Sizing a Pump: A Practical Problem

Let's size a pump to move water from a reservoir to an elevated tank. This is a classic problem you'll face on the job and in interviews.

The scenario

• Flow rate required: 200 gallons per minute (gpm)
• Static lift (elevation change): 60 ft
• Total friction losses in piping: 15 ft of head
• Fluid: water at room temperature

Step 1: Total Dynamic Head (TDH)

The pump must overcome both the elevation change and friction:

TDH = static lift + friction losses
TDH = 60 ft + 15 ft = 75 ft

Step 2: Hydraulic power

Convert flow and head into required hydraulic power:

P (hp) = (Q × H × SG) / 3960
P = (200 × 75 × 1.0) / 3960 ≈ 3.79 hp

Where Q is flow in gpm, H is head in ft, SG is specific gravity, and 3960 is the unit constant for US customary units.

Step 3: Account for efficiency

Real pumps aren't perfect. Assuming 70% pump efficiency:

Brake horsepower = 3.79 / 0.70 ≈ 5.4 hp

You'd select a motor at the next standard size up (likely 7.5 hp) to provide margin.

Key lessons

• Always separate static head from friction head.
• Never forget efficiency - it can add 30-40% to your power requirement.
• Round up to standard motor sizes.

Practice more with the Unit Converter when working across unit systems.