 4inches deep (the approix depth heat can travel in the ground in 1 hour --- see the 'daycreek' thread in this group), is about 50degF. So as long as the greenhouse temperature, Trm, is below 50degF, then the ground is a heat source and supplies heat to the greenhouse.
-----------------------------
-SAND Heat capacity 2.5 BTU/(F-sqft-in)
-SAND Resistance 0.083 hr-sqft-F/(BTU-in)
From past calculations and real-world example (daycreek.com), ground heat travels about 4inches per hour, so:
-SAND Heat capacity 2.5 BTU/(F-sqft-in) * 4inch = 10 BTU/F-sqft
-SAND Resistance 0.083 hr-sqft-F/(BTU-in) * 4inch = 0.33 hr-sqft-F/Btu
The surface area of the ground is:
7' x 7' = ~50sqft
Hgnd = (50degF - Trm)*50sqft / [0.33 hr-sqft-F/Btu]
Hgnd = (50degF - Trm) * 16.5 Btu/hr
-------------------
Now let's include the ground heat in the heatflow equation which again is:
Hgl = - Hgnd + Hrm + Hair
682 = -(50degF - Trm) * 16.5 + (Trm - 20)* 27 + (Trm - 20) * 5
682 = (+16.5+27+5)*Trm - 50*16.5 - 20* (27 + 5)
682 = 44*Trm - 825 - 660
2167 = 44*Trm
Trm = 2167/44 = 49degF
Now, if the outside temperature is 0degF (instead of 20degF):
682 = (+16.5+27+5)*Trm - 50*16.5 - 0* (27 + 5)
682 = 44*Trm - 825
1597 = 44*Trm
Trm = 1507/44 = 34
If a 400Watt growlite were used, and it was 0degF
Hgl = 1364Btu/hr
Then:
1364 = 44*Trm - 825
2189 = 44*Trm
Trm = 2189/44 = 50degF
If there were not any growlites, and it was 20degF outside, then:
0 = 44*Trm - 825 - 660
1485 = 44*Trm
Trm = 1485/44 = 34degF
The lowest daily low temperature in Las Vegas Nv for the month of July is 40degF
See: https://weatherspark.com/averages/31890/1/Las-Vegas-Nevada-United-States
Daytime temperatures are normally in the 50's and 60's.
Using Outside temperature of 40degF, with a 200W growlight we get:
682 = (+16.5+27+5)*Trm - 50*16.5 - 40* (27 + 5)
682 = 44*Trm - 825 - 1320
2827 = 44*Trm
Trm = 2827/44 = 64degF
Using Outside temperature of 40degF, without a 200W growlight we get:
0 = (+16.5+27+5)*Trm - 50*16.5 - 40* (27 + 5)
0 = 44*Trm - 825 - 1320
2145 = 44*Trm
Trm = 2145/44 = 49degF
So, one might have a thermostat to power the growlights if the temperature dropped below 45degF.... AND Further, put the growlites on a timer to give it 12 hours each day (so the plants can get light) during night-time hours when it is coldest outside.
200W * 1kW/1000W * $0.11/kW-hr * 12hr/day = 0.26cents/day
--> $8/month
--> $48/Winter (6-months)
Toby
Greetings,
My son and I are building a 80" x 80" x 8'(tall) 'winter' Greenhouse made of pallets screwed together and stuffed with straw for insulation (R10?). It has plastic stapled to the inside which should prevent heat loss by convection. The Roof has masonite siding on it in case of rain. We will have 200 of grow lights which will also serve as the primary heat source. The ground will be either a heat source or a heat sink (not sure yet.....). The goal is to keep the temperature above freezing so the seedlings don't die.
The surface area of the 4 walls, and ceiling would be:
A = 4*7*8 + 7*7 = 273 sqft
The heat from the growlights
Hgl = 200W * 3.41 Btu/1Watt-hr = 682 Btu/hr
Hgnd = ground heat will be called:
For now, we will assume that heat from the growlites will enter the ground.
Hrm = heat leaving the room of Temperature, Trm, thru the 273sqft of R10 walls and ceiling to the 20degF outside is:
Hrm = (Trm - 20)degF*273sqft / R10 hr-sqft-degF/Btu] = (Trm - 20)* 27 Btu/hr
If the room gets new air each hour equivalent to it's volume, then the air-exchange heat loss for the amount of heat the air in the 400cuft (8'*7'*7') room absorbs to go from 20degF to Trm is:
Hair = (Trm - 20)degF * 1/55 Btu/F /cuft * 400cuft ~= (Trm - 20) * 5 Btu/hr
The heatflow equation then, is:
Hgl = Hgnd + Hrm + Hair
682 = Hgnd + (Trm - 20)* 27 + (Trm - 20) * 5
For now, let's assume the heat flow into or out of the ground is 0, i.e.:
Hgnd = 0
682 = (Trm - 20)* 27 + (Trm - 20) * 5
682 = 33*Trm - 20* (27 + 5)
682 = 33*Trm - 660
1342 = 33*Trm
Trm = 1342/33 = 41degF
If the outside temperature was 0degF, then
Trm = 682/33 = 21degF
--- daid seedlings
So we need to look at the ground temperature.
In the above calc's, Trm is between 20 and 40degF. The Ground temperature, for the southwest (NEvada, Utah, Arizona), 4inches deep (the approix depth heat can travel in the ground in 1 hour --- see the 'daycreek' thread in this group), is about 50degF. So as long as the greenhouse temperature, Trm, is below 50degF, then the ground is a heat source and supplies heat to the greenhouse.
-----------------------------
-SAND Heat capacity 2.5 BTU/(F-sqft-in)
-SAND Resistance 0.083 hr-sqft-F/(BTU-in)
From past calculations and real-world example (daycreek.com), ground heat travels about 4inches per hour, so:
-SAND Heat capacity 2.5 BTU/(F-sqft-in) * 4inch = 10 BTU/F-sqft
-SAND Resistance 0.083 hr-sqft-F/(BTU-in) * 4inch = 0.33 hr-sqft-F/Btu
The surface area of the ground is:
7' x 7' = ~50sqft
Hgnd = (50degF - Trm)*50sqft / [0.33 hr-sqft-F/Btu]
Hgnd = (50degF - Trm) * 16.5 Btu/hr
-------------------
Now let's include the ground heat in the heatflow equation which again is:
Hgl = - Hgnd + Hrm + Hair
682 = -(50degF - Trm) * 16.5 + (Trm - 20)* 27 + (Trm - 20) * 5
682 = (+16.5+27+5)*Trm - 50*16.5 - 20* (27 + 5)
682 = 44*Trm - 825 - 660
2167 = 44*Trm
Trm = 2167/44 = 49degF
Now, if the outside temperature is 0degF (instead of 20degF):
682 = (+16.5+27+5)*Trm - 50*16.5 - 0* (27 + 5)
682 = 44*Trm - 825
1597 = 44*Trm
Trm = 1507/44 = 34
If a 400Watt growlite were used, and it was 0degF
Hgl = 1364Btu/hr
Then:
1364 = 44*Trm - 825
2189 = 44*Trm
Trm = 2189/44 = 50degF
If there were not any growlites, and it was 20degF outside, then:
0 = 44*Trm - 825 - 660
1485 = 44*Trm
Trm = 1485/44 = 34degF
The lowest daily low temperature in Las Vegas Nv for the month of July is 40degF
See: https://weatherspark.com/averages/31890/1/Las-Vegas-Nevada-United-States
Daytime temperatures are normally in the 50's and 60's.
