Question

Air flows through a converging-diverging nozzle/diffuser. Assuming isentropic flow, air as an ideal gas, and constant specific heats determine the state at several locations in the system 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)= 366 Inlet pressure: P1 (kPa) = 496 Inlet Velocity: V1 (m/s) = 99 Area at inlet (cm^2) = 8.7 Mach number at the exit = 1.7 a) Determine the mass flow rate (kg/s) through the nozzle Your Answer=94056 Correct! Exact Answer=0.4067 +/- 1.7E-03 b) Determine the Mach number at the inlet. Your Answer = 0.2723 Incorrect c) Determine the stagnation temperature (K) at the inlet. Your Answer=3707 Correct! Exact Answer=370.88 +/- 1.4E+00 d) Determine the stagnation pressure (kPa) at the inlet. Your Answer=910.48 Correct! Exact Answer=519.53 +/- 2.4E+00 e) Determine the temperature (K) at the throat. Your Answer = 300.00 Correct! Exact Answer=309.0655 +/- 1.1E+00 f) Determine the pressure (kPa) at the throat. Your Answer = sun ona Incorrect. g) Determine the velocity (m/s) at the throat. Your Answers Correct! Exact Answer=352.40 +/- 1.3E+00 h) Determine the area (cm2) at the throat. Your Answer = 2.16 Incorrect. i) Determine the temperature (K) at the exit. Your Answer = 22.00 Correct! Exact Answer=235.0308 +/- 8.1E-01 j) Determine the pressure (kPa) at the exit. Your Answer = 1062 Correct! Exact Answer= 105.25 +/- 3.9E-01 k) Determine the velocity (m/s) at the exit. Your Answer = 52240 Correct! Exact Answer=522.42 +/- 2.1E+00 1) Determine the area (cm^2) at the exit. Your Answer = 4.30 Correct! Exact Answer=4.99 +/- 1.9E-02

Answer 1

Given Inlet temperature, T, = 366 K Inlet pressure, P, = 496 kPa Inlet velocity, V, = 99 m/s Inlet area, Az = 8.7 cm² = 8.7 x 10-4 m2 Mach number at exit, M, = 1.7 (a) Using the ideal gas law density of air at the inlet of the nozzle/diffuser can be obtained as P1 = P. 496 RT 0.287 x 366 = 4.7219 kg/m3 Mass flow rate through the nozzle/diffuser can be computed as m = P2 A, V1 = 4.7219 x 8.7 x 10-4 x 99 = 0.40669 kg/s Mass flow ratet hrough the mozzle = 0.4067 kg/s (b) Mach number at the inlet can be calculated as V M V KRT C1 Where, c is the sonic velocity Substituting the known values M = 99 V1.4 x 287 x 366 0.25816 Mach number at the inlet 0.258

(c) Stagnation temperature at the inlet can be computed as k-1 T01 = T (1+ M 2 Substituting the known values 1.4 - 1 T01 = 366 (1+ x 0.258162 2 - 370.8785 K Stagnation temperture at the inlet = 370.88 K (d) We know k/k-1 Po1 Р. G) Or To1 k/k-1 Poi = P2 Substituting the known values Po1 = 496 1.4 370.8785 1.4-1 = 519.5279 kPa 366 Stagnation pressure at the inlet - 519.53 K

(e) Since flow accelerate from a subsonic velocity to supersonic velocity, Mach number for the flow at throat would be 1. Therefore, the properties at the throat will be reference properties (T*,P*,p*,M") We know T* 2 k +1 T. Therefore T* = = 671) T. Substituting the known values T* = 2 370.8785 = 309.0655 K (1 + 1.4) Temperature at the throat = 309.065 K (1) Pressure at the throat can be obtained from the relation rk/k-1 2 k/k-1 P P. Or 2 k/k-1 P* = P. Substituting the known values 1.4 1.4-1 2 P* = 519.5279 11.4 +1 = 274.45713 kPa Pressure at throat = 274.457 kPa

(g) Since the Mach number is unity at the throat the velocity of the flow will be sonic velocity at the throat. Therefore V = V* = VRT* Substituting the known values V = V1.4 x 287 x 309.0655 = 352.3954 m/s Velocity at the throat 352.4 m/s (h) Density of the flow at the throat can be calculated as p* P* 274.45713 RT* 0.287 x 309.0655 3.09416 kg/m3 Since the mass flow rate remains constant throughout the nozzle m = P14, V1 = p*A*V* On rearranging the above equation A* = m p*V Substituting the known values A* = 0.4067 -= 3.7299 x 10-4 m2 = 3.7299 cm? 3.09416 X 352.3954 Area of the throat 3.73 cm?

(i) Since, the flow is isentropic, the stagnation properties remain constant throughout the flow. Therefore To2 = Toi = T, = 370.8785 K From the isentropic flow relation, the temperature at the exit can be computed as T2 TO2 k-1 2 + -M2 Substituting the known values 370.8785 T, 235.0308 K 1.4 - 1 x 1.72) 2 Temperature at the exit 235.031 K (1) Since the stagnation properties remain constant Po2 = Poi = P, = 519.5279 kPa Using the isentropic flow relations, the pressure at the exit can be computed s T2 k/k-1 P2 = Poz Toz Substituting the known values P2 = 519.5279 1.4 235.0308 1.4-1 = 105.2529 kPa 370.8785 Pressure at the exit = 105.253 kPa

(k) Velocity the exit can be obtained as V2 = M,C2 = M2 VRT, Substituting the known values V2 = 1.7 X V1.4 x 287 x 235.0308 = 522.4158 m/s Velocity at the exit – 522.416 m/s (1) Density of the flow at exit can be computed using the ideal gas law as follows P2 P2 105.2529 RT 0.287 x 235.0308 1.5604 kg/m3 Since the mass flow rate remains constant m = P1AV = P2A, V2 Therefore m Az 0.4067 31.5604 x 522.4158 4.9891 x 10-4 m2 = 4.9891 cm2 PzV2 Area at the exit = 4.989 cm