1. **State the problem:** We want to find the pipe diameter needed to carry a discharge \(Q = 400\, \text{l/s} = 0.4\, \text{m}^3/\text{s}\) with an upstream head of at least 10 m over a 15 m pipe length, considering the roughness coefficient \(n = 0.014\). Then determine the real head upstream for that diameter. 2. **Known: ** - Discharge: \(Q = 0.4\, \text{m}^3/\text{s}\) - Upstream head \(H_1 \geq 10\, \text{m}\) - Downstream head \(H_2 = 5\, \text{m}\) - Pipe length \(L = 15\, \text{m}\) - Roughness \(n = 0.014\, \text{s}/\text{m}^{1/3}\) 3. **Apply Manning equation for discharge in a circular pipe:** \[ Q = \frac{1}{n} A R^{2/3} S^{1/2} \] where - \(A = \frac{\pi D^2}{4}\) is cross-sectional area, - \(R = \frac{A}{P} = \frac{D}{4}\) is hydraulic radius for full pipe (since wetted perimeter \(P = \pi D\)), - \(S\) is the slope of the energy line (head loss per length): \(S = \frac{H_1 - H_2}{L} = \frac{10-5}{15} = \frac{5}{15} = \frac{1}{3} \). 4. **Substitute \(A\) and \(R\) into Manning:** \[ Q = \frac{1}{n} \times \frac{\pi D^2}{4} \times \left(\frac{D}{4} \right)^{2/3} \times \left(\frac{1}{3} \right)^{1/2} \] Simplify: \[ Q = \frac{\pi}{4n} D^{2} \times D^{2/3} 4^{-2/3} \times \frac{1}{\sqrt{3}} = \frac{\pi}{4n \sqrt{3}} D^{(2 + 2/3)} 4^{-2/3} \] \[ Q = \frac{\pi}{4n \sqrt{3}} D^{\frac{8}{3}} \times 4^{-\frac{2}{3}} \] Calculate \(4^{-2/3}\): \[ 4^{-2/3} = (2^2)^{-2/3} = 2^{-4/3} = \frac{1}{2^{4/3}} \] 5. **Rearranged to solve for \(D\):** \[ D^{8/3} = Q \times \frac{4 n \sqrt{3} \times 2^{4/3}}{\pi} \] 6. **Calculate numeric value:** \[ \text{Constants: } n=0.014, Q=0.4 \] Calculate \[ C = \frac{4 \times 0.014 \times \sqrt{3} \times 2^{4/3}}{\pi} \] \[ \sqrt{3} \approx 1.732, \quad 2^{4/3} = 2^{1 + 1/3} = 2 \times 2^{1/3} \approx 2 \times 1.26 = 2.52 \] Calculate numerator: \[ 4 \times 0.014 = 0.056 \] \[ 0.056 \times 1.732 = 0.097 \quad (approx.) \] \[ 0.097 \times 2.52 = 0.244 \] Divide by \(\pi \approx 3.1416\): \[ C = \frac{0.244}{3.1416} \approx 0.078 \] Then \[ D^{8/3} = Q / C = 0.4 / 0.078 = 5.13 \] 7. **Solve for \(D\):** \[ D = (5.13)^{3/8} \] Calculate exponent: \[ \frac{3}{8} = 0.375 \] \[ D = 5.13^{0.375} \approx e^{0.375 \times \ln(5.13)} \] \[ \ln(5.13) \approx 1.637 \] \[ D = e^{0.375 \times 1.637} = e^{0.6139} \approx 1.847\, \text{m} \] 8. **Answer part (a):** The required pipe diameter is approximately \(1.85\, \text{meters}\). 9. **Part (b) Determine the real upstream head:** We check the head using the diameter found by Manning formula to see if upstream head is indeed 10 m or needs adjustment. Rearranging Manning formula for \(H_1\): \[ S = \frac{H_1 - H_2}{L} = \left( \frac{n Q}{A R^{2/3}} \right)^2 \] Calculate \(A\) and \(R\) with \(D=1.847\): \[ A = \frac{\pi (1.847)^2}{4} \approx \frac{3.1416 \times 3.412}{4} = 2.68\, m^2 \] \[ R = \frac{D}{4} = \frac{1.847}{4} = 0.462\, m \] Calculate slope \(S\): \[ S = \left( \frac{0.014 \times 0.4}{2.68 \times (0.462)^{2/3}} \right)^2 \] Calculate \(0.462^{2/3}\): \[ 0.462^{2/3} = e^{\frac{2}{3} \ln(0.462)} = e^{\frac{2}{3} \times (-0.772)} = e^{-0.515} = 0.597 \] Compute numerator: \[ 0.014 \times 0.4 = 0.0056 \] Denominator: \[ 2.68 \times 0.597 = 1.599 \] Slope: \[ S = (0.0056/1.599)^2 = (0.0035)^2 = 1.23 \times 10^{-5} \] Compute head loss \(H_1 - H_2 = S \times L\): \[ = 1.23 \times 10^{-5} \times 15 = 0.0001845\, m \] Because this is very small compared to 5 m assumed earlier, actual head difference is negligible, meaning with diameter of 1.85 m the upstream head could be as low as approximately \(5 + 0.00018 \approx 5.00018\, m\). 10. **Interpretation:** The choice of 1.85 m diameter is larger than necessary for a 5 m head loss but not for 10 m; thus, to maintain a minimum head of 10 m upstream, a smaller diameter pipe or different flow conditions may be investigated. **Final answers:** - a) Required diameter \(\approx 1.85\, \text{m}\) for discharge 400 l/s with upstream head 10 m and downstream 5 m. - b) With that diameter, actual upstream head will be just slightly above 5 m, so 10 m upstream head is more than sufficient.