Question

need help with letter j,k,l,m,n,o just those i solved the top ones just the last 6 thanjs you

mine the temperature (K) at the throat. Your Answer=289.711 Correct! Exact Answer=289.71 +/- 1.0E+00 e) Determine the velocity (m/s) at the throat. Your Answer 341.18 Correct! Exact Answer 341.183 +/- 1.3E+00 f) Determine the pressure (kPa) at the throat. Your Answer=314.09 Correct! Exact Answer=314.10 +/- 1.2E+00 g) Determine the mass flow rate (kg/s) through the nozzle Your Answer=1.0826 Correct! Exact Answer= 1.08264 +/- 9.9E-02 h) Determine the Mach number (Mx) just upstream of (before) the shock. Your Answer=1.5594 Correct! Exact Answer=1.5595 +/- 6.0E-03 i) Determine the maximum velocity (m/s) in the nozzle. Your Answer=478.04 Correct! Exact Answer=478.07 +/- 1.8E+00 1) Determine the Mach number (My) just downstream of (after the shock. Your Answer k) Determine the stagnation pressure (kPa) after the shock. Your Answer 1) Determine the pressure (kPa) at the exit. Your Answer m) Determine the temperature (K) at the exit. Your Answer n) Determine the area (cm^2) at the exit. Your Answer-364.68 Incorrect. ) Determine the velocity (m/s) at the exit. Your Answer = 4.3 Incorrect

Answer 1

Solution - downstream a sy Mac Px Inlet T = 340k = 148. pa Exit M = 0.16 P = 550kpla V=124 m/sec throat A=8.4 cm² Shock M = ? Solution ) Moch number just downstream shock (My ) • (My) from the Normal Shock table for is entropic flow- we have at Mx = 1.5594 21:56 My = 0. 6809 Also it can be found by using the relation between upstecam and downs tecăm properties of of shock- M? = (K-1) M} +2 (1-4-1) (1056)+ 2 2.97344 2km 2 (k-1) 2 (1-4) (1-56)2-(0.4) 6.41408 Mr = 0.46358 må 0-6809 after Pa Mya k) stagnation pressure the shock (Poy) = ? from the normal shock table- at Mx=1.56 and M. My -0.6809 = 0.2743, 541.378 kPa Poy I loy = 541-378 kPa e) Pressure at exit (P2) - from isentropic flow table at M = 0.16 Poy 148.5 0.2743

P2 Pz 0.982 3 Poz Poy P = 0.98 2 3 x 541-378 P2 = 531.796 kPa m). Temperatwu at exit (TG) - From Isentropic flow table- at M = 0.16 Tz - 0.9949 Тоъ = 0.9949 (To) During a shock- stagnation nation temperatur (as well as enthalpy) . Remains constant thro throughout therefore Tog= Toi = sqrth 347.65k T angangaragsru 0.99498347-65k T, 2 345.876k n) Area at exit- From isentopic flow table- aut My = 0.16 Az = 3.6 72 A* A2 = 3:6727 X 804 cm² A2 = 30.8.5cm 9) velocity at exit Y = M₂ 5YR TE 20.16 1.4x 287 X 345,876 2 V₂ = 59.646 m/s