Correction factors for given velocity profile

Reading of the figure (assumed): flow depth $D = 2.4\ \text{m}$. The top 0.4 m moves at a uniform $0.4\ \text{m/s}$. Below it the velocity increases linearly with depth from $0.4\ \text{m/s}$ to $1.4\ \text{m/s}$ at the bed (over the remaining 2.0 m). Take unit width.

Let $s$ be the depth below the surface:

• $0 \le s \le 0.4$: $u = 0.4$
• $0.4 \le s \le 2.4$: $u = 0.4 + 0.5\,(s - 0.4)$, so $ds = du/0.5$

Mean velocity (area under the profile divided by the depth)

$$\begin{aligned} \int u\,ds &= 0.4(0.4) + \frac{0.4 + 1.4}{2}(2.0) = 0.16 + 1.80 = 1.96 \\ V &= \frac{1.96}{2.4} = 0.8167\ \text{m/s} \end{aligned}$$

Integrals of $u^2$ and $u^3$

$$\int u^2 ds = 0.4^2(0.4) + \frac{1.4^3 - 0.4^3}{3 \times 0.5} = 0.064 + 1.7867 = 1.8507$$ $$\int u^3 ds = 0.4^3(0.4) + \frac{1.4^4 - 0.4^4}{4 \times 0.5} = 0.0256 + 1.9080 = 1.9336$$

Correction factors

$$\begin{aligned} \alpha &= \frac{\int u^3\,ds}{V^3 D} = \frac{1.9336}{0.8167^3 \times 2.4} = \frac{1.9336}{1.3073} = 1.479 \\ \beta &= \frac{\int u^2\,ds}{V^2 D} = \frac{1.8507}{0.8167^2 \times 2.4} = \frac{1.8507}{1.6007} = 1.156 \end{aligned}$$

Answer: Energy correction factor $\alpha \approx 1.48$; momentum correction factor $\beta \approx 1.16$.