Engine burns a stoichiometric mixture of fuel

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Reference no: EM13943194

Reaction engineering problem-

Question 7 - A rocket engine burns a stoichiometric mixture of fuel ( liquid hydrogen ) in oxidant ( liquid oxygen) . The combustion chamber is cylindrical , 75 cm long and 60 cm in diameter , and the combustion process produces 108 kg/s of exhaust gases. if combustion is complete, find the rate of reaction of hydrogen and of oxygen. Solution - We want to evaluate - (-Rh₂ = -rh2 ) =1/V*dNh2/dt And , - ro₂ =1/V* d No2/dt Let us evaluate terms.

The reactor volume and the volume in which reaction takes place are identical thus- V = ?/4 * ( 60/100)^2(75/100 ) = 0.79 *( 0.6)^2*(0.75)= 0.2121 m^₂ The reaction - H₂ + ½ O₂ ----(H₂0 Molecular weight H₂ = 2 g O₂ = 32 g ½ O₂ =32/2 =16 g H₂O = 18 g So, H₂O PRODUCED /s = 108 /18 = 6 kg-mole /s H₂ used = 6 k mols/s O₂ usec = 3 kmol/s The rate of the reaction -rh₂ = - 1/0.2121*6 = 2.829 *10^4 moles used/m3 of rocket -s -ro₂ = -1/0.2121*3 =1.415 *10^4 moles /m3 -s -ro₂ = -1/0.2121*3 =1.415 *10^4 moles /m³ -s A rocket engine burns a stoichiometric mixture of fuel ( liquid hydrogen ) in oxidant ( liquid oxygen) . The combustion chamber is cylindrical , 75 cm long and 60 cm in diameter , and the combustion process produces 108 kg/s of exhaust gases .

if combustion is complete, find the rate of reaction of hydrogen and of oxygen. Solution - We want to evaluate - (-Rh₂ = -rh₂ ) =1/V*dNh2/dt And , - ro₂ =1/V* d No2/dt Let us evaluate terms. The reactor volume and the volume in which reaction takes place are identical thus- V = ?/4 * ( 60/100)^2(75/100 ) = 0.79 *( 0.6)^2*(0.75)= 0.2121 m² The reaction - H₂ + ½ O₂ ----(H₂O Molecular weight H₂ = 2 g O₂ = 32 g ½ O₂ =32/2 =16 g H₂O = 18 g So, H₂O PRODUCED /s = 108 /18 = 6 kg-mole /s H₂...

Reference no: EM13943194

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