HW 5 - p 6.35 - Scrubbing a Concentrated Cl2-Air Mixture with Water in a Packed Tower- 8 pts
Exit gas from a chlorinator consists of a mixture of 20 mol% chlorine in air. This concentration is to be reduced to 1% chlorine by water absorption in a packed column to operate isothermally at 20oC and atmospheric pressure. Calculate for 100 kmol/h feed gas.
(a) The minimum flow rate in kilograms per hour.
(b) NOG for twice the minimum flowrate.
(a) The minimum flow rate in kilograms per hour.
(b) NOG for twice the minimum flowrate.
posted by Dr. B at 2:39 PM
3 Comments:
6? Really? That's not what I got and I don't think I did anything wrong...you said in office hours to mention this to you again.
Dr. B
Some people think cucumbers taste better pickled, and I think this problem would taste better if you gave us the right answer...6? I can't believe it, and neither can the pickle.
My cucumber comes out to about 1.35, not 6. Discuss.
pickle & cucumber:
I got NOG = 5.7 when I used just the equilibrium data points (and {0,0}) to compte 1/(Y-Y*). This is not great because the plot of 1/(Y-Y*) is quite curved at lower values of y. Some other students picked evenly spaced points in terms of y (like we did in class) and they got NOG ~ 4. I looked at their work and I think this method is superior to what I did and I now think that the correct answer is closer to 4 than it is to 5.7.
But, I cannot understand how cucmber got 1.35 for NOG. You need to plot 1/(Y-Y*) as a function of Y. The first point on this graph is at Y = 0.0101 and at this point I got 1/(Y-Y*) = 99.0. The area of the FIRST trapezoid alone was 4.56. So, I cannot guess what cucmber did to get NOG = 1.35. If anyone else has an idea, I would be happy if you posted it here.
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