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PHYSICS[100% CORRECT ANSWERS]

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The following data are given for a certain rocket unit: thrust, 8896 N; propellant consumption, 3.867 kg/sec; velocity of vehicle, 400 m/sec; energy content of propellant, 6.911 MJ/kg. Assume 100% c ... ombustion efficiency. Determine (a) the effective velocity; (b) the kinetic jet energy rate per unit flow of propellant; (c) the internal efficiency; (d) the propulsive efficiency; (e) the overall efficiency; (f) the specific impulse; (g) the specific propellant consumption. Answers: (a) 2300 m/sec; (b) 2.645 MJ/kg; (c) 38.3%; (d) 33.7%; (e) 13.3%; (f) 234.7 sec; (g) 0.00426 sec−1. F = 8896 N; m = 3.867 kg/s; u = 400 m/s; Qr = 6.911 MJ/kg; Ncomb = 1.0 (a) C = F/m = 8896/3.867 = 2300 m/s (b) Pjet = ½*m*c^2 = ½*3.867*2300^2 = over 10 million… or Kjet = ½*c^2 = ½*2300^2 = 2.645 MJ/kg…somehow (c) Nint = Pjet/Qr*Ncomb = 2.65/(6.911*1) = 0.383 = 38.3% (d) Np = (2*(u/c))/(1 + (u/c)^2) = (2*(400/2300))/(1+(400/2300)^2) = (2*0.174)/(1+0.03) = 0.348/1.03 = 0.337 = 33.7% (e) N = Nint * Np = 0.383*0.337 = 0.129 = 12.9% (f) c = Is*g so Is = c/g = 2300/9.8 = 234.7 (g) Specific Propellant Consumption refers to the reciprocal of the specific impulse SPC = 1/Is = 1/234.7 = 0.004 [Show More]

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