(Solved) : Use Runge Kutta Method Matlab Solve Following Problem 120 Tubular Reactor Homogeneous Reac Q42700300 . . .

Use Runge kutta method in matlab to solve the following problem

120. A tubular reactor for a homogeneous reaction has the following dimensions: L = 2 m, R = 0.1 m. The inlet reactant concen

Use Runge kutta method in matlab to solve the following problem 120. A tubular reactor for a homogeneous reaction has the following dimensions: L = 2 m, R = 0.1 m. The inlet reactant concentration is cho = 0.03 kmol/m3, and the inlet temperature is T. = 700 K. Other data are as follows: -AH = 104 kJ/kmol, Cp = 1 kJ/(kg K), Ea = 100 kJ/kmol, P = 1.2 kg/m>, u. = 3 m/s and ko = 5 sec-1. The material and energy balances for this reactor result in the following two differential equations after normalization. dy/dz = -Da y exp [8 (1 – 1/0)] de/dz = B Da y exp [8 (1 – 1/0)] – Hw (0 – 0) z = x/L y = c/c. = T/T 0 < z y (z=0) (z=0) < = = 1 1 1 Ow= Tw/T. Da= Lko/uo 8 = Ea/RgT. B = co(-AH) /popt Hv= 2h/R (L/PCpu) = 1. = 3.333 = 0.0172 = 0.357 = 0. a) solve this system of equations if the convective heat transfer coefficient, h, is 0. b) Now assume the overall heat transfer coefficient between the wall and the reactor contents is h= 70 J/(m2 s K) and the wall temperature of the reactor Tw = 700 K. This makes Hy= 2h/R (L/pcpu) = 0.778. Solve this system of equations to determine the outlet reactant concentration and to find the location of the hottest temperature on the reactor wall. Show transcribed image text Use Runge kutta method in matlab to solve the following problem
120. A tubular reactor for a homogeneous reaction has the following dimensions: L = 2 m, R = 0.1 m. The inlet reactant concentration is cho = 0.03 kmol/m3, and the inlet temperature is T. = 700 K. Other data are as follows: -AH = 104 kJ/kmol, Cp = 1 kJ/(kg K), Ea = 100 kJ/kmol, P = 1.2 kg/m>, u. = 3 m/s and ko = 5 sec-1. The material and energy balances for this reactor result in the following two differential equations after normalization. dy/dz = -Da y exp [8 (1 – 1/0)] de/dz = B Da y exp [8 (1 – 1/0)] – Hw (0 – 0) z = x/L y = c/c. = T/T 0

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