API gravity; evaporator with feed bypass: 23-Chem-B6 Petroleum Refining and Petrochemicals, December 2019, Q2

Question 2 — API gravity; evaporator with feed bypass (10 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.

(a) API gravity of crude oil

Definition. API gravity is the American Petroleum Institute’s standard inverse-density scale for petroleum liquids, defined from the specific gravity measured at 60/60 °F: $^{{\circ}}\text{API}=\dfrac{{141.5}}{{\mathrm{{SG}}}}-131.5$. By construction water (SG = 1.000) is exactly 10 °API.

Heavy vs. light. A heavy crude is dense (high specific gravity), and because °API varies inversely with SG, a heavy crude has a low API gravity (typically below ≈22 °API, and below 10 °API it is denser than water). Light, more valuable crudes have high API gravity (> 31 °API). Thus API gravity is a quick proxy for crude quality: higher °API generally means a lighter, sweeter, more easily refined and higher-priced crude.

(b) Evaporator with feed bypass

Given. Feed $F=100\ \text{mol/h}$ salt solution at $x_F=0.10$ salt. Desired product $x_P=0.35$ salt; single-pass evaporator concentrate limited to $x_C=0.50$ salt; the bypass stream is raw feed. Only pure water leaves the evaporator as vapour.

Stream	Salt fraction	Note
Feed F	0.10	100 mol/h
Evaporator concentrate C	0.50	single-pass maximum
Bypass	0.10	raw feed, un-evaporated
Product P	0.35	target

Find. (1) water evaporation rate; (2) bypass fraction; (3) concentrated-product rate; plus a labelled flowsheet.

Approach. Salt is conserved (only water evaporates), so the overall salt balance fixes the product rate and the evaporated water; a salt balance across the final mixer then fixes the bypass split.

(1) Product rate and water evaporated (overall balances). Salt in feed = salt in product: $$F\,x_F=P\,x_P\ \Rightarrow\ P=\frac{{100(0.10)}}{{0.35}}=\boxed{{28.6\ \text{{mol/h}}}}.$$ By the overall total balance the balance of the feed leaves as water vapour, $$\dot W=F-P=100-28.6=\boxed{{71.4\ \text{{mol/h water evaporated}}}}.$$
(2) Bypass fraction (salt balance at the mixer). The product is the evaporator concentrate C (50% salt) mixed with bypass B (10% salt): $C+B=P$ and $0.50\,C+0.10\,B=0.35\,P$. Eliminating $C=P-B$, $$0.50(P-B)+0.10B=0.35P\ \Rightarrow\ 0.40B=0.15P,$$ $$B=0.375\,P=10.7\ \text{{mol/h}}\ \Rightarrow\ \frac{{B}}{{F}}=\frac{{10.7}}{{100}}=\boxed{{0.107\ (10.7\%)}}.$$
(3) Concentrated-product (evaporator concentrate) rate. From $C=P-B$, $$C=28.6-10.7=\boxed{{17.9\ \text{{mol/h at 50\% salt}}}}.$$ The evaporator feed is $E=F-B=89.3\ \text{{mol/h}}$; its salt balance checks: $0.10(89.3)=8.93=0.50(17.9)$, and the evaporator water $E-C=89.3-17.9=71.4\ \text{{mol/h}}$ equals $\dot W$.

Feed splits into an evaporator feed (89.3 mol/h) and a bypass (10.7 mol/h). The evaporator boils off 71.4 mol/h of water, leaving 17.9 mol/h of 50% concentrate, which remixes with the bypass to give 28.6 mol/h of 35% product.

Quantity	Value
(1) Water evaporation rate $\dot W$	71.4 mol/h
(2) Fraction of feed bypassing the evaporator	0.107 (10.7%)
(3) Concentrated product rate P	28.6 mol/h at 35% salt
Evaporator concentrate C / bypass B	17.9 / 10.7 mol/h

Concept & Theory

Topic. API gravity as a density scale; steady-state material balances on a bypass (splitter–unit–mixer) flowsheet with a conserved solute.

Key relations. $^{{\circ}}\text{API}=141.5/\mathrm{SG}-131.5$; overall total balance $F=\dot W+P$; overall salt balance $Fx_F=Px_P$ (water-only vapour); mixer salt balance $x_CC+x_FB=x_PP$ with $C+B=P$.

Why this works. Because the evaporator can only reach 50% but the target is 35%, part of the dilute feed is diverted around the unit and blended back — a classic bypass. Since salt neither enters nor leaves except in the streams, an overall solute balance fixes the product before any internal stream is known; the mixer balance then apportions the split.

Common pitfall. Trying to over-concentrate in one pass (the evaporator physically cannot exceed 50%), or forgetting that the bypass carries salt too — it is raw 10% feed, not pure water, so the mixer balance must include $x_FB$.

Source. Felder & Rousseau, Elementary Principles ch. 4 (bypass and purge balances); Fahim et al., Fundamentals of Petroleum Refining ch. 2 (API gravity).

Question 2 of 5: API gravity; evaporator with feed bypass

Question 2 — API gravity; evaporator with feed bypass (10 marks)

(a) API gravity of crude oil

(b) Evaporator with feed bypass