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section - 1P_{1}=77 \mathrm{KPa}(a b s)T_{1}=268 Kv_{1}=205 \mathrm{~m} / 6here P_{1}=e_{1} R T_{1} \& P_{2}=e_{2} R T_{2}\begin{aligned}\rho_{1} & =\frac{P_{1}}{R T_{1}} \& \rho_{2}=\frac{P_{2}}{R T_{2}} \\\frac{\rho_{1}}{\rho_{2}} & =\frac{P_{1}}{R T_{1}} \times \frac{R T_{2}}{P_{2}} \\& =\frac{P_{1}}{P_{2}} \times \frac{T_{2}}{T} \\\frac{\rho_{1}}{\rho_{2}} & =\frac{77}{45} \times \frac{240}{268} \\\frac{\rho_{1}}{\rho_{2}} & =1.532\end{aligned}here \dot{m}_{1}=\dot{m}_{2} \Rightarrow d_{1} A_{1} v_{1}=P_{2} A_{2} v_{2}here A_{1}=A_{2}=\frac{\pi}{4} D^{2}\begin{array}{l}v_{2}=v_{1} \times \frac{e_{1}}{\rho_{2}} \\=205 \times 1.532 \\\frac{v_{2}=314.06 \mathrm{~m} / \mathrm{s}}{\mathrm{T}} \\\end{array}average air velocity at section 2 . ...