Question 1 of 6: Steam Reforming, Carburetion and an Evaporator with Bypass

Question 1: Steam Reforming, Carburetion and an Evaporator with 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) Reforming reactions from the methane–steam reaction (4 marks)

Steam methane reforming (SMR) reacts methane with steam over a nickel catalyst. The strongly endothermic reforming reaction is followed by the mildly exothermic water–gas shift, which converts the CO to further hydrogen:

$$CH_4 + H_2O \rightleftharpoons CO + 3H_2 \qquad (\Delta H^\circ_{298} = +206\ \text{kJ/mol})$$$$CO + H_2O \rightleftharpoons CO_2 + H_2 \qquad (\Delta H^\circ_{298} = -41\ \text{kJ/mol})$$

Adding the two gives the overall reforming reaction:

$$CH_4 + 2H_2O \rightleftharpoons CO_2 + 4H_2 \qquad (\Delta H^\circ_{298} = +165\ \text{kJ/mol}).$$

Thus one mole of methane yields up to four moles of hydrogen. In practice a little methane is also burned (autothermal/secondary reforming) to supply the endothermic heat, but the two equations above are the reforming reactions proper.

(b) Effect of pressure and temperature (2 marks)

The primary reforming reaction $CH_4 + H_2O \rightleftharpoons CO + 3H_2$ is endothermic and proceeds with an increase in the number of moles (2 → 4). By Le Chatelier’s principle:

• Temperature: higher temperature raises both the equilibrium constant and the rate, driving methane conversion forward. Industrial reformers therefore run hot (700–900 °C).
• Pressure: because the forward reaction increases moles, low pressure thermodynamically favours hydrogen. In practice reformers still run at 15–30 bar for downstream (PSA, ammonia/methanol) and compression economy; the conversion penalty of high pressure is offset with higher temperature and excess steam.

The water–gas shift is mildly exothermic and mole-neutral, so it is favoured by lower temperature and is insensitive to pressure — hence the shift section sits downstream and cooler than the reformer.

(c) Fuel characteristics that influence carburetion (4 marks)

Carburetion is the formation of a combustible air–fuel mixture by vaporising liquid gasoline into the intake air. The physical fuel properties that govern it are:

• Volatility (distillation curve): the dominant property. Front-end volatility (light ends, low 10% point) controls cold starting; the mid-range (50% point) governs smooth acceleration and drivability; the tail (90%/end point) affects mixture distribution and crankcase dilution.
• Vapour pressure (Reid vapour pressure): too high causes vapour lock and evaporative loss; too low hinders cold starting.
• Viscosity and surface tension: govern atomisation — the droplet size formed at the jet, hence the rate of evaporation and mixing.
• Density and latent heat of vaporisation: set the air–fuel ratio metered by a fixed jet and the charge cooling on evaporation.
• Cleanliness / gum-forming tendency: gums and deposits foul the jets and passages and alter metering — the reason detergent additives (part d) are dosed.

(d) Evaporator with a feed bypass

Given. A single evaporator pass concentrates the 15% detergent feed to 65%. The desired product is 50% detergent, and the dilute feed rate is 5000 lbmol/hr. Only water is evaporated (detergent is non-volatile).

Find. (i) how to reach 50% in a single pass; (ii) the production rate $P$ of the 50% product.

(i) What to do — split the feed and bypass part of it

A single evaporator pass overshoots to 65%, which is richer than the 50% wanted. The remedy is a feed bypass: split the dilute feed, send only part ($F_E$) through the evaporator to 65%, and blend the 65% concentrate with the un-evaporated bypass ($F_B$, still 15%). Choosing the split so the blend lands at exactly 50% delivers the target concentration in one pass. Graphically this is the lever rule on a concentration line: the product point (50%) lies between the bypass point (15%) and the concentrate point (65%), and the split ratio is read off as the ratio of the two segment lengths.

(ii) Production rate of the 50% product

Approach. Detergent is conserved (only water leaves as vapour), so an overall detergent balance on the whole unit fixes the product rate independently of the split; the lever rule then gives the individual streams.

1. Overall detergent balance. All the detergent entering with the feed leaves in the product:$$x_F\,F = x_P\,P \;\Rightarrow\; P = \frac{x_F\,F}{x_P} = \frac{0.15(5000)}{0.50} = \boxed{1500\ \text{lbmol/hr}}.$$
2. Split from the blend lever rule. The mixer blends concentrate $C$ (65%) with bypass $F_B$ (15%) to give $P$ (50%), with $P = C + F_B$:$$\frac{C}{P} = \frac{x_P - x_F}{x_C - x_F} = \frac{0.50-0.15}{0.65-0.15}=0.70 \;\Rightarrow\; C = 1050,\quad F_B = 450\ \text{lbmol/hr}.$$
3. Feed to the evaporator (check). The evaporator concentrates $F_E$ (15%) to $C$ (65%); a detergent balance gives $F_E = x_C C / x_F = 0.65(1050)/0.15 = 4550$, and $F_E + F_B = 4550 + 450 = 5000$ lbmol/hr — the feed closes exactly.