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[(-252M0.9) - (200)(0.98)]/ (-262 - 200) = (-422.8)/(-462) = 0.935

3,4 [(200 x 0.7) - 422.8J/(200 - 452) = (-282.8)/( - 252) = 1.122

5. Connect the xmt points with the intersections of the g-line and equilibrium line. Via Eq. (2.40), determine minimum reflux for each stream (Fig. 2.156).

J = 2: R = [0.98 - (0.9)(-252/200)1/0.58 - 0 - 1 = 2.114/0.58 - 1 = 2.64

J = 3: R = [2.114 - 0.7(200/200)1/0.395 - (0.75 - 1)200/200 - 1 = 1.414/0.395 + 0.25 - 1 = 2.83

J = 4:R = 1.414/0.365 + 0.25 - [{-IK-200)/200] - 1 = 1.414/0.365 - 0.75 - 1 = 2.12

J = 5: R = [1.414 - 0.5(500/200)1/0.05 - 0.75 - 1 = 0.164/0.05 - 0.75 - 1 = 1.53

J = 6: K = [0.164 - 0.2(300/200)l/(-0.08) - 0.75 - (-0.5)300/200 - 1 = - 0.136/(-0.08) - 0.75 + 0.75 - 1 = 0.7

A comparison of the minimum reflux ratio for each section shows the highest value to be 2.83 for section J = 3. Therefore, the minimum for the column is 2.83, or 2.83 x 200 = 566 lb-mol/h.

6. For operation at 25 percent above minimum LID, find L for J = 3 from Eq. (2.35) at a minimum reflux: LA = 2.83 x 200 + (-252 x 1) + (200 x 0.75) = 464 lb-mol/h.

To operate at 25 percent above minimum LID, the reflux rate required » (464 x 0.25) + (2.83 x 200) = 682 lb-mol/h. And the reflux ratio = 682/ 200 = 3.41.

7. Find the number of stages by means of the McCabe-Thiele diagram. The component balance lines are constructed using the (Step 4) and the y(o) coordinate [from Eq. (2.40)], using the operating reflux ratio, as shown by the following relationships:

» E / \ 1.414 - 0.5(500/200) 0.164 n no J = 5:y(0) = 3.41 + 1 + 0.75 = 5Ä6 = 003

3.41 + 1 + 0.75 + (-0.5X300/200) "0136 = -0.03