Question 6.4: In the enclosure of an indoor pool, the air needs to be kept......
In the enclosure of an indoor pool, the air needs to be kept in the following conditions: T = 26°C, Φ = 0.65, and maximum chlorine concentra-tion = 1.5 mg/m³. The water temperature of the pool is 24°C, its free surface A = 450 m², and its depth (constant) h = 2 m. The pressure inside the enclosure is the same as the ambient pressure P_{0}=1018 mbar and the ambient temperature and humidity are T_{0} = 5°C and \phi_{0}=0.9. Determine the mass flow rate of renewal air (airflow taken from outside), knowing that
(a) The evaporation rate of the pool water can be calculated by the formula: \ \dot m=25.5\,A(p_{s}^{}-p_{v})/(p_{0}\,-p_{v}); with \dot m expressed in kg/h and A in m², and with p_{v} being the partial pressure of the vapour in the air, and p_{s} the saturation pressure at the water temper-ature of the pool.
(b) The concentration of chlorine in the pool water, kept constant, is 0.15 g/m³, and assuming that in 24 h the total chlorine content in the water is released to the environment, and the maximum concentration of \mathrm{Cl} in the air of the pool is 1.5\cdot10^{-6}\,\mathrm{kg/m^{3}}.
(c) Due to the estimated occupancy in such an installation, at least 1.3 renovations per hour of the air in the enclosure, whose volume is V = 4290 m³, are considered necessary.
(a) From the vapour tables we have p_{s}(24^{\circ}{\mathrm{C}})=30.8 mbar, so
p_{\nu}=\phi p_{s}(26^{\circ}C)=22.55\,m b a rSubstituting values in the given formula, we get that the evaporation rate of the water in the pool is
{\dot{m}}_{e v}=25.5A~{\frac{p_{s}-p_{v}}{p_{0}-p_{v}}}=95.10~{\frac{\mathrm{kg}}{\mathrm{h}}}For maintaining the state of the air inside the pool enclosure, a renewal of airflow would be needed such that if \dot{m}_{a} is the mass flow rate of dry air, we have
{\dot{m}}_{a}\omega_{0}+{\dot{m}}_{e\nu}={\dot{m}}_{a}\omega_{i}where \omega_{0} and \omega_{i} are the absolute humidities of the external and internal air respec-tively. As p_{s}(5^{\circ}{\mathrm{C}})=8.72 mbar
\omega_{0}=0.622{\frac{p_{s}(5^{\circ}{\mathrm{C}})}{{\frac{p_{0}}{\phi_{0}}}-p_{s}(5^{\circ}{\mathrm{C}})}}=4.83~{\frac{{\mathrm{g}}}{{\mathrm{kg~d~a}}}}and for the indoor humidity, as p_{s}(26^{\circ}{\mathrm{C}})=34.69 mbar
\omega_{i}=0.622{\frac{34.69}{{\frac{1018}{0.65}}-34.69}}=14.10~{\frac{\mathrm{g}}{\mathrm{kg~d~a}}}Substituting these values gives
{\dot{m}}_{a}={\frac{{\dot{m}}_{e\nu}}{\omega_{i}-\omega_{0}}}=10,274\ {\frac{\mathrm{kg}}{{\mathrm{h}}}}(b) Calculating the airflow needed to maintain that maximum \mathrm{Cl} concentration in the pool air.
V_{p o}=A.h=900\;m^{3}The exit mass flow rate of the \mathrm{Cl} of the pool water is
{\dot{m}}_{C l}={\frac{V_{p o}c_{C l}}{24}}={\frac{900\cdot0.15\cdot10^{-3}}{24}}=5.62\cdot10^{-3}~{\frac{\mathrm{kg}}{\mathrm{h}}}Undertaking a balance in the \mathrm{Cl}, the following must be true
{\dot{m}}_{a}+{\dot{m}}_{\mathrm{Cl}}={\dot{m}}_{a}\bigl(1+y_{\mathrm{Cl}}^{m a x}\bigr)where y_{Cl} is the mass fraction of \mathrm{Cl} in the pool air. If we know the air density of the pool, the mass fraction for the maximum concentration condition will be y_{\mathrm{Cl}}^{m a x}=1.5\cdot 10^{-6}/\rho_i. Calculating the density of the indoor air, per unit mass of dry air
\nu_{i}=(R_{a}+\omega_{i}R_{\nu})\frac{T_{i}}{p_{i}}=\left(\frac{8.314}{28.9}+0.014\,\frac{8.314}{18}\right)~\frac{299}{1.018\cdot10^{5}}=0.864~\frac{{\mathrm{m}}^{3}}{{\mathrm{kg\,da}}} \\ \varrho_{i}={\frac{1}{\nu_{i}}}=1.157\ {\frac{\mathrm{kg\,d\,a}}{{\mathrm{m}}^{3}}}so the maximum mass fraction of \mathrm{Cl} in the pool air is
y_{\mathrm{Cl}}^{m a x}=1.30\cdot10^{-6}therefore giving
{\dot{m}}_{a}={\frac{{\dot{m}}_\mathrm{CI}}{y_{\mathrm{Cl}}^{m a x}}}=4,323\ {\frac{\mathrm{kg}}{\mathrm{h}}}(c) Let us now see the demands of the renewal air
\dot{V}_{r e}=1.3\cdot V=5,577\,\frac{m^{3}}{h}It is clear that the higher requirements correspond to the need for maintaining humidity in the pool air, so the dry airflow required is \dot{m}_{a}\,=\,10,274~\mathrm{kg/h}. Since the specific volume of the air in the external conditions is
\nu_{0}=(R_{a}+\omega_{0}R_{\nu}){\frac{T_{0}}{p_{0}}}=0.794~{\frac{{\mathrm{m}}^{3}}{{\mathrm{kg~d~a}}}}the airflow is
\dot{V}=\dot{m}_{a}\nu_{0}=8,157\ \frac{{\mathrm{m}}^{3}}{\mathrm{h}}