Feed-Gas Flow Rates to a Steam–Methane Reformer Heater: 23-Chem-A1 Process Balances and Chemical Thermodynamics, May 2014, Q2

Question 2: Feed-Gas Flow Rates to a Steam–Methane Reformer Heater (Part A — 20%)

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. A heater brings a steam + methane feed to 450 °C; both streams are at 1.6 MPa. The molar ratio is $\dot n_{H_2O}:\dot n_{CH_4} = 3:1$, and the specific enthalpies at inlet and outlet states are supplied, so no property look-up is needed.

Species	$H_{in}$ (kJ/mol)	$H_{out,450^\circ C}$ (kJ/mol)	$\Delta H$ (kJ/mol)
H$_2$O (in at 210 °C)	40.571	60.587	20.016
CH$_4$ (in at 30 °C)	14.551	26.393	11.842

Find. $\dot n_{CH_4}$ and $\dot n_{H_2O}$ (mol/h) that make the heater duty equal 2.5 kW.

Figure 2 — the two feed streams (steam at 210 °C and methane at 30 °C, 3:1 molar) enter the heater, absorb Q = 2.5 kW, and leave together at 450 °C.

Approach

Write an energy balance on the heater (no shaft work), express both molar flows through the single unknown $\dot n_{CH_4}$ using the 3:1 ratio, and solve for the duty of 2.5 kW.

Energy balance on the heater. With no work and steady flow, the duty equals the enthalpy rise of the two streams: $$\dot Q = \dot n_{H_2O}\,\Delta H_{H_2O} + \dot n_{CH_4}\,\Delta H_{CH_4}.$$
Impose the 3:1 feed ratio. Let $\dot n_{CH_4}=n$; then $\dot n_{H_2O}=3n$, so $$\dot Q = 3n\,(20.016) + n\,(11.842) = n\,(60.048 + 11.842) = 71.89\,n\ \ \mathrm{kW\ per\ (mol/s)}.$$
Solve for the methane flow. With $\dot Q = 2.5$ kW, $$n = \frac{2.5}{71.89} = 0.03478\ \mathrm{mol/s}.$$
$\dot n_{CH_4} = 0.0348\ \mathrm{mol/s} = 125.2\ \mathrm{mol\ CH_4/h}$.
Steam flow from the ratio. $$\dot n_{H_2O} = 3n = 0.1043\ \mathrm{mol/s}.$$
$\dot n_{H_2O} = 375.6\ \mathrm{mol\ H_2O/h}$.

Quantity	Value
Methane feed $\dot n_{CH_4}$	125.2 mol/h (0.0348 mol/s)
Steam feed $\dot n_{H_2O}$	375.6 mol/h (0.1043 mol/s)
Total feed	500.8 mol/h

Concept & Theory

Topic. Sensible-heat energy balance on a two-component heater with a fixed feed ratio.

Key relations. $\dot Q=\sum_i \dot n_i\,\Delta H_i$; the stream enthalpies are referenced to a common basis so only differences (out−in) matter.

Why this works. Because the enthalpies are already tabulated at the actual inlet and outlet states, the balance collapses to a single linear equation once the 3:1 constraint removes one unknown. No heat capacities or phase changes need to be integrated.

Common pitfall. Mismatching the reference state of each enthalpy, or forgetting that the steam enters at 210 °C (not 25 °C) so its rise to 450 °C is only 20.0 kJ/mol. Watch the unit conversion 1 mol/s = 3600 mol/h.

Source. Felder & Rousseau, Elementary Principles of Chemical Processes, ch. 8 (balances on non-reactive processes).

Species	\(H_{in}\) (kJ/mol)	\(H_{out,450^\circ C}\) (kJ/mol)	\(\Delta H\) (kJ/mol)
H\(_2\)O (in at 210 °C)	40.571	60.587	20.016
CH\(_4\) (in at 30 °C)	14.551	26.393	11.842

Quantity	Value
Methane feed \(\dot n_{CH_4}\)	125.2 mol/h (0.0348 mol/s)
Steam feed \(\dot n_{H_2O}\)	375.6 mol/h (0.1043 mol/s)
Total feed	500.8 mol/h

Question 2 of 8: Feed-Gas Flow Rates to a Steam–Methane Reformer Heater

Question 2: Feed-Gas Flow Rates to a Steam–Methane Reformer Heater (Part A — 20%)

Approach