Mole balances steady-state

The model is described in Sec. 4.3.3. The steady-state balances are written in terms of moles. The Ideal Gas Law is used to calculate the volumetric flow rate from the molar flow at each point in the reactor. This gives also the possibility of considering the influence and temperature or pressure profiles along the tube. 

Assuming that the stoichiometric coefficient of the reactant is — 1 and that there is no change in the number of moles on reaction, the material balance on the volume element at steady state can be written as ... 

The problem is to calculate the steady-state concentration of dissolved phosphate in the five oceanic reservoirs, assuming that 95 percent of all the phosphate carried into each surface reservoir is consumed by plankton and carried downward in particulate form into the underlying deep reservoir (Figure 3-2). The remaining 5 percent of the incoming phosphate is carried out of the surface reservoir still in solution. Nearly all of the phosphorus carried into the deep sea in particles is restored to dissolved form by consumer organisms. A small fraction—equal to 1 percent of the original flux of dissolved phosphate into the surface reservoir—escapes dissolution and is removed from the ocean into seafloor sediments. This permanent removal of phosphorus is balanced by a flux of dissolved phosphate in river water, with a concentration of 10 3 mole P/m3. 

Attempts to define operationally the rate of reaction in terms of certain derivatives with respect to time (r) are generally unnecessarily restrictive, since they relate primarily to closed static systems, and some relate to reacting systems for which the stoichiometry must be explicitly known in the form of one chemical equation in each case. For example, a IUPAC Commission (Mils, 1988) recommends that a species-independent rate of reaction be defined by r = (l/v,V)(dn,/dO, where vt and nf are, respectively, the stoichiometric coefficient in the chemical equation corresponding to the reaction, and the number of moles of species i in volume V. However, for a flow system at steady-state, this definition is inappropriate, and a corresponding expression requires a particular application of the mass-balance equation (see Chapter 2). Similar points of view about rate have been expressed by Dixon (1970) and by Cassano (1980). 

To develop E(B) for two CSTRs in series, we use a slightly different, but equivalent, method from that used for a single CSTR in Section 13.4.1.1. Thus, consider a small amount (moles) of tracer M, nMo = F,dt, where Ft is the total steady-state molar flow rate, added to the first vessel at time 0. The initial concentration of M is cMo = nMo/(V/2). We develop a material balance for M around each tank to determine the time-dependent outlet concentration of M from the second vessel, cM2(l). 

Equation (9.27) defines the so-called axial dispersion coefficient Dax as a model parameter of mixing. Nd is the dispersion flow rate, c the concentration of the tracer mentioned earlier, and S the cross-sectional area of the column. The complete mole flow rate of the tracer consists of an axial convection flow and the axial dispersion flow. The balance of the tracer amount at a cross section of the extractor leads to second-order partial differential equations for both phase flows at steady state. For example, for continuous liquids ... 

This equation says that if the advance corresponds to 1 mole of oxygen, (2/4) moles of vacancies arrived at the reaction front (4 - (A, B) O - (A, B)304). As long as k attains a constant (= steady state) value, the balance of B cations yields (with respect to the local reaction A1 NoB O = (1 —A)-A, BnO+(A/4)-AB204 + A2+ions)... 

The basic equation for a tubular reactor is obtained by applying the general material balance, equation 1.12, with the plug flow assumptions. In steady state operation, which is usually the aim, the Rate of accumulation term (4) is zero. The material balance is taken with respect to a reactant A over a differential element of volume 8V, (Fig. 1.14). The fractional conversion of A in the mixture entering the element is aA and leaving it is (aA + SaA). If FA is the feed rate of A into the reactor (moles per unit time) the material balance over 8V, gives ... 

When a series of stirred-tanks is used as a chemical reactor, and the reactants are fed at a constant rate, eventually the system reaches a steady state such that the concentrations in the individual tanks, although different, do not vary with time. When the general material balance of equation 1.19 is applied, the accumulation term is therefore zero. Considering first of all the most general case in which the mass density of the mixture is not necessarily constant, the material balance on the reactant A is made on the basis of FA moles of A per unit time fed to the first tank. Then a material balance for the rth tank of volume V (Fig. 1.17) is, in the steady state ... 