Using Outside temperature of 40degF, with a 200W growlight we get:
682 = (+16.5+27+5)*Trm - 50*16.5 - 40* (27 + 5)
682 = 44*Trm - 825 - 1320
2827 = 44*Trm
Trm = 2827/44 = 64degF
Using Outside temperature of 40degF, without a 200W growlight we get:
0 = (+16.5+27+5)*Trm - 50*16.5 - 40* (27 + 5)
0 = 44*Trm - 825 - 1320
2145 = 44*Trm
Trm = 2145/44 = 49degF
So, one might have a thermostat to power the growlights if the temperature dropped below 45degF.... AND Further, put the growlites on a timer to give it 12 hours each day (so the plants can get light) during night-time hours when it is coldest outside.
200W * 1kW/1000W * $0.11/kW-hr * 12hr/day = 0.26cents/day
--> $8/month
--> $48/Winter (6-months)
Toby
zoe_lithoi
Dec 25, 2014, 12:53:43 PM12/25/14
to
Greetings,
I took some temperature readings with a usb-type data logger for 2 days. one day had some 200Watt grow lights on, while the other did not. The 2 loggers, unfortunately were not very accurate because to start with, I had them both inside a room next to each other, and one read 80degF while the other read 82degF. I put one outside, and the other inside the greenhouse. I'm pasting the spreadsheet data here, and am not sure how it will appear when it's processed by Google.
No Lights 200W Lights Notes
Tmp-I Tmp-O Tmp-I Tmp-O
Date Time Inside Outside Inside Outside
12/22/14 1616 80 82 Both Temp Probes
In House
12/22/14 1816 63
12/23/14 0 57 56
12/23/14 700 52 53 low Tmp-I & Tmp-O
Equalization: No Heat
Flow into or out of
Greenhouse
12/23/14 830 52 55
12/23/14 1306 82 High Tmp-O
Heat into Ground
12/23/14 1420 62 69 High Tmp-I
12/23/14 1545 62 64
12/24/14 0 50 35
12/24/14 300 30
12/24/14 500 43 30
12/24/14 545 43 30
I estimate the 'ground temperature' equals 53degF by noting that with the lights off on 12/23/14 for a period around 700 (7am), the outside temperature and inside temperature were about equal. I call this equalization. There was no heat flowing into the greenhouse from the outside, and there was no heat flowing into or out of the greenhouse through the ground. This was another way to confirm my estimate in the previous posting on this thread where I said the ground temperature about 4inches deep was about 50degF. IT's not quite that simple. The temperatures 7 hours earlier at 0am on 12/23/14 show heat flowing from the greenhouse to the outside. This heat is being supplied by the ground. So what has happenned is that the ground temperature had heated up (charged up) prior to that, and now this thermal capacitor was discharging. The ground temperature had to have been greater than the greenhouse temp (57). What this tells me is that the ground temperature cycles on that day from about 53 to 58F.
Let's make a better estimate of the overall thermal resistance of the Greenhouse by looking at the temperatures around 1420 to 1545 on 12/23/14. The greenhouse temperature, Trm, was 62F, and the outside air temperature was about 65 to 66F (taking into account the 2deg temperature error mentioned above. In the last posting, I estimated that it had an R10 "R-Value" over a 273sqft surface area (walls and ceiling). Lets call this Rgv
Hrm = heat entering the greenhouse room of Temperature, Trm, from the outside air of temperature To, thru the 273sqft of Rgv walls and ceiling
Hrm = (To - Trm)degF*273sqft / Rgv hr-sqft-degF/Btu]
Hrm = (To - Trm)*273/Rgv Btu/hr
Hgnd = ground heat entering the greenhouse room of Temperature, Trm, thru the 50sqft of R0.33 dirt with temperature Tg
Hgnd = (Tg - Trm)*50sqft / [0.33 hr-sqft-F/Btu]
Hgnd = (Tg - Trm)*150 Btu/hr
If the room gets new air each hour equivalent to it's volume, then the air-exchange heat loss for the amount of heat the air in the 400cuft (8'*7'*7') room absorbs to go from To degF to Trm is:
Hair = (To - Trm)degF * 1/55 Btu/F /cuft * 400cuft
Hair = (To - Trm)* 5 Btu/hr
And lastly, there is another source of heat flow radiative in nature, Hrad, which for now we will assume is 0.
Kierkoff's Current (Heat) flow equation is:
Hrm + Hgnd + Hair + Hrad = 0
(To - Trm)*273/Rgv + (Tg - Trm)*150 + (To - Trm)*5 + 0 = 0
(To - Trm)*(273/Rgv + 5) + (Tg - Trm)*150 = 0
Now let's look at the data at 1420 to 1545 on 12/23/14 as stipulated:
Trm = 62F
To = 66F (The temperature range wa sfrom 69 to 64, but remember that this temperature probe recorded a 2degF higher temperature at the same location and time as the other probe, so the temp range was really 67 to 62. AT 62, it would be the same temperature as the other probe. So we will look at the 66F.)
Tg = 57F (in reality it could be anywhere between 55 to 58F, but since it's at the hottest part of the day and still charging up, 57F is a reasonable estimate IMO.)
(66 - 62)*(273/Rgv + 5) + (57 - 62)*150 = 0
(4)*(273/Rgv + 5) - (5)*150 = 0
(4)*(273/Rgv + 5) - (5)*150 = 0
1092/Rgv + 20 - 750 = 0
1092/Rgv = 730
Rgv = 1092/730 = 1.5
huh. I would have expected more....
I'll have to relook at this and look at the data better.... and look at the radiative heat
Toby
I took some temperature readings with a usb-type data logger for 2 days. one day had some 200Watt grow lights on, while the other did not. The 2 loggers, unfortunately were not very accurate because to start with, I had them both inside a room next to each other, and one read 80degF while the other read 82degF. I put one outside, and the other inside the greenhouse. I'm pasting the spreadsheet data here, and am not sure how it will appear when it's processed by Google.
