Question

Air flow is induced in an insulated tube of 7.16 mm diameter by a vacuum pump. Air is drawn from a room where stagnation pressure is 101 kPa (abs) and stagnation temperature 23 °C, through a smoothly contoured converging nozzle. At station (1), where the converging nozzle joins the constant area tube, the static pressure is 98.5 kPa (abs). At section (2), located some distance downstream in the constant area tube, the air temperature is 14 °C. Determine a) the mass flow rate b) the local stagnation pressure at section (2) c) the frictional force on the duct wall between section (1) and (2) d) Mach number at the section (2) (Hint: Mass flow rate and stagnation temperature remains constant) To = 296 K Po = 101 kPa abs) Flow D=7.16 mm P1 = 98.5 kPa(abs) T2 = 287 K

Answer 1

Palm To=2966 Po= 10ba 0 P= 98.5 hPa T2 = 2814 dirice, there is no any hea so, process should be adiabatic nere so no any heat rejection. at (0) to (1) (4 ***)/v 04/14 => Ti = 2968 23:57 | 101 Ti = 293-887 k and Energy at 'o' and 'l' are same. 0o To - Top 296= (293.887) + V V = (2-11212)X 2X1003'e (v. = 65-165 m 2xlp

@ maso flow rate in- A XVI XS - den xV, X. P. 4. RTL - I (0.007168 x 65.1658 98.5 0.297 x 29311 = 13.064x10 3 kglo 6 Stagnation pressure at (27. 7/4-1 02 LOL T2 0414 => pos = Pix 240) 0 To = Toz 296 - Tot i => V, 2 = (296-287) X2X1005 Y = 134.5 mln

maso How rate (2) is name is (1) in- & AX VJ X S, A x V x 82 - Arya AxVxe » S = Visi V2 134.5 - P - 65.165 x P x T₂ 134.5 P = 46.6 hPa So, from equahin Py = 46-6X (29654 Por = 51.92 hPa 287 ② frictional force = ? change in momentum between I and 2. F+P-P) A = [V, -V] - 3.064 X10x1345-65.165)"

46.6 F = 0.2124 - (98.5 - SEE) X1Dx (0.00716/2 - – 1.8770 F = 1.887 n/ V = 134-5 m/s M= V₂ Jr. R.Tz = 134.5 J 1.48287 x 287 Me = 0.396