1. (25 pts) For the mixture of nC6 (H) and nCs (C), using the y-x and T-y-x diag

Question

1. (25 pts) For the mixture of nC6 (H) and nCs (C), using the y-x and T-y-x diagrams at 101 kPa in the figures below, determine (1) the ratio of mole fraction in Vapor and Liquid phase: (2) determine the equation for q-line and plot the q-line in the y-x Figure; and (3) the compositions of the vapor and liquid phases at 225 °F and ZH-0.5 275 135 1.0 0.9Equilibrium curve 0.8 250 Vapor121. S 0.7 0.6 0.5 0.4 0.3 0.2 225 1072 200 93.3 45° line 175 79.4 Liquid 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 Mole fraction n-hexane in liquid, x 150 65.6 0 0.1 0.2 03 04 0.5 0.6 0.7 0.8 0.9 10 Mole fraction a-hexane, x or y

Explanation / Answer

Plotted in the figure below is the Tie-line at temp 225 F. Also, we plot a vertical line at mole fraction Z= 0.5 representing the feed concentration.

(1) Ratio of moles in vapor:liquid phase : given by Lever rule as the ratio of distances AB/BC

Horizontal distance AB = 0.5 - 0.175 =0.325

Horizontal distance BC = 0.555 - 0.5 = 0.055

Ratio of vapor phase to liquid phase = 0.325/0.055 = 5.91

(3) Compositions in both liquid phases can be directly obtained by the x coordinates of points A and C

Composition in liquid phase (point A): hexane= 0.175, octane = 0.825

Composition in vapor phase (point C): hexane = 0.555, octane = 0.445

(2) We define quality q = fraction of liquid in feed

Here, we have vapor: liquid = 5.91

Thus, q = 1 /(5.91 + 1) = 0.1447

Slope of the Q-line = q/(q-1) = -0.1692

It passes through point (zf,zf) i.e. (0.5,0.5)

Thus, equation is: y-0.5 = -0.1692(x-0.5)

or Equation of Q-line: y = -0.1692x + 0.5846

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