I have a room at my place where I work out a little. The room is south oriented and in the summer the temperature in the room when I come back home around 5pm is more than 40ºC and less than 10% humidity.
I want to calculate the power needed to cool the room in 10 minutes the time I dedicate to warm up and stretch the muscles. After that I will choose a solar generation solution to make the system “green”.
Dimensions: 4,9 x 2,7 x 2,3 = 30,429 m3
Altitude: 850 m
Tmax = 50ºC
HRmin = 10%
Emax = 30,429 m3 x 0,9814 Kg/m3 x 72,37 KJ/Kg = 2161,1868 KJ
Tcomfort = 20ºC
HRcomfort = 40%
Ecomfort = 30,429 m3 x 1,083Kg/m3 x 36,34 KJ/Kg = 1197,5704 KJ
Calculation method justification: Of course this is not the standard method to calculate the cooling power. The reason is that the room is closed almost all the day, where the sun
Entalphy difference = 2161,1868 KJ – 1197,5704 KJ = 963,61 KJ
P = 963,61 KJ / 600 secs = 1,6 KW
Assuming a class A device, P = 1,6 KW x 0,55 = 883 W or 883 / 1,16 = 761 Kcal (FG)
Ok, we know that the air conditioner must provide 761FG and the electricity power is 883W, lets find a solar solution capable to produce the energy. I am looking for do-it-yourself kits consisting in panels, inverter, cables but no batteries at all. This means that you can plug the system directly to a wall socket.
Please note that because the system has no batteries, if there is no sun when the air conditioner is switched on, the energy will be consumed from the grid network. Therefore, the best solution is to program the device to work at peak sun hours in order to maintain the thermal jump lower. Don’t worry about the energy produced where the thermostat stops the air conditioner, because in summer the fridge works more hours than normal.
Find products yourself by googeling “domestic self solar kit”
Me and my circumstance