Question

One of the olf columns in your plant failed. You have been putting together jury-rigged system to separate propane fron n-butane. The column you have found is empty, so within reason you can put in as many stages as needed. It is equipped with a partial condenser and a total reboiler. The column is restricted to a maximum pressure of 800,0 kPa, which means that refrigerator is needed. You have a source of refrigerator, but output is limited to 6000.0 kW. Freed is a saturated liquid at 800.0 kPa that is 56.0 mol% propane. Unless stated otherwise distillate should be 99.9 mol% propane, and bottoms should be 1.0 mol% propane or less. You have more feed available that can be processed.

a) generate a y-x graph at 800.0 kPa from the DePriester chart. From the depriester chart find the boiling points of propane at 800.0 kPa (set Kp=1) and n-butane at 800.0 kPa (set K(n-b)=1). The boiling range for the complete y-x diagram lies between these two temperature. Pick a temperature between the two boiling points, find Kp and K(n-b), calculate X(n-b)=(Kp-1)/(Kp-Kb), Xp=1-X(n-b), and Y(n-b)-K(n-b)X(n-b). Doing this for five to ten temperatures generates the y-x graph.

b) if you built a system with a very large number of stages (like N greater than 100) estimate the kmol/h of 99.9mol% distillate that can be produced and kmol/h of feed processed (bottoms less or equal to 1 mol% propane).

c) Repeat part b for 90.0mol% pure propane distillate and 1% or less propane in the bottoms for a saturated liquid feed.

d) if the required purity of the bottoms stream is changed to 10.0 mol% propane, repeat part b (distillate is 99.9% mol% propane).

e) repeat part b for a 99.9 mol% pure propane distillate and 1% or less propane in the bottoms if the feed is a saturated vapor that is 56.0 mol% propane.

f) epeat part b for a 99.9 mol% pure propane distillate and 1% or less propane in the bottoms if the feed is a saturated vapor that is 66.0 mol% propane.

Data: Heat of vaporization of propane at 800 kPa is 18,800 KJ/Kmol, and for butane at 800 kPa is 22,600 KJ/Kmol

Answer 1

Solution:

Given:

Propane (Kp = 1) = 15.4⁰C (From Depriester chart)

n-butane (Kp=1) = 70.2⁰C (from Depriester chart)

So K values from Dpriester chart between these temperatures can be written as:

T (0C)	Kn (n-butane) = K1	Kp (propane)=K2
15.4	-	1
20	0.34	1.25
25	0.38	1.40
30	0.46	1.5
35	0.5	1.75
40	0.55	2
45	0.65	2.75
50	0.70	2.80
55	0.8	3.125
60	0.9	3.8
65	0.95	4.1
70.2	1	-

Now calculate Y-X from above data as:

K_i = Y_i/X_i

So Y and X can be calculated as:

T (0C)	Kn (n-butane) = K1	Kp (propane)=K2	X1	Y1
15.4	-	1	0	0
20	0.34	1.25	0.274725275	0.093406593
25	0.38	1.40	0.392156863	0.149019608
30	0.46	1.5	0.480769231	0.221153846
35	0.5	1.75	0.6	0.3
40	0.55	2	0.689655172	0.379310345
45	0.65	2.75	0.833333333	0.541666667
50	0.70	2.80	0.857142857	0.6
55	0.8	3.125	0.913978495	0.731182796
60	0.9	3.8	0.965517241	0.868965517
65	0.95	4.1	0.984126984	0.934920635
70.2	1	-	1	1

(b) Given:

Saturated liquid feed = q = 1

So slope of feed line is infinite and it is a vertical line through (Z_f,Z_f) that is (0.56,0.56)

Large number of stages required.

X_D = 0.999; X_W = 0.01

So x_D / R_m+1 = 0.56

From where we will get R_m = 0.7839 = L/D

F = D+W

0.56F = 0.999D+0.01W

L'-L=qF = F

L' = L+F

Let W = 100 kmol/h

So, F = 225 kmol/h and F= 125 kmol/h

Similarly we can solve other parts.