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Question 11.187E: Ambient air is at a condition of 14.7 lbf/in.^2, 95 F, 50% r...

Ambient air is at a condition of 14.7 lbf / in .^{2}, 95 F, 50% relative humidity. A steady stream of air at 14.7 lbf / in .^{2} , 73 F, 70% relative humidity, is to be produced by first cooling one stream to an appropriate temperature to condense out the proper amount of water and then mix this stream adiabatically with the second one at ambient conditions. What is the ratio of the two flow rates? To what temperature must the first stream be cooled?

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\begin{aligned}&P_{1}=P_{2}=P_{5}=14.7   lbf / in ^{2} \\&T_{1}=T_{2}=95  F \\&\phi_{1}=\phi_{2}=0.50, \phi_{4}=1.0 \\& T _{5}=73   F , \phi_{5}=0.70\end{aligned} \begin{aligned}&P_{ v 1}= P _{ v 2}=0.5 \times 0.8246=0.4123, w _{1}= w _{2}=0.622 \times \frac{0.4123}{14.7-0.4123}=0.0179 \\& P _{ v 5}=0.7 \times 0.4064=0.2845 \quad \Rightarrow \quad w _{5}=0.622 \times \frac{0.2845}{14.7-0.2845}=0.0123\end{aligned}

MIX: Call the mass flow ratio  r = m _{ a 2} / m _{ a 1}

Conservation of water mass:  w _{1}+ r w _{4}=(1+ r ) w _{5}

Energy Eq.:

\begin{aligned}h _{ a1 }+ w _{1} h _{ v1 }+ rh _{ a 4}+ rw _{4} h _{ v 4} &=(1+ r ) h _{ a 5}+(1+ r ) w _{5} h _{ v 5} \\\rightarrow & 0.0179+ rw _{4}=(1+ r ) 0.0123\end{aligned}

or

\begin{gathered}r =\frac{0.0179-0.0123}{0.0123- w _{4}}, \text { with } \quad w _{4}=0.622 \times \frac{ P _{ G 4}}{14.7- P _{ G 4}} \\\begin{aligned}0.24 \times 555+0.0179 \times 1107.2+ r \times 0.24 \times T _{4}+ rw _{4} h _{ v 4} \\=(1+ r ) \times 0.24 \times 533+(1+ r ) \times 0.0123 \times 1093.3\end{aligned}\end{gathered}

or        r \left[0.24 \times T _{4}+ w _{4} h _{ G 4}-141.4\right]+11.66=0

Assume  T _{4}=40   F \quad \rightarrow P _{ G 4}=0.12166   psia , h _{ G 4}=1078.9   Btu / lbm

\begin{aligned}& w _{4}=0.622 \times \frac{0.12166}{14.7-0.12166}=0.0052 \\&\frac{ m _{ a 2}}{ m _{ a1 }}=\frac{0.0179-0.0123}{0.0123-0.0052}= 0 . 7 8 8 7\end{aligned}

0.7887[0.24×500 + 0.0052×1078.9 – 141.4] + 11.66 = -0.29 ≈ 0 OK
=>       T _{4} = 40 F

 

187

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