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Question 11.182E: An indoor pool evaporates 3 lbm/h of water, which is removed...

An indoor pool evaporates 3 lbm/h of water, which is removed by a dehumidifier to maintain 70 F, \Phi = 70% in the room. The dehumidifier is a refrigeration cycle in which air flowing over the evaporator cools such that liquid water drops out, and the air continues flowing over the condenser, as shown in Fig. P11.107. For an air flow rate of 0.2 lbm/s the unit requires 1.2 Btu/s input to a motor driving a fan and the compressor and it has a coefficient of performance, \beta= Q _{ L } / W _{ C }=2.0. Find the state of the air after the evaporator, T_{2}, \omega_{2}, \Phi_{2} and the heat rejected. Find the state of the air as it returns to the room and the compressor work input.

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The unit must remove 3 lbm/h liquid to keep steady state in the room. As water condenses out state 2 is saturated.

1:   70 F,   70%    \Rightarrow P _{ g 1}=0.363   psia , h _{ g 1}=1092.0   Btu / lbm ,

P _{ v 1}=\phi_{1} P _{ g 1}=0.2541   psia , \quad w _{1}=0.622 P _{ v1 } /\left( P _{ tot } -P _{ v1 }\right)=0.01094

CV 1 to 2:    \dot{ m }_{ liq }=\dot{ m }_{ a }\left( w _{1}- w _{2}\right) \Rightarrow w _{2}= w _{1}-\dot{ m }_{ liq } / \dot{ m }_{ a }

\begin{aligned}& q_{L}=h_{1}-h_{2}-\left(w_{1}-w_{2}\right) h_{f 2} \\w_{2}=& 0.01094-3 /(3600 \times 0.2)=0.006774 \\P_{v 2}=& P_{g 2}=P_{t o t}   w_{2} /\left(0.622+w_{2}\right)=\frac{14.7 \times 0.006774}{0.628774}=0.1584   psia\end{aligned}

Table F.7.1:    T _{2}= 4 6 . 8 ~ F \quad h _{ f 2}=14.88   btu / lbm , h _{ g 2}=1081.905   Btu / lbm

\begin{aligned}q _{ L } &=0.24(70-46.8)+0.01094 \times 1092-0.006774 \times 1081.905 \\&-0.00417 \times 14.88=10.12   Btu / lbm \text {   dry air }\end{aligned}

 

\dot{ W }_{ c }=\dot{ m }_{ a } q _{ L } / \beta= 1   B tu / s

 

CV Total system :

\begin{aligned}\tilde{ h }_{3}-\tilde{ h }_{1} &=\dot{ W }_{ el } / \dot{ m }_{ a }-\left( w _{1}- w _{2}\right) h _{ f }=1.2 / 0.2-0.062=5.938 Btu / lbm \text { dry air } \\&= C _{ p  a }\left( T _{3}- T _{1}\right)+ w _{2} h _{ v 3}- w _{1} h _{ v 1}\end{aligned}

Trial and error on T _{3}

\begin{aligned}3: w _{3}= w _{2}, h _{3} \Rightarrow T _{3}= 1 1 2  F , \quad P _{ g 3}=1.36  psia , P _{ v 3}= P _{ v 2}=0.1584 \\\phi_{3}= P _{ v 3} / P _{ g 3}=0.12 \quad \text { or } \quad \phi_{3}= 1 2 \%\end{aligned}

 

 

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