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Question 11.179E: A flow of moist air from a domestic furnace, state 1 in Figu...

A flow of moist air from a domestic furnace, state 1 in Figure P11.100, is at 120 F, 10% relative humidity with a flow rate of 0.1 lbm/s dry air. A small electric heater adds steam at 212 F, 14.7 psia generated from tap water at 60 F. Up in the living room the flow comes out at state 4: 90 F, 60% relative humidity. Find the power needed for the electric heater and the heat transfer to the flow from state 1 to state 4.

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State 1:       F.7.1:        P _{ g 1}=1.695   psia , h _{ g 1}=1113.54   Btu / lbm

\begin{aligned}& P _{ v 1}=\phi P _{ g 1}=0.1 \times 1.695=0.1695   psia \\& w _{1}=0.622 \frac{ P _{ v 1}}{ P _{\text {tot }}- P _{ v 1}}=0.622 \frac{0.1695}{14.7-0.1695}=0.00726\end{aligned}

Starte 2:        h _{ f }=28.08   Btu / lbm;              State 2a:          h _{ g 212}=1150.49   Btu / lbm

State 4:          P _{ g 4}=0.699   psia ,    h _{ g 4}=1100.72   Btu / lbm

\begin{aligned}& P _{ v 4}=\phi P _{ g 4}=0.6 \times 0.699=0.4194   psia \\& w _{4}=0.622 \frac{ P _{ v 4}}{ P _{ tot }- P _{ v 4}}=0.622 \frac{0.4194}{14.7-0.4194}=0.0183 \\&\dot{ m }_{ liq }=\dot{ m }_{ a }\left(\omega_{1}-\omega_{4}\right)=0.1(0.0183-0.00726)=0.0011   lbm / s\end{aligned}

Energy Eq. for heater:

\begin{aligned}\dot{ Q }_{\text {heater }} &=\dot{ m }_{ liq }\left( h _{ out }- h _{ in }\right)=0.0011(1150.49-28.08) \\&= 1 . 2 3 5   Btu / s = 1 . 1 7   k W\end{aligned}

Energy Eq. for line (excluding the heater):

\begin{aligned}\dot{ Q }_{\text {line }}=& \dot{ m }_{ a }\left( h _{ a 4}+ w _{4} h _{ g 4}- h _{ a 1}- w _{1} h _{ g 1}\right)-\dot{ m }_{\operatorname{liq}} h _{ g 212} \\=& 0.1[0.24(90-120)+0.0183 \times 1100.72-0.00726 \times 1113.54] \\& \quad-0.0011 \times 1150.49 \\=&- 0 . 7 8   Btu / s\end{aligned}

 

 

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