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% combustion efficiency.

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% combustion efficiency ...

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% combustion 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 A certain rocket has an effective exhaust velocity of 7000 ft/sec; it consumes 280 lbm/sec of propellant mass, each of which liberates 2400 Btu/lbm. The unit operates for 65 sec. The rated flight velocity equals 5000 ft/sec. Calculate (a) the specific impulse; (b) the total impulse; (c) the mass of propellants required; (d) the volume that the propellants occupy if their average specific gravity is 0.925. Answers: (a) 217.5 sec; (b) 3,960,000 lbf-sec; (c) 18,200 lbm; (d) 315 ft3. c = 7000 f/s or 2133.6 m/s; m = 280 lbm or 8.7 slug (a) Is = c/g = 2133.6/9.8 = 217.7 sec (b) It = F*t; c = F/m so F = c*m F = 7000*8.7 = 60900 It = 60900 * 65 = 3,958,000 lbf-sec (c) Burns 280 lbm/s for 65 sec – 280*65 = 18200 lbm (d) SG = Psubstance/Ph20; Psubstance = SG*Ph20 = 0.925*1 = 0.925 0.925 = Density of object/62.4 = 57.72lbs/ft^3 P = m/v; V = p/m v = 57.72/18200 https://www.coursehero.com/file/14633907/HW-1/ This study resource was shared via CourseHero.com A Russian rocket engine (RD-110) consists of four nonmoveable thrust chambers supplied by a single turbopump. The exhaust from the turbine of the turbopump then drives four vernier chamber nozzles (which can be rotated to provide some control of the flight path). Using the information below, determine the thrust, effective exhaust velocity, and mass flow rate of the four vernier thrusters Individual thrust chambers (vacuum): F = 73.14 kN, c = 3279 m/sec Overall engine with verniers (vacuum): F = 297.93 kN, c = 3197 m/sec Answers: 5.358 kN, 1351 m/sec, 3.965 kg/sec F = m*c; 73.14 = 3279m; m = 22 kg/s F = m*c; 297.93 = 3197m; m = 93 kg/s 73.14*4 = 292.56; 297.93-292.56 = 5.37 kN 22*4 = 88; 93-88 = 5 kg/s F = m*c; 5.37 = 5c; c = 1074 m/s A certain rocket engine has a specific impulse of 250 sec. What range of vehicle velocities (u, in units of ft/sec) would keep the propulsive efficiencies at or greater than 80%. Also, how could rocket–vehicle staging be used to maintain these high propulsive efficiencies for the range of vehicle velocities encountered during launch? Answers: 4021 to 16,085 ft/sec; design upper stages with increasing I s. Np = (2*(u/c))/(1+(u/c)^2) 0.8 = (2*(u/2450))/1+(u/2450)^2 ? = 2 ? ? 1 + ( ? ? ) 2 0.8 = 2 ? 2450 1 + ( ? 2450) 2 ? = 1225 ?? ? = 4900 m/s (4019 to 16076 ft/s) Certain experimental results indicate that the propellant gases of a liquid oxygen– gasoline reaction have a mean molecular mass of 23.2 kg/kg-mol and a specific heat https://www.coursehero.com/file/14633907/HW-1/ This study resource was shared via CourseHero.com ratio of 1.22. Compute the specific heat at constant pressure and at constant volume, assuming a perfect gas. Given Cp / Cv = 1.22 where Cp and Cv are specfic heats at constant pressure and constant volume respectively. and Cp - Cv = 8.314 J/ mole K 1.22 Cv - Cv = 8.314 Thus Cv = 8.314/0,.22 = 37.79 J/mole K ans Cp = 46.104 J/mole K The actual conditions for an optimum expansion nozzle operating at sea level are given below. Calculate v2, T2, and CF . The mass flow m˙ = 3.7 kg/sec; p1 = 2.1 MPa; T1 = 2585 K;M = 18.0 kg/kg-mol; and k = 1.30. R = 8314.3/18 = 461.9 M = (A*p1)/T1 * square root (((KM)/R)*(2/(k+1))^((k+1)/(k-1)) 3.7 = (A(2.1x10^6))/2585 * square root ((1.3/461.9)*(2/(1.3+1))^(2.3/0.3) A = 1.336 M = p/RT*M*square root (KRT)*A 3.7 = (2.1*10^6)/(461.9*2585) *M*square root (1.3*461.9*2585)*1.336 M = 0.00126 pv = MRT v2 = (MRT)/p = (3.7*461.9*2585)/(2.1*10^6) v2 = 2.1 T0/T2 = 1+((k-1)/2)*m^2 2585/T2 = 1+ (0.3/2) * 0.00126^2 T2 = 2584.99 CF = Nitrogen at 500 degrees C (k=1.38, molecular mass is 28.00) flows at a Mach number of 2.73. What are its actual and its acoustic velocity? a=√kRT R= 8314.3/28= 296.939 a=√(1.38*296.939*500)= 452.645 =acoustic velocity T_0=T[1+1/2 (k-1) M^2 ] =T_0=500[1+1/2 (1.38-1) 〖2.73〗^2 ]= 2124.0765 M=√(2/(k-1) (T_0/T-1) )= M=√(2/(1.38-1) (1208.0255/500-1) )=2.73 M=v/a=1536.6/562.856=2.73 v=a*M=562.856*2.73=1536.6 m/s =actual velocity