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5 m^(3) of air with a relative humidity of 22% at 15^(@)C and 3 m^(3) of air with a relative humidity of 46% at 28^(@)C have been mixed together. The total volume of the mixture is 8m^(3). Find the relative humidity of the mixture. |
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Answer» Solution :Pind first the mass of moisture in each volume of air, i.E. the absolute humidity of the volumes which are mixed. We have in 5 m of air `m_(1)=f_(1)V_(1)=rho_(1)^("sat")B_(1)V_(1)=12.8xx0.22xx5=14.1g` In `3m^(2)` of air `m_(2)=rho_(2)^("sat")B_(2)V_(2)=27.2xx0.46xx3=37.5g` Next find the absolute humidity of the mixture: `f=(m_(1)+m_(2))//(V_(1)+V_(2))=6.45g//m^(2)` To find the relative humidity we MUST find the temperature of the mixture. Neglecting the vapour mass we may write the equation of heat balance in the form `rho_(0)V_(1)e_(0)(t-15)=rho_(0)V_(2c_(0))(28-t)` where the subscripts O SHOW that the density and the specific heat refer to air. We have`t= 20^(@)C`. Now it is easy to compute the rolative bumidity. |
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