Calculating vs. Estimating Pipe Velocity

During our last Pump School session, one attendee asked how much error there would be if the velocity head and gauge elevation correction were not accounted for. While the most obvious answer perhaps would be “It depends,” we nevertheless went over a specific example with numbers to illustrate, as presented below. Such examples are numerous for relatively basic fluid transfer from one tank to another, or from a tank to a process. Let’s take a typical single-stage pump, operating as shown on a performance curve in Image 1.

Operating flow in this example is 70 gallons per minute (gpm) of water, with suction gauge reading 5 pounds per square inch gauge (psig), and discharge gauge reading 80 psig. The suction pipe is 1.5 inches, and the discharge pipe is 1 inch. The suction gauge is 1 foot above pump centerline. The discharge gauge is 6 feet above pump centerline. When doing a quick calculation in the field, it is not uncommon to disregard velocity head correction and gauge elevation correction. Usually this gets reasonably close to a “practical” answer, although there are cases when the error could be substantial. It is important to know when you can use a shortcut, and when to go the long way. For a quick calculation, all we need are gauge pressures: Head ~ (80-5) x 2.31 / 1.0 (specific gravity for water) = 173 feet But let’s also do it the long way: What is the flow velocity in a 1-inch pipe when pumping water through it at 70 gpm? velocity in a pipe = flow / area For simplicity, assume the net pipe flow opening is the same as the pipe nominal diameter. Discharge side: V_pipe = flow_(gpm) x 0.321 / area_(in2)(0.321 is a units conversion constant) A_pipe = π / 4 x 12 = 0.79 in² V_pipe = 70 gpm x 0.321 / 0.79 in² = 28.4 ft/sec Suction pipe area is 3.14 / 4 x 1.5² = 1.8 in² Suction line velocity is 70 x 0.321 / 1.8 = 12.4 ft/sec, and Discharge side velocity head = V_Pipe² / 2g = 28.4² / 64.4 = 12.5 ft (g = 32.2, gravitational constant) Velocity head at the suction pipe is 12.42 / 64.4 = 2.5 ft In absolute units, suction pressure is 5 + 14.7 = 19.7 psia (19.7 x 2.31 / 1.0 = 45.5 ft) In absolute units, discharge pressure is 80 + 14.7 = 94.7 psia (94.7 x 2.31 / 1.0 = 218.8 ft) Suction head (including pressure and velocity head components) = 45.5 + 2.5 = 48 ft Discharge head = 218.8 + 12.5 = 231 ft Gauge elevation difference = 6 – 1 = 5 ft Pump head (corrected by gauges elevations) = 231 – 48 + 5 = 188 ft And so, the error between an “on-the-fly” estimate and a more exact calculation is 188 / 173 = 9 percent. Is this error big and significant? Let me know of a good example where you would consider this error not significant and one when ignoring it would be a big problem. We will use your feedback and examples at my next Pump School session.