Sizing a pump for a borehole supply
A farm must raise water from a borehole to a storage tank. The water surface in the borehole is 38 m below ground level. The tank stands 12 m above ground, and water is delivered into the top of the tank. The required flow rate is 15 litres per minute. A 12 V motor drawing 18 A drives the pump, and the manufacturer states the pump unit is 60 % efficient. The farmer plans to fit a tap at the bottom of the tank, where the water will be 2.0 m deep, and to use a hose rated at 50 kPa. A helper says the 12 V, 18 A supply is more than enough because '216 W is a lot of power'. The installer warns that pipe friction is equivalent to an extra 5 m of lift and that the motor current will rise if the pipe is narrowed. Neither has calculated the actual power needed.
1. Calculate the useful power needed to lift the water, with and without the friction allowance.
Read the worked answer
Convert the flow rate: 15 litres per minute = 15 kg per minute = 15/60 = 0.25 kg/s.
Lift = 38 m from the water surface to ground level + 12 m from ground to the tank top = 50 m. Water delivered into the top of the tank must reach that height, so 50 m is the correct head.
Useful power = (mass per second) × g × h = 0.25 × 10 × 50 = 125 W.
With the 5 m friction head the effective lift is 55 m: 0.25 × 10 × 55 = 137.5 W.
Using g = 9.8 N/kg instead of 10 N/kg gives 122.5 W and 134.8 W, about 2 % lower. The friction allowance is not optional: it is real energy that leaves the system as heat in the pipe, and it must be supplied by the motor.
2. Compare the available electrical power with the power needed and judge whether the 12 V, 18 A supply is adequate.
Read the worked answer
Electrical input available = V × I = 12 × 18 = 216 W.
At 60 % efficiency the mechanical power delivered to the water is 0.60 × 216 = 130 W.
The calculation to compare against is 125 W without friction and 137.5 W with the 5 m friction head. So the supply is just sufficient in the ideal case (130 W > 125 W, a margin of only 4 %), but insufficient once pipe friction is included (137.5 W needed against 130 W available).
The helper's reasoning is wrong because 216 W is the input, not the output: 40 % of it (about 86 W) is lost in the motor and pump. If the farm tries to run at 15 L/min as installed, the motor will draw more than 18 A, run hot and possibly overheat, or the flow will simply be below 15 L/min.
Practical fixes: use a larger-bore delivery pipe to cut the friction head, reduce the flow rate, raise the supply voltage/current, or choose a more efficient pump. The correct specification is an input of about 137.5/0.60 ≈ 229 W, not 216 W.
3. Calculate the water pressure at the tap and judge whether the 50 kPa hose is suitable.
Read the worked answer
The pressure at the tap is due to the depth of water above it: p = ρgh = 1000 × 10 × 2.0 = 2.0 × 10⁴ Pa = 20 kPa.
This is the gauge pressure (above atmospheric), which is the quantity a hose rating refers to, so 20 kPa is the number to compare with 50 kPa. The hose rating is 2.5 times the static pressure, so it is suitable for the static head.
Two cautions. First, if the tap is closed suddenly, the moving water is stopped and a surge (water hammer) can briefly produce a pressure well above 20 kPa; the 50 kPa rating gives some margin for that. Second, the pump delivers into the tank, so the tap does not see pump pressure directly - but if the pump were ever connected straight to the hose, the pressure would be far higher and the 50 kPa hose would be unsafe. A 25 kPa hose would be marginal even for the static head.