A double-suction split case pump (18-in suction nozzle size), similar to the one shown in Figure 1, moves 8000-gpm at 400-ft head in a paper mill. Before reading further, stop to make a rough guess on something: how much energy per year (7/24/365) is wasted if its wear rings are worn to double their initial clearance? Assume, for simplicity, continuous operation.

Let's see how close your guess was by using some pump selection programs that are free and readily available on the Internet. After logging in and entering your operating parameters, you'll quickly get a performance curve showing head-capacity curves at various diameters, power, and NPSHr. This also shows, although not generally recognized or appreciated as a good source of information, data at the top right corner of the chart (see Figure 2) that can tell you more about the pump. Let's use this information to answer our question about energy.

What drives the leakage across the wear rings is differential pressure across the ring. On one side of the ring is suction pressure; on the other side, the pressure is close to discharge pressure, but reduced somewhat due to a parabolic pressure distribution caused by vortex motion between the impeller and casing walls. There are ways to calculate this pressure precisely, but as a rough rule of thumb, this pressure decreases as a ratio of a ring diameter to impeller diameter.

Notice that in the upper right-hand corner of the legend on the chart, the impeller eye diameter size for this 16 X 18 - 30G pump is 179.0-in2, from which we calculate the eye diameter. 179.0 = π/4 x (Deye2 - Dshaft2), which yields Deye = 15.4-in, assuming a 3-in shaft. If you know the shaft/sleeve size more exactly, you'll get a more accurate answer. A ring diameter is a little larger, and assuming the ring is 0.25-in thick, we get Dring ~ 16-in - close enough to begin the leakage calculation.

First, we estimate pressure (head) at the ring diameter. As noted above, it varies (approximately) with the diameter ratio: Hring ~ 400 x (16/29.5) = 217-ft, where 29.5-in is the impeller diameter found from the performance curve (see Figure 2) for the specified operating conditions. 217-ft of head throttles by the wear ring, acting essentially as an orifice, for which (check your books on basic hydraulics for an orifice) the velocity of leakage is:

V = 0.61 x √(2gH) = 0.61 x √(2 x 32.2 x 217) = 72-ft/sec, where "g" is gravitation constant

Looking up a standard ring (diametral) clearance for a 16-in ring, we find 0.028-in, and the ring area then calculates as:

Aring = π x 16 x (0.028/2) = 0.7-in2

From the ring area and velocity, we can now calculate ring flow: Qring = V x Aring ÷ 0.321 = 158-gpm, or 316-gpm leakage accounting for both rings of the impeller. When rings are worn to double the initial value, their area doubles and flow leakage doubles, i.e. an additional 316-gpm are lost, or 316 ÷ 8000 ~ 4 percent of rated pump flow.

The rated power, corresponding to the rated flow, is approximately 900-hp (from Figure 2) and thus 4 percent of it is 36-hp, or 27-kW. Running non-stop for one year at \$0.10 per kW-hr, consumes:

27 x (24 x 365) x 0.1 = \$23,652

There are many things I can think of doing with \$23,652 - and still have a lot left over after paying for the replaced new rings, with proper clearance, during the next repair.

Note: Any pump manufacturer's software should provide similar results.

As always, we cannot part without a quiz! As I am writing this article, our repair shop manager is considering replacing the wear rings for a 12NK vertical turbine pump for a customer, as part of the overhaul. Using the procedure outlined above, plus any insights you may have on a 12NK design and its typical applications (hint: it is a single stage, pumping water), quantify your answer to the question: Does it make sense for a pump user to have the clearances restored during the on-going repair?

Be creative! The answer may surprise you. The best three answers will get you a free admission to PumpTec-2007 in Atlanta.

Pumps & Systems, March 2007