Question 16. An ideal rocket has the following characteristics: Chamber pressure Nozzle exit pressure Specific heat ratio : Average molecular mass Chamber temperature 27.2 atm 3 psia 1.20 21.0 lbm/lb-mol 4200 °F Determine the critical pressure ratio, the gas velocity at the throat, the expansion area ratio, and the theoretical nozzle exit velocity. Answers: 0.5645; 16. An ideal rocket has the following characteristics: Chamber pressure Nozzle exit pressure Specific heat ratio : Average molecular mass Chamber temperature 27.2 atm 3 psia 1.20 21.0 lbm/lb-mol 4200 °F Determine the critical pressure ratio, the gas velocity at the throat, the expansion area ratio, and the theoretical nozzle exit velocity. Answers: 0.5645; 3470 ft/sec; 14.8; and 8570 ft/sec.

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Transcribed Image Text: 16. An ideal rocket has the following characteristics: Chamber pressure Nozzle exit pressure Specific heat ratio : Average molecular mass Chamber temperature 27.2 atm 3 psia 1.20 21.0 lbm/lb-mol 4200 °F Determine the critical pressure ratio, the gas velocity at the throat, the expansion area ratio, and the theoretical nozzle exit velocity. Answers: 0.5645; 3470 ft/sec; 14.8; and 8570 ft/sec.

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An:-(1)\begin{array}{l}P_{0,}=27 \cdot 2 a \mathrm{tss} \\T_{01}=4200^{\circ} \mathrm{F}\end{array}(1)\begin{array}{l}\text { Thoot } \\M=1\end{array} \quad \begin{array}{l}P_{2}=3 \text { psia } \\P_{2}=0.204 \text { atm } \\\gamma=1.2 \text { (2) } M=21 \mathrm{lbm} / 1 \mathrm{lmot}\end{array}Coiticel pressure Retio:- Retio of pressure at the section where Sonic velocity is attained to the inlet pressure of a nozzle is celled the criticel presswre Reto.So, for M=1 u r=1.2 from Isentropic table\frac{P}{P_{0}}=0.5645gas velocity at throat (M=1)\begin{array}{c}M=V / a \\V=a=\sqrt{\gamma R T} \\T_{01}=4200^{\circ} \mathrm{F}=2588.106 \mathrm{~K} \\R=\frac{8314.3 \mathrm{~J} / \mathrm{Kmol}}{21 \mathrm{~kg} / \mathrm{Kmol}} \\R=396 \mathrm{~J} / \mathrm{kgK}\end{array}for M=1 from Isentropic teble\begin{array}{l}\frac{T_{t}}{T_{01}}=0.90909 \\T_{t}=2353.367 \mathrm{~K} \\V_{t}=\sqrt{1.2 \times 396 \times 2353.367} \\V_{t}=1057.5065 \mathrm{~m} / \mathrm{s} \\V_{t}=3469.5095 \mathrm{Ht} / \mathrm{s} \\\end{array}Expansion Ratio:\frac{P_{2}}{P_{01}}=\frac{0.204}{27.2}=0.0075for \frac{P_{2}}{P_{01}} from Isentropic table\frac{A_{2}}{A_{t}}=14.8and and M_{2}=3.55for M_{2}=3.55 from Isentropic toble\begin{aligned}\frac{T_{2}}{T_{02}} & =0.4424 \\T_{2} & =1145.24 \mathrm{~K} \\V_{2} & =M_{2} \sqrt{\gamma R T_{2}} \\& =3.55 \sqrt{1.2 \times 396 \times 1145.24} \\V_{2} & =2618.88 \mathrm{~m} / \mathrm{s} \\V_{2} & =8592.126 \mathrm{ft} / \mathrm{s}\end{aligned} ...