Straight equilibrium line

Equation 28 and its liquid-phase equivalent are very general and valid in all situations. Similarly, the overall mass transfer coefficients may be made independent of the effect of bulk fiux through the films and thus nearly concentration independent for straight equilibrium lines ... 

The concept of a mass-transfer unit was developed many years ago to represent more rigorously what happens in a differential contactor rather than a stagewise contactor. For a straight operating line and a straight equilibrium line with an intercept of zero, the equation for calculating the number of mass-transfer units based on the overall raffinate phase N r is identical to the Kremser equation except for the denominator when the extraction factor is not equal to 1.0 [Eq. (15-23)]. 

The relationship of the overall gas-phase mass transfer coefficient KQG to the individual film coefficients may be found from equations 4 and 5, assuming a straight equilibrium line ... 

When temperature and pressure are constant, the equilibrium values corresponding to a constant are expressed graphically by a straight equilibrium line passing through the origin with a slope of or H. 

With a straight operating line and a straight equilibrium line (Henry s Law for example), where the entering absorption liquid is pure, i.e., xao = 10, Nqg as shown in Eq. (48), it now becomes... 

Hog and Nog aie called the overall gas-phase height of a transfer unit and the number of overall gas-phase transfer units, respectively. In the case of a straight equilibrium line, f 0G °ften nearly concentration-independent as explained earlier. In such cases, use of equation 47 is especially convenient because NQG, as opposed to NG, can be evaluated without solving for the interfacial concentrations. In all other cases, HQG must be retained under the integral and its value calculated from / h and M at different points of the equilibrium line as... 

Although it was derived for a straight operating line and straight equilibrium lines, Eqs. will be... 

Solutions for the foregoing involve graphical or numerical integration of the expressions for /itotai- Such calculations are obviously complicated. There are, however, some simplified cases that can be used if we have dilute gases with straight equilibrium lines. 

These definitions are somewhat arbitrary since, in llie case of a truly cocurrent operation such as that of Fig. 5.7, it would be impossible to obtain a leaving concentration in phase E higher than Y or one in phase R lower than Xg. They are nevertheless useful, as will be developed in later chapters. The two Murphree efficiencies are not normally equal for a given stage, and they can be simply related only when the equilibrium relation is a straight line. Thus, for a straight equilibrium line of slope w = (KJ — Y /iX — X it can be shown that... 

Simplified Design Procedure for Linear Equilibrium and Operating Lines. A straight operating line occurs when the concentrations are low such that and remain essentially constant. (The material balance is obtained from equation 35.) In cases where the... 

For dilute solutions in which both the operating and the equilibrium lines are straight and in which heat effects can be neglected, the integral term in Eq. (14-27) is... 

Equations (14-168) and (14-170) have been developed for binaiw mixture separations and hold for cases where the operating hne and equilibrium line are straight. Thus, when there is curvature, the equations should be used for sections of the column where hnearity can be assumed. When the eqiiihbriiim line and operating line have the same slope, HETP = Hog and Nog = (theoretical stages). 

The end points of the operating line on an XY plot (Fig. 15-13) are X., Y, andXy, Y., and the number of theoretical stages can be stepped off graphically. The equilibrium curve is taken from the Hand type of correlation shown earlier (Fig. 15-9). When the equilibrium line is straight, its intercept is zero, and the operating line is straight, the number of theoretical stages can be calculated with one of the Kremser equations [Eqs. (l5-14) and (15-15)]. When the intercept of the eqnihbrinm line is not zero, the value of YJK, should be used... 

Note that if the problem of accurate graphical representation occurs in the rectification end of the diagram, the corresponding relation to use to calculate the balance of the trays, assuming straight line operating and equilibrium lines in the region is [59] ... 

Figure 9-69. Mass transfer diagrams. The number of transfer units can be determined by the difference in concentration or vapor pressure, particularly over ranges where the equilibrium line is essentially straight. Used by permission of Czermann, J. J., Gyokheqyi, S. L, and Hay, J. J., Petroleum Refiner, V. 37, No. 4 (1958) p. 165 all rights reserved.

If the operating and equilibrium lines are straight, and they usually can be taken as such when the concentrations are small, the number of stages required can be calculated using the equations given by Robinson and Gilliland (1950). 

For the idealised situation where the operating and equilibrium lines are straight, the overall column efficiency and the Murphree plate efficiency are related by an equation derived by Lewis (1936) ... 

Partial pressure in the exit gas at 95 per cent recovery = 60.8 x 0.05 = 3.04 mm Hg Over this range of partial pressure the equilibrium line is essentially straight so Figure 11.40 can be used to estimate the number of stages needed. 

The equilibrium data from Table 10.2 are plotted in Figure 10.6a. It can be seen that this does not form a straight-line relationship overall. Figure 10.6a shows the operating line of maximum slope that touches the equilibrium line at xR = 0.1576 kg AA/kg Water. The slope of this line is the ratio of feed to extraction flowrates. If the liquids are immiscible, then the flowrate of solute-free feed (F) is equal to the flowrate of the... 

The equilibrium data given are represented by a straight line of slope m = 1.20. As shown in Problem 12.12, the equation for Nog may be integrated directly when the equilibrium line is given by ye = mx to give ... 

The operating line may be fixed by trial and error as it passes through the point (0.001, 0), and 9 theoretical plates are required for the separation. Thus it is a matter of selecting the operating line which, with 9 steps, will give X2 = 0.001 when Xt = 0.06. This is tedious but possible, and the problem may be better solved analytically since the equilibrium line is straight. 

Show that regardless of the orientation of a straight dislocation line and its Burgers vector, there will exist a stress system that will convert the dislocation line into a helix whose axis is along the position of the original dislocation when the point-defect concentration is at the equilibrium value characteristic of the stress-free crystal. Use the simple line-tension approximation leading to Eq. 11.12. 

The simplest possible case occurs when (1) both the operating and equilibrium lines are straight (i.e., the solutions are dilute) (2) Henry s law is valid (y"/x = yjx, = m) and (3) absorption heat effects are negligible. Under these conditions, the integral term in Eq. (14-21) may be computed by Colburn s equation [Tran.s. Am. Inst. Chem. Eng., 35,211(1939)] ... 

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