Posted by zoe_lithoi on November 30, 2014, 5:28 pm
Straw-infilled-Pallet Winter Greenhouse.
Greetings,
My son and I are building a 80" x 80" x 8'(tall) 'winter' Greenhouse made o
f pallets screwed together and stuffed with straw for insulation (R10?). I
t has plastic stapled to the inside which should prevent heat loss by conve
ction. 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 groun
d 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 B
tu/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 B
tu/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 th
is 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 t
ravels 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/Bt
u
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 4
0degF
See: https://weatherspark.com/averages/31890/1/Las-Vegas-Nevada-United-Stat
es
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 * $.11/kW-hr * 12hr/day = 0.26cents/day
--> $/month
--> $8/Winter (6-months)
Toby
Posted by zoe_lithoi on December 25, 2014, 5:53 pm
Greetings,
I took some temperature readings with a usb-type data logger for 2 days. on
e day had some 200Watt grow lights on, while the other did not. The 2 logge
rs, 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 pa
sting 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 l
ights off on 12/23/14 for a period around 700 (7am), the outside temperatu
re 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 an
other way to confirm my estimate in the previous posting on this thread whe
re 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 su
pplied 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 wa
s discharging. The ground temperature had to have been greater than the gre
enhouse 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 Green
house by looking at the temperatures around 1420 to 1545 on 12/23/14. The g
reenhouse temperature, Trm, was 62F, and the outside air temperature was ab
out 65 to 66F (taking into account the 2deg temperature error mentioned abo
ve. In the last posting, I estimated that it had an R10 "R-Value" over a 27
3sqft surface area (walls and ceiling). Lets call this Rgv
Hrm = heat entering the greenhouse room of Temperature, Trm, from the ou
tside 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, thr
u 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 th
e 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 reasonabl
e 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
On Sunday, November 30, 2014 9:28:27 AM UTC-8, zoe_lithoi wrote:
> 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 con
vection. The Roof has masonite siding on it in case of rain. We will have 2
00 of grow lights which will also serve as the primary heat source. The gro
und will be either a heat source or a heat sink (not sure yet.....). The go
al 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 groun
d.
>
> Hrm = heat leaving the room of Temperature, Trm, thru the 273sqft of R
10 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 ai
r-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 dept
h 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 th
e 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-St
ates
>
> 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 temperatur
e dropped below 45degF.... AND Further, put the growlites on a timer to giv
e it 12 hours each day (so the plants can get light) during night-time hour
s when it is coldest outside.
>
> 200W * 1kW/1000W * $.11/kW-hr * 12hr/day = 0.26cents/day
> --> $/month
> --> $8/Winter (6-months)
>
> Toby
Posted by zoe_lithoi on December 25, 2014, 7:03 pm
Greetings,
So, I'm refining my calculations by taking account the heatflow thru the do
or 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 thr
u 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 1
2/23/14 as stipulated:
Trm = 62F
To = 66F
Tg = 57F
(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
On Thursday, December 25, 2014 9:53:43 AM UTC-8, zoe_lithoi wrote:
> 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 log
gers, unfortunately were not very accurate because to start with, I had the
m both inside a room next to each other, and one read 80degF while the othe
r 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 tempera
ture and inside temperature were about equal. I call this equalization. The
re 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 w
here 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 12/23/1
4 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 temperatur
e 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 g
reenhouse temp (57). What this tells me is that the ground temperature cycl
es on that day from about 53 to 58F.
>
> Let's make a better estimate of the overall thermal resistance of the Gre
enhouse 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 a
bove. 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, t
hru 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 ai
r-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, Hra
d, 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 th
is temperature probe recorded a 2degF higher temperature at the same locati
on and time as the other probe, so the temp range was really 67 to 62. AT 6
2, 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 reasona
ble 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 t
he radiative heat
>
> Toby
>
> On Sunday, November 30, 2014 9:28:27 AM UTC-8, zoe_lithoi wrote:
> > Straw-infilled-Pallet Winter Greenhouse.
> > Greetings,
> >
> > My son and I are building a 80" x 80" x 8'(tall) 'winter' Greenhouse ma
de of pallets screwed together and stuffed with straw for insulation (R10?)
. It has plastic stapled to the inside which should prevent heat loss by c
onvection. 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 g
round 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 gro
und.
> >
> > 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 temperatu
re, for the southwest (NEvada, Utah, Arizona), 4inches deep (the approix de
pth heat can travel in the ground in 1 hour --- see the 'daycreek' thread i
n this group), is about 50degF. So as long as the greenhouse temperature, T
rm, 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 he
at 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 temperat
ure dropped below 45degF.... AND Further, put the growlites on a timer to g
ive it 12 hours each day (so the plants can get light) during night-time ho
urs when it is coldest outside.
