By C.F. ‘Chubb’ Michaud, CWS-VI

In previous chapters, it was pointed out that a filtration system design is an exercise in engineering. Although we started with flows through pipes, fittings and valves and then moved on to lateral and eductor design, a proper sequence is to start with the design of the filter media itself, then the tank it goes into and work backwards. We hope that the feed pipe and water pressure are adequate. If not, a booster pump may be necessary.

What is a pump?
A pump is a device that imparts energy to a fluid and causes it to flow. One of the more common types of pumps used in water treatment is the centrifugal pump. These are rotary pumps that work by accelerating fluid from their center intake to the edge, thus increasing the energy of the fluid by centrifugal force. Since pumps are all part of hydraulic science, this chapter is devoted to determining the power requirements and calculating pump sizes. Instruction is offered on the reading and interpretation of pump curves.

Figure 17. The impeller of a centrifugal pump

Sizing pumps
Pumps impart energy to water to increase both flow and pressure. If a pump is rated at 50 gpm and 50 psi, it usually means that it will do 50 gpm into an atmospheric tank (at zero head pressure) and 50 psi against a closed valve (at zero flow). Having it do 50 gpm and 50 psi is quite another story!

Determining the boost needed
A fluid system can be characterized by an operating curve similar to the curves shown in Part 2, Figure 2 and is simplistically represented here in Figure 18.

This curve shows how much pressure has to be supplied (vertical axis) to achieve a given flowrate (horizontal axis).1 Unfortunately, this curve is not in any book nor is it part of the manual you might receive with new equipment. It has to be determined for the complete  treatment system design, which is unique to your installation. It includes head losses through the pipe, filters and fittings and elevation changes. As we would expect, it takes an increased pressure to achieve an increased flowrate. Next we look at a pump performance curve (Figure 19).

The relationship of pump flow and pressure
Pump curves supplied by the manufacturer contain a wealth of information—almost too much information. The simplest of curves provides pressure information on the vertical axis and flow information on the horizontal axis, as shown in Figure 19. Remember, pump curves are almost always shown as feet of head and have to be converted to psi or other pressure units (refer to Table 1, Part 1) to utilize it. Here, I am showing psi.

This curve shows a specific pump at a stated speed with a given impeller size that can deliver 75 psi of  head pressure and a flow up to 100 gpm. Note that at the full pressure, the flow is zero and at the full flow, the pressure is zero. Can this pump deliver 60 gpm at 50 psi (green dot)? Yes it can. In fact, it can deliver 70 gpm at 50 psi or 60 gpm at 57 psi. A throttling valve can be used to tweak the flow and pressure up to the limits of the red line curve. Can this pump deliver 70 gpm at 60 psi (red dot)? No. That point is outside the performance curve. Knowing the operating needs of pressure and flow will allow you to ‘ballpark’ your pump needs by simply referring to a set of charts.

What if we change the impeller size? A smaller impeller, for instance, would turn just as fast but it would move less water. In other words, it does less work and requires less energy (and for a given job may be less expensive to operate). Figure 20 shows three impeller sizes ranging from six to 10 inches. The blue curves that intersect the pump curves show the pump efficiencies at various pressures and delivered flows. Generally, the best operating point for a pump will be close to that of the best efficiency.2

Relating what we have to what we need
By superimposing our system curve (Figure 18) onto the pump curve (Figure 20) we can see a guide developing (Figure 21). Where the green system curve intersects the pump curves (called the operating point or duty point) is where we choose our pump.3 In this case, we pick the 10-inch impeller model because it has the best efficiency (lowest power consumption for the work delivered).

Pump efficiency
Pump efficiency is defined as the ratio of the power imparted on the fluid by the pump in relation to the power supplied to drive the pump. Its value is not fixed for a given pump but is a function of the discharge and therefore, the operating head. For centrifugal pumps, the efficiency tends to increase with flowrate up to a point midway through the operating range (peak efficiency) and then declines as the flowrates rise further. Pump performance data, such as this, are usually supplied by the pump manufacturer to help you make your selection. Pump efficiencies tend to decline over time due to wear (e.g., increasing clearances as impellers reduce in size due to wear).

Like most machines, it is not possible to achieve 100-percent efficiency. To be conservative, I generally use 50-percent efficiency for approximating pump size. I do the calculations for just the pump and then double it (50-percent efficiency) to account for head  losses in piping and filters. This only represents a best guess but usually allows you to move forward. Our goal is to select a pump that operates as close to its peak as possible. Note that the pump efficiency decreases if the flow is too high or too low. As a rule, if the pump efficiency is below about 60 percent, you should look at other pumps (as might be the case above).

