Question

V-3. Air flows at steady state through a horizontal, well-insulated, constant-area duct whose diameter is 0.75 in. At the inlet, p1 = 55.9 lbf/in.2, T1 = 7200R, and the velocity is 9 ft/s. The temperature of the air leaving the duct is 700°R. Determine (a) the velocity at the exit, in ft/s. (b) the pressure at the exit, in lbf/in.2 (c) the magnitude of the net horizontal force exerted by the duct wall on the air, in lbf. In which direction does the force act?

Answer 1

Given Dala: Air flows at steady state through a Horizontal Well Insulated constant area duct whole Dia meter = 0.25 in = 0.01905 m fressure Pl=55.9 lbflin = 385.41kpa Thrup Ti = 720 R = 400K ; T2= 700 R = 388.8898 Velocity v = 94/see = 2.743 ou/sec @ To Deterneine The velouty at the exit Air velocity in duets is a really critical factor in how well duets do their job of eficiently and quietly Moving the right amount of air from one place. V = RTV/p, = 287 x 400 - 297.86 xíó 385.41X10² = 0.299 melkg Mass Flow rate using M= Avila - 1/4 'V) - 17 x(0.01905)X2743 Lń = 2.632 kg see 0-292 Steady flow energy Equation is - a= Wt in (h2h)+%m (V2 - v tmg (22-Z) V = y,'+200,- 12) = Va= Jv+(400-42€6.b) a) Va=J(2:74372 (401.3-389.6Jxrom Na - 151.15 Mise cor) (up = 495.4 ft/sec) To find the Pressure exit At stage2 in = AV2/02 AVI - AV.2 VaR vs - 0.2978951.15 2.743 1.636mplis P2 P2 =RT2 P2 = 283 x 388.89 = 68222,145 1.636 P2 = 68.22 kpa = 9.89 lbflin

% To calculate the magnitude of the net houzo antal force exerted by the duet wall on the aus & also finding direction. FTP, A, -P2 A2 = rns (V2-4) Foń (12-v) + ALP2-P) Fa 2.612 (151.15-2.743)+(TV 10.01905)*(68-23-385)x18) € 390.607 – 90.28 f = 300.32 KN (or) If = 67.44 ebf] Net Honzontal force exerted by the duet W all on the air is acting Right side.