However in this case the line mistakenly crosses over

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Unformatted text preview: efore, instead, the operating line is drawn tangent to the equilibrium curve from the point x = xD = 0.675 as shown on the diagram by a solid line. The slope of the operating line = L/V = 0.467. From Eq. (7-27), Rmin = (L/D)min = 0.467/(1-0.467) = 0.876. The operating reflux ratio = R = 1.5Rmin = 1.5(0.876) = 1.314. From Eq. (7-7), L/V = R/(1+R) = 1.314/(1+1.314) = 0.568. On a set of three McCabe-Thiele diagrams on the next page, the rectifying section operating line has this slope and passes through the point, y=0.675, x=0.675. the stripping section operating line passes through the point, y=0.0023, x=0.0023 and intersects the vertical q-line at the point where the rectifying section operating line intersects the q-line. As seen, the equilibrium stages are stepped off starting at the top, with a switch from the rectifying section to the stripping section to minimize the number of stages and, thus, locating the optimal feed stage. To achieve accuracy, one diagram covers the high-concentration region, one the mid...
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