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 small amount of methane may also be partially oxidised (autothermal/secondary reforming) to supply reaction 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 increases both the equilibrium constant and the rate, driving conversion of methane forward. Industrial reformers therefore run hot (700–900 °C).
• Pressure: because the forward reaction increases moles, low pressure thermodynamically favours hydrogen production. However, reformers are actually run at 15–30 bar for downstream (PSA, ammonia/methanol synthesis) and compression economy; the conversion penalty of high pressure is offset by operating at higher temperature and with 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 is placed 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 and warm-up; 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, altering metering — which is why detergent additives (part d) are used.

Of these, volatility is the single most important characteristic: the fuel must vaporise readily and completely in the intake to form a homogeneous, ignitable mixture across the operating range.

(d) Evaporator with bypass to hit an intermediate concentration

(i) What to do, and why. (5 marks) A single pass of the evaporator raises the detergent concentration from 12% straight to 58% — it cannot stop at 45%. To obtain an intermediate concentration of 45% in one pass, split the feed and bypass part of it around the evaporator: send only a portion of the 12% feed through the evaporator (producing 58% concentrate) and blend that concentrate with the bypassed 12% feed. Because 45% lies between 12% and 58%, a suitable split gives exactly 45%. Graphically this is the lever rule on a concentration line: the product (45%) sits between the two mixing streams, and the required fraction of concentrate is the ratio of the near arm to the whole span.

On the composition line, the fraction of the product that must come from the 58% concentrate is

$$f_C = \frac{x_P - x_F}{x_C - x_F} = \frac{0.45 - 0.12}{0.58 - 0.12} = 0.717,$$

so about 72% of the product stream is evaporator concentrate and 28% is bypassed feed.

(ii) Production rate of 45% product.

Given. Dilute feed $F = 2000$ lbmol/hr at $x_F = 12\%$ detergent; single-pass concentrate $x_C = 58\%$; desired product $x_P = 45\%$. Detergent is non-volatile — the evaporator removes only water, so all detergent that enters leaves in the product.

Find. The production rate $P$ of the 45% product stream.

Approach. Detergent is conserved through the whole splitter–evaporator–mixer system, so an overall detergent balance fixes the product rate directly, independent of the split.

1. Detergent fed to the system. On the total 2000 lbmol/hr feed,$$\dot n_{\text{det,in}} = x_F\,F = 0.12(2000) = 240\ \text{lbmol/hr}.$$
2. Overall detergent balance. No detergent leaves with the water vapour, so all 240 lbmol/hr reports to the 45% product:$$x_P\,P = x_F\,F \;\Rightarrow\; P = \frac{0.12(2000)}{0.45} = \boxed{533.3\ \text{lbmol/hr}}.$$
3. Consistency of the split (lever rule). With $f_C = 0.717$, the concentrate and bypass streams are$$C = f_C P = 382.6,\qquad B = P - C = 150.7\ \text{lbmol/hr}.$$A detergent balance across the evaporator alone gives the feed actually evaporated, $A = x_C C / x_F = 0.58(382.6)/0.12 = 1849.3$ lbmol/hr, and $A + B = 2000$ lbmol/hr closes the total feed. The water vapour removed is $V = A - C = 1466.7$ lbmol/hr.