Urgent
Solved 1 Answer
See More Answers for FREE
Enhance your learning with StudyX
Receive support from our dedicated community users and experts
See up to 20 answers per week for free
Experience reliable customer service
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 . ...