Question

A nozzle is designed to deliver a supersonic air flow, R = 287 J/Kg/K, of Mach M = 2.19

The reservoir has a pressure of p0 = 648kPa and T0= 300K.

The nozzle exit has an area of 0.233 m^2.

The nozzle flow exits into an environment that is kept at constant pressure p_b which matches the exit pressure of the nozzle. As long as there are no shock waves appearing in – or outside the nozzle, the complete flow process through the nozzle and into the environment can be considered isentropic.

1. Determine the exit pressure of the nozzle and the mass flux going through the nozzle.

At a later instant, with T0 and p0 being the same, the pressure of the environment into which the nozzle flow exits exists has increased by 539 kPa.

2. Will the mass flux be the same as in part a

• If not what is the new mass flux and what is the value of the Mach number at the throat
• If yes what pressure must exist at the nozzle exit if one wants to reduce the mass flux to 80% of the mass flux in part a? the nozzle cannot be modified. What is, in this case, the Mach number at the throat.

Answer 1

that, R Sal:- It is given 287 J/kg ZK. Exit mach number = Me= 2.19 lo = 648 kPa To 300 R Ae. 0.233 m? Po exttpMX Pe 2 . isentropic flow properties For Me = 2.19 from for Me = 2.19 2.19, we get Po po Pb Now, 10.5296 = pe Po Po le pe Ph 648 10.5296 x Pos Р. ХР. Now, fo le Pe= Pt = 61.54 kPa reservice density exit air density Po 6-487-to Po RT 287X 300 3 Р. = 7.52 Kg dim? from isentropic flow properties chart for Me=2.19

تهیه = 5.374 be Now, foloxle 7.52 = 5.374x de Pe le 1.4 kg/m3 Also from. same chart, for the M= 2.19 To 1.9592 Te Ә) Те 1S3.12K 300 1.9592 Te Mass flow rate. m Pe AeVe Here, Ve = mele. merrte m mefe Ae VERTE m = 1.4x0.233 X 2.19 *J1-4x287x153.12 m 177.19 kg / sec. flux going through the nozzle, Hence, man m 177.19 760.47 H 0.233 man flux. = 760.47 kg /m?s