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Five percent excess air is used in burning the reformer
Chapter 13, Problem 13.2(choose chapter or problem)
Five percent excess air is used in burning the reformer fuel; it is drawn into the system at \(30^{\circ} \mathrm{C}\) and 70% relative humidity. Estimate the average molecular weight of the air. Why does it differ from the value of 29 determined in Example 3.3-4 even though the ratio of nitrogen to oxygen is the same? Determine the flow rate of this stream \(\left(\mathrm{kmol}, \mathrm{m}^{3}\right)\) per kmol of natural gas burned.
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QUESTION:
Five percent excess air is used in burning the reformer fuel; it is drawn into the system at \(30^{\circ} \mathrm{C}\) and 70% relative humidity. Estimate the average molecular weight of the air. Why does it differ from the value of 29 determined in Example 3.3-4 even though the ratio of nitrogen to oxygen is the same? Determine the flow rate of this stream \(\left(\mathrm{kmol}, \mathrm{m}^{3}\right)\) per kmol of natural gas burned.
ANSWER:Step 1 of 7
Given data regarding the burning of five percent excess air in the reformer fuel:
\(\begin{aligned}P&=1\mathrm{atm}\\ \text{ R.H. }&=\frac{p_{\mathrm{H}_2\mathrm{O}}}{\dot{p_{\mathrm{H}_2\mathrm{O}}}}\times100=70\%\\ p_{\mathrm{H}_2\mathrm{O}}&=0.0424\mathrm{\ bar}\\ p_{\mathrm{H}_2\mathrm{O}}^*&=0.02968\mathrm{\ bar}\end{aligned}\)
Basis: 1 mole dry air (\(79\%\mathrm{\ N}_2(\mathrm{mol}\%)\) and \(21\%\mathrm{\ O}_2(\mathrm{mol}\%)\)).
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