Question 5 of 6: Gasoline Properties and an Evaporator with Feed Bypass

Question 5: Gasoline Properties and an Evaporator with Feed Bypass (20 marks — 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.

(a)(i) Meaning of the three properties (4 marks)

• Reid vapour pressure (RVP): the absolute vapour pressure of the gasoline measured at 100 °F (37.8 °C) in a standard bomb (ASTM D323) — an index of the fuel’s front-end (light-end) volatility.
• Boiling range: the span of temperatures over which the gasoline distils (ASTM D86 distillation curve, from initial boiling point through the 10/50/90% points to the final boiling point).
• Antiknock characteristics: the fuel’s octane quality (RON, MON and the pump AKI) — its resistance to engine knock.

(a)(ii) Why they matter (4 marks)

• RVP: too high causes vapour lock, high evaporative (VOC) emissions and a safety/venting concern; too low hampers cold starting. It is capped seasonally and geographically.
• Boiling range: the front end governs cold starting, the mid-range controls warm-up and acceleration/drivability, and the tail affects complete combustion, deposits and crankcase-oil dilution — so the whole curve must be tailored.
• Antiknock: sets the highest compression ratio and spark advance the engine can use, hence its power and thermal efficiency, and prevents the destructive pressure spikes of knock.

(b) Evaporator with a feed bypass

Given. The evaporator overshoots the target concentration in a single pass, so part of the dilute feed is bypassed and re-blended to land on 50%.

Find. (i) schematic; (ii) product rate $P$; (iii) water evaporated $V$; (iv) bypass fraction.

(i) Process schematic

(ii) Production rate of the 50% product

Approach. Detergent is non-volatile, so all of it entering with the feed leaves in the product — a single overall detergent balance fixes $P$.

1. Overall detergent balance. $x_F F = x_P P$, hence$$P = \frac{x_F F}{x_P} = \frac{0.12(1000)}{0.50} = \boxed{240\ \text{lb/h}}.$$

(iii) Rate of water evaporation

1. Overall mass balance. Only water leaves as vapour $V$: $F = P + V$, so$$V = 1000 - 240 = \boxed{760\ \text{lb/h of water evaporated}}.$$

(iv) Fraction of feed that bypasses

Approach. A detergent balance on the evaporator alone (all 760 lb/h of water is removed there) gives the feed split, since the bypass carries no water out.

1. Evaporator detergent balance. Let $F_1$ enter the evaporator (12%) and leave as concentrate $C$ (60%) after losing all the water $V$: $x_F F_1 = x_C(F_1 - V)$, i.e. $0.12F_1 = 0.60(F_1 - 760)$, giving$$F_1 = \frac{0.60(760)}{0.60-0.12} = \frac{456}{0.48} = 950\ \text{lb/h}.$$
2. Bypass stream and fraction. $F_2 = F - F_1 = 1000 - 950 = 50$ lb/h, so$$\text{bypass fraction} = \frac{F_2}{F} = \frac{50}{1000} = \boxed{0.05\ (5\%)}.$$
3. Check. Concentrate $C = F_1 - V = 190$ lb/h at 60% → 114 lb/h detergent; bypass detergent $=0.12(50)=6$; product detergent $=114+6=120=0.12(1000)$ ✓, and $C+F_2 = 190+50 = 240 = P$ ✓.