Dr. Nelik (aka “Dr. Pump”) is president of Pumping Machinery, LLC, an Atlanta-based firm specializing in pump consulting, training, equipment troubleshooting and pump repairs. Dr. Nelik has 30 years of experience in pumps and pumping equipment. He can be contacted at www.pump-magazine.com. To learn more and register for his upcoming PUMPTEC conference, visit www.pumpconference.com.

In the August 2013 issue of Pumps & Systems, I presented a summer pump quiz about cavitation. This column includes the solutions to the cavitation quiz.

### The Questions & Solutions

**What is a tank water level above the suction connection? **

The tank water level height (H0) can be found from the definition of the net positive suction head available (NPSHa):

a. NPSHa = H_{suction} – Hvapor = H_{0} + H_{a} – H_{vapor} = 43.2 feet (given)

b. Atmospheric pressure = H_{a} = 34 ft (14.7 psia) on top of the open tank. For water, vapor pressure at ambient temperature is 0.34 psia = 0.8 ft

c. Then H_{0} = 43.2 + 0.8 – 34 = 10 ft

**How long is the pipe if it has a 3-inch diameter? How long is the pipe if it has a 12-inch diameter? Note: assume a typical flow through a pump of this size. Keep a practical perspective. Use your experience and judgment when lacking exact data.**

We will make an estimate of a 3-inch diameter suction pipe length, resulting in an additional 20 feet of hydraulic friction losses, assuming that the pipe is straight—with no kinks or bends. There are several ways to calculate this. Existing charts that show friction losses (typically in feet of losses per 100 feet of pipe length) can be used. For viscous fluids, charts also present values of losses for different viscosities. In this case (water), viscosity corrections do not have to be considered. The other method is to use the formula on which the charts are based:

hf = f x (L/D) x V^{2}/2g

Where:

L and D = the pipe length and diameter

V = the velocity of fluid inside the pipe

g = 32.2 ft^{2}/sec (gravitational constant)

The main preference of using the charts instead of the formula is that the friction coefficient is unknown and depends on many factors—including pipe roughness, Reynolds number and flow regime (laminar, transition, turbulent). However, in most cases, the practical variations of it are between 0.02 to 0.04, and taking a 0.03 value is a good starting point for a reasonable engineering answer—not entirely scientific, but practical.

However, another practical dilemma is determining which flow to use. To counter the skeptics, we will measure the flow, ask the operators or make a rough estimate for a typical 3-inch pump performance curve. While not all 3-inch pumps are alike, they are much different in flow than a 12-inch pump.

Looking at several 3-inch pumps’ catalog curves, the flow varies substantially—depending on the speed, impeller diameter and the system in which they are installed. Some pumps are throttled by the operator to nearly deadhead conditions. Others operate significantly outside their curve. A good estimate of average flow is approximately 300 gallons per minute (gpm).

V_{(3-inch pipe)} = 300 x 0.321/(3.14 x 3^{2})/4 = 13.6 ft/sec

Correcting for the pump suction versus discharge piping at this point can be set aside. Right now, we are just trying to get a rough estimate.

If:

h_{f} = f x (L/D) x V^{2}/2g = 20 = 0.03 x (L/3) x 13.6^{2}/2g

Then:

L = 696 inches = 58 feet

For a 12-inch pump, 5,000 gpm will be a starting point.

V_{(12-inch pipe)} = 5,000 x 0.321/(3.14 x 12^{2})/4 = 14.2 ft/sec

Why do you think the velocity was so similar to a much smaller pump?

If: h_{f} = f x (L/D) x V^{2}/2g = 20 = 0.03 x (L/12) x 14.2^{2}/2g Then: L= 2,552 inches = 211 feet With several estimates, there is now a starting point.

**In Figure 3, assume that no kinks or turns are present in the piping. Then modify your answer by estimating and adding the pipe losses shown.**

Again, we will use a practical approach to the kinks. The 90-degree pipe bend results in a loss of one velocity head. For non 90-degree bends, the loss would be less or possibly more, but we will use the 90-degree bend as a good starting point. Approximately five bends are shown in Figure 3, and each one takes the following:

h_{bend 3-inch} = V^{2}/2g =13.6^{2} /2g = 3 feet

h_{bend 12-inch} = V^{2}/2g = 14.2^{2} /2g = 3 feet

For five bends, 15 feet of losses are added. That would leave only 5 feet (20 – 5) for friction losses (initially assumed as 20 feet), and it would require recalculation, with the resulting pipe length much shorter to produce the same overall friction loss. As shown in this example, it is not the length—but the kinks—that kill pumps in cavitation. In conclusion, keep your pipes straight. Email your comments or join the debate at my next pump school: www.pumpingmachinery.com/pump_school/pump_school.htm.