Saturation material balances

Equation 1 is normally integrated by graphical or numerical means utilizing the overall material balance and the saturated air enthalpy curve. 

Note that ai will gradually increase during the course of the reaction and will reach its saturation value, agjKu, when B is depleted. Dropping the accumulation term for ai i) represents a form of the pseudo-steady hypothesis. Since component B is not transferred between phases, its material balance has the usual form for a batch reactor ... 

Pure component in gas phase and saturated liquid phase First-order kinetics in A In this case, the gas phase is a pure component A (CAG is constant) and the liquid phase is considered to be saturated with A (C is constant). Furthermore, the intrinsic rate is considered to be of fust order with respect to A -rm = /cmCAS, per unit mass of catalyst. Under these conditions, the material balances for the gas component A in the gas and liquid phases (eqs. (3.365) and (3.367)) are not needed CAL is constant and equal to CAl q = CAG/HA. The same analysis is valid for reactions of fu st order for both components, if Cbl -> > Cal and thus CBS = const (pseudo-first order) (Smith, 1981). Then (eq. (3.369))... 

Figure 9.15. Relations in a packed continuous How air-water contactor, (a) Sketch of the tower with differential zone over which the enthalpy and material balances are made, (b) Showing equilibrium and operating lines from which the integrand 1 fths — h) can be found as a function of liquid temperature T. (c) Showing interfacial conditions as determined by the coefficient ratio km/kh when this value is large, interfacial and saturation temperatures are identical.

A mixture containing 12 mol % water is to be separated by distillation into products with 99.5 and 0.5 mol % butanol. The accompanying flowsketch of a suitable process utilizes two columns with condensing-subcooling to 40°C. The 53% saturated solution is refluxed to the first column, and the 98% is fed to the second column. The overhead of the second column contains a small amount of butanol that is recycled to the condenser for recovery. The recycle material balance is shown with the sketch. 

This kind of software information is now also available to the wider marketplace through, for example, electronic books that provide an infinitely customizable, all-in-one package, combining technical manual calculations for all common water treatment areas, displays for system material balance, saturation index tables, report generator, etc. An example of this type of software is the Water Treatment Handbook On-Disk from LXF Inc. 

SimSim fills the gap between material balance techniques and complex reservoir simulation yet keeping the simplicity and speed of the material balance but providing reservoir simulation like results, i.e. pressure, saturation, hydrocarbons in place and fluid flux distribution within the reservoir. 

The porosity of a solid is most easily determined by the imbibition method. A sample of solid is dried, weighed while dry, then saturated with a wetting liquid (such as water or heptane), under vacuum. The accessible pore volume is calculated by a material balance ... 

Samples were considered for further work only when the material balance of saturates, aromatics, and resins was within 1% of total recovery. 

Material Balance Equation for a Reservoir Producing Above the Saturation Pressure. When a reservoir exists at a pressure above its bubble point oil can be produced by expansion of the reservoir fluid as the pressure is reduced to the saturation pressure. This process is shown for a constant volume reservoir in Figure 95. The pressure declines from the original reservoir pressure to a pressure which is equal to or greater than the saturation pressure. 

Prediction of Oil Reservoir Behavior. By combining the material balance equation, the reservoir saturation equation, llie producing gas-oil ratio equation, and the equation relating the cumulative gasoil ratio to the producing gas-oil ratio, it is possible to compute the... 

Let us first do the degree-of-freedom analysis. Three unknown variables appear on the chart— rtt, /i2 2nd y. Since only two species are involved in the process, we can only write two independent material balances, leaving us one equation short. If we fail to observe that the gas at the condenser outlet is saturated with water, solution of the problem would be impossible however, the saturation condition supplies the needed third equation, Raoult s law. 

There are three unknowns on the chart m, x, m2). We will assume that the solution leaving the crystallizer is saturated at 40 C Accordingly, the value of jc may be determined from the known solubility of KNO3 at that temperature, and the remaining two variables may be determined from material balances. From Figure 6.5-1, the solubility at 40 C is roughly 63 kg KNO3/IOO kg H O. The calculations follow. 

The solution of material balance problems involving partial saturation, condensation, and vaporization will now be illustrated. Remember the drying problems in Chap. 2 They included water and some bone-dry material, as shown at the top of Fig. 3.19. To complete the diagram, we add the air that is used to remove the water from the material being dried. 

You can analyze material balance problems involving water vapor in air in exactly the same feshion as you analyzed the material balance problems for the drying of leather (or paper, etc.), depending on the information provided and sought. (Humidity and saturation problems that include the use of energy balances and humidity charts are discussed in Chap. 4.)... 

Before discussing fractional flow analysis, we first need to derive the moving velocity of a saturation discontinuity or shock. Figure 2.13 shows a saturation shock from 8 2 to S i. 8 2 moves from Xj to X2 during the time interval At = t2 - h. The total injection rate, qt, is constant, but the water cut changes from f i to f 2, which corresponds to 8 i and 8 2, respectively. Therefore, during the time interval. At, the total incremental water injected into the block from Xi to X2 is (q,)(At)(f 2-f i). Meanwhile, this incremental water injected results in the increase in saturation from 8 1 to 8 2. The material balance of water gives... 

The material balance already has been made on the concentrations. The calculated saturations (82) using Eqs. 7.53 and 7.54 should be close to each other. 

Side Draws The 7-1 inc of a side draw passes through the intersection of the operating lines above and below the draw tray. A liquid side draw from stage j is a saturated liquid with composition Xj. The liquid draw composition is the same as the draw tray liquid composition, and therefore the liquid draw 7-line is a vertical line passing through X. A material balance on a tray with liquid draw 5 T is written as... 

In the irreversible case, the concentration wave entering the particle will saturate each spherical layer before it penetrates further. That is, ct cannot exceed zero at a particular radius r = r, until q, at that r has become equal to the saturation value qm. With c/C0 = q/qm = 1 — (ri/rp)3 by material balance over one particle, Eq. (90) leads to ... 

For both polymerization and rubber processing it is important to know the type and amount of unsaturation. In this connection IR spectroscopy is perhaps the most useful technique, although it requires a preliminary evaluation of band absorptivities by means of suitable standards, analyzed by an independent technique. A number of methods are available to analyze these standards NMR spectroscopy, the iodine-monochloride method, gas-chromatographic analysis of pyrolysis products, materials balance determination, etc.12 Provided the termo-nomer content of the standard is not too low, the most useful technique is NMR, based on the ratio between the signal from all saturated protons (generally with chemical shifts of 2S from TMS) and the signal from olefinic protons, observed at lower fields (45-5.5 5). 

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