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Q1: Air Flows steadily between two sections in a long, straight portion of4-in inside diameter. The uniformly distributed temperature and pressure at each section are given. If the average air velocity (non-uniform velocity distribution) at section 2 is 1000ft/s, calculate the average air velocity at Section (l). Given: Air behaves like an ideal gas D1- D2- 4 in; pl- 100psia, p2-18.4psia T1 5400R, T2 453 R. v2 1000ft/s
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Answer #1

\rho 1A1V1 = \rho 2A2V2

m1 = m2

Assuming under these conditions that air behaves as ideal gas, from ideal gas equation

\rho 2/\rho 1 = p2T1/p1T2

V1 = P2T1V2 / P1T2 = 18.4 * 813 *1000 / (100 * 726) = 206.049 ft/s

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