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Question 11.178E: Two moist air streams with 85% relative humidity, both flowi...

Two moist air streams with 85% relative humidity, both flowing at a rate of 0.2 lbm/s of dry air are mixed in a steady flow setup. One inlet flowstream is at 90 F and the other at 61 F. Find the exit relative humidity.

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CV mixing chamber.

Continuity Eq. water:              \dot{ m }_{\text {air }} w _{1}+\dot{ m }_{\text {air }} w _{2}=2 \dot{ m }_{\text {air }} w _{ ex };

Energy Eq.:                              \dot{ m }_{\text {air }} \tilde{ h }_{1}+\dot{ m }_{\text {air }} \tilde{ h }_{2}=2 \dot{ m }_{\text {air }} \tilde{ h }_{ ex }

Properties from the tables and formulas

\begin{aligned}& P _{ g 90}=0.699 ; P _{ v 1}=0.85 \times 0.699=0.594   psia \\& w _{1}=0.622 \times 0.594 /(14.7-0.594)=0.0262 \\& P _{ g 61}=0.2667 ; P _{ v 2}=0.85 \times 0.2667=0.2267   psia \\& w _{2}=0.622 \times 0.2267 /(14.7-0.2267)=0.00974\end{aligned}

Continuity Eq. water:      w _{ ex }=\left( w _{1}+ w _{2}\right) / 2=0.018;

For the energy equation we have      \tilde{ h }= h _{ a }+ wh _{ v }           so:

2 \tilde{ h }_{ ex }-\tilde{ h }_{1}-\tilde{ h }_{2}=0=2 h _{ a  ex }- h _{ a  1}- h _{ a  2}+2 w _{ ex } h _{ v  ex }- w _{1} h _{ v 1}- wh _{ v  2}

we will use constant heat capacity to avoid an iteration on T _{ ex }.

\begin{aligned}& C _{ p  \text { air }}\left(2 T _{ ex }- T _{1}- T _{2}\right)+ C _{ p  H 2 O }\left(2 w _{ ex } T _{ ex }- w _{1} T _{1}- w _{2} T _{2}\right)=0 \\& T _{ ex }=\left[ C _{ p \text { air }}\left( T _{1}+ T _{2}\right)+ C _{ p  H 2 O }\left( w _{1} T _{1}+ w _{2} T _{2}\right)\right] /\left[2 C _{ p  \text { air }}+2 w _{ ex } C _{ p  H 2 O }\right] \\&\quad=[0.24(90+61)+0.447(0.0262 \times 90+0.00974 \times 61] / 0.4961 \\&\quad=75.7   F\end{aligned}

 

\begin{aligned}& P _{ v  ex }=\frac{ w _{ ex }}{0.622+ w _{ ex }} P _{ tot }=\frac{0.018}{0.622+0.018} 14.7=0.413   psia \\& P _{ g  ex }=0.445   psia \Rightarrow \quad \phi=0.413 / 0.445= 0 . 9 3 \text {   or   } 9 3 \%\end{aligned}

 

 

 

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