Name: Nguyen Huu Cong
ID: Cttt 10210113
ECH 158B: Process Design
Homework #3
Problem 1:
The temperature of water at exit:

𝑚
𝑐
𝑐
𝑝,𝑐(Tc,2 –Tc,1) =
𝑚
ℎ
𝑐
𝑝,ℎ(Th,1 –Th,2)
Or 1260000)(Tc,2 - 90) = 0.4160000)(230-130) = 6.4∙106 BTU/hr
So Tc,2= 114.6 oF
Then, the log mean temperature difference is:
LMTD ∆Tm =
𝑇
1−
𝑇
2
𝑙𝑛
𝑇
1
𝑇
2=
𝑇
ℎ,2−
𝑇
𝑐,1
𝑇
ℎ,1−
𝑇
𝑐,2
𝑙𝑛
𝑇
ℎ,2−
𝑇
𝑐,1
𝑇
ℎ,1−
𝑇
𝑐,2= (130− 90)−(230− 114.6
𝑙𝑛(130−90 (230−114.6

= 71.2 oF
Using F-Method, we estimate the F-Factor:
R =
Th,1 –Th,2
Tc,2 –Tc,1 =
𝑚
𝑐
𝑐
𝑝,𝑐
𝑚
ℎ
𝑐
𝑝,ℎ=
1∙(260000
0.4∙(160000
= 4.06
P =
Tc,2 –Tc,1
Th,1 –Tc,1 =
114.6−90
230−90= 0.18
So from the chart F-Factor = 0.85 is acceptable
From q= UAF∆Tm( UA =
𝑞
F∆Tm=
𝑚
𝑐
𝑐
𝑝,𝑐(Tc,2 –Tc,1
F∆Tm=
1∙(260000)(114.6 − 90)
0.85∙71.2

= 105750 BTU/hroF
The area of tube: Di = 0.902 in so the Area Ai = 0.639 in2 and (Perry’s 8th, Table 11-12)
area/ft= 0.236 ft2/ft
Therefore, 238 20” tubes will have area: As = 20x0.236x238 = 1123 ft2
The overall heat coefficient: U = 105750/1123 = 94.2 BTU/hrft2F
Let calculated the heat transfer coefficient for inside of tube:
G
𝑚
𝑐
𝐴
𝑠=
160000(238/2)∙(0.639/144= 302990 lb/hrft2
Reynold Re =
𝐺
𝐷
𝑖
𝜇=
302990∙(0.902/12
0.9= 25306
Prandal Number Pr =
𝑐
𝑝∙𝜇
𝑘
0.4∙0.9
0.2= 1.8
Then,
ℎ
𝑖
𝐷
𝑖
𝑘= 0.023∙Re0.8∙Pr0.33 = 0.023x253060.8x1.80.33 = 93.2
So hi = (93.2x0.2)/(0.902/12) = 247.98 BTU/hrft2F
For outside of tubes










So,




Problem 3:
i) The first step is to plot the equilibrium diagram and on it erect vertivcals at xB, xF , xB. There shoud be extended to the diagonal of the diagram.






The second step is to draw the feed line. Here, f = 0, and the feed line is vertical and is a continuation of line x = xF.
XD =
97
32
97
32
3
18= 0.95, XF =
40
32
40
32
60
18 = 0.27, XB =
3
32
3
32
97
18 = 0.1
The third step is to plot the operating lines. The intercept of the rectifying line on the y axis is 0.95/(3.5 + 1) = 0.211. From the intersection of this operating line and the feed line the stripping line is drawn.
The fourth step is to draw the rectangular steps between the two operating lines and the equilibrium curve. In drawing the steps, the transfer from the rectifying line to the stripping line is at the seventh step. By counting steps it is found that, besides the reboiler, 11 ideal plates are needed and feed should be introduced on the seventh plate from the top
ii) Physical properties of methanol: Molecular weight is 32, normal boiling point is 65°C,and the density of vapor is

The density of liquid methanol is 810 kg/m3 at 0°C and 792 kg/