Let's break down the problem step by step: **Given:** - Average speed of water, \( V = 4 \, \text{m/s} \) - Distance to tank, \( L = 6 \, \text{km} = 6000 \, \text{m} \) - Pipe diameter, \( d = 0.8 \, \text{m} \) - Friction factor, \( f = 0.01 \) - Constant head to be maintained, \( h = 10 \, \text{m} \) **Required:** Absolute value of the pressure recorded by a pressure gauge at the exit of the pump (in bar). --- ### Step 1: Calculate Head Loss due to Friction (\( h_f \)) The Darcy-Weisbach equation: \[ h_f = f \frac{L}{D} \frac{V^2}{2g} \] - \( f = 0.01 \) - \( L = 6000 \) - \( D = 0.8 \) - \( V = 4 \) - \( g = 9.81 \) \[ h_f = 0.01 \times \frac{6000}{0.8} \times \frac{4^2}{2 \times 9.81} \] \[ h_f = 0.01 \times 7500 \times \frac{16}{19.62} \] \[ h_f = 75 \times 0.8155 = 61.16 \, \text{m} \] --- ### Step 2: Total Head Required Total head = Head to be maintained + Head loss due to friction: \[ H_{\text{total}} = 10\, \text{m} + 61.16\, \text{m} = 71.16\, \text{m} \] --- ### Step 3: Convert Head to Pressure \[ P = \rho g H \] - \( \rho \) (density of water) = \( 1000 \, \text{kg/m}^3 \) - \( g = 9.81 \, \text{m/s}^2 \) - \( H = 71.16 \) \[ P = 1000 \times 9.81 \times 71.16 = 698,089.6 \, \text{Pa} \] Convert to bar: \[ 1 \, \text{bar} = 100,000 \, \text{Pa} \] \[ P_{\text{bar}} = \frac{698,089.6}{100,000} = 6.98 \, \text{bar} \] --- ### Step 4: Choose the Closest Answer - 10 bar - 8 bar - 9 bar - **7 bar** (closest to 6.98 bar) --- ## **Final Answer** **7 bar** is the closest answer. So, **option 4: 7 bar** is correct.