HW 2 - p 3.40 - Overall Mass Transfer Coefficient for a Packed Tower- 6 pts
A new type of cooling-tower packing is being tested in a laboratory column. At two points in the column, 0.7 ft apart, the following data have been taken. Calculate the overall volumetric mass-transfer coefficient Kya that can be used to design a large, packed-bed cooling tower, where a is the mass-transfer area, A per unit volume, V, of tower.
Bottom Top
Water temperature, oF 120 126
Water vapor pressure, psia 1.69 1.995
Mole fraction H2O in air 0.001609 0.0882
Total pressure, psia 14.1 14.3
Air rate, lbmol/h 0.401 0.401
Column area, ft2 0.5 0.5
Water rate, lbmol/h 20 20
posted by Dr. B at 2:20 PM
5 Comments:
Does the (yai-yab)Lm = ((yai-yao)-(yai-yab))/ ln((yai-yao)/(yai-yab))
And yao=0, Na=Air rate*volume pack bed ?
Anon 4:11 PM:
(yai-yab)Lm = ((yai,1-yab,1)-(yai-yab,2))/ ln((yai,2-yab,1)/(yai-yab,2))
where:
yab,1 = mole fraction of A (water) in the bulk gas phase at the bottom of the 0.7 ft section of the column that we are analyzing.
yab,2 = mole fraction of A (water) in the bulk gas phase at the top of the 0.7 ft section of the column that we are analyzing.
yai,1 = mole fraction of A (water) in the gas phase at water interface at the bottom of the 0.7 ft section of the column that we are analyzing.
yab,2 = mole fraction of A (water) in the gas phase at water interface at the top of the 0.7 ft section of the column that we are analyzing.
dr. phil 5:02 PM:
Yes, the mole fractioon of water in the bulk gas is different at the top and the bottom of the column. See my previous comment to see how this effects your calculations.
It is also true that (1-yA)LM is different at the top and bottom of the 0.7 ft section of the column that we are analyzing. In this case the thing to do is to evaluate (1-yA)LM at the top and again at the bottom and use the average of these two values.
NA = nA / Amt, where nA = rate at which moles of a are transferred in our section of the column. You can evaluate nA from a water mole balance. Amt is the surface area for matt transfer. This, you cannot evaluate. Use a = Amt / Vcol. Then, solve for (Ky a).
I hate to correct dr. b but his equation didn't make much sense with the log means that we have used before. I tried it anyway and it disagreed with the answer that he gave in the assignment sheet. I experimented and put it how it would be if it were like all the other log means that we have done and got the right answer. So the equation I used is:
(yai-yab)Lm= ((yai2-yab2)-(yai1-yab1))/ln((yai2-yab2)/(yai1-yab1))
with the same coordinate system except that the second yab,2 is now yai2 because that was a mistake I believe.
gwydion 9:23 PM:
There is definitely a typo in my formula. It should be:
(yai-yab)Lm = ((yai,1-yab,1)-(yai,2-yab,2))/ ln((yai,1-yab,1)/(yai,2-yab,2))
This is equivalent to what gwydion posted.
Thank you for straightening me out !
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