Lever arm rule

An example of a binary eutectic system AB is shown in Figure 15.3a where the eutectic is the mixture of components that has the lowest crystallisation temperature in the system. When a melt at X is cooled along XZ, crystals, theoretically of pure B, will start to be deposited at point Y. On further cooling, more crystals of pure component B will be deposited until, at the eutectic point E, the system solidifies completely. At Z, the crystals C are of pure B and the liquid L is a mixture of A and B where the mass proportion of solid phase (crystal) to liquid phase (residual melt) is given by ratio of the lengths LZ to CZ a relationship known as the lever arm rule. Mixtures represented by points above AE perform in a similar way, although here the crystals are of pure A. A liquid of the eutectic composition, cooled to the eutectic temperature, crystallises with unchanged composition and continues to deposit crystals until the whole system solidifies. Whilst a eutectic has a fixed composition, it is not a chemical compound, but is simply a physical mixture of the individual components, as may often be visible under a low-power microscope. 

Step 2 Locate point M on a straight line connecting L and V using the inverse lever-arm rule ... 

Thus, if the feed and solvent compositions and flow rates are known, the mixture point M can be determined (Figure 11.4). The solvent and feed points Lq and 2jv+i are first plotted at their known compositions, and a straight line is drawn through them. The mixture point, M, is located on the basis of the lever arm rule ... 

The difference point can be located on the basis of either side of this equation by passing a line through either pair and extending it to A to satisfy the lever arm rule ... 

Figure 3.20 Control volume used to explain lever-arm rule.

When applied to extraction problems, the two feed streams V and O are equivalent to the incoming feed and solvent streams. The stream F would represent a two-phase mixture, which would separate into the raffinate and extract phases. The component A is usually the solute and the component B is usually the diluent, although the lever-arm rule will work no matter how the axes of the diagram are arranged. When solving extraction problems graphically, it is really useful to remember equations ... 

The lever-arm rule is a graphical alternative to solving a mass balance. 

A special case of the lever-arm rule, which renders it applicable to extraction analysis, is an equilibrium-limited stage for a three-phase system (Figure 3.22). Everything stated for the lever-arm rule still applies here since the mass balances around the control volume (equilibrium stage) are still the same. The compositions of the three streams will still lie on a straight line, and stream ratios can still be calculated as before. 

We know that M must lie on a line connecting 0 and Vn, and Ojv+i and Vb (these lines are tie-lines). Using the lever-arm rule, we can locate M. In other words, we can write ... 

Figure 5.13 To plot the delta point 1 Locate the points corresponding to the compositions of Vq (solvent), Ojv+iAfVb (feed), and Oi (exit raffinate). 2 Find point M from a component balance or the lever-arm rule. 3 Make line A/ Vq. Extend line OiM to the saturation curve (this is point VV). 4 Extend Ojv+i Vf/ and Oi Vo the intersection is at A.

Note that in Figure 5.13, this ratio decreases as the point M moves toward the point Vo (lever-arm rule). This causes a decrease in the solute concentration of Vjv. When the line AVoDjv+i changes such that it falls exactly on a tie-line (equilibrium and operating lines... 

Now all of the streams are located, and their compositions can be read from the graph. Using the lever-arm rule to find the amounts of the streams ... 

Show that the lever-arm rule is valid for one inlet and two outlet streams for a contact stage. 

The quantities of extract and raffinate can be computed from the lever-arm rule, or by the materia] balance for C ... 

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