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Question 11.180E: Atmospheric air at 95 F, relative humidity of 10%, is too wa...

Atmospheric air at 95 F, relative humidity of 10%, is too warm and also too dry. An air conditioner should deliver air at 70 F and 50% relative humidity in the amount of 3600 ft ^{3} per hour. Sketch a setup to accomplish this, find any amount of liquid (at 68 F) that is needed or discarded and any heat transfer.

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CV air conditioner. Check first the two states, inlet 1, exit 2.

In:    P _{ g 1}=0.8246   psia , h _{ g 1}=1102.9   Btu / lbm , \quad h _{ f , 68}=36.08   Btu / lbm,

P _{ v 1}=\phi_{1} P _{ g 1}=0.08246   psia , \quad w _{1}=0.622 P _{ v 1} /\left( P _{\text {tot }}- P _{ v 1}\right)=0.0035

Ex:    P _{ g 2}=0.36324   psia , \quad h _{ g 2}=1092   Btu / lbm

P _{ v 2}=\phi_{2} P _{ g 2}=0.1816   psia , \quad w _{2}=0.622 P _{ v 2} /\left( P _{ tot } -P _{ v 2}\right)=0.00778

Water must be added  \left( w _{2}> w _{1}\right) .  Continuity and energy equations

\begin{aligned}&\dot{ m }_{ A }\left(1+ w _{1}\right)+\dot{ m }_{ liq }=\dot{ m }_{ A }\left(1+ w _{2}\right) \quad \& \quad \dot{ m }_{ A } h _{1 mix }+\dot{ m }_{ liq } h _{ f }+\dot{ Q }_{ CV }=\dot{ m }_{ A } h _{2 mix } \\&\dot{ m }_{ tot }= P \dot{ V }_{ tot } / R T =14.7 \times 3600 \times 144 / 53.34 \times 529.67=270   lbm / h \\&\dot{ m }_{ A }=\dot{ m }_{ tot } /\left(1+ w _{2}\right)=267.91   lbm / h \\&\dot{ m }_{ liq }=\dot{ m }_{ A }\left( w _{2}- w _{1}\right)=267.91(0.00778-0.0035)= 1 . 1 4 7   lbm / h\end{aligned} \begin{aligned}\dot{ Q }_{ CV }=& \dot{ m }_{ A }\left[ C _{ p  a }\left( T _{2}- T _{1}\right)+ w _{2} h _{ g 2}- w _{1} h _{ g 1}\right]-\dot{ m }_{ liq } h _{ f , 68} \\=& 267.91[0.24(70-95)+0.00778 \times 1092-0.0035 \times 1102.9] \\& \quad-1.147 \times 36.08 \\=&- 4 0 6 . 8   Btu / h\end{aligned}

 

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