Mass and Energy Balance, Contaminant Partitioning and Microbiology: 18-Env-A1 Principles of Environmental Engineering, May 2017, Q1

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) Steady-State Phosphate Mass Balance on the Lake

Given. A completely-mixed lake of volume $V = 1\times10^{5}\ \text{m}^3$, fed by a river of flow $Q_i = 1\times10^{4}\ \text{m}^3/\text{yr}$ carrying $C_i = 10\ \text{mg/L}$ PO4 from an upstream sewage-plant discharge; net evaporation from the lake surface is $Q_e = 1\times10^{4}\ \text{m}^3/\text{yr}$ (pure water loss — no PO4 leaves with it); the outflow $Q_o$ is unknown; PO4 decays in the lake as a first-order reaction with $K = 0.10\ \text{yr}^{-1}$; steady state applies.

Find. The outflow PO4 concentration $C_o$.

Approach. First close the water (volume) balance on the lake to find $Q_o$, then write the steady-state PO4 mass balance on the same completely-mixed control volume and solve for $C_o$.

Figure 1. Control volume for the lake mass balance: river inflow ($Q_i$, $C_i$), evaporative loss ($Q_e$, PO4-free), and the outflow ($Q_o$, $C_o$) being solved for.

Close the water balance. At steady state the lake volume is constant, so total inflow equals total outflow. Evaporation removes only water (no PO4), so it is a separate loss term from the liquid outflow: $$Q_i = Q_o + Q_e \;\;\Rightarrow\;\; Q_o = Q_i - Q_e = 1\times10^{4} - 1\times10^{4} = \boxed{0\ \text{m}^3/\text{yr}}.$$ Physically, the river inflow is exactly balanced by evaporative loss, so this lake has no net liquid outflow — it behaves as a terminal (closed) water body for mass-transport purposes.
Write the unsteady-state PO4 balance on the lake. Treating the lake as a single completely-stirred reactor (CSTR) with first-order decay, accumulation equals mass in minus mass out minus decay: $$V\frac{dC}{dt} = Q_iC_i - Q_oC_o - KVC_o.$$
Apply the steady-state condition. With $dC/dt = 0$: $$0 = Q_iC_i - Q_oC_o - KVC_o \;\;\Rightarrow\;\; C_o = \dfrac{Q_iC_i}{Q_o + KV}.$$
Substitute $Q_o = 0$ and evaluate. With no liquid outflow, the only PO4 sink is in-lake decay, so the balance reduces to $Q_iC_i = KVC_o$: $$C_o = \dfrac{Q_iC_i}{KV} = \dfrac{(1\times10^{4})(10)}{(0.10)(1\times10^{5})} = \dfrac{1\times10^{5}}{1\times10^{4}} = \boxed{10\ \text{mg/L}}.$$ The outflow (in-lake) concentration equals the inflow concentration for this particular combination of numbers — a coincidence of the given data ($Q_i = KV$ numerically), not a general result; halving $K$ or doubling $V$ would double $C_o$ relative to $C_i$.

Quantity	Value
Outflow (liquid) rate $Q_o$	0 m³/yr
Steady-state PO4 concentration $C_o$	10 mg/L

Check: assumes the lake is a single completely-mixed (CSTR) control volume with uniform PO4 concentration equal to the outflow concentration $C_o$, and that the decay rate $K$ is a true first-order constant independent of season/temperature. With $Q_o = 0$, the lake has effectively zero hydraulic flushing by the liquid pathway (only evaporative loss), so all PO4 removal is by in-lake decay — a case worth noting since it means any inaccuracy in $K$ has an amplified effect on the predicted $C_o$ (there is no dilution safety margin from flushing).

(ii) Environmental Factors Affecting Air–Water Partitioning of Polar Chemicals

Henry’s Law, $p = K_H\,C$ (or, on a dimensionless basis, $K_H^{\prime} = C_{\text{air}}/C_{\text{water}}$), predicts the equilibrium split of a chemical between the gas and dissolved-liquid phases. For a genuinely polar, non-ionizable compound such as propanol, $K_H$ is intrinsically small (propanol is highly water-soluble and has low vapour pressure relative to its solubility), so it strongly favours the aqueous phase; two environmental factors that shift that split further are:

