Question is attached Q 5. Methanol (denoted as Me) is partially condensed out of a gas stream containing 60 mole% methanolvapor and the balance nitrogen. Process specifications lead to the flowchart shown below;QU/s)4 mol Me(v)/sN2(v)22°C, 6 atm100 mol/sCONDENSERMe(v) (60 mol%)N2(v) (40 mole%)70°C, 1 atmMe(l)22°C, 6 atmThe process operates at steady state. Calculate the required cooling rate.(Felder and Rousseau, 2005)

Answered: Q 5. Methanol (denoted as Me) is…

Introduction to Chemical Engineering Thermodynamics

8th Edition

ISBN:9781259696527

Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart

Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart

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Problem 1.1P

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Q 5. Methanol (denoted as Me) is partially condensed out of a gas stream containing 60 mole% methanol
vapor and the balance nitrogen. Process specifications lead to the flowchart shown below;
QU/s)
4 mol Me(v)/s
N2(v)
22°C, 6 atm
100 mol/s
CONDENSER
Me(v) (60 mol%)
N2(v) (40 mole%)
70°C, 1 atm
Me(l)
22°C, 6 atm
The process operates at steady state. Calculate the required cooling rate.
(Felder and Rousseau, 2005)

Expert Solution

Step 1

Given that: (from the images shown)

Feed rate = 100 mol/s

Methanol (Me) = 60%

Nitrogen (N₂) = 40%

Hence,

Initial methanol present in feed = 100 mol/s ×0.60 = 60 mol/s

Temperature of stream = 70⁰C

Pressure of stream= 1 atm

Temperature inside the chamber = 22⁰C

Pressure inside the chamber = 6 atm

The product feed has, methanol Me = 4 mol/s

Hence, methanol condensed = 60 mol/s – 4 mol/s = 56 mol/s

In this process, the feed is condensed and then compressed adiabatically ( keeping heat constant) to increase the pressure from 1 atm to 6 atm in the product feed.

For ethanol, the specific heat ratio (n) =1.20 (from data)

Hence, the adiabatic temperature relation is shown as:

$\frac{T_{2}}{T_{1}} = {(\frac{P_{2}}{P_{1}})}^{[(n - 1) / n]}$

Here, T₂= final temperature = 22⁰C, P₂ = Final pressure= 6 atm, T₁ = condensation temperature

P₁ = initial pressure = 1 atm, n = specific heat ratio = 1.20

Putting values:

$\frac{22 ⁰ C}{T_{1}} = {(\frac{6 a t m}{1 a t m})}^{[(1.20 - 1) / 1.20]} = 6^{0.166} \frac{22 ⁰ C}{T_{1}} = 1.3464 T_{1} = 16.34 ⁰$

Now, Cp of components should be taken at average teparature. Initial feed was 70⁰C.

So,

$a v e r a g e t e m p e r a t u r e = \frac{16.34 ⁰ C + 70 ⁰ C}{2} = \frac{86.34 ⁰ C}{2} = 43.17 ⁰ C$

Hence, average temperature = 43.17⁰+ 273 K = 316.17 K

At this temperature ( from standard data of handbook)

Cp ( methanol) =45.355 J/⁰C

Cp (nitrogen) = 29.141 J/⁰C

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