Question

2. Air flows through a converging-diverging nozzle. There is a normal shock in the divergmy section of the nozzle. The areas at throat, section 1. and section 2 are given in the figure below The sections 1 and 2 are located immediately upstream and downstream of the normal respectively. Therefore, the sections 1 and 2 are assumed to have the same area. There temperature and pressure are given as 300 K and 110 kPa, respectively. The entire isentropic except for the portion across the normal shock. (a) Find the local temperature (5 pts) and local pressure (5 pts) at section 1, (b) Find the local temperature (5 pts) and local pressure (5 pts) at section 2 (c) Find the total temperature (5 pts) and total pressure (5 pts) at section 3. (d) Draw a control volume for the portion of flow across the normal shock (10 pts). Normal Shock Reservoir To = 300 K Po = 110 kPa Throat At = 0.001 m2 A1 = A2 = 0.002 m A1 = A2 = 0.002 m2 Exit Exit

Answer 1

Normal Shock Exit throat To =300k Po = 116 kpon At=0.com 0 0 0 A = A₂ = 0.002 m² 0.002 = 2 0-101 2, from ixentropic tablex, corresponding mach number in 2.2. & M, = 2-2 a :@ M,=2-2, from isentropic tablex = 0.5031 - Ti = BDO X0-5081 - 152-43K o »po= 152-43K = 0.0935 P = 110x0.0935 = 10.285kpa →/P, = 10.285 kpa] b) From Normal shock tables, @m, = 2.2, M2 = 0.5471, P2 = 5:48 I 2 = 1.8569 845 P₂ = 10.285 X 5-48 = 56.361 kpa [P2 = 56-361 kPa

T₂ = 152.43 X 1-8569 T2 = 283.04% Poz = 110 x0:628) Por = 69.091 kpa stagnation Pressure is same through out the isentropic flow. . Poz = Poz = 69.091 kpa. temperature is some isentropic throughout the stagnation flow ... Toi = To = To3 = 300k. (d) T2 =283-04K 7, -13243K m, =2-2 m₂ =0.547 control volume Normal a cross Theashock. r. There is an Entropy difference in