To refresh air in a room, a counterflow heat exchanger is mounted in the wall, as shown in Fig. P11.122. It draws in outside air at 33 F, 80% relative humidity and draws room air, 104 F, 50% relative humidity, out. Assume an exchange of 6 lbm/min dry air in a steady flow device, and also that the

Question

To refresh air in a room, a counterflow heat exchanger is mounted in the wall, as shown in Fig. P11.122. It draws in outside air at 33 F, 80% relative humidity and draws room air, 104 F, 50% relative humidity, out. Assume an exchange of 6 lbm/min dry air in a steady flow device, and also that the room air exits the heat exchanger to the atmosphere at 72 F. Find the net amount of water removed from room, any liquid flow in the heat exchanger and [latex](T, \phi)[/latex] for the fresh air entering the room.

Accepted Answer

State 3: [latex]P _{ g 3}=1.0804 psia , h _{ g 3}=1106.73 Btu / lbm[/latex] ,

[latex]P _{ v 3}=\phi_{3} P _{ g 3}=0.5402, \quad w _{3}=0.622 P _{ v 3} /\left( P _{ ext {tot }}- P _{ v 3} ight)=0.02373[/latex]

The room air is cooled to [latex]72 F < T _{ ext {dew } 1}=82 F [/latex] so liquid will form in the exit flow channel and state 4 is saturated.

4: [latex]72 F , \phi=100 \% \Rightarrow P _{ g 4}=0.3918 psia , h _{ g 4}=1092.91 Btu / lbm[/latex] ,

[latex] w _{4}=0.017, h _{ f 4}=40.09 Btu / lbm[/latex]

1: [latex]33 F , \phi=80 \% \Rightarrow P _{ g 1}=0.0925 psia , h _{ g 1}=1075.83 Btu / lbm[/latex],

[latex]P _{ v 1}=\phi_{1} P _{ g 1}=0.074 psia , \quad w _{1}=0.00315[/latex]

CV 3 to 4: [latex]\dot{ m }_{ liq , 4}=\dot{ m }_{ a }\left( w _{3}- w _{4} ight)=6(0.02373-0.017)= 0 . 0 4 lbm / min[/latex]

CV room:

[latex]\begin{aligned}\dot{ m }_{ v , out } &=\dot{ m }_{ a }\left( w _{3}- w _{2} ight)=\dot{ m }_{ a }\left( w _{3}- w _{1} ight) \&=6(0.02373-0.00315)= 0 . 1 2 3 5 lbm / min\end{aligned}[/latex]

CV Heat exchanger: [latex]\dot{ m }_{ a }\left( ilde{ h }_{2}- ilde{ h }_{1} ight)=\dot{ m }_{ a }\left( ilde{ h }_{3}- ilde{ h }_{4} ight)-\dot{ m }_{ liq } h _{ f 4}[/latex]

[latex]\begin{gathered}C _{ p a }\left( T _{2}- T _{1} ight)+ w _{2} h _{ v 2}- w _{1} h _{ v 1}= C _{ p a }\left( T _{3}- T _{4} ight)+ w _{3} h _{ v 3}- w _{4} h _{ v 4}-\left( w _{3}- w _{4} ight) h _{ f 4} \0.24\left( T _{2}-33 ight)+ w _{2} h _{ v 2}-3.3888=0.24(104-72)+26.2627-18.5795-0.2698 \0.24 T _{2}+0.00315 h _{ v 2}=26.402 btu / lbm\end{gathered}[/latex]

Trial and error on [latex]T _{2}: \quad T _{2}= 9 5 . 5 F , \quad P _{ g 2}=0.837 psia , P _{ v 2}= P _{ v 1}[/latex]

[latex]\phi= P _{ v 2} / P _{ g 2}=0.074 / 0.837=0.088 ext { or } \phi=9 \%[/latex]

Question 11.183E: To refresh air in a room, a counterflow heat exchanger is mo...

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