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23-Chem-A1 Process Balances and Chemical Thermodynamics · May 2013

Question 2 of 7: Methanol Condensed in a Distillation Accumulator

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — May 2013 — 04-Chem-A1 Process Balances and Chemical Thermodynamics. Three-hour, open-book exam; any non-communicating calculator permitted. Format: seven questions in three parts — answer one of Q1–Q2 (Part A), one of Q3–Q4 (Part B) and two of Q5–Q7 (Part C); four questions of equal value constitute a complete paper. All seven are solved below for completeness. Where property data are needed they are stated explicitly in each Given block. Units follow the paper (mixed SI and US customary).

Reference texts: Felder, Rousseau & Bullard, Elementary Principles of Chemical Processes (4th ed., Wiley) — material & energy balances; Smith, Van Ness, Abbott & Swihart, Introduction to Chemical Engineering Thermodynamics (8th ed., McGraw-Hill) — VLE, residual properties, reaction equilibrium; supporting data from Perry's Chemical Engineers' Handbook (9th ed.) and the NIST Chemistry WebBook.

Question 2: Methanol Condensed in a Distillation Accumulator (equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Overhead vapour $V$ (95.0% methanol) is condensed into the accumulator. Over the 12-h test: product to storage $P=7500$ lbm; reflux $R=0.80\,V$; drum level fell by 3.5 in at 50 gal/in; the drum liquid is the 95% methanol condensate. Density of the condensate is taken as that of methanol, $\rho\approx 0.792\ \text{g/mL} = 6.61\ \text{lb/gal}$ (property data, open-book).

Find. The mass of methanol condensed during the test (i.e. the methanol content of the total condensed vapour $V$).

CondenserAccumulatorDrum (5 psig)Overhead vapour V95% methanolReflux R = 0.8 VProduct 7500 lbm(to storage)
Figure 2 — Accumulator-drum control volume: condensed vapour $V$ in; reflux $R=0.8V$ and product $P$ out; the falling level supplies the shortfall.

Approach. Write an unsteady total mass balance on the accumulator drum over the test — condensed vapour in, reflux and product out, and the measured inventory change — then convert $V$ to methanol using the 95% purity.

  1. Convert the level drop to an inventory change. The level fell 3.5 in at 50 gal/in, so the liquid inventory dropped by $$\Delta \mathcal{V} = -(3.5)(50) = -175\ \text{gal}\;\Rightarrow\; \Delta m = -175\times 6.61 = -1157\ \text{lbm}.$$ The minus sign says the drum supplied 1157 lbm of stored liquid during the test.
  2. Unsteady total balance on the drum. Accumulation = in − out, with the vapour condensing in and both reflux and product leaving: $$\Delta m = V - (P + R) = V - P - 0.80V = 0.20\,V - P.$$
  3. Solve for the total vapour condensed. Substituting $\Delta m=-1157$ lbm and $P=7500$ lbm, $$0.20\,V = P + \Delta m = 7500 - 1157 = 6343\;\Rightarrow\; V=\frac{6343}{0.20}=\boxed{31{,}700\ \text{lbm condensed}}.$$
  4. Methanol content of the condensate. The overhead vapour (and thus the condensate) is 95.0% methanol, so $$m_{\text{MeOH}} = 0.950\,V = 0.950(31{,}700)=\boxed{3.01\times10^{4}\ \text{lbm methanol}}.$$ This is the methanol that condensed over the 12-h test; the balance (≈ 1580 lbm) is water.
QuantityResult
Inventory change (level drop)$-175$ gal $\approx -1157$ lbm
Total vapour condensed, $V$$\approx 31{,}700$ lbm
Methanol condensed$\approx 3.01\times10^{4}$ lbm (≈ 30,100 lbm)

Check: the condensate density is taken as pure-methanol (6.61 lb/gal); at 95 wt% methanol and the drum temperature it may run 1–3% higher, shifting the inventory term by only ≈ ±30 lbm and $V$ by ≈ ±150 lbm (<0.5%). The 5 psig drum pressure is not needed for the liquid balance.