5 ms·

Can a tea light heat a greenhouse?

24 points by firebaze 3y ago

aaron695 3y ago
[dead]
Obscurity4340 3y ago
Isn't the answer to these "type" questions almost always "no"?
- kylebenzle 3y ago
  Maybe, but in this case it is 100% yes. I used to have an expensive kerosene heater in my greenhouse it was a huge pain, now single candle does the same job if I hear it's going to get much below freezing. Never had an issue yet
nullc 3y ago
The water freezing outside may buffer temperatures right at the freezing point, so it may not require much heat to prevent frost on the inside.
sgarland 3y ago
I thought I'd calculate the greenhouse's innate heating capability from solar radiation for fun. Let the greenhouse have dimensions of 3 m x 2 m, with an eaves height of 2m, and a roof slope of 1:2, giving a ridge height of 2.5 m. This yields a surface area for the walls up to the eaves of (2 * (3 m * 2 m)) + (2 * (2 m * 2 m)) = 20 m^2, and to the ridge of (2 * ((2 m * 0.5 m) / 2) = 1 m^2. The roof has a surface area of (2 * (3 m * sqrt(1^2 + 0.5^2)m)) ~= 6.71 m^2, for a total of ~= 27.7 m^2. Assume 95% (arbitrarily decided) is glass, so 26.3 m^2. Short-wave transmissivity of typical single-pane glass is 88% [0] at a 0-degree angle of incidence. Plain glass has a refractive index of approximately 1.5, and air is 1.0. Using Snell's Law and Fresnel's Laws, we can calculate the transmissivity for realistic angles. The geographic center of the continental U.S. [1] will see a maximum solar elevation of 73.61 deg during the summer solstice, and a maximum of 26.76 deg during the winter solstice. Using Snell's Law, solving for the refracted angle yields arcsin(sin((90 - 73.61) deg)/1.5) ~= 10.84 deg and arcsin(sin((90 - 26.76) deg)/1.5) ~= 36.53 deg, respectively. These are for the sides of the building, and I assume that each side gets equal amounts of light at complementary angles throughout the day. Re-doing these calculations for the roof's slope yielded 29.21 deg and 0.13 deg for refracted angles. Using Fresnel's Laws (or a calculator for them [2]) we get 95.9% and 96.0% for light entering the roof, and 77.4% and 95.9% for light entering the walls. I'll use the average of each of these for yearly, so 95.95% for the roof, and 86.65% for the walls. An online solar insolation calculator [3] with all losses turned off yielded an average daily maximum of 8.708 kWh/m^2/day, or 0.36283 kW/m^2, and an average daily minimum of 2.437 kWh/m^2/day, or 0.10154 kW/m^2. The average of these is thus 0.23219 kW/m^2. The roof has a glass area of 6.38 m^2, and the walls have 19.92 m^2. Summing the product of insolation, area, and transmissivity together, we get 5.438 kW entering the greenhouse. The greenhouse has a volume of (3 m * 2 m * 2 m) + (3 m * (2 m * 0.5 m / 2)) = 13.5 m^3. Assuming dry air at 20 C, this is a delta-T/sec of (5438 W * 1 sec) / (13.5 m^3 * 1.2 kg/m^3 * 1005 J/(kg * K)) = 0.334 degC/sec. Standard greenhouse glass seems to be 3mm thick, giving it a heat transfer coefficient (assuming no coatings) of 6.3 W/(m^2 * delta-degC). Thus, the combined surface area is losing 165.69 W * delta-degC. Assuming the outside temperature was 0 C, and the internal temperature was 20 C, that's 3312 W. The two forces on their own would reach equilibrium at 5438 W / 165.69 W * delta-degC = 32.82 delta-degC. This ignores transpiration from the plants inside, heat absorption characteristics of various things inside the greenhouse, etc. This paper [4] has quite a bit of detail on parts I elided. [0]: https://greenhouse.hosted.uark.edu/Unit03/Section02.html https://greenhouse.hosted.uark.edu/Unit03/Section02.html [1]: https://geohack.toolforge.org/geohack.php?pagename=Geographic_center_of_the_United_States&params=39_50_N_98_35_W_region:US-KS_type:landmark&title=1918+geographic+Center+of+the+Contiguous+United+States https://geohack.toolforge.org/geohack.php?pagename=Geographi... [2]: https://www.lasercalculator.com/fresnel-reflection-and-transmission-calculator/ https://www.lasercalculator.com/fresnel-reflection-and-trans... [3]: https://www.fabhabs.com/solar-insolation-calculator https://www.fabhabs.com/solar-insolation-calculator [4]: https://faculty.ksu.edu.sa/sites/default/files/lmhdr_lthlth_wlrb.pdf https://faculty.ksu.edu.sa/sites/default/files/lmhdr_lthlth_...
danielfoster 3y ago
Smudge pots and heaters are also sometimes used in wine vineyards to prevent frost. The idea is not that the heat itself will prevent frost, but that the generation of heat will create air circulation that prevents frost from forming. Warm air rises and cool air falls, so anything that promotes circulation-- even a a fan-- should have the same effect. So this really isn't about the heat generated or the insulation of the greenhouse.
- mikecoles 3y ago
  When doing field work near an orchard, "wind mills" were noticed scattered across the fields. I asked what the wind mills were for, thinking they were pumping water for irrigation. I was told they keep the fruits safe from frost. It took me many years to understand how that worked.