> >
> > 200W * 1kW/1000W * $.11/kW-hr * 12hr/day = 0.26cents/day
> > --> $/month
> > --> $8/Winter (6-months)
> >
> > Toby
Posted by zoe_lithoi on December 25, 2014, 7:23 pm
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 p
erhaps, TflaF
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 1
2/23/14 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
On Thursday, December 25, 2014 11:03:42 AM UTC-8, zoe_lithoi wrote:
> 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 t
hru 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 14sq
ft 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
>
> (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
>
> On Thursday, December 25, 2014 9:53:43 AM UTC-8, zoe_lithoi wrote:
> > 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 l
oggers, unfortunately were not very accurate because to start with, I had t
hem both inside a room next to each other, and one read 80degF while the ot
her 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 whe
n 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 t
he lights off on 12/23/14 for a period around 700 (7am), the outside tempe
rature and inside temperature were about equal. I call this equalization. T
here was no heat flowing into the greenhouse from the outside, and there wa
s no heat flowing into or out of the greenhouse through the ground. This wa
s 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. I
T'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 bein
g supplied by the ground. So what has happenned is that the ground temperat
ure had heated up (charged up) prior to that, and now this thermal capacito
r was discharging. The ground temperature had to have been greater than the
greenhouse temp (57). What this tells me is that the ground temperature cy
cles on that day from about 53 to 58F.
> >
> > Let's make a better estimate of the overall thermal resistance of the G
reenhouse by looking at the temperatures around 1420 to 1545 on 12/23/14. T
he greenhouse temperature, Trm, was 62F, and the outside air temperature wa
s 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 th
e 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, H
rad, 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 loca
tion 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 a
t the 66F.)
> > Tg = 57F (in reality it could be anywhere between 55 to 58F, but sinc
e it's at the hottest part of the day and still charging up, 57F is a reaso
nable 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
> >
> > On Sunday, November 30, 2014 9:28:27 AM UTC-8, zoe_lithoi wrote:
> > > 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 ha
ve 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.....). Th
e 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 g
round.
> > >
> > > 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 th
e 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 tempera
ture, 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 t
o 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-sqf
t-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 agai
n 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 Jul
y is 40degF
> > > See: https://weatherspark.com/averages/31890/1/Las-Vegas-Nevada-Unite
d-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 temper
ature 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 * $.11/kW-hr * 12hr/day = 0.26cents/day
> > > --> $/month
> > > --> $8/Winter (6-months)
> > >
> > > Toby
Posted by zoe_lithoi on December 25, 2014, 7:51 pm
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 p
er hour. Heat travels about 4" through the ground in 1 hour. (see the appe
ndix 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 re
calculating the heat through the floor:
Hgnd = ground heat entering the greenhouse room of Temperature, Trm, thr
u 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 1
2/23/14 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"
On Thursday, December 25, 2014 11:23:10 AM UTC-8, zoe_lithoi wrote:
> Greetings,
>
> I am continuing to refine my calculations. This time by taking into accou
nt the radiative heat flow.
>
> The room Temperature, Trm is 62F. The surface temperature of the floor is
perhaps, TflaF
>
> 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 greenhou
se)
> 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
>
> On Thursday, December 25, 2014 11:03:42 AM UTC-8, zoe_lithoi wrote:
> > Greetings,
> >
> > So, I'm refining my calculations by taking account the heatflow thru th
e 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 14
sqft 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
> >
> > (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
> >
> > On Thursday, December 25, 2014 9:53:43 AM UTC-8, zoe_lithoi wrote:
> > > Greetings,
> > >
> > > I took some temperature readings with a usb-type data logger for 2 da
ys. 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 w
hen 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 tem
perature 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 thre
ad 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 be
ing supplied by the ground. So what has happenned is that the ground temper
ature had heated up (charged up) prior to that, and now this thermal capaci
tor was discharging. The ground temperature had to have been greater than t
he 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 mention
ed above. In the last posting, I estimated that it had an R10 "R-Value" ove
r 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 ceilin
g
> > > 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, Tr
m, 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 th
e 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 tha
t this temperature probe recorded a 2degF higher temperature at the same lo
cation 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 si
nce it's at the hottest part of the day and still charging up, 57F is a rea
sonable 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
> > >
> > > On Sunday, November 30, 2014 9:28:27 AM UTC-8, zoe_lithoi wrote:
> > > > Straw-infilled-Pallet Winter Greenhouse.
> > > > Greetings,
> > > >
> > > > My son and I are building a 80" x 80" x 8'(tall) 'winter' Greenhous
e made of pallets screwed together and stuffed with straw for insulation (R
10?). 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. T
he 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 d
ie.
> > > >
> > > > 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 273sqf
t of R10 walls and ceiling to the 20degF outside is:
> > > >
> > > > Hrm = (Trm - 20)degF*273sqft / R10 hr-sqft-degF/Btu] = (Trm - 2
0)* 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 - 2
0) * 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 tempe
rature, for the southwest (NEvada, Utah, Arizona), 4inches deep (the approi
x depth heat can travel in the ground in 1 hour --- see the 'daycreek' thre
ad in this group), is about 50degF. So as long as the greenhouse temperatur
e, 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), groun
d 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-s
qft-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 ag
ain 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 J
uly is 40degF
> > > > See: https://weatherspark.com/averages/31890/1/Las-Vegas-Nevada-Uni
ted-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 ge
t:
> > > >
> > > > 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 temp
erature 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-tim
e hours when it is coldest outside.
> > > >
> > > > 200W * 1kW/1000W * $.11/kW-hr * 12hr/day = 0.26cents/day
> > > > --> $/month
> > > > --> $8/Winter (6-months)
> > > >
> > > > Toby
> Greetings,
>
> My son and I are building a 80" x 80" x 8'(tall) 'winter' Greenhouse made