Sewer for 200 ha catchment at 0.6 depth

Step 1: Peak runoff (rational method)

$t_c=55$ min. No rainfall formula is given, so $i=\dfrac{760}{t+10}$ mm/hr is assumed:

$$\begin{aligned} i&=\frac{760}{55+10}=11.69\ \text{mm/hr}\\ Q&=\frac{CiA}{360}=\frac{0.45\times11.69\times200}{360}=2.923\ \text{m}^3/\text{s} \end{aligned}$$

Step 2: Hydraulic elements at $d/D=0.6$

For $d/D=0.6$: $\theta = 2\cos^{-1}(1-2d/D)=203.07^\circ$, $A/A_f=\dfrac{\theta-\sin\theta}{2\pi}=0.6265$, $R/R_f=1-\dfrac{\sin\theta}{\theta}=1.1106$, so $v/v_f=(R/R_f)^{2/3}=1.0724$ and $Q/Q_f=(A/A_f)(v/v_f)=0.6718$ (n taken constant).

Step 3: Diameter

Here the velocity at peak flow is fixed ($v=1.5$ m/s), so the flow area is

$$A_{p}=\frac{Q}{v}=\frac{2.923}{1.5}=1.949\ \text{m}^2$$ $$A_f=\frac{A_p}{A/A_f}=\frac{1.949}{0.6265}=3.111\ \text{m}^2\ \Rightarrow\ D=\sqrt{\frac{4A_f}{\pi}}=1.990\ \text{m}$$

Adopt $D=2.10$ m.

Step 4: Slope

$R=\dfrac{R}{R_f}\times\dfrac{D}{4}=1.1106\times\dfrac{1.990}{4}=0.5525$ m. From Manning's formula $v=\dfrac{1}{n}R^{2/3}S^{1/2}$:

$$S=\left(\frac{v\,n}{R^{2/3}}\right)^2=\left(\frac{1.5\times0.013}{0.5525^{2/3}}\right)^2=0.00084\approx\frac{1}{1192}$$

(For the adopted $D=2.10$ m, the slope needed for the same 1.5 m/s is $1$ in 1281.)

Answer: Diameter about 1.990 m (adopt 2.10 m), slope about 1 in 1192 (0.6 depth, v = 1.5 m/s, Q = 2.923 m³/s).