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Question 11.177E: To save on natural gas use in the previous problem it is sug...

To save on natural gas use in the previous problem it is suggested to take and cool the mixture and condense out some water before heating it back up. So the flow is cooled from 200 F to 120 F as shown in Fig. P11.83 and the now dryer mixture is heated to 1500 F. Find the amount of water condensed out and the rate of heating by the natural gas burners for this case.

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Eq.11.25:          P _{ v }=\phi P _{ g }=0.70 \times 11.53   psia =8.07   psia

Eq.11.28:          \omega=0.622 \frac{ P _{ v }}{ P _{ tot }- P _{ v }}=0.622 \times \frac{8.07}{14.7-8.07}=0.757

Flow:

\begin{aligned}&\dot{ m }_{ tot }=\dot{ m }_{ a }+\dot{ m }_{ v }=\dot{ m }_{ a }(1+\omega) \quad so \\&\dot{ m }_{ a }=\frac{\dot{ m }_{ tot }}{1+\omega}=19.92   lbm / s ; \quad \dot{ m }_{ v }=\dot{ m }_{\text {tot }} \frac{\omega}{1+\omega}=15.08   lbm / s\end{aligned}

Now cool to 120 F:      P _{ g }=1.695   \text { psia }< P _{ v 1} \quad \text {   so } \quad \phi_{2}=100 \%, P _{ v 2}=1.695   psia

\begin{aligned}&\omega_{2}=0.622 \frac{ P _{ v }}{ P _{ tot }- P _{ v }}=0.622 \times \frac{1.695}{14.7-1.695}=0.08107 \\&\dot{ m }_{ liq }=\dot{ m }_{ a }\left(\omega_{1}-\omega_{2}\right)=19.92(0.757-0.08107)=13.464   lbm / s\end{aligned}

The air flow is not changed so the water vapor flow for heating is

\dot{ m }_{ v 2}=\dot{ m }_{ v1}-\dot{ m }_{ liq }=15.08-13.464=1.616   lbm / s

Now the energy equation becomes

\begin{aligned}\dot{ Q } &=\dot{ m }_{ a }\left( h _{3}- h _{2}\right)_{ a }+\dot{ m }_{ v 2}\left( h _{3}- h _{2}\right)_{ v } \\&=19.92(493.61-157.96)+1.616(12779-996.45) / 18.015 \\&=7743   Btu / s\end{aligned}

Comment: If you solve the previous problem you find this is only 47% of the heat for the case of no water removal.

………………………………………

Eq.11.25 :  \phi=\frac{P_{v}}{P_{g}}=\frac{\rho_{v}}{\rho_{g}}=\frac{v_{g}}{v_{v}}

Eq.11.28 :  \omega=0.622 \frac{P_{v}}{P_{a}}=0.622 \frac{P_{v}}{P_{\text {tot }}-P_{v}}

 

177

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