No Lights 200W Lights Notes
Tmp-I Tmp-O Tmp-I Tmp-O
Date Time Inside Outside Inside Outside
12/22/14 1616 80 82 Both Temp Probes
In House
12/22/14 1816 63
12/23/14 0 57 56
12/23/14 700 52 53 low Tmp-I & Tmp-O
Equalization: No Heat
Flow into or out of
Greenhouse
12/23/14 830 52 55
12/23/14 1306 82 High Tmp-O
Heat into Ground
12/23/14 1420 62 69 High Tmp-I
12/23/14 1545 62 64
12/24/14 0 50 35
12/24/14 300 30
12/24/14 500 43 30
12/24/14 545 43 30
I estimate the 'ground temperature' equals 53degF by noting that with the lights off on 12/23/14 for a period around 700 (7am), the outside temperature and inside temperature were about equal. I call this equalization. There was no heat flowing into the greenhouse from the outside, and there was no heat flowing into or out of the greenhouse through the ground. This was another way to confirm my estimate in the previous posting on this thread where I said the ground temperature about 4inches deep was about 50degF. IT's not quite that simple. The temperatures 7 hours earlier at 0am on 12/23/14 show heat flowing from the greenhouse to the outside. This heat is being supplied by the ground. So what has happenned is that the ground temperature had heated up (charged up) prior to that, and now this thermal capacitor was discharging. The ground temperature had to have been greater than the greenhouse temp (57). What this tells me is that the ground temperature cycles on that day from about 53 to 58F.
Let's make a better estimate of the overall thermal resistance of the Greenhouse by looking at the temperatures around 1420 to 1545 on 12/23/14. The greenhouse temperature, Trm, was 62F, and the outside air temperature was about 65 to 66F (taking into account the 2deg temperature error mentioned above. In the last posting, I estimated that it had an R10 "R-Value" over a 273sqft surface area (walls and ceiling). Lets call this Rgv
Hrm = heat entering the greenhouse room of Temperature, Trm, from the outside air of temperature To, thru the 273sqft of Rgv walls and ceiling
Hrm = (To - Trm)degF*273sqft / Rgv hr-sqft-degF/Btu]
Hrm = (To - Trm)*273/Rgv Btu/hr
Hgnd = ground heat entering the greenhouse room of Temperature, Trm, thru the 50sqft of R0.33 dirt with temperature Tg
Hgnd = (Tg - Trm)*50sqft / [0.33 hr-sqft-F/Btu]
Hgnd = (Tg - Trm)*150 Btu/hr
If the room gets new air each hour equivalent to it's volume, then the air-exchange heat loss for the amount of heat the air in the 400cuft (8'*7'*7') room absorbs to go from To degF to Trm is:
Hair = (To - Trm)degF * 1/55 Btu/F /cuft * 400cuft
Hair = (To - Trm)* 5 Btu/hr
And lastly, there is another source of heat flow radiative in nature, Hrad, which for now we will assume is 0.
Kierkoff's Current (Heat) flow equation is:
Hrm + Hgnd + Hair + Hrad = 0
(To - Trm)*273/Rgv + (Tg - Trm)*150 + (To - Trm)*5 + 0 = 0
(To - Trm)*(273/Rgv + 5) + (Tg - Trm)*150 = 0
Now let's look at the data at 1420 to 1545 on 12/23/14 as stipulated:
Trm = 62F
To = 66F (The temperature range wa sfrom 69 to 64, but remember that this temperature probe recorded a 2degF higher temperature at the same location and time as the other probe, so the temp range was really 67 to 62. AT 62, it would be the same temperature as the other probe. So we will look at the 66F.)
Tg = 57F (in reality it could be anywhere between 55 to 58F, but since it's at the hottest part of the day and still charging up, 57F is a reasonable estimate IMO.)
(66 - 62)*(273/Rgv + 5) + (57 - 62)*150 = 0
(4)*(273/Rgv + 5) - (5)*150 = 0
(4)*(273/Rgv + 5) - (5)*150 = 0
1092/Rgv + 20 - 750 = 0
1092/Rgv = 730
Rgv = 1092/730 = 1.5
huh. I would have expected more....
I'll have to relook at this and look at the data better.... and look at the radiative heat
Toby
zoe_lithoi
Dec 25, 2014, 2:03:42 PM12/25/14
to
Greetings,
So, I'm refining my calculations by taking account the heatflow thru the door which I had just reckoned as part of the straw-infill greenhouse.
The ~2'x7' plywood 'door'has an Rvalue of about R0.5 hence the heatflow thru the door
Hdr = (To-Trm)*14sqft/R0.5sqft-degF-hr/Btu
Hdr = (To-Trm)* 28 Btu/hr
The surface area of the walls and ceiling can now be reduced by this 14sqft as well, from 273 to 259sqft:
Hrm = (To - Trm)*259/Rgv Btu/hr
Kierkoff's Current (Heat) flow equation is now :
Hrm + Hdr + Hgnd + Hair + Hrad = 0
(To - Trm)*259/Rgv + (To-Trm)* 28 + (Tg - Trm)*150 + (To - Trm)*5 + 0 = 0
(To - Trm)*(259/Rgv + 28 + 5) + (Tg - Trm)*150 = 0
(To - Trm)*(259/Rgv + 33) + (Tg - Trm)*150 = 0
Now let's look at the data, as in the previous posting at 1420 to 1545 on 12/23/14 as stipulated:
(66 - 62)*(259/Rgv + 33) + (57 - 62)*150 = 0
(4)*(259/Rgv + 33) - (5)*150 = 0
(4)*(259/Rgv + 33) - (5)*150 = 0
1092/Rgv + 132 - 750 = 0
1036/Rgv = 618
Rgv = 1036/618 = 1.7
It 'still' should be more.
If the ground temp is 58 instead of 57.
(4)*(259/Rgv + 33) - (4)*150 = 0
1092/Rgv + 132 - 600 = 0
1036/Rgv = 468
Rgv = 1036/468 = 2.2
Toby
So, I'm refining my calculations by taking account the heatflow thru the door which I had just reckoned as part of the straw-infill greenhouse.
The ~2'x7' plywood 'door'has an Rvalue of about R0.5 hence the heatflow thru the door
Hdr = (To-Trm)*14sqft/R0.5sqft-degF-hr/Btu
Hdr = (To-Trm)* 28 Btu/hr
The surface area of the walls and ceiling can now be reduced by this 14sqft as well, from 273 to 259sqft:
Hrm = (To - Trm)*259/Rgv Btu/hr
Kierkoff's Current (Heat) flow equation is now :
Hrm + Hdr + Hgnd + Hair + Hrad = 0
(To - Trm)*259/Rgv + (To-Trm)* 28 + (Tg - Trm)*150 + (To - Trm)*5 + 0 = 0
(To - Trm)*(259/Rgv + 28 + 5) + (Tg - Trm)*150 = 0
(To - Trm)*(259/Rgv + 33) + (Tg - Trm)*150 = 0
Now let's look at the data, as in the previous posting at 1420 to 1545 on 12/23/14 as stipulated:
Trm = 62F
To = 66F
Tg = 57F
To = 66F
(66 - 62)*(259/Rgv + 33) + (57 - 62)*150 = 0
(4)*(259/Rgv + 33) - (5)*150 = 0
(4)*(259/Rgv + 33) - (5)*150 = 0
1092/Rgv + 132 - 750 = 0
1036/Rgv = 618
Rgv = 1036/618 = 1.7
It 'still' should be more.
If the ground temp is 58 instead of 57.
