How Much Water is Needed to Cool Solar Panels via Evaporation?

[said.alomairi]

In July the Hanze University and the Rijks University Groningen participate in the Frisian Solar Challenge . This is world race for solar boats. The solar cells we use are so-called B50 cells with an efficiency of 21%.


The cells perform worse if the temperature rises. The way we cool the solar cell is by means of evaporation of water.

Assume that the solar panel is 60 degrees Celsius. The radiation of the sun is 1000W/m2.
Twenty one percent of the 1000 W is converted in electricity and does not have to be cooled.

We assume that the approximately 800 W will be converted into heat. What is the amount of water needed in kg/s to cool the panel by making use of a phase change. The phase change form liquid to gas.

Assumptions:
• All the water will evaporate
• The temperature of the water is 18 degrees of Celsius
• The surface of the solar panel is 1 m2

Calculate the flux?

i tried to calculate this but always i got it wrong

so please anyone can help me on this as it is my final assignment before i start my final project

 

[NateD]

The amount of water needed in kg/s to cool the panel by making use of a phase change can be calculated using the following formula: Flux = (Heat energy / (Latent heat of vaporization x Temperature difference)) Flux = (800 W / (2.256 MJ/kg x (60 - 18) °C)) Flux = 0.0819 kg/s