Choosing a Separation Sequence on Lowest Depreciation: 23-Chem-A5 Chemical Plant Design and Economics, May 2018, Q2

Question 2: Choosing a Separation Sequence on Lowest Depreciation (20 marks)

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.

Quantity	Value
Feed rate	100 kg/min
Composition (mass)	pentane 90 %, hexane 8 %, heptane 2 %
Component flows	90 / 8 / 2 kg/min
Cost rule	annual (capital/depreciation) cost ∝ unit input capacity (kg/min)
Energy rule	energy cost ∝ energy consumed

Find. The depreciation index (sum of unit input capacities) of each scheme, hence which layout is cheaper, with an energy cross-check.

Approach

Depreciation follows capital, and capital is proportional to each unit's throughput; so the decision metric is simply the sum of the input capacities of the four units in each scheme. Trace the component flows through both trains, add up the four unit feeds, and compare. A sensible-heat energy proxy confirms the ranking.

Figure 2.1 — Scheme 1: the light, most-plentiful component (pentane, 90 kg/min) is removed first, so only 10 kg/min reaches the second stage. Unit feeds: 100, 100, 10, 10 kg/min.

Figure 2.2 — Scheme 2: the heavy, scarce component (heptane, 2 kg/min) is removed first, so 98 kg/min must still be processed by the second stage. Unit feeds: 100, 100, 98, 98 kg/min.

Split the feed into component flows. On the 100 kg/min basis, $\dot m_{C_5}=0.90(100)=90$, $\dot m_{C_6}=0.08(100)=8$, $\dot m_{C_7}=0.02(100)=2$ kg/min (check: $90+8+2=100$).
Scheme 1 unit capacities (remove the light key first). Separator 1 at 50 °C flashes off the pentane, so its bottoms carry only hexane+heptane, $8+2=10$ kg/min, to the second stage. The four unit input flows are therefore heater 1 = 100, separator 1 = 100, heater 2 = 10, separator 2 = 10: $$\Sigma_1 = 100+100+10+10 = 220 \text{ kg/min}$$ $\boxed{\text{Scheme 1 depreciation index }\Sigma_1 = 220\ \text{kg/min}}$
Scheme 2 unit capacities (remove the heavy key first). Separator 1 at 85 °C drops only the heptane (2 kg/min) as bottoms; its overhead, pentane+hexane $=90+8=98$ kg/min, must be cooled and separated again. The four unit input flows are heater 1 = 100, separator 1 = 100, cooler = 98, separator 2 = 98: $$\Sigma_2 = 100+100+98+98 = 396 \text{ kg/min}$$ $\boxed{\text{Scheme 2 depreciation index }\Sigma_2 = 396\ \text{kg/min}}$
Compare depreciation. Because cost (and hence depreciation) scales with capacity, $$\frac{\Sigma_2}{\Sigma_1} = \frac{396}{220} = 1.80,$$ so Scheme 2 carries 80 % more installed capacity — and 80 % more depreciation — than Scheme 1. $\boxed{\text{Scheme 1 has the lower depreciation and is the superior layout}}$
Energy cross-check (sensible-heat proxy). Taking the feed at 20 °C and equal specific heat per unit mass, an energy index $\sum \dot m\,\lvert\Delta T\rvert$ gives, for Scheme 1, $100(50-20)+10(85-50)=3000+350=3350$; for Scheme 2, $100(85-20)=6500$ of heating (plus $98(85-50)=3430$ of cooling that Scheme 1 avoids entirely). Scheme 2 needs $6500/3350 = 1.94\times$ the heating and a cooling duty, because it heats the whole feed to 85 °C and then cools 98 kg/min back down — a wasteful heat-then-cool. The energy comparison reinforces the depreciation decision.

Metric	Scheme 1 (light first)	Scheme 2 (heavy first)
Unit input capacities (kg/min)	100, 100, 10, 10	100, 100, 98, 98
Depreciation index Σ (kg/min)	220	396
Relative depreciation	1.00 (base)	1.80 (+80 %)
Heating index (kg·°C/min)	3350	6500 (+ 3430 cooling)
Decision	Scheme 1 is superior on both depreciation and energy.

Check: energy proxy The energy cross-check assumes equal mass-specific heat for all three cuts, no latent-heat term (a “gas/liquid separator” is treated as a flash, not a full reboiled column), and a 20 °C feed. These simplifications do not affect the depreciation ranking, which rests only on the input-capacity rule the question supplies; they simply confirm the same winner on operating cost.

Concept & Theory

Topic. Separation-sequence selection by capital/depreciation, and the “remove the most plentiful component first” heuristic.

Key relations. Depreciation ∝ capital ∝ Σ(unit throughput); energy proxy $\sum \dot m\,c_p\,\Delta T$; heuristic ordering of separators to minimize the load carried downstream.

Why this works. When each unit's cost scales linearly with the flow it handles, the cheapest train is the one that shrinks the downstream flows earliest. Removing pentane first (90 % of the feed) means the second heater and separator handle only 10 kg/min; removing heptane first (2 %) leaves 98 kg/min still to process, so the second stage is nearly full-size. This is the classical heuristic “do the largest split first” / “remove the most plentiful component early,” and here it also happens to be the easiest split (pentane's boiling point is far from the other two).

Common pitfall. Comparing only the number of units (both schemes have four) instead of their capacities, or forgetting that Scheme 2 adds a cooling duty on top of a larger heating duty. Another trap is confusing depreciation (a capital-linked charge, here the capacity sum) with energy cost (a throughput×ΔT charge) — the question asks for the former.

Source. Turton et al., Ch. 11 (separation-train synthesis heuristics); Peters, Timmerhaus & West, Ch. 6–9 (depreciation & capital cost as a function of size).

Question 2 of 6: Choosing a Separation Sequence on Lowest Depreciation

Question 2: Choosing a Separation Sequence on Lowest Depreciation (20 marks)

Approach