# Problem Overview: Total Head of a Francis Turbine Given Data: - Head to center of spiral casing, \( H_{c} = 3.8\, \text{m} \) - Inlet velocity of water, \( V_{in} = 5\, \text{m/s} \) - Discharge, \( Q = 2.1\, \text{m}^3/\text{s} \) - Hydraulic efficiency, \( \eta_h = 87\% = .87 \) - Overall efficiency, \( \eta_o = 84\% = .84 \) - Velocities at inlet and exit of draft tube are 1 m below the center line - Tailrace level is 3 m from the top of the draft tube - Neglect whirl and leakage losses --- # Step 1: Determine the Power Developed by the Turbine ### Hydraulic Power (\( P_h \)) \[ P_h = \rho g Q H_{h} \] Where: - \( \rho = 100\, \text{kg/m}^3 \) (density of water) - \( g = 9.81\, \text{m/s}^2 \) - \( Q = 2.1\, \text{m}^3/\text{s} \) - \( H_{h} \) = Total head (unknown, to be found) --- ### Step 2: Find the Hydraulic Power Using Hydraulic Efficiency Hydraulic efficiency relates the power developed at the turbine to the flow energy: \[ \eta_h = \frac{\text{Power at the turbine shaft}}{\text{Flow energy input}} \] Flow energy input per second: \[ \text{Flow energy} = \frac{1}{2} \rho V_{in}^2 Q \] Calculate flow energy: \[ E_{flow} = \frac{1}{2} \times 100 \times (5)^2 \times 2.1 = .5 \times 100 \times 25 \times 2.1 = .5 \times 100 \times 52.5 = 26250\, \text{W} \] Hydraulic power: \[ P_h = \eta_h \times E_{flow} = .87 \times 26250 \approx 22838\, \text{W} \] --- ### Step 3: Find the Total Head \( H \) Total head: \[ H = \frac{P_h}{\rho g Q} \] Calculate denominator: \[ \rho g Q = 100 \times 9.81 \times 2.1 = 100 \times 20.6 = 20600\, \text{W/m} \] Calculate total head: \[ H = \frac{22838}{20600} \approx 1.11\, \text{m} \] **Note:** This is the effective head contributing to power, but we need to account for other head components. --- # Step 4: Find the Total Head \( H_{total} \) The total head includes: - Head to the center of the spiral casing: \( H_{c} = 3.8\, \text{m} \) - Velocity head at inlet: \( \frac{V_{in}^2}{2g} \) - Velocity head at draft tube exit - Tailrace level ### Velocity head at inlet: \[ h_{v1} = \frac{V_{in}^2}{2g} = \frac{25}{2 \times 9.81} \approx 1.275\, \text{m} \] ### Velocity head at draft tube exit: - Since velocities at inlet and exit are 1 m below the center line, the velocity heads are similar. --- ### Step 5: Calculate the Total Head Total head \( H_{total} \) is sum of: - Head to the center of spiral casing: 3.8 m - Velocity head at inlet: 1.275 m - Head loss due to velocity change in draft tube (assuming negligible whirl and leakage) - Head difference from tailrace level (3 m below top of draft tube) **Assuming the head at tailrace level is the reference, the total head is:** \[ H_{total} = H_{c} + \text{velocity head} + \text{additional head losses} \] Since velocities at inlet and exit are given, and the overall efficiency is 84%, the total head delivered to the turbine is: \[ H_{total} = \frac{P_{shaft}}{\rho g Q} \] From earlier, shaft power: \[ P_{shaft} = \eta_o \times P_{input} \] But since the earlier calculation for power was based on flow energy, and efficiencies are given, the total head is approximately: \[ H_{total} \approx 3.8\, \text{m} + 1.275\, \text{m} + \text{additional head losses} \] Considering the head losses are small or included in the efficiency, the final total head: \[ \boxed{ H_{total} \approx 3.8 + 1.275 \approx 5.075\, \text{m} } \] --- # **Final Answer:** **The total head on the turbine is approximately \(\boxed{5.1\, \text{m}}\).**