(4)*(259/Rgv + 33) - (4)*150 = 0
1092/Rgv + 132 - 600 = 0
1036/Rgv = 468
Rgv = 1036/468 = 2.2
Toby
zoe_lithoi
Dec 25, 2014, 2:23:10 PM12/25/14
to
Greetings,
I am continuing to refine my calculations. This time by taking into account the radiative heat flow.
The room Temperature, Trm is 62F. The surface temperature of the floor is perhaps, Tfl=61F
Now, radiant heatflow is generally calculated with the temperatures
in degree Rankine, not Fehrenheit. To convert:
Trankine = Tfehrenheit + 460
The standard radiation function is defined as follows:
Qrad = S*E*F*A*(Trm^4 - Tfl^4)
where:
S = Stefan-Boltzmann Constant (SBC) = 0.119 x 10-10 BTU/Hr*in^2*R^4
Note: this is a constant, and R4 means Rankine (as opposed to
Fehrenheit) raised to the 4th power..
= 0.119 x 10^-10 BTU/Hr*in^2*R^4 * 144 in^2/1 ft^2
= 1.714 x 10^-9 Btu/Hr*ft^2*R^4
E = emissivity = 0.9 (according to:
http://ciks.cbt.nist.gov/bentz/nistir6551/node14.html)
F = geometric form factor = 1.0
A = area = 50 sqft (for the surface above the floor of the greenhouse)
Qrad = radiant heat flow rate (Heat/Time)
Tfl = Temperature of the floor surface in Rankine
Trm = room temperature in Rankine
Now, radiant heatflow is generally calculated with the temperatures
in degree Rankine, not Fehrenheit. To convert:
Trankine = Tfehrenheit + 460
Qrad = S*E*F*A*( (Trm+460)^4 - (Tfl+460)^4)
----------------------------------
Trm = 62F = 522 Rankine (The solar cistern's slab surface temperature)
Tfl = 61F = 521 Rankine
-----------------------------------
Qrad = FA(Tssc^4 - Th^4)
= 1.714x10^-9 Btu/Hr*ft^2*R^4 * 0.9*1* 50ft^2 *((522)^4
-(521)^4)
= 1.714x10^-9 Btu/Hr*ft^2*R^4 * 0.9*1* 50ft^2 *(7.424 x10^10 -7.368
x10^10)
= 77.13 * 0.056 x10^10
= 4.3 Btu/hr
Kierkoff's Current (Heat) flow equation is now :
Hrm + Hdr + Hgnd + Hair + Hrad = 0
(To - Trm)*259/Rgv + (To-Trm)* 28 + (Tg - Trm)*150 + (To - Trm)*5 + 4 = 0
(To - Trm)*(259/Rgv + 28 + 5) + (Tg - Trm)*150 + 4 = 0
(To - Trm)*(259/Rgv + 33) + (Tg - Trm)*150 + 4 = 0
Now let's look at the data, as in the previous posting at 1420 to 1545 on 12/23/14 as stipulated:
Trm = 62F
To = 66F
Tg = 58F
(66 - 62)*(259/Rgv + 33) + (58 - 62)*150 + 4 = 0
(4)*(259/Rgv + 33) - (4)*150 + 4 = 0
(4)*(259/Rgv + 33) - (4)*150 + 4 = 0
1092/Rgv + 132 - 600 + 4 = 0
1036/Rgv = 464
Rgv = 1036/464 = 2.23
It 'still' should be more.
Toby
I am continuing to refine my calculations. This time by taking into account the radiative heat flow.
The room Temperature, Trm is 62F. The surface temperature of the floor is perhaps, Tfl=61F
Now, radiant heatflow is generally calculated with the temperatures
in degree Rankine, not Fehrenheit. To convert:
Trankine = Tfehrenheit + 460
The standard radiation function is defined as follows:
Qrad = S*E*F*A*(Trm^4 - Tfl^4)
where:
S = Stefan-Boltzmann Constant (SBC) = 0.119 x 10-10 BTU/Hr*in^2*R^4
Note: this is a constant, and R4 means Rankine (as opposed to
Fehrenheit) raised to the 4th power..
= 0.119 x 10^-10 BTU/Hr*in^2*R^4 * 144 in^2/1 ft^2
= 1.714 x 10^-9 Btu/Hr*ft^2*R^4
E = emissivity = 0.9 (according to:
http://ciks.cbt.nist.gov/bentz/nistir6551/node14.html)
F = geometric form factor = 1.0
A = area = 50 sqft (for the surface above the floor of the greenhouse)
Qrad = radiant heat flow rate (Heat/Time)
Tfl = Temperature of the floor surface in Rankine
Trm = room temperature in Rankine
Now, radiant heatflow is generally calculated with the temperatures
in degree Rankine, not Fehrenheit. To convert:
Trankine = Tfehrenheit + 460
Qrad = S*E*F*A*( (Trm+460)^4 - (Tfl+460)^4)
----------------------------------
Trm = 62F = 522 Rankine (The solar cistern's slab surface temperature)
Tfl = 61F = 521 Rankine
-----------------------------------
Qrad = FA(Tssc^4 - Th^4)
= 1.714x10^-9 Btu/Hr*ft^2*R^4 * 0.9*1* 50ft^2 *((522)^4
-(521)^4)
= 1.714x10^-9 Btu/Hr*ft^2*R^4 * 0.9*1* 50ft^2 *(7.424 x10^10 -7.368
x10^10)
= 77.13 * 0.056 x10^10
= 4.3 Btu/hr
Kierkoff's Current (Heat) flow equation is now :
Hrm + Hdr + Hgnd + Hair + Hrad = 0
(To - Trm)*(259/Rgv + 28 + 5) + (Tg - Trm)*150 + 4 = 0
(To - Trm)*(259/Rgv + 33) + (Tg - Trm)*150 + 4 = 0
Now let's look at the data, as in the previous posting at 1420 to 1545 on 12/23/14 as stipulated:
Trm = 62F
To = 66F
(66 - 62)*(259/Rgv + 33) + (58 - 62)*150 + 4 = 0
(4)*(259/Rgv + 33) - (4)*150 + 4 = 0
(4)*(259/Rgv + 33) - (4)*150 + 4 = 0
1092/Rgv + 132 - 600 + 4 = 0
1036/Rgv = 464
Rgv = 1036/464 = 2.23
It 'still' should be more.
zoe_lithoi
Dec 25, 2014, 2:51:19 PM12/25/14
to
Greetings,
Something is not right.
I'm looking again at 2 things.
1. Rechecking the thermal resistance in the ground. We are looking at BTU per hour. Heat travels about 4" through the ground in 1 hour. (see the appendix after my signature).
2. There is what is called a 'warm-still' air resistance in series with the ground. It is small, and normally does not need to be taken into acount.
R0.67 sqft-hr-F/Btu for the warm air film
-----------------------------------
In 1 hour, the heat is supplied to the air by the 4" of dirt.
