Question

2. A variable mesh screen produces a linear control volg and axisymmetrie velocity profile as indicated in Figure which is given by U2 = Umax () in the air flow through a 0.6 m diameter (D) circular cross section duct. Umax is given 46 m/s. as The static 14 pressures upstream and downstream of the screen are 1.4 kPa and I kPa and are uniformly distributed over the flow cross section area. Neglecting the force exerted by the duct wall on the flowing air, calculate the drag foree (R). (p-1.23 kg/m', and d1-2xrdr). a. Write the Linear Momentum equation in integral form b. Simplify it c. Find an algebraic equation that gives R. & Compute R. Good Luck!

Answer 1

Control volume , Umar = 46 m/s U22 Umayr Dz0-6 DzObm H = 1.23 kalm3 P = 1.4kpa P, z Ikkia V, - 30 ml axial component of liviear momentum the flow : Application equation of to 0 – V, PV,A, + ( us t u aarder = P, A - Rx - Pz Az - fueAit (u2² anodo = P, A, - l Az-Rx & Ra= tvira Di? _ 247 Š 42 ²8 do + P 1 2 ² 2 2 2 Green Us z Umowl ha) » hovering the team 152.95* * From 0 (1) 2n = Tu, I D, ² - 2 + + (53.33 27r dr + Pi 2 D, 2 P = 2 2 ² Rez 17 2 - 2 1 8 x 153.33² 1 2 1 2 3 4 Pi Dia Pere her tv - PAT2 133 55°(2:34) + ID?P27/02 Re: 1.23 x 30°,1 x 0.62-2011.23x153-3560,57; +(4001000) X, = 312.8582 -** *4 + 118.04 Hoe Rx 258.138N 367.74