Balancing depth with service bank and free board

Given: $B = 8$ m, full supply depth $D = 1.5$ m, cutting slope $n_1 = 1$ (1:1), banking slope $n_2 = 1.5$ (1.5:1), top width of ordinary bank $b_1 = 1.8$ m, top width of service bank $b_2 = 5.0$ m, free board $F = 0.6$ m.

Set-up. Let $d$ = depth of the bed below ground level (balancing depth). The water surface is $D - d$ above ground, and the top of the bank is at $D + F - d$ above ground:

$$h = D + F - d = 2.1 - d$$

Cutting area: $A_c = (8 + d)\,d$

Filling area (one ordinary bank and one service bank, slope 1.5:1 on both faces):

$$A_f = h(b_1 + 1.5h) + h(b_2 + 1.5h) = h(6.8 + 3h)$$

Equate with $h = 2.1 - d$:

$$\begin{aligned} d^2 + 8d &= (2.1-d)(6.8 + 6.3 - 3d)\\ d^2 + 8d &= 3d^2 - 19.4d + 27.51\\ 2d^2 - 27.4d + 27.51 &= 0 \end{aligned}$$ $$d = \frac{27.4 - \sqrt{27.4^2 - 4\times2\times27.51}}{4} = \frac{27.4-\sqrt{530.68}}{4} = 1.091\ \text{m}$$

(The larger root, $d>2.1$ m, is rejected because it gives a negative bank height.)

Check: $h = 1.009$ m; $A_c = 9.917$ m$^2$ and $A_f = 9.917$ m$^2$.

Answer: balancing depth (bed below ground level) = 1.09 m; bank height above ground = 1.01 m.