Conduit partly full with C = K R^n

Given data

$d=0.9$ m, $L=3600$ m, bed slope $S_0=1/1500$, normal depth $y=0.6$ m, $Q=0.322$ m³/s, $C=KR^{1/6}$.

(i) Value of $K$ (partly full)

Geometry for $y/d=0.6/0.9=0.667$. Angle subtended at the centre by the wetted arc:

$$\cos\frac\theta2=1-\frac{2y}{d}=1-\frac{2\times0.6}{0.9}=-0.3333\;\Rightarrow\;\frac\theta2=109.47^\circ,\ \theta=218.94^\circ=3.8213\ \text{rad}$$ $$A=\frac{d^2}{8}(\theta-\sin\theta)=\frac{0.81}{8}(3.8213-\sin218.94^\circ)=0.4505\ \text{m}^2$$ $$P=\frac{d\theta}{2}=\frac{0.9\times3.8213}{2}=1.7196\ \text{m},\qquad R=\frac AP=0.2620\ \text{m}$$

Chezy's equation $Q=CA\sqrt{RS_0}$ gives

$$C=\frac{Q}{A\sqrt{RS_0}}=\frac{0.322}{0.4505\sqrt{0.2620/1500}}=54.08\ \text{m}^{1/2}/\text{s}$$ $$K=\frac{C}{R^{1/6}}=\frac{54.08}{0.2620^{1/6}}=67.60$$

(ii) Discharge when flowing full

Full pipe: $A_f=\dfrac\pi4(0.9)^2=0.6362$ m², $R_f=\dfrac d4=0.225$ m.

$$C_f=KR_f^{1/6}=67.60\times0.225^{1/6}=52.72$$

When full and the reservoirs differ by 4.5 m (losses at entry and exit neglected), the energy (friction) slope is

$$S_f=\frac{h_f}{L}=\frac{4.5}{3600}=0.00125$$ $$Q_f=C_fA_f\sqrt{R_fS_f}=52.72\times0.6362\times\sqrt{0.225\times0.00125}=0.5625\ \text{m}^3/\text{s}$$

Answer: (i) $K=67.6$ (with $C=54.1$ for the partly full flow); (ii) $Q_{full}=0.562$ m³/s.