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Solution Data:- m=700 \mathrm{bbl} / \mathrm{hr}\text { S.G. }=0.8524Air fuel ratio f(x)=20Average specific heat of stock =2.268 \mathrm{kj} / \mathrm{lg}{ }^{\circ} \mathrm{C}Length (L)=6 \mathrm{~m}dicono D=6.5 \mathrm{~cm}=0.065 \mathrm{~m}Total prajected Area =150 \mathrm{~m}^{2}. (conversion =200 \mathrm{lb} / \mathrm{bbl} )\Rightarrow Total Heat Liberated (Q)=M_{\text {fuel }} \times \mathrm{NHV}\begin{aligned}& =40.5 \times 10^{3} \times 1700 \\\therefore Q & =6.88 \times 10^{7} \mathrm{kj} / \mathrm{hr}\end{aligned}\rightarrow projected Area of one tube (L * D)=6 \times 0.065=0.39 \mathrm{~m}^{2}\rightarrow No. of tubes =\frac{150}{0.39}=385 Tubes\Rightarrow \therefore \alpha \cdot A_{p}=0.88 \times 385 \times(2 \times 0.065 \times 6)=264.26 \mathrm{~m}^{2}\begin{aligned}\Rightarrow \text { Heat absorption } \%(R) & =\frac{1}{1+\frac{G \sqrt{Q / \alpha A P}}{S}} \times 100 \\& =\frac{1}{1+\frac{20 \sqrt{6.88 \times 10^{7} / 264.26}}{14200}} \times 100 \\\therefore \%(R) & =58.18 \%\end{aligned}\sim outlet temperatune of the stoek:-\begin{array}{l}\therefore Q=m C_{P} \Delta t \\\therefore 0.5818 \times 6.88 \times 10^{7}=700 \times 200 \times 0.8524 \times 2.268 \times \Delta t \\\therefore \Delta t=147.89\end{array}so, the outlet temperature is equal to 147.89+230=377.89^{\circ} \mathrm{C} ...