Essay on Modelling in Chemical Engineering Sample

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Two phases fluidize bed reactor mass balance:
Fluidized bed reactor is widely used for carrying out multiphase reaction. One example of application of fluidized bed reactor is the production of polyethylene and polypropylene [1]. The following is a simplified schematic of the bed reactor used in polymerization of polyethylene [1].

The two phases in the fluidized bed reactor are the bubble phase and the dense phase. The bubble phase contains the essential reactant for the polymerization-ethylene (or any olefin). The monomer then is transferred to the emulsion (dense) phase by mean of mass transfer. The carrier gas hydrogen is not necessary an inert gas. The polymerization only takes place in the emulsion phase.
A simplified mass balance for generic isothermal fluidized bed reactor is shown by JOHN ROBERT LIGON [2]. A simple block diagram for the system is shown below.

Subscript E means dense phase, subscript B means bubble phase. In order to simplify the problem, the following assumptions have been made: 1. The bubble phase is always at a quasi-steady state. 2. Reaction is first order and only occurs in dense phase. 3. Volume of both phase are constant. 4. Constant flow rate though both phases. 5. Although the overall mass transfer coefficient is related to bubble size and number, thus related to volumetric flow rate. However, here for simplicity, it is assumed to be constant. 6. The system is isothermal thus the reaction coefficient r is constant.
Mass balance for the bubble phase:
Mass in-Mass out=Mass accumulation dMbdt=Fb*Cbo-Cb+k(Ce-Cb) Since V is constant,
VbdCbdt=Fb*Cbo-Cb+k(Ce-Cb)
dCbdt=FbVb*Cbo-Cb+kVb(Ce-Cb)
Let k/Vb=K, then we has dCbdt=FbVb*Cbo-Cb+K(Ce-Cb) Mass balance for the dense phase: dMedt=Fe*Ceo-Ce+kCb-Ce-R*Ve*Ce Since V is constant
VedCedt=Fe*Ceo-Ce+K*VbCb-Ce-R*Ve*Ce
Divide both sides by Ve: dCedt=FeVe*Ceo-Ce+K*VbVeCb-Ce-R*Ce Up to this point both equations can be solved simultaneous to determine the concentration in both dense phase
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