Question 2 of 5: Pulse-Tracer Analysis of a River — Tracer Quantity and Reach Volume

Question 2: Pulse-Tracer Analysis of a River — Tracer Quantity and Reach Volume (25 marks: a 18, b 7)

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. Volumetric flow $Q = 6000\ \text{m}^3/\text{s}$; a triangular response curve $C(t)$ at the dam with $C=0$ for $t<20$ d, a linear rise to $C_{max}=10^{-6}$ units/m$^3$ at $t=35$ d, and a linear fall to $C=0$ at $t=125$ d. The half-life exceeds 10 years, so radioactive decay over the $\sim100$-day passage is negligible (the tracer is conserved).

Find. (a) total units of tracer injected; (b) volume of the river reach.

Approach. The dam records the exit-age (pulse) response, so the quantity injected is the flow times the area under $C(t)$ (zeroth moment), and the mean residence time is the centroid of $C(t)$ (first moment); the reach volume follows from $V = Q\bar t$.

1. Conservation of the injected tracer (part a). Every unit dumped upstream must pass the dam (no decay), so $M = \displaystyle\int_0^\infty C\,Q\,dt = Q\!\int_0^\infty C\,dt = Q\,(\text{area under the }C\text{-}t\text{ curve}).$
2. Area of the triangular pulse. Base $= 125-20 = 105$ days, height $=10^{-6}$ units/m$^3$: $$\int C\,dt = \tfrac12 (105\ \text{d})(10^{-6}) = 52.5\times10^{-6}\ \tfrac{\text{units}\cdot\text{d}}{\text{m}^3}.$$
3. Convert to SI time and multiply by flow. With $1\ \text{d} = 86{,}400$ s, the area is $52.5\times10^{-6}\times86{,}400 = 4.536\ \text{units}\cdot\text{s}/\text{m}^3$, hence $$M = (6000)(4.536) = \boxed{2.72\times10^{4}\ \text{units}.}$$
4. Mean residence time (part b). For a pulse response, $\bar t = \dfrac{\int t\,C\,dt}{\int C\,dt}$, which is the time-centroid of the curve. The centroid of a triangle is the average of its three vertex times: $$\bar t = \frac{20 + 35 + 125}{3} = 60\ \text{days} = 5.184\times10^{6}\ \text{s}.$$
5. Reach volume. $$V = Q\,\bar t = (6000)(5.184\times10^{6}) = \boxed{3.11\times10^{10}\ \text{m}^3\ (\approx 31\ \text{km}^3).}$$ Check: mean velocity $=400{,}000/5.184\times10^{6} = 0.077$ m/s, cross-section $=Q/v = 7.8\times10^{4}$ m$^2$, and $V = 7.8\times10^{4}\times400{,}000 = 3.1\times10^{10}$ m$^3$ — consistent.