Question

Engineers are conducting a sea level test of the space shuttle main engine. They are suspicious that the nozzle is not performing up to specifications due to the total pressure losses. To aid in determining its operational characteristics, the engineers have installed a series of wall static pressure taps at the nozzle exit station. Test parameters include: 467 kg/sec of gas leaving the combustion chamber operating at a total pressure and total temperature of 20.7 MPa and 4080 K respectively, and a nozzle exit area equal to 77 times that of the throat area. Assuming the nozzle is running full (without any flow separation) and adiabatically, with a molecular weight of 13.0 and a gamma of 1.25, determine

1. The nozzle throat area (assuming isentropic flow to the throat)

2. The Mach number at the nozzle exit where the average wall static pressure is measured as 0.016MPa

3. The total pressure recovery to the nozzle exit.

Answer 1

I have solved part 1 and 2 for you, i didn't solve the 3 part because the only way to calculate pressure recovery known to me is through the gas table but the gas table of gamma 1.25 is not available to me. if you provide the gas table of gamma 1.25 then i can solve part 3 for you.

throat mlor exit Now, the given parameters are. in = 467 kg/S. P, = 20.7 MPa. T, = 4000K A = 774 å = 0 M =13 8 = 1.25 [ Aez area exit , A = throat area Cadiabatic process) The experiment is conducting at sea level, so sea level pressuse o 101325 10.10 1325 mpa will be the back pressure Now, P = 0.101325 mPa. Pressure at throat

e [ 2 1.25 +1 1.25 1.25 + 20.7mpa (1.25+1) P = 0.93665 * 20.7 mPa. Pt = 19.3888 MPa с. е. 7 г. - поzz, чJ bя сhооtiq , Now, for choaked condition, maximum velocity at throat = sonoic velocity I v = VRTE Now, throat Area, A m v A = m for choaked condition, throat temp. T = ( 2 Ja To • 62 Ju yoso V. 25+1 T = 3626. 667K

putting values in 0 Ą: 467 kg x y 8.314x3626.667 I knop T 30 J 12 molk kg 19.3888x10° Pax $1.25 = 1.03752 K10°m°f the rJJ 4 = 10.35 cod = 10.375 cm han Im? (checking for unit) of Area sound F ERT .4Звук бяв at exit, Velocity of sound - ų = J KRT Now flow through nozzle is issentropic Sor 1.25-1 1, : 4000 [ 0.01675 20.7

T = 973.39 K bon So, velocity of sound at T = 973.39 k ų = 51.4+ 287x973.39 - 625.386 m/s Now, Area at exit Ag = 77*A= 0.07988 m² Now, Aspire ou 0.07988 - 467 kg x 85364 x 973.39 UK 0.076 x 10 Pa rum 0.07988 = 290717.0002 ING 0.016 X106 Nas y = 18.1698 m/s 0.07988 o = 227.46 m/s Mach No. = velocity of velocity of fluid sound in fluid

so, Mach No.- 227.46 625 386 Mach No, M = 0.3637 & Subsonic flow. The stagnation pressure will be [ 2 1.25 5.20.1[14 (1.25-1)=(0-363) = 22.468 MPa Now, we have a р" > Р, > Pore 22.468 > 0.101385 > 0.016. The thuid which aclieved subsonic in diverging section velocity at throat cowlnus accelerating the supersone velocity in diverging

section comes to a sudden stop. A normal shock develops at a sechon between the throat & exit plane which tad cause a sudden drop in velouty to subsonic level & sut sudden increase in pressure Flow through shock become highly irreversible. 3. The total pressure recovery is the ratio of total average pressure at the exit of the inlet to the free stream total pressure & total pressure - 22.468 MPa recovery