What is condensate lift?


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In a furnace, 2 chemical reactions are happening – 1 mole of solid carbon reacts with 1 mole of oxygen gas to generate 1 mole of carbon dioxide gas; 1 mole of solid carbon reacts with 0.5 mole of oxygen gas to generate 1 mole of carbon monoxide gas. In a given process, 100 kmol of carbon is burned in a furnace. (a) Calculate the theoretical oxygen gas needed by assuming that all the carbon is burned completely to carbon dioxide gas. (b) Calculate the theoretical air needed by assuming that all the carbon is burned completely to carbon dioxide gas and there is only 21 % of oxygen gas. (c) Determine the amount of air required (in kmol) if 50 % excess oxygen gas must be satisfied for (a) and (b). (d) It has latter been found that 20 % of the carbon undergoes incomplete combustion resulting to carbon monoxide gas production. The rest of the carbon undergoes complete combustion. Calculate the total oxygen gas required stoichiometrically based on the actual process.

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CHEMICAL FLUID MECHANIC - EXAMPLE 3.2 : The terminal velocity of a falling object, v is given by v = sqrt [ 4g (R - r) D / (3Cr) ] where sqrt is the square root of, g = 9.81, D = 0.000208, R = 1800, r = 994.6, m = 0.000893. The Reynold number, L is given by L = rD (v) / m. The C for various conditions are : C = 24 / L for L < 0.1; C = 24 (1 + 0.14 L^0.7) / L for 0.1 <= L <= 1000; C = 0.44 for 1000 < L <= 350000; C = 0.19 - 80000 / L for 350000 < L. Find the value of v for the situation above by trial and error, ^ is power, <= is less than or equal to.

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MASS TRANSFER - EXAMPLE 4.3 : According to Adolf Eugen Fick (1829 - 1901) : rate of diffusion v increases with less wall thickness t, increased area A and decreased molecular weight of a fluid M. The diffusion constant D decreased with increasing M. (a) By assuming v, t, dP, A, M and D changes proportionally of each other, find the equation of v as a function of t, dP, A and D. (b) The ratio of self diffusion constant D, at T = 273 K and P = 0.1 MPa, for gases B and C are 1.604 : 0.155. If only 2 gases exist in such a system : hydrogen and nitrogen, find the type of gas for B and C with reference to their molecular weights M. (c) By using the equation of kinetic energy 0.5 MV = constant where V = square of v, find the ratio of V for B and V for C, or V(B) / V(C), as a function of M(B) and M(C), where M(B) is molecular weight of B and M(C) the molecular weight of C : Graham's Law of Diffusion.

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