Question

Air at a local pressure of 650 kpa and local temperature of 280 K enter a frictinless adiabatic convergent- Divergent nozzle with velocity of 160m/s. the throat area of nozzle is 3.2cm^2 and the exit area is 14cm^2 and the shock occurs where the nozzle area is 8cm^2 fined the following:

1- Back pressure.

2- The flow velocity on both side of the shock. 3-What is the flow rate.

4- Exit Mach number.

5- Mach number on both side of the shock

Answer 1

" To = 280k At=312 come A = 14 cm² 2 Solution Given Pia = 6 sokpa V = 160m/sec A shock - 8 cm² = (1) Calculate the back pressure We know that o = JYRT 57.4x 287 x 280 - 835.416 m/sec 160 V. M = Mach Number = -0.477 835.416 for + M = 0.477 -0.48 Pi -0.8538 (by interpolation) Po - P Po 650 0.8538 0.8538 > = ) Po = 761.3k la 280 0.9568 -o Or 9 568 - To То. 1 To =292.64k 4.375 O 0262 Ae 2 Pe. Po + 14. 8.2 1) AB Pe 0.0262X 761.3 = .19.97 kra And 19, 97k la Back pressure (Pe) (2) Calculate the flow velocity on both side 2.5 - ' 4 A 8 3.2 2.44 fox A Mx - 2.5 2 Aby

=JORTY 1.4 x 287 X 278.4 C = 3340 45 m/sec & velocity after the shock Vy My C = 0.52 x 334.45 Vy = 473.92 m/sec / Ang Calculate flow rate - mass flow in Chock- レ * p* V A SURT RT* RT* for .M=1 p* 481-90x108 287X243.77 X3.2 x 10 x = 0.528x761.3 p* = 401. g6k pa 1.4x287*24377 = 0.833 To T* = 243.77K 20.5754 kg/sec | Ang (4) Calculate exit Mach Number A e = 14cm2 before shock (Ax) = 8 cm2 Ae Ae / At Ax 1.2886 Ae / A = 14 Ax / A - Ae/Aě 2.25505 Mexit 0.263 21255 for. Ae/A

5.) Mach Number before the shock I 2.44 Mach Number after the shock = 0.52