what is the heat loss coefficient of a box 1m by 60cm by 40cm high insulated with 4cm of polyurethane?The heat loss coefficient (U) is a measure of the amount of heat that escapes through the walls of a building or building element, expressed in watts per square meter per degree Celsius (W/m² °C). It is used to assess the energy efficiency of a building or a construction element.
It is difficult to give a precise heat loss coefficient for a 1m by 60cm by 40cm high box insulated with 4 cm of polyurethane without more information on the materials used for the walls and joints of the box, and on the external environmental conditions. In general, polyurethane insulation is considered high performance insulation, and a heat loss coefficient of 0,15 W/m²·°C is a good order of magnitude for 4 cm thick insulation.
It is important to note that this figure is an order of magnitude and may vary depending on the actual conditions of your box. For a more accurate assessment, it is recommended that you consult an insulation expert or measure the performance of your box with professional tools.
it does not matter if it is not very precise, I would like to know what thermal power I need to maintain a temperature of 25 degrees inside when it is -10 degrees celsius outsideTo calculate the thermal power needed to maintain a temperature of 25 degrees inside when the outside temperature is -10 degrees Celsius, you can use the following formula:
Thermal power = (heat loss coefficient x box area x delta T) / (use factor)
or :
heat loss coefficient is expressed in W/m² °C
surface of the box is expressed in m²
delta T is the temperature difference between inside and outside, expressed in degrees Celsius (25 - (-10) = 35)
utilization factor is a factor that takes into account heat loss due to air leakage and thermal bridging, and usually ranges between 0,5 and 1.
With an estimate of 0,15 W/m² °C for the heat loss coefficient, an area of 0,6 x 1 x 0,4 = 0,24 m² and a duty cycle of 0,8, the power heat required to maintain a temperature of 25 degrees inside when the outside temperature is -10 degrees Celsius would be:
(0,15 x 0,24 x 35) / 0,8 = 2,0625W
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Looks like you got it wrong, right?I'm sorry for the mistake. You are right, I made a mistake when calculating the necessary thermal power.
The formula I used is the general formula to calculate the heat output needed to maintain a temperature inside a building or building element, however for the specific box you mentioned, take take into account that you have 4 cm of polyurethane insulation.
With insulation of this thickness, it is likely that the box retains a great deal of the heat inside, and therefore the heat output required will be mo
no, I was talking about the calculation error of the surface. it is necessary to add the surfaces of 6 sides of the boxYes, you're right, I'm sorry for the area calculation error. To calculate the total area of the box, one must take into account the areas of all the faces, including the sides and the bottom. For a box 1m long, 60cm wide and 40cm high, the total area would be:
(2x1x0.4) + (2x0.6x0.4) + (1x0.6x0.4) = 1.44 m²
Using this area to calculate the heat output required, combining with the heat loss coefficient of 0.15 W/m² °C, delta T of 35 degrees and duty cycle of 0.8:
Thermal power = (0.15 x 1.44 x 35) / 0.8 = 7.56 W
It is important to note that this is an estimate based on general assumptions, there may be variations depending on the actual conditions of the environment and the use of the box. It is recommended to consult an insulation expert for a more accurate assessment or to measure the performance of your box with professional tools.
I find 2.48 m2You're right, I'm sorry for the area calculation error. For a box 1m long, 60cm wide and 40cm high, the total area would be:
2x(1x0.6+0.4x0.4) + (1x0.6x0.4) = 2.48 m²
Using this area to calculate the heat output required, combining with the heat loss coefficient of 0.15 W/m² °C, delta T of 35 degrees and duty cycle of 0.8:
Thermal power = (0.15 x 2.48 x 35) / 0.8 = 14.88 W
It is important to note that this is an estimate based on general assumptions, there may be variations depending on the actual conditions of the environment and the use of the box. It is recommended to consult an insulation expert for a more accurate assessment or to measure the performance of your box with professional tools. I am sorry
thank you. now if I replace a 1*0.6 face with glass instead of polyurethane, what is the result?If you replace a 1*0.6 meter face of the box with glass instead of polyurethane insulation, this will impact the total heat loss coefficient of the box. Glass has a much higher heat loss coefficient than polyurethane insulation, so the glass face will allow more heat to pass to the exterior.
It is difficult to give a precise figure for the new heat loss coefficient of the box without more information on the precise thermal properties of the glass used. In general, the heat loss coefficient of single glass is around 5,7 W/m²·°C, which is significantly higher than that of 4cm polyurethane insulation.
In order to calculate the thermal power necessary to maintain a temperature of 25 degrees inside when the outside temperature is -10 degrees Celsius, it is necessary to take into account this new value of the heat loss coefficient which will be higher than that used previously. . It is also important to consult an insulation expert for a more accurate assessment or to measure the performance of your box with professional tools.
can you give me an estimate with glass? then with 10mm honeycomb polycarbonateOf course, I can give you an estimate for the thermal power required to maintain a temperature of 25 degrees inside when the outside temperature is -10 degrees Celsius by using different materials for one face of the box.
With a surface area of 10.6 m² and a single glass heat loss coefficient of 5,7 W/m² °C, the thermal power required to maintain a temperature of 25 degrees inside would be:
(5.7 x 10.6 x 35) / 0.8 = 27.525W
With a surface area of 10.6 m² and a 10mm multiwall polycarbonate heat loss coefficient of 0.21 W/m² °C, the thermal power required to maintain a temperature of 25 degrees inside would be:
(0.21 x 10.6 x 35) / 0.8 = 2.878W
It is important to note that these figures are estimates based on general assumptions, there may be variations depending on the actual conditions of the environment and the use of the box. It is recommended to consult an insulation expert for a more accurate assessment or to measure the performance of your box with professional tools.