Dirt has an Rvalue of R0.083/inch
R0.083/inch * 4 = R0.33 sqft-hr-F/Btu for 4 inches of dirt
plus
R0.67 sqft-hr-F/Btu for the warm air film
= R1 sqft-hr-F/Btu
This is 3 times greater than my earlier value I used which was R0.33. So recalculating the heat through the floor:
Hgnd = ground heat entering the greenhouse room of Temperature, Trm, thru the 50sqft of R1 dirt with temperature Tg
Hgnd = (Tg - Trm)*50sqft / [R1 hr-sqft-F/Btu]
Hgnd = (Tg - Trm)*50 Btu/hr
Kierkoff's Current (Heat) flow equation is now :
Hrm + Hdr + Hgnd + Hair + Hrad = 0
(To - Trm)*259/Rgv + (To-Trm)* 28 + (Tg - Trm)*50 + (To - Trm)*5 + 4 = 0
(To - Trm)*(259/Rgv + 28 + 5) + (Tg - Trm)*50 + 4 = 0
(To - Trm)*(259/Rgv + 33) + (Tg - Trm)*50 + 4 = 0
Now let's look at the data, as in the previous posting at 1420 to 1545 on 12/23/14 as stipulated:
Trm = 62F
To = 66F
Tg = 58F
(66 - 62)*(259/Rgv + 33) + (58 - 62)*50 + 4 = 0
(4)*(259/Rgv + 33) - (4)*50 + 4 = 0
1092/Rgv + 132 - 200 + 4 = 0
1036/Rgv = 64
Rgv = 1036/64 = 16.2
That is much more realistic!!!...
My original estimation was R10.... Maybe 16.2 is too large.
Let's go back to the earlier posting which used Tg as 57.
(66 - 62)*(259/Rgv + 33) + (57 - 62)*50 + 4 = 0
(4)*(259/Rgv + 33) - (5)*50 + 4 = 0
1092/Rgv + 132 - 250 + 4 = 0
1036/Rgv = 114
Rgv = 1036/114 = R9
This seems the best fit.
Let's look at even more data.
Toby
Toby
Appendix: Heat travels about 4inches into the ground in 1 hour.
Let's look at the depth, D, below 1 sqft of slab surface, As, heat
will travel in 1 hr, t:
t = time = 1hr
As = 1 sqft
tr = thermal resistivity = 1.7 hr*ft*F/Btu
(fig 11-1 of my earth-coupled heat transfer book)
Cv = 30 Btu/F/cuft
td = thermal diffusivity = 1/tr*C
-*-*-*-*-*-*-*-*-*
C = 30 Btu/F/cuft * D*1sqft = 30D Btu/F
td = 1/[(1.7 hr*ft*F/Btu)*(30D Btu/F)] = 1/51D sqft/hr
D = (1/51D sqft/hr)/(D ft) * 1 hr
D = 1/[51*D^2]
D^3 = 1/51 = 0.31 ft
D = (1/51)^(1/3) = 0.307' = 3.7"
Something is not right.
I'm looking again at 2 things.
1. Rechecking the thermal resistance in the ground. We are looking at BTU per hour. Heat travels about 4" through the ground in 1 hour. (see the appendix after my signature).
2. There is what is called a 'warm-still' air resistance in series with the ground. It is small, and normally does not need to be taken into acount.
R0.67 sqft-hr-F/Btu for the warm air film
-----------------------------------
In 1 hour, the heat is supplied to the air by the 4" of dirt.
Dirt has an Rvalue of R0.083/inch
R0.083/inch * 4 = R0.33 sqft-hr-F/Btu for 4 inches of dirt
plus
R0.67 sqft-hr-F/Btu for the warm air film
= R1 sqft-hr-F/Btu
This is 3 times greater than my earlier value I used which was R0.33. So recalculating the heat through the floor:
Hgnd = ground heat entering the greenhouse room of Temperature, Trm, thru the 50sqft of R1 dirt with temperature Tg
Hgnd = (Tg - Trm)*50sqft / [R1 hr-sqft-F/Btu]
Hgnd = (Tg - Trm)*50 Btu/hr
Kierkoff's Current (Heat) flow equation is now :
Hrm + Hdr + Hgnd + Hair + Hrad = 0
(To - Trm)*(259/Rgv + 28 + 5) + (Tg - Trm)*50 + 4 = 0
(To - Trm)*(259/Rgv + 33) + (Tg - Trm)*50 + 4 = 0
Now let's look at the data, as in the previous posting at 1420 to 1545 on 12/23/14 as stipulated:
Trm = 62F
To = 66F
Tg = 58F
(4)*(259/Rgv + 33) - (4)*50 + 4 = 0
1092/Rgv + 132 - 200 + 4 = 0
1036/Rgv = 64
Rgv = 1036/64 = 16.2
That is much more realistic!!!...
My original estimation was R10.... Maybe 16.2 is too large.
Let's go back to the earlier posting which used Tg as 57.
(66 - 62)*(259/Rgv + 33) + (57 - 62)*50 + 4 = 0
(4)*(259/Rgv + 33) - (5)*50 + 4 = 0
1092/Rgv + 132 - 250 + 4 = 0
1036/Rgv = 114
Rgv = 1036/114 = R9
This seems the best fit.
Let's look at even more data.
Toby
Toby
Appendix: Heat travels about 4inches into the ground in 1 hour.
Let's look at the depth, D, below 1 sqft of slab surface, As, heat
will travel in 1 hr, t:
t = time = 1hr
As = 1 sqft
tr = thermal resistivity = 1.7 hr*ft*F/Btu
(fig 11-1 of my earth-coupled heat transfer book)
Cv = 30 Btu/F/cuft
td = thermal diffusivity = 1/tr*C
-*-*-*-*-*-*-*-*-*
C = 30 Btu/F/cuft * D*1sqft = 30D Btu/F
td = 1/[(1.7 hr*ft*F/Btu)*(30D Btu/F)] = 1/51D sqft/hr
D = (1/51D sqft/hr)/(D ft) * 1 hr
D = 1/[51*D^2]
D^3 = 1/51 = 0.31 ft
D = (1/51)^(1/3) = 0.307' = 3.7"
zoe_lithoi
Dec 25, 2014, 6:10:28 PM12/25/14
to
Straw-infilled-Pallet Winter Greenhouse.
Greetings,
I've taken all the postings, and condensed all the refinements into this one posting....
My son and I are building a 80" x 80" x 8'(tall) 'winter' Greenhouse made of pallets screwed together and stuffed with straw for insulation which I estimate would be about R10, but which, based on temperature measurements, turns out to be about R8.8. It has plastic stapled to the inside which should prevent heat loss by convection. The Roof has masonite siding on it in case of rain. We will have 200W of grow lights which will also serve as the primary heat source. The 200W grow lights, based on the temperature data, yield only about 90W of heat, while the rest of the electrical energy is converted to light. The ground will be a heat source when the outside temperature is lower than about 53degF, and the ground is a heat sink when the outside temperature is greater than 53degF. The goal is to keep the temperature above freezing so the seedlings don't die.
