Question

A pump with an efficiency of 80% is used to draw groundwater from a well for irrigating a flat field of area 108 hectares. The base period and delta for paddy crop on this field are 120 days and 144 cm, respectively. Water application efficiency in the field is 80%. The lowest level of water in the well is 10 m below the ground. The minimum required horse power (h.p.) of the pump is _________ (round off to two decimal places).

Hint:

The minimum horsepower required for a pump depends on the volume of water, head, efficiency, and the time over which the irrigation is to be done.

Answer 1

Correct Answer: 30

To find the minimum required horsepower, we first calculate the volume of water required using the formula: \[ \text{Volume of water} = \text{Area} \times \text{Delta} \times \frac{\text{Base Period}}{\text{Time period}} \] where: - Area = 108 hectares = 108 \times 10,000 m\(^2\), - Delta = 144 cm = 1.44 m. Thus, the volume of water required for irrigation is: \[ \text{Volume} = 108 \times 10^4 \times 1.44 \times \frac{120}{365} = 489,856.16 \, \text{m}^3. \] Next, the work done by the pump is given by: \[ \text{Work} = \text{Volume of water} \times \text{Head} \times \rho \times g, \] where:
- \( \rho = 1000 \, \text{kg/m}^3 \) (density of water),
- \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity),
- Head = 10 m (elevation of water level).
Now, the required power is calculated as: \[ \text{Power} = \frac{\text{Work}}{\text{Time}} = \frac{489,856.16 \times 10 \times 1000 \times 9.81}{120 \times 3600} = 31.23 \, \text{kW}. \] Since the pump efficiency is 80%, the input power is: \[ \text{Input Power} = \frac{31.23}{0.8} = 39.03 \, \text{kW}. \] Convert this to horsepower: \[ \text{Horse Power} = \frac{39.03 \times 1000}{746} = 52.3 \, \text{h.p.}. \] Thus, the minimum required horse power is: \[ \boxed{32.0 \, \text{h.p.}}. \]