Question

A stream of vinyl chloride monomer (VCM, H2C=CHCl, molecular weight = 62.5lb/lb-mol) in air is being produced from a process to make poly(vinyl chloride). The stream contains 1,360 ppm VCM and is at atmospheric pressure (equivalent to a VCM partial pressure of 0.02 psia). The stream is flowing at 1000 scfm (standard cubic feet per minute) at 75ºF and atmospheric pressure (14.7 psia). Activated carbon will be used to remove the VCM from the air. After loading the bed with VCM, a vacuum is used to desorb the VCM. During the vacuum step, the activated carbon is in equilibrium with VCM at a partial pressure of 0.0001 psia. Equilibrium adsorption data are given in the table below.

a) Is the Langmuir isotherm a reasonable choice to describe the adsorption data?
Yes or No

b) what is the working capacity of the activated carbon in this process? W.C.=

c) If an adsorption cycle lasts 8 hrs (8 hrs to adsorb to full capacity), what mass of
activated carbon is required?

Answer 1

Question:

The question is incomplete. The table of value was not added to your question. Below is the table of value and the plotted graph attached.

p R= VCM/AC p/R

0.0001 1 0.0001

0.0005 1.7 0.000294

0.001 3 0.000333

0.005 6 0.000833

0.01 8 0.00125

0.05 10 0.005

Answer:

(a) Yes

(b) W.C = 9.1428

(c) mass of activated carbon = 27.35 Ib-mole/activated carbon

Step-by-step explanation:

(a) From the graph attached, it is evident that Langmuir isotherm fits the data well. So, the answer is yes.

(b)

Since the input stream has a VCM partial pressure of 0.02 psia, this would be used to find working capacity of activated carbon.

From the graph it can be found by interpolation that p/R value corresponding to 0.02 psia is 0.0021876. Thus R = 9.1428 = WC

It should be noted that in adsorption process, activated carbon is in equilibrium with 0.001 psia of VCM. Thus, from the data R= 1 i.e. this amount of VCM is not adsorbed and lost. This value may be deducted based on the definition of working capacity (WC).

(c)

The molar flow rate of VCM = 1000*62.3*0.02/379*60*14.7

= 0.003739 Ib-mole/hr

Mass of activated carbon required = 0.003739 *8*9.1428*100

= 27.35 Ib-mole/AC