The surface area of the 4 walls, and ceiling would be:
A = 4*7*8 + 7*7 = 273 sqft
==========================
Wall and Ceiling Rvalue (Straw-Infilled Pallets)
Wood Door Rvalue
==========================
The Rvalue of a strawbale is somewhere around R50 for a 24inch wide or so bale. So a 5inch wide straw-infilled palet might be 1/5th of this, i.e. R10. In this case, it would be over a 273sqft surface area (walls and ceiling). Lets call this Rgv. Really, though, the 2'x7', i.e. 14Sqft door, has an Rvalue of about R0.5. Leaving the Wall area = 273 - 14 = 259sqft
Hdr = heat entering the greenhouse room of Temperature, Trm, from the outside air of temperature To, thru the 14sqft of R0.5 wood door
Hdr = (To - Trm)degF*14sqft / R0.5 hr-sqft-degF/Btu]
Hdr = (To - Trm)*28 Btu/hr
Hrm = heat entering the greenhouse room of Temperature, Trm, from the outside air of temperature To, thru the 259sqft of Rgv walls and ceiling
Hrm = (To - Trm)degF*259sqft / Rgv hr-sqft-degF/Btu]
Hrm = (To - Trm)*259/Rgv Btu/hr
=============================
Ground Heat and the 'Warm Air Still Resistance'
=============================
Heat travels about 4" through the ground in 1 hour. (see the appendix after my signature). In 1 hour, the heat is supplied to the air by the 4" of dirt. There is what is called a 'warm-still' air resistance in series with the ground. It is small, and normally does not need to be taken into acount, but because the ground also has a 'small' R-value, we need to take it into account.
If the room gets new air each hour equivalent to it's volume, then the air-exchange heat loss for the amount of heat the air in the 400cuft (8'*7'*7') room absorbs to go from To degF to Trm is:
Hair = (To - Trm)degF * 1/55 Btu/F /cuft * 400cuft
Hair = (To - Trm)* 5 Btu/hr
(To - Trm)*259/Rgv + (To-Trm)* 28 + (Tg - Trm)*50 + (To - Trm)*5 = 0
(To - Trm)*(259/Rgv + 28 + 5) + (Tg - Trm)*50 = 0
(To - Trm)*(259/Rgv + 33) + (Tg - Trm)*50 = 0
Now plugging in the temperature values just given above...
(66 - 62)*(259/Rgv + 33) + (57 - 62)*50 = 0
(4)*(259/Rgv + 33) - (5)*50 = 0
1092/Rgv + 132 - 250 = 0
1036/Rgv = 118
Rgv = 1036/118 = R8.8
Kierkoff's Current (Heat) flow equation is now :
Hrm + Hdr + Hgnd + Hair = 0
(To - Trm)*259/Rgv + (To-Trm)* 28 + (Tg - Trm)*50 + (To - Trm)*5 = 0
(To - Trm)*(259/R8.8 + 28 + 5) + (Tg - Trm)*50 = 0
(To - Trm)*(29 + 33) + (Tg - Trm)*50 = 0
(To - Trm)*(62) + (Tg - Trm)*50 = 0
=============================================
Now, let's add a 200W growlite.
Kierkoff's Current (Heat) flow equation is now :
Hrm + Hdr + Hgnd + Hair + Hgl = 0
(To - Trm)*259/Rgv + (To-Trm)* 28 + (Tg - Trm)*50 + (To - Trm)*5 + 682 = 0
(To - Trm)*(259/R8.8 + 28 + 5) + (Tg - Trm)*50 + 682 = 0
(To - Trm)*(29 + 33) + (Tg - Trm)*50 + 682 = 0
(To - Trm)*(62) + (Tg - Trm)*50 + 682 = 0
The data shows that Tg varies from 53 to 57degF, 53 being the coldest part of the night. If the outside air, To, is 30degF, what would Trm be?
(30 - Trm)*(62) + (53 - Trm)*50 + 682 = 0
1860 - Trm*62 + 2650 - Trm*50 + 682 = 0
5192 = 112*Trm
Trm = 5192/112 = 46degF
The data between 5am and 5:40am on 12/24/14 shows that Trm = 43degF when To = 30.
The 'theory' now matches the real data fairly well.
What could be the source of the 3degF difference (46degF - 43degF)?
Perhaps, it is our grow lites are not producing 200W of heat.
Let's make Trm=43, and solve for the grow lite Heat, Hgl
1860 - Trm*62 + 2650 - Trm*50 + Hgl = 0
4510 - Trm*(62 + 50) + Hgl = 0
4510 - 43*112 + Hgl = 0
4510 - 4816 + Hgl = 0
Hgl = 306 Btu/hr
Hgl = 306 Btu/hr * 1 Watt/3.41Btu/hr = 90watts
So, of the 200W electrical input, 90Watts make heat, and 110 Watts go towards making light. So, from a heat standpoint, 90/200 = 45% efficiency.
From a light perspective, 110/200 = 55% efficiency.
Hgl really equals:
Hgl = 0.45 * 200W *3.41 Btu/1Watt-hr = 306 Btu/hr
Now, redoing the model:
Kierkoff's Current (Heat) flow equation is now :
Hrm + Hdr + Hgnd + Hair + Hgl = 0
(To - Trm)*259/Rgv + (To-Trm)* 28 + (Tg - Trm)*50 + (To - Trm)*5 + 306 = 0
(To - Trm)*(259/R8.8 + 28 + 5) + (Tg - Trm)*50 + 306 = 0
(To - Trm)*(29 + 33) + (Tg - Trm)*50 + 306
(To - Trm)*(62) + (Tg - Trm)*50 + 306 = 0
If the outside air, To, is 30degF, and the Ground Temp, Tg = 53degF, what would Trm be?
(30 - Trm)*(62) + (53 - Trm)*50 + 306 = 0
1860 - Trm*62 + 2650 - Trm*50 + 306 = 0
4816 = 112Trm
Trm = 4816/112 = 43degF
If the outside air, To, is 20degF, and the Ground Temp, Tg = 53degF, what would Trm be?
(20 - Trm)*(62) + (53 - Trm)*50 + 306 = 0
1240 - Trm*62 + 2650 - Trm*50 + 306 = 0
4196 = 112Trm
Trm = 4816/112 = 37 degF
What would be the coldest outside air, To, such that the room temp, Trm = just above freezing, i.e. 33DegF? ( Ground Temp, Tg = 53degF)
(To - 33)*(62) + (53 - 33)*50 + 306 = 0
62To - 2046 + 1000 + 306 = 0
To = 740/62
To = 12 degF
If the coldest we want the room to be = 45degF, and suppose the coldest it gets outside is 0degF, then what wattage of grow lights would we need?
(To - Trm)*(62) + (Tg - Trm)*50 + Hgl = 0
(0 - 45)*(62) + (53 - 45)*50 + Hgl = 0
-2790 + 400 +Hgl = 0
Hgl = 2390 Btu/hr
Hgl = 2390 Btu/hr * 1Watt/3.41 Btu/1Watt = 700W
Since the grow lights are 45% efficient from a heat perspective.
