A 500 m2 light-colored swimming pool is located in a normal suburban site, where the measured wind speed at 10 m height is 3 m/s. The water temperature is 25°C, the ambient air temperature is 17°C, and relative humidity is 60%. There are no swimmers in the pool, the temperature of the make-up water is 22°C, and the solar irradiation on a horizontal surface for the day is 20.2 MJ/m2-d. How much energy must the solar system supply (gss) to the pool to keep its temperature to 25°C?
The energy balance of the pool is given by qe + qr + qc + q*
The velocity at 0.3 m above the pool surface is 0.3 X 3 = 0.9 m/s. The partial pressures for air and water are given by Eqs. (5.20) and (5.21). The saturation water vapor pressure at air temperature, ta, is also given by Eq. (5.21); therefore,
Ps = 1GG(G.GG4516 + G.GGG7l78t, + 6.944 X 1G~713) = 1GG(G.GG4516 + G.GGG7178 X 17
2.649 X 1G~612
Saturation water vapor pressure can also be obtained from Eq. (5.21) by using tw instead of ta. Therefore,
From Eq. (5.18), evaporation heat losses are qe = (5.64 + 5.96v0.3)(Pw - Pa) = (5.64 + 5.96 X 0.9)(3.166 - 1.162)
From Eq. (5.22), radiation heat losses are 24 X 3600 4 4
= 0.0864 X 0.95 X 5.67 X 10-8(2984 - 263.74) = 14.198MJ/m2-d
From Eq. (5.26), convection heat losses are qc = 0.0864(3.1 + 4.1v)(tw - ta) = 0.0864(3.1 + 4.1 X 0.9)(25 - 17)
From steam tables, hfg, the latent heat of vaporization of water at 25°C is equal to 2441.8 MJ/kg. Therefore, the daily evaporation rate is given by Eq. (5.28):
qc 4.693 X 103 , „„„, , 2a m = =-= 1.922 kg/m2-d evp hfg 2441.8
From Eq. (5.27), the heat losses due to the make-up water are qmuw = mevp€p (tmuw - tw) = 1.922 X 4.18(22 - 25)
= 24.10MJ/m2 -d (the negative sign is not used, because all the values are losses)
From Eq. (5.29), solar radiation heat gain is qs = aHt = 0.85 X 20.2 = 17.17MJ/m2-d Therefore, the energy required by the solar system to keep the pool at 25°C is qss = Qe + Qr + qc + qmuw - qs = 22.052 +14.198 + 4.693 + 24.1 -17.17
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