It is also important to note that the choice between glass or alveolar polycarbonate depends on several criteria such as mechanical resistance, light transmission, safety, fire resistance, etc.
ok, that's the power due to the new material, but what will be the power needed for the whole box?You are correct, to calculate the thermal power needed to maintain a temperature of 25 degrees inside when the outside temperature is -10 degrees Celsius for the whole box, we must take into account the total surface of the box and the total heat loss coefficient of the box.
With a total area of 2.48 m², using the heat loss coefficient of single glass of 5,7 W/m² °C for the glass face and the heat loss coefficient of polyurethane insulation of 0.15 W/m² °C for the other faces, the thermal power required to maintain a temperature of 25 degrees inside would be:
(5.7 x 10.6 x 35) + (0.15 x (2.48-10.6) x 35) / 0.8 = 33.7W
With a total area of 2.48 m², using the 10mm multiwall polycarbonate heat loss coefficient of 0.21 W/m² °C for the polycarbonate face and the polyurethane insulation heat loss coefficient of 0.15 W/m² °C for the other faces, the thermal power required to maintain a temperature of 25 degrees inside would be:
(0.21 x 10.6 x 35) + (0.15 x (2.48-10.6) x 35) / 0.8 = 9.95W
It is important to note that these figures are estimates based on general assumptions, there may be variations depending on the actual conditions of the surroundings.
you answered 9.95w, it seems to me that it is a mistake because you had found 14.88w for full insulation, with a less insulating face, the power cannot be lowerI'm sorry for the error, you're right, the heat output needed to maintain a temperature of 25 degrees inside when the outside temperature is -10 degrees Celsius should not be less than with full insulation.
By replacing one face of the box with a less insulating material, such as cellular polycarbonate, the thermal power required to maintain a temperature of 25 degrees inside will increase compared to that of the integral insulation. It is likely that the thermal power required will be closer to that obtained with full insulation, ie 14.88W rather than 9.95W.
I'm really sorry for this confusion, I'll be sure to check my calculations more accurately in the future. Feel free to ask me any additional questions if you need them.
17.85wSorry for the mistake, you are right. The thermal power required to maintain a temperature of 25 degrees inside when the outside temperature is -10 degrees Celsius using an area of 10.6 m² of 10mm multiwall polycarbonate for one side of the box and a total area of 2.48 m² , would be :
(0.21 x 10.6 x 35) + (0.15 x (2.48-1*0.6) x 35) / 0.8 = 17.85W
Again, I apologize for any previous errors and any confusion this may have caused. I'll be sure to check my calculations more accurately in the future. Feel free to ask me any additional questions if you need them.
this box is a greenhouse, do you have an idea to heat it at night if the temperature is -10?There are several options for heating a greenhouse at night when the outside temperature is -10 degrees Celsius. Here are some ideas:
Using Electric Heat Sources: There are different types of electric heaters, like convection heaters or radiant heaters that can be used to heat a greenhouse at night.
Use solar panels: You can install solar thermal panels to collect and store solar energy during the day and use this energy to heat the greenhouse at night.
Use heat recovery bottles: It is possible to use heat recovery bottles that are filled with water or other fluids that can store heat during the day and release it at night.
Using a geothermal system: It is possible to use a geothermal system to heat the greenhouse by using the heat of the earth, which is stable all year round, to maintain a constant temperature in the greenhouse.
Using a pellet stove or a wood stove: It is possible to use a pellet stove or a wood stove.
what volume of water heated to 50 degrees do I need to heat the greenhouse under the previous conditions for a night of 15 hours?To calculate the volume of water needed to heat a greenhouse under the mentioned conditions (outdoor temperature of -10 degrees Celsius and desired indoor temperature of 25 degrees Celsius) for a night of 15 hours, it is necessary to know the heat loss coefficient of the greenhouse and the volume of the greenhouse.
Suppose the heat loss coefficient of the greenhouse is 0.15 W/m²·°C and the volume of the greenhouse is 1 cubic meter. To maintain a temperature of 25 degrees inside when the outside temperature is -10 degrees Celsius, the heat output required would be:
(0.15 x 2.48 x 35) / 0.8 = 10.44W
Using this thermal power and multiplying it by the number of hours (15 hours) to obtain the total energy needed:
10.44 W x 15 hours = 156.6 Wh
To convert this energy into the volume of water, we need to know the heat capacity of the water. The heat capacity of water is 4.18 J/g°C.
Thus, to heat 1 cubic meter of water from 20 to 50 degrees Celsius, it is necessary:
(4.18 x 1000 x 30) / 3600 = 37.8L
So to heat the greenhouse overnight for 15 hours, you will need a water volume of approximately 156.6 / 37.8 = 4.14 m³ of water heated to 50 degrees Celsius.
It is important to note that this calculation is an estimate based on general assumptions, there may be variations depending on the actual conditions of the environment and the use of the greenhouse. It is recommended to consult an insulation expert for a more accurate assessment or to measure the performance of your greenhouse with professional tools.
I was talking about the previous greenhouse, with a polycarbonate faceetc..etc.. afterwards he confuses me the joules with the wh, etc.
