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Step 1Given data:Heat duty =3 \times 10^{6} \mathrm{Btu} / \mathrm{hr}Efficiency =75 \%Furnace is fired with gaseous fuel at a rate of 17 \mathrm{~b} air//b fuel (\mathrm{NHV}=17000 \mathrm{Btu} / \mathrm{b} )Tubes are arranged in 2 rows and of 5 in OD, 40 \mathrm{ft} length and 2 \times OD spacingHeat rate =35000 \mathrm{Btu} / \mathrm{hr}.Step 2% heat absorbed in the radiation section ( R \%)We know,\% heat absorbed in radiant section, R=\left(\begin{array}{l}1 \\1\end{array}\right)G=a i r / f u e l ratio =17 l b air / lb fuel\alpha=0.986 for two rows at spacing 2 O DSince, Q in Btu/hr, S=4200Area of wall having tubes in front of it, A_{c p}=L N \frac{C}{12}\begin{array}{l}L=\text { length }=40 \mathrm{ft} \\C=\text { center to center spacing } \\N=\text { number of tube per row } \\n=\text { no. of rows } \\A=L n N \frac{D}{12} \\A=n A_{c p} \frac{D}{C} \\R Q=A q\end{array}q= rate of heat absorption per square foot of projected tube areaSo, Substituting all these in\left(\begin{array}{l}1 \\1\end{array}\right)\begin{array}{l}q\left(\frac{D}{C} \frac{n}{a}\right)=\frac{(1-R)^{2}}{R}\left(\frac{s}{G}\right)^{2} \text { For most commercial case, } D / C=5 / 10 \\35000\left(\frac{5}{10} \frac{2}{0.986}\right)=\frac{(1-R)^{2}}{R}\left(\frac{4200}{17}\right)^{2} 0.581=(1-R)^{2} \Rightarrow R=47.55 \%\end{array}Heat absorbed in radiation zone =R \times Q=0.4755 \times 30 \times 10^{6}=14265000 \mathrm{Btu}2 Heat absorbed in the convection section: ...