When a system design includes a centrifugal pump, an important issue in its design is matching the head loss-flow characteristic with the pump so that it operates at or close to the point of its maximum efficiency. Pump efficiency is an important aspect of pumps and can change over time (as impellers wear) and with changes in the operating conditions (temperature, for instance). On larger systems, pump efficiency should be tested regularly. Thermodynamic pump testing is one method.4

Thermodynamic pump testing is a form of testing where only the temperature rise and power consumed need to be measured to determine the efficiency of the pump. This is measured by means of temperature and pressure probes fitted to tapping points on the pump’s inlet and outlet. In this case, the instrumentation has to be very accurate, as the temperature increases may only be fractions of a degree. This method is so accurate that the flowrate for a pump can be derived from the pump efficiency measurement and electrical power supplied to the pump. In other words, by comparing how much work is being put into the pump (electrical consumption) and how much work is being generated by the pump (temperature and pressure), it is possible to derive the flowrate.

Figure 22 provides water horsepower (WHP) requirements for sizing a pump.13 Pumps are not 100-percent efficient, however, so the question is: “How big a pump motor is required to deliver the water horsepower found on the curve?” Pumps typically run at 50- to 70-percent efficiency. In our example above, where the question is posed as to what size pump would deliver 50 gpm (note logarithmic curve) at 50 psi (115 ft head), we see from Figure 22 (locate the small red circle) that we need about 1.5 WHP. You can calculate that using Equation 5.

Equation 5. Calculating water horsepower
WHP = Q H/3960 where:1
WHP = water horsepower
Q = flow rate in gpm
H = the pressure head in feet ( psi x 2.31 to get pressure in feet
WHP = 50 x 50 x 2.31/3960 = 1.46 WHP

Taking into account the efficiency, we can double that to 3 HP as a reasonable starting point. This is for a 1,750-rpm pump operating on 60 Hz power. If this pump is run on 50 Hz, it will run slower. There will be a 17-percent loss in rpm (50/60 = 0.83 or 83 percent) and a 30-percent loss of power (50/60)2 = 0.694 or 69 percent of the power (an approximately 30-percent loss).

Cavitation
Pumps need adequate feed pressure. Otherwise, they can produce so much suction that they boil the intake water (create vapor) and cavitate. This can be very damaging to the moving parts of the pump. If the pump works too hard to pull water in, it can create vapor and produce bubbles at the eye of the impeller. As these bubbles are carried over to the discharge side, they are compressed back into a liquid. Cavitation often sounds like the pump is passing gravel! This change can be so violent that it can actually chew away material from the impeller, leading to a failure of the pump.3 For this reason, pump intakes are often larger than the outlets.

Keep in mind that a pump cannot ‘suck’ from the basement to the roof even if the pump could create the perfect vacuum. It can only draw water the equivalent of one atmosphere or about 33 feet (at most). That’s because pumps don’t actually suck; atmospheric pressure pushes fluid in to fill the vacuum. If you fill a clear 35-foot PVC pipe with water and cap one end while standing the open end vertically in a bucket of water, the fluid will drop down, creating a void, to an equivalent head representing the outside or atmospheric condition. The void space that forms is a near-perfect vacuum. This the principle by which the barometer works; it is filled with mercury so it only has to be 36 inches long instead of 36 feet, due to mercury weighing 13.6 times that of water.

You will often note a separate curve at the bottom of the pump curve display and shown as net positive suction head required (NPSHR). This is the feed pressure (in feet of head) required to keep the pump happy.3 If the pump has to do work to satisfy its own intake requirements, this will subtract power from the pressure side of the delivery. This value will depend on the size of the pump but is generally only a few feet of head (but never zero). Feeding a pump from a supply tank is generally not a problem, as long as the needed NPSHR is taken into account with a low-level shutoff in the tank that is at least equal to the intake (NPSHR) requirements of the pump.

How do deep wells supply water to the surface?
Pumps have a limited suction head but a nearly unlimited pressure head. Well pumps are located in the well and they work by pushing the water to the surface. Even with the old iron hand pumps located on the farm, the pump portion was down in the well pushing water out of the spout.

What are the final considerations for selecting a pump size?
Let’s take the case of a businessman who decided to buy a farm and live the life of the gentleman farmer on weekends. The farm had a quaint old windmill that pumped water from a shallow table and stored it in a nearby 2,500-gallon plastic tank. The businessman tore down the old farm house and built a new 9,000-square-foot cottage at the top of a nearby hill with a gorgeous view. He calls the local water guy and says he needs to move the water from the storage tank to his cottage. It’s just him and the missus, and a few dozen close friends on weekends. You’re okay with the 2,500-gallon storage (given the constant wind) and considering the reliability of the old (but well functioning) windmill, but you have to size a pump.