Water pH (for ionizable polar chemicals). Henry’s Law strictly applies to the neutral, un-ionized form of a compound — the ionized (charged) form is essentially non-volatile because it is stabilized in solution by hydration. For weak acids and bases (e.g., phenols, chlorinated phenoxy herbicides, ammonia), the fraction present as the neutral species is set by the Henderson–Hasselbalch relationship relative to the compound’s $\text{p}K_a$. As pH moves the equilibrium toward the ionized form (e.g., pH above $\text{p}K_a$ for a weak acid), the effective Henry’s constant for the bulk (ionized + neutral) chemical drops sharply, because only the shrinking neutral fraction can volatilize. A simple neutral alcohol like propanol has no acid/base equilibrium in the environmental pH range, so pH itself has little direct effect on propanol specifically — but the mechanism is the dominant control for the many polar organics (weak acids/bases) that are ionizable, which is why pH is listed as a governing environmental factor for polar-chemical partitioning generally.
Water salinity (the “salting-out” effect). Dissolved ions (e.g., in seawater or brackish estuarine water) compete with an organic solute for water molecules of hydration, effectively reducing the solute’s solubility and raising its activity coefficient in the aqueous phase. This is quantified by the Setschenow equation, $\log(K_{H,\text{salt}}/K_{H,\text{fresh}}) = K_s\,[\text{salt}]$, and it increases the effective Henry’s constant — i.e., a polar chemical such as propanol partitions somewhat more readily to the gas phase (or to a NAPL/sorbed phase) as salinity rises, because it is being “pushed” out of the increasingly occupied aqueous phase. The effect is generally modest for very polar, highly soluble compounds like short-chain alcohols compared with sparingly-soluble nonpolar compounds, but the direction is the same.

(Soil/sediment organic-carbon content is the third factor named in the question and governs the water–soil partitioning coefficient $K_d = f_{oc}K_{oc}$ rather than the water–air split; it is mentioned here only to distinguish it from the two air–water factors selected above.)

(iii) The Chick-Watson Law of Disinfection

The Chick-Watson law is the standard rate expression for microbial inactivation by a chemical disinfectant. It states that the rate of inactivation (die-off) of microorganisms is first order in the concentration of surviving organisms and depends on the disinfectant concentration raised to a coefficient of dilution, giving the integrated form

$$\dfrac{N}{N_0} = \exp\!\left(-k^{\prime}\,C^{\,n}\,t\right),$$

where $N_0$ is the initial number (or concentration) of viable organisms, $N$ is the number surviving at time $t$, $C$ is the disinfectant concentration (assumed constant over the contact time), $t$ is the contact (exposure) time, $k^{\prime}$ is the coefficient of specific lethality (a measure of the disinfectant’s potency against the particular organism), and $n$ is the coefficient of dilution, which reflects how sensitive the kill rate is to disinfectant concentration versus contact time ($n=1$ for the simplest, and most commonly assumed, case, giving the familiar $Ct$ (concentration×time) product used in regulatory disinfection credit calculations).

A key underlying assumption of the Chick-Watson law is that the microbial population is homogeneous with respect to susceptibility — every organism in the population is assumed to have identical resistance to the disinfectant, so a single rate constant $k^{\prime}$ describes the whole population’s die-off. In practice, real populations contain sub-populations with a range of resistances (e.g., spores versus vegetative cells, or organisms embedded in flocs/particles that shield them from contact), which is why real disinfection curves commonly show an initial rapid kill followed by a slower “tailing” phase that the simple Chick-Watson exponential does not capture.

Concept & Theory

Topic. Steady-state mass balance on a well-mixed water body, environmental controls on Henry’s-Law air–water partitioning, and first-order disinfection kinetics.

Key relations. Water balance $Q_i = Q_o + Q_e$; steady-state CSTR pollutant balance $0 = Q_iC_i - Q_oC_o - KVC_o \Rightarrow C_o = Q_iC_i/(Q_o+KV)$; Henry’s Law $p=K_H\,C$ (neutral species only) with pH-driven ionization and salinity-driven salting-out as the two governing environmental controls illustrated here; Chick-Watson $N/N_0 = \exp(-k^{\prime}C^n t)$.

Why this works. All three sub-parts reduce to a competition between a supply/removal rate and a governing coefficient: the lake balance weighs hydraulic flushing against first-order decay, Henry’s Law partitioning weighs the neutral (volatile) fraction against ionization or salting-out effects, and Chick-Watson weighs disinfectant potency and contact time against the population’s (assumed uniform) resistance.

Common pitfall. Forgetting to close the water balance before writing the pollutant balance (missing that $Q_o$ can be zero, or even negative if evaporation exceeds inflow, which would signal an error in the given data); applying Henry’s Law to the ionized form of a weak acid/base; and assuming the Chick-Watson $Ct$ product transfers unchanged to a different concentration than the one at which $k^{\prime}$ was measured, when in fact $k^{\prime}$ (for $n\ne1$) is concentration-dependent.

Source. Davis & Cornwell, Introduction to Environmental Engineering, mass-balance and disinfection-kinetics chapters; Sawyer, McCarty & Parkin, Chemistry for Environmental Engineering and Science, Henry’s Law and Setschenow-equation treatment.

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

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

(i) Steady-State Phosphate Mass Balance on the Lake

(ii) Environmental Factors Affecting Air–Water Partitioning of Polar Chemicals

(iii) The Chick-Watson Law of Disinfection