Dear Mr. Phukan,         If we check the psychrometric charts, we will find that 1 kg of air at 35 C, and 50% humidity carries 18 gm of water. If we succeed to cool this air to 10 C, then we will be able to condense 10 gm of out of the 18 gm. 1 kg of air at 35 C occupies a volume of 870 Liter (0.87 m3). The specific heat for air is 1.01 KJ/Kg K, which means the heat needed to be extracted from air = 1.01 X (35 - 10) = 25.25 KJ/kg.     Let us consider the low average daily consumption per person to be 240 Lit/day = 10 Lit/hr. If 10 Lit/ 10 gm = 1000, then you need to circulate 1000 kg of air/hr, that is you need to cool 870 m3 of air by 25 C in 1 hr. Your heat load will be 25250 KJ/hr /0.5 (Efficiency) = 14 KW, and your mechanical load will be 870 m3 of air/hr = 1 KW, then your needed electrical supply is equal to 15 KWhr, which will need 90 - 100 m2 of solar panels to generate 10 lit/hr of water to cover the restricted water needs of one person????       I do not want to comment????

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Dear Mr. Phukan,

   

    If we check the psychrometric charts, we will find that 1 kg of air at 35 C, and 50% humidity carries 18 gm of water. If we succeed to cool this air to 10 C, then we will be able to condense 10 gm of out of the 18 gm. 1 kg of air at 35 C occupies a volume of 870 Liter (0.87 m3). The specific heat for air is 1.01 KJ/Kg K, which means the heat needed to be extracted from air = 1.01 X (35 - 10) = 25.25 KJ/kg.

    Let us consider the low average daily consumption per person to be 240 Lit/day = 10 Lit/hr. If 10 Lit/ 10 gm = 1000, then you need to circulate 1000 kg of air/hr, that is you need to cool 870 m3 of air by 25 C in 1 hr. Your heat load will be 25250 KJ/hr /0.5 (Efficiency) = 14 KW, and your mechanical load will be 870 m3 of air/hr = 1 KW, then your needed electrical supply is equal to 15 KWhr, which will need 90 - 100 m2 of solar panels to generate 10 lit/hr of water to cover the restricted water needs of one person????

      I do not want to comment????