From the water (always get an analysis), you determine that the owner needs 1) an iron filter, 2) a softener and 3) a nitrate filter. From the fixture count and other fictitious input, you determine that you should design for a true 10-gpm demand flowrate and a minimum delivered pressure of 50 psi at the entrance of the cottage (but after any water treatment). You factor in a well tank of sufficient size with a high pressure of 80 psi and a minimum of 60 psi.

The ground level of the two-storey cottage is 60 feet above the storage tank base and we add an additional 15 feet to reach the upstairs baths and hot tubs. The horizontal run is 200 feet. The vertical is 75 feet and the hypotenuse of that triangle is 213.6 feet (see Part 1). You opt to transfer the water up the hill, using 1.5-inch pipe (refer to Part 2) to minimize the pressure losses. To get there, you have to do six right turns (six elbows) (see Part 3). Then you reduce the pipe down to one inch to feed the cottage.

You select three 14-inch diameter filters (cross-sectional surface area = 1.07 sq ft ~ 1) using 1.25-inch control valves (Cv=9.2) (refer to Parts 4 and 5) and since the eductors are built into the valves, you can skip re-reading Part 6. At full design flow, you will have 10 gpm/sq ft on all three filters. From the media data sheets, you see that the iron filter has a ΔP of 1.3 psi per foot of bed depth, the softener is 0.9 psi/ft and the nitrate filter is 1.0 psi per foot (see Parts 7 and 8). You are confident in your filter sizes for the water temperature and contaminant challenge (see Part 9). Now, the easy part.

Pulling together what you have learned
The pressure drop through 213 feet of 1.5-inch pipe at 10 gpm is five feet of head per 100 feet or 10.75 feet. The six elbows add the equivalent of 21 more feet of pipe (plus one foot of head). Total head loss through the pipe is 11.75 feet plus the loss from the 75 feet of elevation gain (total now is 86.75 feet). Bushing down from 1.5-inch to 1.25-inch and from there to 1.0-inch is an additional three feet of pipe or 0.15 feet. So we have 86.9 feet thus far.

From Figure 22, we can see that a 0.5 WHP pump will do the job (trace the intersection of the 10-gpm flow line to the brown 250 foot of head line). Using my fudge factor of 50-percent efficiency, we have a 1 HP pump. And you thought it was going to be bigger!

Since the well tank feeds a storage tank there is no energy boost provided by the well tank. Note: a well pump supplying only 60 psi (138.6 feet of head) would still be able to deliver 10 gpm to the cottage (required head = 123.3 feet). However, there would only be a residual working pressure of 138.6 – 123.3 = 15.3 feet or about 7 psi. Our well tank tops off at 80 psi (185 feet of head) so there is a working pressure of 185 – 123 = 62 feet of head or 27 psi even before the pump kicks in.

In the old days before calculators and the above example, we are trying to determine how much work is required to deliver 10 gpm of water against a head requirement of 238.8 feet. Ten gallons of water weighs 83.4 pounds. To move it against a 238.8-foot head means we have to calculate 83.4 lbs/min x 238.8 ft = 19,916 ft lbs/min. We know that one horsepower is equal to 33,000 ft lbs/min so we can calculate 19,916/33,000 = 0.6 HP. Using a more generous 60-percent pump efficiency, we have 0.6HP/60% = 1 HP. This example does not try to calculate the working pressure losses from within the cottage. The original problem was to deliver a 50-psi working pressure to cover the internal needs.

Conclusions
Pumps supply energy to a fluid in terms of an increased pressure, which in turn promotes higher flow. There are, perhaps, fewer issues that will develop from a pump that is slightly large than from one that is slightly small. Many downstream filters rely on adequate flow and pressure (think eductor, Part 6) and that is often supplied by a booster pump. I would invite readers knowledgeable on the subject of variable-drive pumps to enlighten us all on the merits of same.

References

1. The Engineering Toolbox, www.engineeringtoolbox.com/pump-system/curves.
2. System Curves, www.gouldpumps.com/cpf.
3. Pump Performance, Cotton CRC Water Team, NSW Dept of Primary Industries.
4. Wikipedia.com. Thermodynamic Pump Test.
5. McNally Institute, www.mcnallyinstitute.com, Calculating Water Horsepower.