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Advanced cmbustion home work no1 edition 2

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Sabir Ahmed
Sabir Ahmed

This document summarizes the solutions to several problems regarding rocket performance calculations for a rocket with a thrust of 8896 N, propellant consumption of 3.867 kg/s, and flight velocity of 400 m/s. It includes calculations of effective exhaust velocity, kinetic energy of the jet, internal efficiency, propulsive efficiency, overall efficiency, specific impulse, and specific propellant consumption. It also includes calculations of specific power assuming a dry mass of 80 kg and duration of 3 minutes. Finally, it describes plotting the variation of thrust and specific impulse against altitude using atmospheric pressure data and engine parameters from tables.

Engineering
1 of 4
β€«Ψ§Ω„Ψ±Ψ­ΩŠΩ…β€¬ ‫الرحمن‬ ‫هللا‬ ‫بسم‬ Sudan University of Science &Technology College Of Post Graduates Studies PhD Program in Mechanical Engineering by Courses and Dissertation GE 721 - Advanced Combustion Homework No. (1) Prepared by student: Sabir Abushousha Ahmed Abushousha Supervisor: Dr. Mohammed Hassan Mohammed Abuuznien February 2015
Question 2 A rocket has a thrust of 8896 N and propellant consumption of 3.867 kg/sec. The vehicle flies at a velocity of 400 m/sec and the propellant specific energy content (heat of combustion) is 16.911 megajoule/kg (from Sutton). Find the following: a. Effective exhaust velocity b. Kinetic energy of the jet for 1 kg of fuel c. Internal efficiency ( d. Propulsive efficiency e. Overall efficiency f. Specific impulse ) g. Specific propellant consumption Given the following from the problem statement: F = 8896 N π‘šΜ‡ = 3.867 kg sec 𝑒 = 400 m sec QR =6.911β‹…106 J/ kg a. Effective exhaust velocity: 𝐢 = F mΜ‡ = 8896 3.867 = 2300.5 m sec b. Specific kinetic energy of the jet: 𝐾𝐸𝑗𝑒𝑑 = 0.5 C2 = 0.5 βˆ— (2300.5 )2 = 2.646130 Γ—106 J /kg c. Internal efficiency πœ‚π‘–π‘›π‘‘ = 𝐾𝐸 𝑗𝑒𝑑 QR πœ‚π‘–π‘›π‘‘ = 2.646130 Γ— 106 6.911βˆ—106 = 0.38288 = 38.3 d. Propulsive efficiency The speed ratio 𝑣 = u c = 400 2300.5 = 0.1739
πœ‚ 𝑝 = 2 . u 1 + u2 = 2 βˆ— 400 1 + (2300.5)2 = 0.3375 = 33.75% e. Overall efficiency: πœ‚ 𝑝 = F . u m .Μ‡ QR = 8896 βˆ— 400 3.867 βˆ— 6.911 βˆ— 106 = 0.1331 = 13.3% f. Specific impulse: 𝐼𝑠𝑝 = c g = 2300.5 9.81 = 234.6 s g. Specific propellant consumption: TSFCW = 1 𝐼𝑠𝑝 = 1 234.5848 = 0.0042629 1 S Question 4 For the rocket in Problem 2, calculate the specific power, assuming a propulsion system dry mass of 80 kg and a duration of 3 min. mΜ‡ = 80 3 βˆ— 60 = 0.4444 𝑃𝑗𝑒𝑑 = 1 2 𝐹𝑔0 𝐼𝑠 = 1 2 𝐹𝑔0 𝐢 𝑔0 = 1 2 𝐹 𝐢 𝑃𝑗𝑒𝑑 = 1 2 βˆ— 𝐹𝑣2 = 1 2 βˆ— 8896 βˆ— 2300.5 = 10232624 𝑀 ∝= 𝑃 𝑗𝑒𝑑 π‘š0 = 10232624 80 =127907.8 w Question 7 Plot the variation of the thrust and specific impulse against altitude, using the atmospheric pressure information given in Appendix 2, and the data for the Minuteman first-stage rocket thrust chamber in Table 11-3. Assume that P2 = 8.66 psia. Solution CONVERT THE UNITS 8.66 psi = 59708.598158924 pascal Assuming a ratio of specific heats to be 1.3 and gas constant to be 345.7 kJ/kg K,
FROM TABLE 11 EXIT VELOCITY 𝑣2 = √ 2π‘˜ π‘˜ βˆ’ 1 𝑅𝑇 [1 βˆ’ ( 𝑝2 𝑝1 ) (π‘˜βˆ’1)/π‘˜ ] Throat area (in 2 ) =164.2 =0.105935272 m2 β€’ Expansion Area Ratio: 10 = 𝐴2 𝐴 𝑑 𝐴2 = 0.105935272 m2 βˆ— 10 = 10.6π‘š2 Mass flow rate =A2*v2/V2 THURST 𝐹 = π‘šΜ‡ 𝑣2 + (𝑝2 βˆ’ 𝑝3)𝐴2 Specific Impulse 𝐼𝑆 = F mΜ‡ g0 All data are tabulated in excel sheet attached to the home work 0.1013 MPa atmospheric pressure which has the value * A A A A e throat exit ο€½ο€½ο₯