Although the solutions are fed separately to the first tank, we may for the purpose of argument consider them to be mixed together just prior to entering the tank as shown in Fig. 1.18. If the ester is denoted by A and the caustic soda by B and x is the number of moles of ester (or caustic soda) which have reacted per litre of the combined solutions, a material balance on the ester over the first tank gives, for the steady state ... 

In the steady state the algebraic sum of the moles entering and leaving the element will be zero. By expanding terms evaluated at (z + Sz) and (r + Sr) in a Taylor series about the points z and r respectively and neglecting second order differences, the material balance equation becomes ... 

In Chapter 2 (Section 2.9.2) the steady-state design of a reactor-stripper process was studied. Now we investigate the dynamic controllability of this process. The dynamic model of the reactor is the same as Eqs. (3.9)—(3.11) except there is a second stream entering the reactor, the recycle stream D (kmol/s) from the column with composition. Vj) (mole fraction A). The reactor effluent is F (kmol/s) with composition z (mole fraction A). The reactor component and energy balances are ... 

Suppose that a 1000 kmol h 1 feed stream, consisting of 30.0% by mole M-pentane and the remainder w-hexane, is to be separated into 95.0% pure pentane and 95.0% pure hexane streams using a distillation column. Determine the flow rates of the output streams through the use of mass balances, assuming steady-state operation. We will assume three digits of significance for this example. 

We are going to formulate the mole balance equation in a CSTR operating at steady state, where nothing changes with time. This means that there is no accumulation inside the reactor and that the inlet, as well as outlet streams are at steady state. Following the general equation the mole balance for a component i in a CSTR is... 

The mass balance equation, expressed in moles, for the catalytic reactor is given by (see Figure 9.16) Input - Output + Production = Accumulation where, in a steady state, the accumulation is zero. 

In this section we apply die general energy balance [Equation (8-22)] to the CSTR and to the tubular reactor operated at steady state. We then present example problems showing how the mole and energy balances are combined to size reactors operating adiabadcally. 

If one were to examine Figure E8-4.2, one would observe that if a parameter were changed slightly, there might be more than one intersection of the energy and mole balance curves. When more than one intersection occurs, there is more than one set of conditions that satisfy both the energy balance and mole balance, and consequently there will be multiple steady states at which the reactor may operate. 

The startup of a fixed volume CSTR under iscrthennal conditions is rare, but it does occur occasionally. Here we want to determine the time necessary to reach steady-state operation. We begin with the general mole balance equation applied to Figure 4-13a ... 

In Example 9-4 we saw how a 500-gal CSTR used for the production of propylene glycol approached steady-state. For the flow rates and conditions (e.g., Tq = 75°F, = 60° ), the steady-state temperature was 138°Fand the corresponding conversion was 75.5%. Determine the steady-state temperature and conversion that would result if the entering temperature were to drop from 75°F to 70°F, assuming that all other conditions remain the same. First, sketch the steady state conversions calculated from the mole and energy balances as a function of temperature before and after the drop in entering temperature occurred. Next, plot the "conversion,"concentration of A, and the temperature in the reactor as a function of time after the entering temperature drops from 75°F to 70°F. 

The steady-state conversions can be calculated from the mole balance. 

The reactant feedstream enters at the top of the reactor and flows rapidly through the reactor relative to the flow of the catalyst through the reactor (Figure 10-29). If the feed rates of the catalyst and the reactants do not vary with time, the reactor operates at steady state that is, conditions at any point in the reactor do not change with time. The mole balance on reactant A over A IT is... 

Comments. At equilibrium no changes in properties occur with time either in the system or in its surroundings. However, under steady state conditions inputs and outputs of the system remain in balance so that the properties of the system are not altered, but changes do occur in the surroundings as a result of such processes. A more scientific characterization is provided in Chapter 6. Number of Independent Components. The least number of chemically distinct species whose mole numbers must be specified to prepare a particular phase. Comments. Due account must be taken of any prevailing chemical equilibria since in such cases the concentrations of the various participating species cannot all be independently altered. The number of independent components may then be determined from the number of distinct chemical compounds present in the system minus the number of chemical equations that specify their interactions. This matter is taken up in Section 2.1. 

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