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In contrast to the liquid rocket, the solid propellant rocket is self-contained and has no entrance ducts. Using a control-volume analysis for the conditions shown, compute the rate of mass loss of the propellant, assuming ideal gas at outlet with R=297 m2/(s2K). Assume the diameter at the exit section to be: De = 25 cm. Note: P.=RT,297) (750) 90000 =0.404 kg/m³. %3D Propellant Exit section D. P.= 90 kPa V, = 1150 m/s T, - 750 K Combustion: 1500 K. 950 kPa Propellant m, = 44.70 kg/s m, =22.81 kg/s m = 32.84 kg/s m. = 11.82 kg/s

In contrast to the liquid rocket, the solid propellant rocket is self-contained and has no entrance ducts. Using a control-volume analysis for the conditions shown, compute the rate of mass loss of the propellant, assuming ideal gas at outlet with R=297 m2/(s2K). Assume the diameter at the exit section to be: De = 25 cm. Note: P.=RT,297) (750) 90000 =0.404 kg/m³. %3D Propellant Exit section D. P.= 90 kPa V, = 1150 m/s T, - 750 K Combustion: 1500 K. 950 kPa Propellant m, = 44.70 kg/s m, =22.81 kg/s m = 32.84 kg/s m. = 11.82 kg/s

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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In contrast to the liquid rocket, the solid propellant rocket is self-contained and has no entrance ducts. Using a control-volume analysis for the conditions shown, compute the rate of mass loss of the propellant, assuming ideal gas at outlet with R=297 m2/(s?K). Assume the diameter at the exit section to be: De = 25 cm. P 90000 RT, (297) (750) = 0.404 kg/m. Note: P. = Propellant Exit section D. Pe = 90 kPa V = 1150 m/s T = 750 K Combustion: 1500 K. 950 kPa Propellant m. = 44.70 kg/s m = 22.81 kg/s m. = 32.84 kg/s m. = 11.82 kg/s %3D
Transcribed Image Text:In contrast to the liquid rocket, the solid propellant rocket is self-contained and has no entrance ducts. Using a control-volume analysis for the conditions shown, compute the rate of mass loss of the propellant, assuming ideal gas at outlet with R=297 m2/(s?K). Assume the diameter at the exit section to be: De = 25 cm. P 90000 RT, (297) (750) = 0.404 kg/m. Note: P. = Propellant Exit section D. Pe = 90 kPa V = 1150 m/s T = 750 K Combustion: 1500 K. 950 kPa Propellant m. = 44.70 kg/s m = 22.81 kg/s m. = 32.84 kg/s m. = 11.82 kg/s %3D
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