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Advanced cmbustion home work no1 edition 2

  • 1. β€«Ψ§Ω„Ψ±Ψ­ΩŠΩ…β€¬ ‫الرحمن‬ ‫هللا‬ ‫بسم‬ Sudan University of Science &Technology College Of Post Graduates Studies PhD Program in Mechanical Engineering by Courses and Dissertation GE 721 - Advanced Combustion Homework No. (1) Prepared by student: Sabir Abushousha Ahmed Abushousha Supervisor: Dr. Mohammed Hassan Mohammed Abuuznien February 2015
  • 2. Question 2 A rocket has a thrust of 8896 N and propellant consumption of 3.867 kg/sec. The vehicle flies at a velocity of 400 m/sec and the propellant specific energy content (heat of combustion) is 16.911 megajoule/kg (from Sutton). Find the following: a. Effective exhaust velocity b. Kinetic energy of the jet for 1 kg of fuel c. Internal efficiency ( d. Propulsive efficiency e. Overall efficiency f. Specific impulse ) g. Specific propellant consumption Given the following from the problem statement: F = 8896 N π‘šΜ‡ = 3.867 kg sec 𝑒 = 400 m sec QR =6.911β‹…106 J/ kg a. Effective exhaust velocity: 𝐢 = F mΜ‡ = 8896 3.867 = 2300.5 m sec b. Specific kinetic energy of the jet: 𝐾𝐸𝑗𝑒𝑑 = 0.5 C2 = 0.5 βˆ— (2300.5 )2 = 2.646130 Γ—106 J /kg c. Internal efficiency πœ‚π‘–π‘›π‘‘ = 𝐾𝐸 𝑗𝑒𝑑 QR πœ‚π‘–π‘›π‘‘ = 2.646130 Γ— 106 6.911βˆ—106 = 0.38288 = 38.3 d. Propulsive efficiency The speed ratio 𝑣 = u c = 400 2300.5 = 0.1739
  • 3. πœ‚ 𝑝 = 2 . u 1 + u2 = 2 βˆ— 400 1 + (2300.5)2 = 0.3375 = 33.75% e. Overall efficiency: πœ‚ 𝑝 = F . u m .Μ‡ QR = 8896 βˆ— 400 3.867 βˆ— 6.911 βˆ— 106 = 0.1331 = 13.3% f. Specific impulse: 𝐼𝑠𝑝 = c g = 2300.5 9.81 = 234.6 s g. Specific propellant consumption: TSFCW = 1 𝐼𝑠𝑝 = 1 234.5848 = 0.0042629 1 S Question 4 For the rocket in Problem 2, calculate the specific power, assuming a propulsion system dry mass of 80 kg and a duration of 3 min. mΜ‡ = 80 3 βˆ— 60 = 0.4444 𝑃𝑗𝑒𝑑 = 1 2 𝐹𝑔0 𝐼𝑠 = 1 2 𝐹𝑔0 𝐢 𝑔0 = 1 2 𝐹 𝐢 𝑃𝑗𝑒𝑑 = 1 2 βˆ— 𝐹𝑣2 = 1 2 βˆ— 8896 βˆ— 2300.5 = 10232624 𝑀 ∝= 𝑃 𝑗𝑒𝑑 π‘š0 = 10232624 80 =127907.8 w Question 7 Plot the variation of the thrust and specific impulse against altitude, using the atmospheric pressure information given in Appendix 2, and the data for the Minuteman first-stage rocket thrust chamber in Table 11-3. Assume that P2 = 8.66 psia. Solution CONVERT THE UNITS 8.66 psi = 59708.598158924 pascal Assuming a ratio of specific heats to be 1.3 and gas constant to be 345.7 kJ/kg K,
  • 4. FROM TABLE 11 EXIT VELOCITY 𝑣2 = √ 2π‘˜ π‘˜ βˆ’ 1 𝑅𝑇 [1 βˆ’ ( 𝑝2 𝑝1 ) (π‘˜βˆ’1)/π‘˜ ] Throat area (in 2 ) =164.2 =0.105935272 m2 β€’ Expansion Area Ratio: 10 = 𝐴2 𝐴 𝑑 𝐴2 = 0.105935272 m2 βˆ— 10 = 10.6π‘š2 Mass flow rate =A2*v2/V2 THURST 𝐹 = π‘šΜ‡ 𝑣2 + (𝑝2 βˆ’ 𝑝3)𝐴2 Specific Impulse 𝐼𝑆 = F mΜ‡ g0 All data are tabulated in excel sheet attached to the home work 0.1013 MPa atmospheric pressure which has the value * A A A A e throat exit ο€½ο€½ο₯