Recent content by StoningTree

It's more like a thought experiment, so we can say that the air and the water never touches/are separated. I'm not sure how i take the humidity in account when using the ideal Gas Law: p*V = n*Ru*T n=p*V/Ru*T (With 180m^3, 25°C and 1.01325 bar) = 7356.6044 mol ? The mole fraction of water...

First i calculated the latent heat of vaporation: h_vap(15°C)= 2465.4 kJ/kg In the next step i calculated how much water is already in the air: x=0.622*j*ps/(p-j*ps)=0.00988 kg/kg where: j=50% ps=0.03171 bar p=1.01325 bar Now that i know that i can calculate the enthalpy of humid air: h_hum =...