Question 5 of 7: Free water in aggregate, and uniform flow in a trapezoidal channel

Question 5: Free water in aggregate, and uniform flow in a trapezoidal channel

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) a wet aggregate sample weighing 310.0 N whose oven-dry weight is 280.0 N, for an aggregate whose absorption is 4.0 percent; (b) a trapezoidal concrete channel 6 m wide at the invert with side slopes of 1 vertical to 2 horizontal, on a longitudinal slope of 3 percent, running at a uniform depth of 3 m.

Find. (a) the free (surface) water in the wet sample, expressed as a percentage; (b) the discharge carried by the channel, expressed as a volume per day.

Approach. For (a), split the total water the sample carries into the part absorbed into the aggregate pores, which is fixed by the absorption and the oven-dry weight, and the remainder, which is free water on the particle surfaces. For (b), compute the section's area and wetted perimeter at the stated depth, form the hydraulic radius, apply Manning's equation for uniform flow, and convert to a daily volume.

1. Part (a) — total water in the sample. Everything the sample loses on oven drying is water, absorbed and free together: $$W_{\text{total}} = W_{\text{wet}} - W_{\text{dry}} = 310.0 - 280.0 = 30.0\ \text{N}$$
2. Absorbed water. Absorption is defined on the oven-dry weight and measures the water needed to bring the aggregate to the saturated surface-dry condition: $$W_{\text{abs}} = \frac{A}{100}\,W_{\text{dry}} = 0.040 \times 280.0 = 11.2\ \text{N}$$
3. Free (surface) water by difference. $$W_{\text{free}} = W_{\text{total}} - W_{\text{abs}} = 30.0 - 11.2 = \boxed{18.8\ \text{N}}$$ This is the water that will end up in the concrete mixture and must be deducted from the batch water; the absorbed 11.2 N is already inside the aggregate and takes no part in hydration or workability.
4. Express it as a percentage. Aggregate moisture is conventionally reported on the oven-dry weight, on which basis the total moisture content is $30.0/280.0 = 10.714$ percent and the free water is $$\text{free water} = \frac{W_{\text{free}}}{W_{\text{dry}}}\times 100 = \frac{18.8}{280.0}\times 100 = \boxed{6.71\ \text{percent}}$$ which is exactly the total moisture content less the absorption, $10.714 - 4.000 = 6.714$ percent, as it must be. The question asks for the free water “in the original wet sample”, so it is worth also giving the figure on the wet weight, $18.8/310.0 = 6.06$ percent; the two bases differ by 0.65 of a percentage point and only the oven-dry basis is additive with the absorption.
5. Part (b) — flow area. For a trapezoidal section of bottom width $b$ and side slope $z$ horizontal to 1 vertical, at depth $y$, $$A = (b + zy)\,y = (6 + 2 \times 3)(3) = 12 \times 3 = 36.0\ \text{m}^2$$ The top width is $T = b + 2zy = 6 + 12 = 18.0$ m.
6. Wetted perimeter and hydraulic radius. Each sloping side has length $y\sqrt{1+z^2}$, so $$P = b + 2y\sqrt{1 + z^2} = 6 + 2(3)\sqrt{5} = 6 + 13.416 = 19.416\ \text{m}$$ $$R_h = \frac{A}{P} = \frac{36.0}{19.416} = 1.8541\ \text{m}$$
7. Mean velocity from Manning's equation. With the flow uniform, the friction slope equals the bed slope, $S_f = S_0 = 0.03$: $$V = \frac{1}{n} R_h^{2/3} S_0^{1/2} = \frac{1}{0.013}(1.8541)^{2/3}(0.03)^{1/2}= 76.923 \times 1.5091 \times 0.17321 = 20.107\ \text{m/s}$$
8. Discharge. $$Q = AV = 36.0 \times 20.107 = \boxed{723.9\ \text{m}^3\text{/s}}$$
9. Convert to a daily volume. The question asks for cubic metres per day, so multiply by the 86 400 seconds in a day: $$\forall = Q \times 86\,400 = 723.86 \times 86\,400 = \boxed{6.254 \times 10^{7}\ \text{m}^3\text{/day}}$$ that is, about 62.5 million cubic metres per day.