24 points by firebaze 3y ago

aaron695 3y ago

[dead]

Obscurity4340 3y ago

Isn't the answer to these "type" questions almost always "no"?

kylebenzle 3y ago
Maybe, but in this case it is 100% yes. I used to have an expensive kerosene heater in my greenhouse it was a huge pain, now single candle does the same job if I hear it's going to get much below freezing. Never had an issue yet

nullc 3y ago

The water freezing outside may buffer temperatures right at the freezing point, so it may not require much heat to prevent frost on the inside.

sgarland 3y ago

I thought I'd calculate the greenhouse's innate heating capability from solar radiation for fun. Let the greenhouse have dimensions of 3 m x 2 m, with an eaves height of 2m, and a roof slope of 1:2, giving a ridge height of 2.5 m. This yields a surface area for the walls up to the eaves of (2 * (3 m * 2 m)) + (2 * (2 m * 2 m)) = 20 m^2, and to the ridge of (2 * ((2 m * 0.5 m) / 2) = 1 m^2. The roof has a surface area of (2 * (3 m * sqrt(1^2 + 0.5^2)m)) ~= 6.71 m^2, for a total of ~= 27.7 m^2. Assume 95% (arbitrarily decided) is glass, so 26.3 m^2. Short-wave transmissivity of typical single-pane glass is 88% [0] at a 0-degree angle of incidence. Plain glass has a refractive index of approximately 1.5, and air is 1.0. Using Snell's Law and Fresnel's Laws, we can calculate the transmissivity for realistic angles. The geographic center of the continental U.S. [1] will see a maximum solar elevation of 73.61 deg during the summer solstice, and a maximum of 26.76 deg during the winter solstice. Using Snell's Law, solving for the refracted angle yields arcsin(sin((90 - 73.61) deg)/1.5) ~= 10.84 deg and arcsin(sin((90 - 26.76) deg)/1.5) ~= 36.53 deg, respectively. These are for the sides of the building, and I assume that each side gets equal amounts of light at complementary angles throughout the day. Re-doing these calculations for the roof's slope yielded 29.21 deg and 0.13 deg for refracted angles. Using Fresnel's Laws (or a calculator for them [2]) we get 95.9% and 96.0% for light entering the roof, and 77.4% and 95.9% for light entering the walls. I'll use the average of each of these for yearly, so 95.95% for the roof, and 86.65% for the walls. An online solar insolation calculator [3] with all losses turned off yielded an average daily maximum of 8.708 kWh/m^2/day, or 0.36283 kW/m^2, and an average daily minimum of 2.437 kWh/m^2/day, or 0.10154 kW/m^2. The average of these is thus 0.23219 kW/m^2. The roof has a glass area of 6.38 m^2, and the walls have 19.92 m^2. Summing the product of insolation, area, and transmissivity together, we get 5.438 kW entering the greenhouse. The greenhouse has a volume of (3 m * 2 m * 2 m) + (3 m * (2 m * 0.5 m / 2)) = 13.5 m^3. Assuming dry air at 20 C, this is a delta-T/sec of (5438 W * 1 sec) / (13.5 m^3 * 1.2 kg/m^3 * 1005 J/(kg * K)) = 0.334 degC/sec. Standard greenhouse glass seems to be 3mm thick, giving it a heat transfer coefficient (assuming no coatings) of 6.3 W/(m^2 * delta-degC). Thus, the combined surface area is losing 165.69 W * delta-degC. Assuming the outside temperature was 0 C, and the internal temperature was 20 C, that's 3312 W. The two forces on their own would reach equilibrium at 5438 W / 165.69 W * delta-degC = 32.82 delta-degC. This ignores transpiration from the plants inside, heat absorption characteristics of various things inside the greenhouse, etc. This paper [4] has quite a bit of detail on parts I elided. [0]: https://greenhouse.hosted.uark.edu/Unit03/Section02.html https://greenhouse.hosted.uark.edu/Unit03/Section02.html [1]: https://geohack.toolforge.org/geohack.php?pagename=Geographic_center_of_the_United_States&params=39_50_N_98_35_W_region:US-KS_type:landmark&title=1918+geographic+Center+of+the+Contiguous+United+States https://geohack.toolforge.org/geohack.php?pagename=Geographi... [2]: https://www.lasercalculator.com/fresnel-reflection-and-transmission-calculator/ https://www.lasercalculator.com/fresnel-reflection-and-trans... [3]: https://www.fabhabs.com/solar-insolation-calculator https://www.fabhabs.com/solar-insolation-calculator [4]: https://faculty.ksu.edu.sa/sites/default/files/lmhdr_lthlth_wlrb.pdf https://faculty.ksu.edu.sa/sites/default/files/lmhdr_lthlth_...

danielfoster 3y ago

Smudge pots and heaters are also sometimes used in wine vineyards to prevent frost. The idea is not that the heat itself will prevent frost, but that the generation of heat will create air circulation that prevents frost from forming. Warm air rises and cool air falls, so anything that promotes circulation-- even a a fan-- should have the same effect. So this really isn't about the heat generated or the insulation of the greenhouse.

mikecoles 3y ago
When doing field work near an orchard, "wind mills" were noticed scattered across the fields. I asked what the wind mills were for, thinking they were pumping water for irrigation. I was told they keep the fruits safe from frost. It took me many years to understand how that worked.