Question 1 of 7: Mass and Energy Balance, Contaminant Partitioning and Microbiology

Question 1: Mass and Energy Balance, Contaminant Partitioning and Microbiology (20 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.

(i) Henry’s Law and the Aqueous–Gaseous Fate of Volatile Contaminants

Henry’s Law states that, for a dilute solution at equilibrium, the partial pressure of a volatile solute in the gas phase above a liquid is directly proportional to its concentration (or mole fraction) in the liquid phase: $p = H \cdot C$ (or $p = H_x \cdot x$), where $H$ is the compound-specific Henry’s constant. A large $H$ means the contaminant strongly favours the gas phase at equilibrium — it is volatile and will tend to strip out of water into air (or, conversely, out of soil pore water into soil gas); a small $H$ means the contaminant stays predominantly dissolved. Environmental engineers use $H$ to predict fate and transport and to design unit processes: a high-$H$ compound (e.g., trichloroethylene, benzene) in groundwater will partition readily into an air stream, which is exactly the mechanism exploited by air-stripping towers and soil-vapour extraction, and the same property means such a compound volatilizes readily from an open tank, aerated basin, or spill, creating an inhalation exposure pathway that a low-$H$ compound (e.g., many metals, most sugars) would not. $H$ also increases with temperature, so volatilization losses (and stripping-system efficiency) are temperature-sensitive.

(ii) Indicator Microorganism for Drinking-Water Treatment Effectiveness

The standard indicator organism used to assess the effectiveness of surface-water treatment for drinking-water production is Escherichia coli (E. coli), a member of the fecal coliform group (total coliforms and enterococci are also used as complementary indicators, but E. coli is the primary regulatory indicator under the Guidelines for Canadian Drinking Water Quality). Four characteristics make an organism suitable as a regulatory indicator:

1. Consistent association with fecal contamination. The organism must be present whenever pathogens of fecal origin (bacteria, viruses, protozoa) are present, and absent when they are not, so a positive test reliably signals a fecal contamination event.
2. Present in greater numbers than the pathogens it represents. This gives the indicator a safety margin — if the indicator is undetectable, pathogen levels are very likely below a level of concern — and makes it easier and cheaper to detect than the pathogens themselves.
3. Comparable or greater resistance to treatment and environmental die-off than the target pathogens. If the indicator survives disinfection/treatment less well than pathogens, a "clean" indicator result could mask surviving pathogens; the indicator should be an equal or more conservative surrogate for treatment barrier performance.
4. Easy, rapid, inexpensive, and reliable to culture/detect using standard, reproducible laboratory methods (e.g., membrane filtration or defined-substrate tests), so that routine, frequent regulatory monitoring is practical for water utilities of any size.

(iii) Steady-State Mass Balance on a CSTR and the Limiting-Rate Cases

Given. A completely stirred tank reactor (CSTR) at steady state: feed concentration $C_{A0}$, outlet/tank concentration $C_A$, volume $V$, constant volumetric flow rate $Q$ in and out, and consumption rate $r_A = -kVC_A$ (mol/s) for the reaction A → B.

Find. The general mass-balance relation for $C_A$, and the value $C_A$ approaches in the two limiting cases $k=0$ and $k \to \infty$.

Approach. Write the unsteady-state species balance (accumulation = in − out + generation, with consumption as negative generation), then apply the given steady-state condition and solve algebraically for $C_A$.

1. Unsteady-state mass balance on A. Accumulation in the tank equals mass flow in, minus mass flow out, minus the reaction consumption: $$V\frac{dC_A}{dt} = QC_{A0} - QC_A - kVC_A.$$
2. Apply the steady-state condition. The problem states feed and outlet concentrations are constant over time, so $dC_A/dt = 0$: $$0 = QC_{A0} - QC_A - kVC_A \;\;\Rightarrow\;\; C_A(Q+kV) = QC_{A0}.$$
3. Solve for the outlet concentration. $$\boxed{C_A = \dfrac{C_{A0}}{1 + kV/Q}}.$$
4. Evaluate the limit $k=0$ (no reaction). With $k=0$ the denominator is 1, so $C_A = C_{A0}$: the reactor behaves as a pure hydraulic pass-through and the outlet concentration equals the feed concentration — nothing is removed.
5. Evaluate the limit $k \to \infty$ (instantaneous reaction). As $k \to \infty$, $kV/Q \to \infty$, so $C_A \to C_{A0}/\infty = 0$: the reaction consumes A far faster than it can accumulate, so essentially complete conversion is achieved and the outlet concentration approaches zero regardless of the specific values of $V$ or $Q$.