Question

Homework Problem HW8 2 Attemps Used 2/20 Due Date - Thu, Jul 30, 2020, 23:59:59 Current Time Thu, Jul 30, 2020, 01:42:09 Air flows through a converging-diverging nozzle/diffuser. A normal shock stands in the diverging section of the nozzle. Assuming sentropic flow, air as an ideal gas, and constant specific heats determine the state at several locations in the system. Solve using equations rather than with the tables Note: The specific heat ratio and gas constant for air are given as k-1.4 and R-0.287 kJ/kg-K respectively. --Given Values Inlet Temperature: T1 (K) - 340 Inlet pressure: PI (kPa) - 550 D Inlet Velocity: VI (m/s) - 124 Throat Area: A (cm/2) - 8.4 Pressure upstream of (before) shock: Px (kPa) - 148.5 Mach number at exit: M -0.16 a) Determine the Mach number at the inlet Your Answer 1061 Incorrect b) Determine the stagnation temperature (K) at the inlet. Your Answer-34765 Correct! Exact Answer: 34765+.12E+00 c) Determine the stagnation pressure (kPa) at the inlet. Your Answer - 569 92 Incorrect d) Determine the temperature (K) at the throat Your Answer 26855 Incorrect. c) Determine the velocity (ms) at the throat Your Answer 125 82 Incorrect :) Determine the pressure (kPa) at the throat, Answer- Incorrect

Answer 1

Pi (1+0:2 Mail] 3. Ti = 340k P, = 550kpa VI 3 124m/s 20 Athout = 8.4cm Menit 2016 I p befose shock = P x = 148.5 kpa Inlet Macken 2. = MART m/s = 364.61 - / 4х287xзцо Ma, vi ai 124 369.61 20.33 stagnation temp :- (24) 2 2x(1005) = 347.643 Tit a z 340 + To = Tit 20p Stognation pren we = 550 $50 (170.23(0.373) Po. 2593.08kpa at Hout M=1 PA -0.528 Po let - 313.14624 kpa To 20.834 * = 290k

mut -CALXV thout = T8R Ishoul 11.4x287X290 ✓ thood - 341.357) Psh xh = Monorrtth 2. Johan = eshsout RT those Cthroat = 3.36 kg/m3 in shrout Ishrat Alhout thout = 3.36 *8.4x16x341.35 Divine in ghosecut 1.07 kg/ mnozzle ? P =COT 550 = Cx0.28 7*340 G=5 = 5.63 164 ? full Mosole zl, Aqva A in AX (1+0.2xMef 13 1.728 Mai i+0:2X60-332 AL (1+04 8.ux104 1.728 x0.23

from normal sock table A, -0.001571 m2 A m, = GIAN 0 42 25.68X0.001571 X 124 m = 1.09 Kgl Pa = 148.5*pa Esentropic flow B 148.5kpa 593.08kpa -0.25038 Po. M = 142 (before the shock trave) Tz То 20.5756 T2= = 200.10K 20.35 M2 = 102 M2=0.25 (k-1) Maz²+2 (0.4x 1.92² +2) 24 Maz?- (k-1) 2x14x1.92²-04 (mock numbers after sock uzove) [2k ma? (k-1) [2+(K-1) maup] (k+1)2 mai 2 [2x1481972–ay (2km? (] = {2+(x-) MB] (214) ²x 1.923 TE Tz (2+04x1.92). 1162

TE T2 -1.62 T3 = 162X Tz -1.62x200.10 [T =324.876 (Temp after shock) Sony point 3-4 (asentropic How) (Paz =Poy) a 1 Tos = toy at Mg = 70.35 [1 +0.2 M32] 3:57 Pane = (1+0.28 037) RA Pz P2 Ktol (2xm;< gr-1] man (2x14x1922-04) = 4.1342 2 2.4 Pet 24.134XB = 4.134X148.5 = 614 kpa Pos = {1+0.2X03847** *61Y = 668.28 Toza [1+02x0.382] XT3 - [1 +0-260362)* 324.87 Toz = 382.821 - Toy AA16 at Me =0.16 3.5 3.5 Pay = (1 +0.2(Me}] - {1+22X(016)] Py = =/656.uukpa = Pe ' Poy=P03 668-28 1.018

To Ти [1+0.263)?] {1+02 (Over] Tu = 33282 liol - 331 IC ko Te -881K Pe=lexRTO 656.44 = lex0.287X331 € 56.4 kg) is 2364.68ms zve Veruit zorte 1.4x287X331 = Ce feve 1.09 = @ 6.9 x AeX364.68 2 Ae = 4.33x104 A = 4,3 cm