A three-cylinder single acting reciprocating pump has a plunger diameter of 100 mm and a stroke length of 300 mm . The pump is installed at a level 5 m above the surface of water reservoir and with a vertical height of 25 m above the pump. Calculate the power required to drive the pump at \( 40 \mathrm{r} / \mathrm{min} \) if the efficiency of the motor is \( 80 \% \).

**Step 1. Calculate the volume displaced by one plunger stroke** The volume displaced by one stroke in one cylinder is given by \[ V = \frac{\pi}{4} D^2 L, \] where \(D = 100 \, \text{mm} = 0.1 \, \text{m}\) (plunger diameter) and \(L = 300 \, \text{mm} = 0.3 \, \text{m}\) (stroke length). Thus, \[ V = \frac{\pi}{4} (0.1)^2 (0.3) = \frac{\pi}{4} (0.01)(0.3) = \frac{\pi}{4} \times 0.003 \approx 0.002356 \, \text{m}^3. \] --- **Step 2. Determine the total flow rate** The pump is a three-cylinder single-acting pump operating at \(40 \, \text{r/min}\). Therefore, the total number of strokes per minute is \[ \text{Total Strokes per Minute} = 3 \times 40 = 120. \] The flow rate per minute is \[ Q_{\text{min}} = 120 \times 0.002356 \, \text{m}^3 \approx 0.2827 \, \text{m}^3/\text{min}. \] Convert this to cubic meters per second: \[ Q = \frac{0.2827}{60} \approx 0.004712 \, \text{m}^3/\text{s}. \] --- **Step 3. Calculate the total head against which the pump works** The pump is installed \(5 \, \text{m}\) above the water reservoir, and the discharge point is \(25 \, \text{m}\) above the pump. Thus, the total vertical lift (head) is \[ H = 5 + 25 = 30 \, \text{m}. \] --- **Step 4. Find the hydraulic power required** The hydraulic power needed to lift water is given by \[ P_h = \rho g Q H, \] where \(\rho = 1000 \, \text{kg/m}^3\) (density of water), \(g = 9.81 \, \text{m/s}^2\) (acceleration due to gravity), \(Q \approx 0.004712 \, \text{m}^3/\text{s}\), and \(H = 30 \, \text{m}\). Substitute the values: \[ P_h = 1000 \times 9.81 \times 0.004712 \times 30. \] First, calculate \(9.81 \times 30 = 294.3\). Then, \[ P_h = 1000 \times 0.004712 \times 294.3 \approx 1387 \, \text{W}. \] --- **Step 5. Account for the motor efficiency** Given that the motor efficiency is \(80\% = 0.8\), the actual input power required is \[ P_{\text{in}} = \frac{P_h}{\eta} = \frac{1387}{0.8} \approx 1734 \, \text{W}. \] Thus, the required power to drive the pump is approximately \[ \boxed{1.73 \, \text{kW}}. \]