700W/0.45 = 1560Watts of Grow Lights are needed.
Since the existing Grow Lights (200W light bulbs) produce 90W of 'heat', one could just get a 610Watt heater.
Greetings,
I've taken all the postings, and condensed all the refinements into this one posting....
My son and I are building a 80" x 80" x 8'(tall) 'winter' Greenhouse made of pallets screwed together and stuffed with straw for insulation which I estimate would be about R10, but which, based on temperature measurements, turns out to be about R8.8. It has plastic stapled to the inside which should prevent heat loss by convection. The Roof has masonite siding on it in case of rain. We will have 200W of grow lights which will also serve as the primary heat source. The 200W grow lights, based on the temperature data, yield only about 90W of heat, while the rest of the electrical energy is converted to light. The ground will be a heat source when the outside temperature is lower than about 53degF, and the ground is a heat sink when the outside temperature is greater than 53degF. The goal is to keep the temperature above freezing so the seedlings don't die.
The surface area of the 4 walls, and ceiling would be:
A = 4*7*8 + 7*7 = 273 sqft
I took some temperature readings with a usb-type data logger for 2 days. one day had some 200Watt grow lights on, while the other did not. The 2 loggers, unfortunately were not very accurate because to start with, I had them both inside a room next to each other, and one read 80degF while the other read 82degF. I put one outside, and the other inside the greenhouse. I'm pasting the spreadsheet data here, and am not sure how it will appear when it's processed by Google.
No Lights 200W Lights Notes
Tmp-I Tmp-O Tmp-I Tmp-O
Date Time Inside Outside Inside Outside
12/22/14 1616 80 82 Both Temp Probes
In House
12/22/14 1816 63
12/23/14 0 57 56
12/23/14 700 52 53 low Tmp-I & Tmp-O
Equalization: No Heat
Flow into or out of
Greenhouse
12/23/14 830 52 55
12/23/14 1306 82 High Tmp-O
Heat into Ground
12/23/14 1420 62 69 High Tmp-I
12/23/14 1545 62 64
12/24/14 0 50 35
12/24/14 300 30
12/24/14 500 43 30
12/24/14 545 43 30
I estimate the 'ground temperature' equals 53degF by noting that with the lights off on 12/23/14 for a period around 700 (7am), the outside temperature and inside temperature were about equal. I call this equalization. There was no heat flowing into the greenhouse from the outside, and there was no heat flowing into or out of the greenhouse through the ground. This was another way to confirm my estimate in the previous posting on this thread where I said the ground temperature about 4inches deep was about 50degF. IT's not quite that simple. The temperatures 7 hours earlier at 0am on 12/23/14 show heat flowing from the greenhouse to the outside. This heat is being supplied by the ground. So what has happenned is that the ground temperature had heated up (charged up) prior to that, and now this thermal capacitor was discharging. The ground temperature had to have been greater than the greenhouse temp (57). What this tells me is that the ground temperature cycles on that day from about 53 to 58F.
Let's make a better estimate of the the thermal resistance of the Greenhouse by looking at the temperatures around 1420 to 1545 on 12/23/14. The greenhouse temperature, Trm, was 62F, and the outside air temperature was about 65 to 66F (taking into account the 2deg temperature error mentioned above.
No Lights 200W Lights Notes
Tmp-I Tmp-O Tmp-I Tmp-O
Date Time Inside Outside Inside Outside
12/22/14 1616 80 82 Both Temp Probes
In House
12/22/14 1816 63
12/23/14 0 57 56
12/23/14 700 52 53 low Tmp-I & Tmp-O
Equalization: No Heat
Flow into or out of
Greenhouse
12/23/14 830 52 55
12/23/14 1306 82 High Tmp-O
Heat into Ground
12/23/14 1420 62 69 High Tmp-I
12/23/14 1545 62 64
12/24/14 0 50 35
12/24/14 300 30
12/24/14 500 43 30
12/24/14 545 43 30
I estimate the 'ground temperature' equals 53degF by noting that with the lights off on 12/23/14 for a period around 700 (7am), the outside temperature and inside temperature were about equal. I call this equalization. There was no heat flowing into the greenhouse from the outside, and there was no heat flowing into or out of the greenhouse through the ground. This was another way to confirm my estimate in the previous posting on this thread where I said the ground temperature about 4inches deep was about 50degF. IT's not quite that simple. The temperatures 7 hours earlier at 0am on 12/23/14 show heat flowing from the greenhouse to the outside. This heat is being supplied by the ground. So what has happenned is that the ground temperature had heated up (charged up) prior to that, and now this thermal capacitor was discharging. The ground temperature had to have been greater than the greenhouse temp (57). What this tells me is that the ground temperature cycles on that day from about 53 to 58F.
==========================
Wall and Ceiling Rvalue (Straw-Infilled Pallets)
Wood Door Rvalue
==========================
The Rvalue of a strawbale is somewhere around R50 for a 24inch wide or so bale. So a 5inch wide straw-infilled palet might be 1/5th of this, i.e. R10. In this case, it would be over a 273sqft surface area (walls and ceiling). Lets call this Rgv. Really, though, the 2'x7', i.e. 14Sqft door, has an Rvalue of about R0.5. Leaving the Wall area = 273 - 14 = 259sqft
Hdr = heat entering the greenhouse room of Temperature, Trm, from the outside air of temperature To, thru the 14sqft of R0.5 wood door
Hdr = (To - Trm)degF*14sqft / R0.5 hr-sqft-degF/Btu]
Hdr = (To - Trm)*28 Btu/hr
Hrm = heat entering the greenhouse room of Temperature, Trm, from the outside air of temperature To, thru the 259sqft of Rgv walls and ceiling
Hrm = (To - Trm)degF*259sqft / Rgv hr-sqft-degF/Btu]
Hrm = (To - Trm)*259/Rgv Btu/hr
=============================
Ground Heat and the 'Warm Air Still Resistance'
=============================
Heat travels about 4" through the ground in 1 hour. (see the appendix after my signature). In 1 hour, the heat is supplied to the air by the 4" of dirt. There is what is called a 'warm-still' air resistance in series with the ground. It is small, and normally does not need to be taken into acount, but because the ground also has a 'small' R-value, we need to take it into account.
R0.67 sqft-hr-F/Btu for the warm air film
Dirt has an Rvalue of R0.083/inch
R0.083/inch * 4 = R0.33 sqft-hr-F/Btu for 4 inches of dirt
plus
R0.67 sqft-hr-F/Btu for the warm air film
= R1 sqft-hr-F/Btu
R0.083/inch * 4 = R0.33 sqft-hr-F/Btu for 4 inches of dirt
plus
R0.67 sqft-hr-F/Btu for the warm air film
= R1 sqft-hr-F/Btu
Hgnd = ground heat entering the greenhouse room of Temperature, Trm, thru the 50sqft of R0.33 dirt with temperature Tg
Hgnd = (Tg - Trm)*50sqft / [R1 hr-sqft-F/Btu]
Hgnd = (Tg - Trm)*50 Btu/hr
If the room gets new air each hour equivalent to it's volume, then the air-exchange heat loss for the amount of heat the air in the 400cuft (8'*7'*7') room absorbs to go from To degF to Trm is:
Hair = (To - Trm)degF * 1/55 Btu/F /cuft * 400cuft
Hair = (To - Trm)* 5 Btu/hr
Now let's look at the data at 1420 to 1545 on 12/23/14 as stipulated:
Trm = 62F
To = 66F (The temperature range wa sfrom 69 to 64, but remember that this temperature probe recorded a 2degF higher temperature at the same location and time as the other probe, so the temp range was really 67 to 62. AT 62, it would be the same temperature as the other probe. So we will look at the 66F.)
