Component balance integral

There are only two possible values for concentration in a CSTR. The inlet stream has concentration and everywhere else has concentration The reaction rate will be the same throughout the vessel and is evaluated at the outlet concentration, SIa = A(ctout,bout, ) For the single reactions considered in this chapter, continues to be related to by the stoichiometric coefficient and Equation (1.13). With SS a known, the integral component balance, Equation (1.6), now gives useful information. For component A,... 

Volume changes also can be mechanically determined, as in the combustion cycle of a piston engine. If V=V(i) is an explicit function of time. Equations like (2.32) are then variable-separable and are relatively easy to integrate, either alone or simultaneously with other component balances. Note, however, that reaction rates can become dependent on pressure under extreme conditions. See Problem 5.4. Also, the results will not really apply to car engines since mixing of air and fuel is relatively slow, flame propagation is important, and the spatial distribution of the reaction must be considered. The cylinder head is not perfectly mixed. 

In the component balance equations, dY]/dt will therefore be zero, whereas dXj/dt may still be quite large. This can obviously cause considerable difficulties in the integration procedure, owing to equation stiffness. 

This equation can be integrated together with the component balances and the reaction kinetic expressions where the kinetics could, for example, be of the form... 

The component balance equations I and II are integrated with respect to distance to give the volumetric flow rate of each component, (y G)z and (yB G)z, at any position Z, along the length of the reactor. 

Solution of the required column height is achieved by integrating the two component balance equations and the heat balance equation, down the column from the known conditions Xi , yout and TLin, until the condition that either Y is greater than or X is greater than Xqui is achieved. In this solution approach, variations in the overall mass transfer capacity coefficient both with respect to temperature and to concentration, if known, can also be included in the model as required. The solution procedure is illustrated by the simulation example AMMON AB. 

The three features outlined in this section have profound implications for a plant s control strategy. Simple examples in this chapter will illustrate the effects of material recycle and component balancing. Chapter 5 contains more details of the effects created by energy integration on the entire plant. 

Component balances can often be quite subtle, but they are particularly important in processes with recycle streams because of their integrating effect. They depend upon the specific kinetics and reaction paths in the system. They often affect what variable can be used to set production rate or reaction rate in the reactor. The buildup of chemical components in recycle streams must be prevented by keeping track of chemical component inventories (reactants, products, and inerts) inside the system. 

The component balances can be integrated in the same way. The initial condition is that the volume in the tank is Vo of pure solvent and the concentration of the solid is zero. To find the analytical solutions to these equations, we specify V[t] and then we use DSolve to simultaneously solve for the concentrations, calling the set of two solutions "a." Two functions are named and then extracted from the solution set and assigned to these names. Finally, the two new functions are placed back into the original differential equations and tested for validity. 

Now we can use this technique to determine how the concentration of dye changes as a function of time in a well-stirred vessel. We need only write and integrate the component balance on the dye to have the answer because the volume of the tank is assumed not to change with time. We will use the following equation for the rate of change of the dye mass ... 

The applicability of the component balance equations with reaction terms is limited. It requires the knowledge of the reaction kinetics and then, the equations are rather part of a more complex mathematical model involving heat and mass transfer equations and the like. In balancing proper, the integral reaction rates W (n) in (4.2.11) can be known only approximately or rather, they are unknown. We will now show how they can be eliminated from the set of balance equations. 

Neglecting the change of accumulation in the nodes, the individual component balances of nodes n e N (units of the system, without the environment node Hq) are of the form "output minus input = integral production rate" for any component C,< see Section 4.2. Here, the integral production rate of C,( (in mass units) in node n equals, by (4.2.11)... 

The integral reaction rates (n) can be computed, if the component balance equations have been solved for the quantities riy. (n), provided the reactions are independent see Remark to Section 4.5. Let R(n) = 1 for simplicity. Omitting the indices r and n, Eq.(4.5.12) reads... 

With (positive) mass flowrates the component balances (4.2.1) read = 1, , K = y[m,+ ylni2 + Wr where Wr is the integral ( extensive ) reaction rate. We have... 

The system of equations (F.l) has been analyzed in Chapter 4 see the fine-printed paragraph that follows after (4.3.3). This set of reactions can be called maximal in the given system of components. Instead of transforming the component balance equations according to Section 4.3, we can also use directly the component balances (4.2.8) for the given node n (reactor) and = 1, , 7 (components C, listed above). Then the number of reactions R n) = 4 for the system (F.l) and the reaction extents (time - integrated reactions rates W ) will represent unknown parameters eliminated in the course of reconciliation. In this case, the parameters are observable cf. Remark to Section 4.5. If adding another admissible reaction, it will be linearly dependent on the system (F.l). 

There is rather little change in Xq during step 3 at constant R and F, thanks to the semibatch character of the process. If Xd is taken as constant, the still path can be calculated at given R, F, and Xp (see Fig. 8). At the end of the second step of the process (at time f2=Afi+Ar2), the actual still composition x,2 can be calculated from a simple component balance. The still path of step 3 can be determined by integrating eq 2 started from Xs=Xt2- The solution is a straight line if Xd is constant. The theoretical end of step 3 is the time when the still path reaches the boundary therefore, x,3 is estimated by the intersection of the still path with the boundary. Once x,2 and X(3 are known, the amount of used solvent, the amount of yielded distillate, and the final recovery can be calculated. 

The solution methodology involves a numerical integration using the method of Sundaram and Evans [1]. The overall material and component balances are given by... 

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