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I've been trying to solve this problem for a while with no success. I think I figured out that $m_1$ is $420 kg/hr$ and that $n_{3,A} = 3 kmol/h$ using the mass balance equation:

$input-output+generation-consumption=accumulation$

where steady state conditions mean there will be no accumulation and the lack of a reaction means that there will be no generation or consumption.

Everything else I've tried to solve for the remaining values has led me to a dead end (420=420, 16=16, etc.) Any help would be greatly appreciated!

Question

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  • $\begingroup$ I think probably you the question is incomplete. There are 5 unknowns in total and only one equation that can be formed. Neither you have given any maximum condition to solve the problem based on optimization approach, like transportation problem. Please check your question. $\endgroup$
    – Roopesh
    May 9 '18 at 10:21
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First, applying your equation to your case, $n_1-(n_2+n_3) + 0 - 0 = 0$ or $n_1 = n_2+n_3$

$$n_{1,a}=n_{2,a}+n_{3,a} \space (1) \\ n_{1,b}=n_{2,b}+n_{3,b} \space (2)\\ n_{1,c}=n_{2,c}+n_{3,c} \space (3) $$

and since there's nothing that separates/filters the components in different proportions, the output will contain exactly the same fraction as input.

$$ f n_{2,a} + (1-f) n_{3,a} = n_{1,a} \space (4)\\ f n_{2,b} + (1-f) n_{3,b} = n_{1,b} \space (5)\\ f n_{2,c} + (1-f) n_{3,c} = n_{1,c} \space (6)\\ $$

where $f$ is the ratio at which the streams are split into output - the same for all components.

Looking at the table, we have $n_{1,A}$ and $n_{2,a}$. From (1) we can quickly find $n_{3,a}$ : 4-1=3.

With these 3 we can find $f$ from (4): 0.25

Remainder of $n$ is trivial as we have column $1$ - multiply the $1$ column value by 0.25 for column $2$ and 0.75 for column $3$, e.g. $n_{2,b} = 10\cdot0.25 = 2.5$

For totals in kg/h, simply take the units given to find the factor by which to multiply:

$$total_{c} = \sum_i^{a,b,c}{MW_i \cdot n_{c,i}} $$

(remembering the 1000 factor of $kmol$ vs $mol$ - multiply $n$ by 1000 first): For column $1$

(4*10 + 10*20 + 6*30)*1,000 = 40+200+180 = 420,000 kg/h.

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  • $\begingroup$ Thank you for your help, I wasn't sure that I could use the same fraction for each component. I also didn't catch the mol to kmol unit in the chart so thanks for that too :) $\endgroup$
    – LostChemE
    May 9 '18 at 21:30
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I'm not going to solve this question numerically, but I believe you've been given enough information. Look at it piece by piece:

  1. You have the composition of stream 1. Assume the divider maintains the composition ratios and just splits up the total flowrates. This means that the ratio of streams S2/S1 = S2_A/S1_A = 1/4. You can use that factor to solve for S2_B and S2_C.

  2. That also tells you that S3/S1 = 3/4. Use that to determine the molar flowrates of S3_A,B,C.

  3. Use the MW values to calculate total mass flows.

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  • $\begingroup$ Thank you! I wasn't sure if I was allowed to assume that the components were divided in the same ratio. That makes it way easier :) $\endgroup$
    – LostChemE
    May 9 '18 at 21:29

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