Tg = 57F (in reality it could be anywhere between 55 to 58F, but since it's at the hottest part of the day and still charging up, 57F is a reasonable estimate IMO.)
Trm = 62F
To = 66F (The temperature range wa sfrom 69 to 64, but remember that this temperature probe recorded a 2degF higher temperature at the same location and time as the other probe, so the temp range was really 67 to 62. AT 62, it would be the same temperature as the other probe. So we will look at the 66F.)
Tg = 57F (in reality it could be anywhere between 55 to 58F, but since it's at the hottest part of the day and still charging up, 57F is a reasonable estimate IMO.)
Kierkoff's Current (Heat) flow equation is now :
Hrm + Hdr + Hgnd + Hair = 0
(To - Trm)*259/Rgv + (To-Trm)* 28 + (Tg - Trm)*50 + (To - Trm)*5 = 0
(To - Trm)*(259/Rgv + 28 + 5) + (Tg - Trm)*50 = 0
(To - Trm)*(259/Rgv + 33) + (Tg - Trm)*50 = 0
Now plugging in the temperature values just given above...
(66 - 62)*(259/Rgv + 33) + (57 - 62)*50 = 0
(4)*(259/Rgv + 33) - (5)*50 = 0
1092/Rgv + 132 - 250 = 0
1036/Rgv = 118
Rgv = 1036/118 = R8.8
Kierkoff's Current (Heat) flow equation is now :
(To - Trm)*259/Rgv + (To-Trm)* 28 + (Tg - Trm)*50 + (To - Trm)*5 = 0
(To - Trm)*(259/R8.8 + 28 + 5) + (Tg - Trm)*50 = 0
(To - Trm)*(29 + 33) + (Tg - Trm)*50 = 0
(To - Trm)*(62) + (Tg - Trm)*50 = 0
=============================================
Now, let's add a 200W growlite.
The heat from the growlights
Hgl = 200W * 3.41 Btu/1Watt-hr = 682 Btu/hr
The problem is, that not all of this 200W is going towards 'heat'. Some of it is going towards light. We will have to compare theory with real data to make a better thermal model.
Hgl = 200W * 3.41 Btu/1Watt-hr = 682 Btu/hr
Kierkoff's Current (Heat) flow equation is now :
(To - Trm)*259/Rgv + (To-Trm)* 28 + (Tg - Trm)*50 + (To - Trm)*5 + 682 = 0
(To - Trm)*(259/R8.8 + 28 + 5) + (Tg - Trm)*50 + 682 = 0
(To - Trm)*(29 + 33) + (Tg - Trm)*50 + 682 = 0
(To - Trm)*(62) + (Tg - Trm)*50 + 682 = 0
The data shows that Tg varies from 53 to 57degF, 53 being the coldest part of the night. If the outside air, To, is 30degF, what would Trm be?
(30 - Trm)*(62) + (53 - Trm)*50 + 682 = 0
1860 - Trm*62 + 2650 - Trm*50 + 682 = 0
5192 = 112*Trm
Trm = 5192/112 = 46degF
The data between 5am and 5:40am on 12/24/14 shows that Trm = 43degF when To = 30.
The 'theory' now matches the real data fairly well.
What could be the source of the 3degF difference (46degF - 43degF)?
Perhaps, it is our grow lites are not producing 200W of heat.
Let's make Trm=43, and solve for the grow lite Heat, Hgl
1860 - Trm*62 + 2650 - Trm*50 + Hgl = 0
4510 - Trm*(62 + 50) + Hgl = 0
4510 - 43*112 + Hgl = 0
4510 - 4816 + Hgl = 0
Hgl = 306 Btu/hr
Hgl = 306 Btu/hr * 1 Watt/3.41Btu/hr = 90watts
So, of the 200W electrical input, 90Watts make heat, and 110 Watts go towards making light. So, from a heat standpoint, 90/200 = 45% efficiency.
From a light perspective, 110/200 = 55% efficiency.
Hgl really equals:
Hgl = 0.45 * 200W *3.41 Btu/1Watt-hr = 306 Btu/hr
Now, redoing the model:
Kierkoff's Current (Heat) flow equation is now :
(To - Trm)*259/Rgv + (To-Trm)* 28 + (Tg - Trm)*50 + (To - Trm)*5 + 306 = 0
(To - Trm)*(259/R8.8 + 28 + 5) + (Tg - Trm)*50 + 306 = 0
(To - Trm)*(29 + 33) + (Tg - Trm)*50 + 306
(To - Trm)*(62) + (Tg - Trm)*50 + 306 = 0
If the outside air, To, is 30degF, and the Ground Temp, Tg = 53degF, what would Trm be?
(30 - Trm)*(62) + (53 - Trm)*50 + 306 = 0
1860 - Trm*62 + 2650 - Trm*50 + 306 = 0
4816 = 112Trm
Trm = 4816/112 = 43degF
If the outside air, To, is 20degF, and the Ground Temp, Tg = 53degF, what would Trm be?
(20 - Trm)*(62) + (53 - Trm)*50 + 306 = 0
1240 - Trm*62 + 2650 - Trm*50 + 306 = 0
4196 = 112Trm
Trm = 4816/112 = 37 degF
What would be the coldest outside air, To, such that the room temp, Trm = just above freezing, i.e. 33DegF? ( Ground Temp, Tg = 53degF)
(To - 33)*(62) + (53 - 33)*50 + 306 = 0
62To - 2046 + 1000 + 306 = 0
To = 740/62
To = 12 degF
If the coldest we want the room to be = 45degF, and suppose the coldest it gets outside is 0degF, then what wattage of grow lights would we need?
(To - Trm)*(62) + (Tg - Trm)*50 + Hgl = 0
(0 - 45)*(62) + (53 - 45)*50 + Hgl = 0
-2790 + 400 +Hgl = 0
Hgl = 2390 Btu/hr
Hgl = 2390 Btu/hr * 1Watt/3.41 Btu/1Watt = 700W
Since the grow lights are 45% efficient from a heat perspective.
700W/0.45 = 1560Watts of Grow Lights are needed.
Since the existing Grow Lights (200W light bulbs) produce 90W of 'heat', one could just get a 610Watt heater.
rakeshm...@gmail.com
Aug 28, 2015, 2:19:43 AM8/28/15
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