Question

Normal Shock Nozzle Exit (4, -6 cm? Back pressure Air from a reservoir at 350 K and 500 kPa, flows through a converging-diverging nozzle. The throat area is 3 cm- and the exit area is 6 cm. A normal shock appears, for which the downstream (region 2) Mach number (M2) is 0.6405. Reservoir Throat (A = 3 cm (a) What is the Mach number (M]) upstream of the shock? 350K, 500 kPa (abs) (b) What is the area where the shock appears? (c) Find the mass flow rate through the nozzle. (d) Find the back pressure (Pb). Region 2 (e) Find the back pressure (Pb) for a design condition where the flow at the exit is supersonic without any shock occurring in the nozzle and outside of the nozzle. Region 1

Answer 1

Then, Toi (2+ - 2 M² ) But, M*=d. - 350k 350K - 1+ 0.2 T* I - T - 350 1.2 = 291.67k Also, for = (1+ za upya Then, 해 0.4. Poi = (at 0.2312) pe = pk - 0.684 Pos Also, Por - Pol R Tol

>> Pei = 4.86 kg/m3 Then, kg/m3 * = 3.14 kg/on? - 3.14 x 3 x 10 tx dx SRT kg/s. - 0.634 x 4.96 on3 Now, mass flow rate, nok q* _ Merry ชา V* from = 0.323 kg/

Now, from isentropic table for M₂ = 0.6405, we get :- A. 2 - 1.14515 (AT), How, Axit Aexit At (A*), Az 12 At AL (A* ) Here, dexit = he Ae CA* 1) ㅗ w/ x 1.145 15 1.3376 froom (i) As - 1.3376 and A=A₂ 1.712 (AR) Froom isentoopic table for Ae 1.112, Me ~ 0.365

from normal shock table, for get: M2.1.7 Poz 0.8557. Pos Them, Por = 0.8557 x 500 kPa | Pog= 427.85 kPa flow in regions 2 is iseotoopic Thurs, Pora Poe Also, Poe مائے (a + 8-22 m² =>۲: = |P6= 390.37 kPa Ry > Design Condition Here, Poi = Poe Also, Ae colm -2 U

7 Here, the flow will be sentropic throughout from isentropic flow table, for Ae -2 > getan A Me ~ 2.2 Now, X N-1 Poe (1+ de me al Pb - 5ookpq = 10.693 Pn - 46.76 kPa } -Design back pressure