What is the surface area of ​​a flat solar collector per ton of water?

Introduction What is the surface area of ​​a flat solar collector per ton of water? A vacuum glass collection tube with a diameter of 4.7 cm and a length of 1.25 m should correspond to 6 liters of water. Of course, some water heaters are. better and more

What is the surface area of ​​a flat solar collector per ton of water?

A vacuum glass collection tube with a diameter of 4.7 centimeters and a length of 1.25 meters should correspond to 6 liters of water. Of course, some water heaters may only fit 5.0 or a few liters for better and faster heat collection. The heat output of a flat panel collector is 50 to 90 liters per square meter when the sunshine is good, which can be calculated according to the actual situation.

How many tons of water can be lost after eight hours of solar water supply?

I will give you a formula for reference. You can calculate how much heat or how many kilowatts. -hours of electricity required to raise the temperature of each ton of water by n degrees< /p>

P total=cm(t2-t1)×Kc÷860kcal/kw (Kc is the heat loss coefficient , 1.05) (860kcal/kw is the heat needed to increase the temperature of each liter of water by 1 degree)

For examplemple, the water temperature (or room temperature) is 5℃ and heated to 55℃

=1kcal/kg.℃×1000kg (1 ton of water)×(55-5) ℃×1.05÷860kcal/kw=61kw, that is, heating a ton of water generally requires 61 kilowatt hours of electricity. , the average annual value of the energy efficiency coefficient COP of air source heat pumps is 3.5, which requires 61/3.5 = 17 kilowatt hours of electricity. If there is a difference between off-peak power and off-peak power in some locations, you can use off-peak power for heating, saving money.

Assuming that 50% of the water is heated by solar energy throughout the year, the electricity consumption per ton of hot water is only 17/2 = 8.5 degrees. Assuming that the annual water consumption is 1,000 tons, the electricity consumption is 1,000 * 8.5 = 8,500 kilowatt hours of electricity. Assuming that the spawns of electricity are 1 yuan/kilowatt hour, the cost is 8,500 * 1 = 8,500 yuan.

It depends on the speed of your water supply (flow). Certainly, if the water pressure is high, the water will be filled quickly and more water will flow out at the same time, and vice versa. According to ordinary water pressure, it takes. approximately 30 minutes to fill a 300 liter water tank. According to this calculation, it will take an hour. The flow rate is 600 liters, eight hours is 8x600=4800 liters and one ton of water is 1000 liters. p>

According to this speed, approximately 4.8 tons of water will be lost in eight hours

I hope to adopt thank you

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