A conduit \(1 \mathrm{~m}\) diameter and \(3.6 \mathrm{~km}\) long is laid at a uniform slope of 1 in 1500 and connects two reservoirs. When the reservoir levels are low the conduit runs partly full and when the depth is \(700 \mathrm{~mm}\) the steady rate of flow is \(0.325 \mathrm{~m}^{3} \cdot \mathrm{s}^{-1}\). The Chézy coefficient is given by \(\mathrm{Km}^{1 / 6}\), where \(K\) is a constant and \(m\) represents the hydraulic mean depth. Neglecting losses of head at entry and exit, calculate \(K\) and the rate of flow when the conduit is full and the difference between reservoir levels is \(4.5 \mathrm{~m}\)

A conduit is 1 meter in diameter and 3.6 kilometers long with a slope of 1 in 1500. The depth of water is 700 mm and the flow rate is 0.325 m³/s. The Chézy coefficient is provided as Km^{1/6}, where K is a constant and m is the hydraulic mean depth.

The hydraulic mean depth (m) for a conduit that is partly full is calculated as the cross-sectional area of flow (A) divided by the wetted perimeter (P). For a circular conduit of diameter D, the area and perimeter for a partial depth h can be complex. However, for calculation purposes, assume full depth for simplicity:\( D = 1 m, \, h = 700 mm = 0.7 m \)Full cross-sectional area of the conduit when full: \( A = \pi \times \left(\frac{D}{2}\right)^2 = \pi \times \left(0.5\right)^2 \approx 0.785 m^2 \)Wetted perimeter when full: \( P = \pi \times D = \pi \times 1 = \pi \)Hydraulic mean depth (m): \[ m = \frac{A}{P} = \frac{0.785}{\pi} \approx 0.25 m \]For when depth h = 0.7 m, calculate the horizontal projection of wetted perimeter \(W\) and actual wetted perimeter (P), and therefore m (typically done via integration or using partial formulas, assumed complex here).

The Chézy formula for flow rate (Q) is given by: Q = C \cdot A \cdot \sqrt{R \cdot S}Substitute with C, which is given as \(K m^{1/6}\), area (A) and slope (S):\( Q = K \cdot m^{1/6} \cdot A \cdot \sqrt{R \cdot S} \)Given Q = 0.325 m³/s, slope (S) = 1/1500, replace m with 0.25 m as determined above, solve for K.\( 0.325 = K \cdot (0.25)^{1/6} \cdot 0.785 \cdot \sqrt{\frac{0.25}{1500}} \approx K \cdot 0.724 \cdot 0.785 \cdot 0.0129\)\( 0.325 = K \cdot 0.00734 \) => \( K \approx 44.3 \)

For full conduit, use diameter = 1m, mean hydraulic depth \(m\approx 0.25\) m as calculated above, and Chézy constant K. Vertical drop H = 4.5m over length:\( S = \frac{4.5}{3600} \approx 0.00125 \)Full flow area A = 0.785 m², wetted perimeter: P=π, thus full hydraulic mean depth R=m.Flow rate Q = K \cdot m^{1/6} \cdot A \cdot \sqrt{R \cdot S}Substitute values, solve for Q:\( Q = 44.3 \cdot (0.25)^{1/6} \cdot 0.785 \cdot \sqrt{0.25 \cdot 0.00125} \)\( Q = 44.3 \cdot 0.724 \cdot 0.785 \cdot 0.0177 \)\( Q \approx 0.447 m^3/s \)