Question

Consider the flow through a rocket engine nozzle. In the combustion chamber, the gas which results from the combustion of the rocket fuel and oxidizer is at a pressure and temperature of 15 atm and 2500K, respectively; the molecular weight and specific heat at constant pressure of the combustion gas are 12 kg/kmol and 4157 J/kg · K, respectively. Assume that the gas flow through the nozzle is an isentropic expansion of calorically perfect gas, with a temperature of 1350K and an area of 50 cm2 at the exit. Calculate the pressure, Mach number, velocity, and mass flowrate at the exit. Gas constants Ř 8314 R= = 10 J/(kg · K) = 692.8 J/(kg · K). Cv = Cp - R = 4157 – 6927.8 J/kg · K = 3464 J/(kg · K) y = Cp - 4157 a = 346A = 1.2

Answer 1

condition firsta soln Let us note the given given At entering in nozzle P = 15 atm Ti= 25ook At exit of hozzle Ta = 1350k A2 = 50 cm2 - molecular mass of gas (m) = 12 kg /kmol - Iscentropic expansion -) (p = 4157.J/kg K R = 692.8 3 1 kg k Cu = 3464 J I kg K r= 12 Unknows or target pressure at exit (P2) = ? velouity at exit (V2) = ? mach no at exit (m2) = ? mass flow rate at exit (ria) = ? now Let us first doow the nozzle y exit entering P2, Tz PIT Iscentropic expansion PL Now from adiabetic equation (Pw=const)

P2 = 0.371 atm y Pressure at exit of nozzle. ' from Ist law of thermodynamics for open system steady flow hit us ho + gzyst of the TV2 +głe & ou 2000 2 DOO hit uit = hat V₂2 2000 VICLLV2 V2=52000 Chi-ha) v2= √ 2000 CP (T.-T2) V2 = 12000 x 4.157 x(2500-1350) [N2 = 3092 mis velouity at exit of hozzle we know that mach number = velouity of air (12) (a) velocity of sound at that temper Velouity of Sound = VYRT2 - 11.2x 692.8x1350 [a= 1059.4 mis

mach number at exit of nozzle (m2) = V m2 = 3092 1059.4 2 = 2.918 - mach no of nozzle at exit now, mass flow rate at exit of nozzle = m2 = A2V2 m2 = A2 V2 = 50x10" X 3092 me = 15.46 m3/s ) mass flow rate at exit of nozzle
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