Let's break down and solve each part step-by-step. --- ## **Given Data** - **Flow rate, \( Q \):** 100 gal/min (in both directions) - **Frictional head loss, \( h_f \):** 30 ft - **Reservoir elevations:** \( Z_1 = 150 \) ft (upper), \( Z_2 = 20 \) ft (lower) - **Duration (day/night):** 12 h each - **Water temperature:** 20°C (density ≈ 100 kg/m³) - **Pump-turbine system** --- ## **Step 1: Unit Conversions** ### **Convert Flow Rate to m³/s** \[ 1\, \text{gal} = 3.785 \times 10^{-3}\, \text{m}^3 \] \[ Q = 100\, \text{gal/min} \times 3.785 \times 10^{-3}\, \text{m}^3/\text{gal} = 3.785\, \text{m}^3/\text{min} \] \[ Q = \frac{3.785}{60}\, \text{m}^3/\text{s} \approx .0631\, \text{m}^3/\text{s} \] ### **Convert Head and Loss to meters** \[ 1\, \text{ft} = .3048\, \text{m} \] \[ \Delta z = Z_1 - Z_2 = 150 - 20 = 130\, \text{ft} \times .3048 = 39.624\, \text{m} \] \[ h_f = 30\, \text{ft} \times .3048 = 9.144\, \text{m} \] --- ## **Step 2: Power Calculations** ### **(a) Power extracted by the turbine during the daytime** **Effective head available to the turbine:** \[ H = \Delta z - h_f = 39.624 - 9.144 = 30.48\, \text{m} \] **Power extracted (turbine):** \[ P_{turbine} = \rho g Q H \] Where: - \( \rho = 100\, \text{kg/m}^3 \) - \( g = 9.81\, \text{m/s}^2 \) - \( Q = .0631\, \text{m}^3/\text{s} \) - \( H = 30.48\, \text{m} \) \[ P_{turbine} = 100 \times 9.81 \times .0631 \times 30.48 \] \[ P_{turbine} = 100 \times 9.81 \times 1.924 \] \[ P_{turbine} = 100 \times 18.887 \] \[ P_{turbine} \approx 18,887\, \text{W} = \boxed{18.9\, \text{kW}} \] --- ### **(b) Power delivered by the pump at night** **Total head pump must overcome:** \[ H_{pump} = \Delta z + h_f = 39.624 + 9.144 = 48.768\, \text{m} \] \[ P_{pump} = \rho g Q H_{pump} \] \[ P_{pump} = 100 \times 9.81 \times .0631 \times 48.768 \] \[ P_{pump} = 100 \times 9.81 \times 3.079 \] \[ P_{pump} = 100 \times 30.201 \] \[ P_{pump} \approx 30,201\, \text{W} = \boxed{30.2\, \text{kW}} \] --- ### **(c) Net Power Produced Over 24 Hours** **Time for each phase:** 12 h **Energy extracted:** \[ E_{turbine} = P_{turbine} \times 12\, \text{h} = 18.9\, \text{kW} \times 12 = 226.8\, \text{kWh} \] **Energy consumed by pump:** \[ E_{pump} = P_{pump} \times 12\, \text{h} = 30.2\, \text{kW} \times 12 = 362.4\, \text{kWh} \] **Net energy:** \[ E_{net} = E_{turbine} - E_{pump} = 226.8 - 362.4 = \boxed{-135.6\, \text{kWh}} \] **Net power is negative.** This means more energy is consumed by the pump than is produced by the turbine. --- ### **(d) Why is this type of system practical?** **Pumped storage** is practical because it allows storing energy during periods of low demand (cheap electricity at night) and generating electricity during periods of high demand (expensive electricity during the day), despite round-trip losses. It helps stabilize the grid and balance supply and demand. --- ## **Summary Table** | Part | Value | |------|------------------------| | a | 18.9 kW (turbine) | | b | 30.2 kW (pump) | | c | -135.6 kWh (net, 24h) | | d | Stores energy for peak demand, grid stability | --- ### **Let me know if you want to see efficiency calculations or more details!**