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Question 11.186E: A large air separation plant takes in ambient air (79% N2, 2...

A large air separation plant takes in ambient air (79% N _{2}, 21% O _{2} by volume) at 14.7 lbf / in .^{2}, 70 F, at a rate of 2 lb mol/s. It discharges a stream of pure O _{2} gas at 30 lbf / in .^{2}, 200 F, and a stream of pure N _{2} gas at 14.7 lbf / in .^{2}, 70 F. The plant operates on an electrical power input of 2000 kW. Calculate the net rate of entropy change for the process.

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\dot{ S }_{ gen }=-\frac{\dot{ Q }_{ CV }}{ T _{0}}+\dot{ n }_{ i } \Delta \overline{ s }_{ i }=-\frac{\dot{ Q }_{ CV }}{ T _{0}}+\left(\dot{ n }_{2} \overline{ s }_{2}+\dot{ n }_{3} \overline{ s }_{3}-\dot{ n }_{1} \overline{ s }_{1}\right)

The energy equation, Eq.4.10 gives the heat transfer rate as

\begin{aligned}\dot{ Q }_{ CV } &= \Sigma \dot{ n } \Delta \overline{ h }_{ i }+\dot{ W }_{ CV }=\dot{ n }_{ O _{2}} \overline{ C }_{ P 0  O _{2}}\left( T _{2}- T _{1}\right)+\dot{ n }_{ N _{2}} \overline{ C }_{ P 0  N _{2}}\left( T _{3}- T _{1}\right)+\dot{ W }_{ CV } \\&=0.21 \times 2 \times[32 \times 0.213 \times(200-70)]+0-2000 \times 3412 / 3600 \\&=+382.6-1895.6=-1513   Btu / s\end{aligned}

Use Eq.6.16 (see page 295 also) for the entropy change

\begin{aligned}\Sigma \dot{n}_{i} \Delta \bar{s}_{i}=& 0.21 \times 2\left[32 \times 0.219 \ln \frac{660}{530}-\frac{1545}{778} \ln \frac{30}{0.21 \times 14.7}\right] \\&+0.79 \times 2\left[0-\frac{1545}{778} \ln \frac{14.7}{0.79 \times 14.7}\right]\end{aligned}

= -1.9906 Btu/R s

\dot{ S }_{ gen }=+\frac{1513}{530}-1.9906= 0 . 8 6 4   B t u / R ~ s

…………………………………………..

Eq.4.10 : \dot{Q}_{ C.V. }+\sum \dot{m}_{i}\left(h_{i}+\frac{ V _{i}^{2}}{2}+g Z_{i}\right)=\sum \dot{m}_{e}\left(h_{e}+\frac{ V _{e}^{2}}{2}+g Z_{e}\right)+\dot{W}_{ C . V. }

Eq.6.16 : s_{2}-s_{1}=C_{p 0} \ln \frac{T_{2}}{T_{1}}-R \ln \frac{P_{2}}{P_{1}}

 

186

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