Question 1 of 7: Contaminant Dilution, Reactor Kinetics & Disinfection

Question 1: Contaminant Dilution, Reactor Kinetics & Disinfection (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.

1(i) — Maximum dischargeable Zn²⁺ (6 marks)

Given. Stream flow above the outfall $Q_s=300\ \text{L/s}$ with upstream $C_s=0\ \text{mg/L}$; industrial discharge $Q_w=100\ \text{L/s}$; allowable in-stream concentration after complete mixing $C_{mix}=0.010\ \text{mg/L}$; steady state, conservative (Zn²⁺ is not consumed over the mixing reach).

Find. The maximum Zn²⁺ concentration $C_w$ permitted in the 100 L/s discharge.

Approach. Write a steady-state mass balance on Zn²⁺ at the point where the discharge fully mixes with the creek, then solve for the unknown effluent concentration.

1. Conservation of mass at the mixing point. Mass in equals mass out for a conservative substance: $$Q_s C_s + Q_w C_w = (Q_s + Q_w)\,C_{mix}$$
2. Isolate the effluent concentration. With $C_s=0$: $$C_w = \frac{(Q_s+Q_w)\,C_{mix} - Q_s C_s}{Q_w} = \frac{(300+100)(0.010) - 0}{100}$$
3. Evaluate. $$\boxed{C_w = 0.040\ \text{mg Zn}^{2+}/\text{L}}$$ The total downstream flow is 400 L/s, so the creek dilutes the effluent four-fold; the discharge may carry four times the in-stream limit.

A quick check confirms the limit is exactly met: $(300\times0 + 100\times0.040)/400 = 0.010\ \text{mg/L}$. This is the classic mixing-zone allocation — note it grants no margin for upstream loading, so a non-zero background would tighten $C_w$ proportionally.

1(ii) — Time for 80% ammonia conversion (6 marks)

Given. Batch (completely mixed) reactor, first-order decay of NH₃; $C_0=200\ \text{mg/L}$, $k=0.1\ \text{h}^{-1}$; target conversion $X=0.80$, so residual $C=(1-X)C_0=40\ \text{mg/L}$.

Find. The reaction time $t$ to reach 80% conversion.

Approach. For a batch reactor the first-order integrated law relates residual fraction to time directly.

1. First-order batch model. Mass balance $\dfrac{dC}{dt} = -kC$ integrates to $$C = C_0\,e^{-kt}.$$
2. Solve for time. Rearranging and inserting $C/C_0 = 0.20$: $$t = \frac{\ln(C_0/C)}{k} = \frac{\ln(200/40)}{0.1} = \frac{\ln 5}{0.1}.$$
3. Evaluate. $$\boxed{t = 16.1\ \text{h}}$$

Because the process is first order, the same 16.1 h would achieve 80% removal from any starting concentration — the time depends on the fractional conversion, not the absolute load. Full nitrification also consumes alkalinity and oxygen, which the aerobic reactor must supply.

1(iii) — Reading the disinfection curves (8 marks)

On this concentration–contact-time (C·t) chart each line is the locus giving 99% kill of one organism; a line lying higher / farther to the right demands a larger C·t product and therefore represents a more resistant organism.

• (a) Most resistant: the Poliomyelitis virus — its curve sits highest, requiring the greatest C·t.
• (b) Least resistant: E. coli — the lowest curve, inactivated at the smallest C·t.
• (c) Is E. coli a good indicator of viral inactivation? No. E. coli is a vegetative bacterium and the least resistant of the four; the enteric viruses (poliomyelitis, Coxsackie A5, adenovirus) all require substantially more C·t. A dose sufficient to give 99% kill of E. coli would leave the viruses under-treated, so absence of the indicator does not guarantee viral safety — it underestimates the disinfection duty. Design must be set by the most resistant pathogen, not the indicator.
• (d) Minimum contact time at HOCl = 0.1 mg/L for 99% kill of all microbes. The governing organism is the most resistant one, so the required time is read off the Poliomyelitis line at $c=0.1\ \text{mg/L}$, giving ≈ 12 minutes.