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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. (727), Rmin = (L/D)min = 0.467/(10.467) = 0.876. The operating reflux ratio = R = 1.5Rmin = 1.5(0.876) = 1.314. From Eq. (77), L/V = R/(1+R) = 1.314/(1+1.314) = 0.568. On a set of three McCabeThiele 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 qline at the point where the rectifying section operating line intersects the qline. 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 highconcentration region, one the mid...
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 Spring '06
 selebi
 Distillation, pH, Mass Transfer

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