A cold feed enters the column at 40 degrees Centigrade. The bubble point of the mixture, containing
66.7 percent benzene and 33.3 percent toluene is 94 degrees Centigrade. Calculate the feed quality, q, if the
following data are given:
Latent heat of vaporization of benzene: 7,566 calories per
mole, Latent heat of vaporization of toluene: 7,912 calories per mole.
Heat capacity of benzene: 32.8 calories per mole per degree
Centigrade, Heat capacity of toluene: 40.5 calories per mole per degree
Centigrade.
Solution:
Quality of feed, q is equal to the ratio of heat required to
vaporize one mole of feed to the latent heat of vaporization of the feed.
Heat required to vaporize one mole of benzene is equal to
the sum of the sensible heat and the latent heat of vaporization. This is equal to 32.8 times (94 minus 40)
plus 7,566. This value is equal to
9,337.2 calories per mole of benzene.
Heat required to vaporize one mole of toluene is equal to
40.5 times (94 minus 40) plus 7,912.
This value is equal to 10,097 calories per mole of toluene.
Plugging in the values of mole fractions of benzene and
toluene in the feed results in the heat requirement for vaporization, HG. It is equal to 0.667 times 9,337.2 plus
0.333 times 10,097. This value is 9,590.2
calories per mole of feed.
Latent heat of vaporization, Hl, is 0.667 times 7,566 plus 0.333 times 7,910 and is equal
to 7,680.6 calories per mole of feed.
Quality of the feed, q, is equal to the ratio of 9590.2 and
7,680.6 and its value is equal to 1.25.