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

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 (T, \phi) for the fresh air entering the room.

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

P _{ v 3}=\phi_{3} P _{ g 3}=0.5402, \quad w _{3}=0.622 P _{ v 3} /\left( P _{\text {tot }}- P _{ v 3}\right)=0.02373

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

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

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

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

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

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

CV room:

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

CV Heat exchanger:      \dot{ m }_{ a }\left(\tilde{ h }_{2}-\tilde{ h }_{1}\right)=\dot{ m }_{ a }\left(\tilde{ h }_{3}-\tilde{ h }_{4}\right)-\dot{ m }_{ liq } h _{ f 4}

\begin{gathered}C _{ p  a }\left( T _{2}- T _{1}\right)+ w _{2} h _{ v 2}- w _{1} h _{ v 1}= C _{ p  a }\left( T _{3}- T _{4}\right)+ w _{3} h _{ v 3}- w _{4} h _{ v 4}-\left( w _{3}- w _{4}\right) h _{ f 4} \\0.24\left( T _{2}-33\right)+ 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}

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

\phi= P _{ v 2} / P _{ g 2}=0.074 / 0.837=0.088 \text {   or   } \phi=9 \%

 

 

 

183

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