Well and Pump Rehabilitation

Part 7: Less common but effective methods

By Ed Butts, PE, CPI

We began a long discussion on well and pump rehabilitation in the June 2017 column. The series has included an outline of various methods available to improve and stabilize well and well pump efficiency and reliability.

This month, as a fitting wrap-up to the series, we outline some other less common but nonetheless effective methods to raising pumping plant efficiency. Examples will demonstrate how rapidly these improvements can recover the initial cost, plus conserve and lower energy consumption.

Methods to improve the efficiency and lower energy consumption and operating costs of a pumping plant are as long and versatile as the combinations of pumps and motors available in the world.

For one, materials used for pump components with an inherently low hydraulic friction as the component passes through water—such as bronze, porcelain or special coatings, lined and stainless steel—will result in lower hydraulic loss within the pump itself. Use of composite materials, Ni-Resist, or ductile iron impellers and bowls instead of bronze or cast iron is often favored for abrasive or corrosive conditions.

Sandblasting and applying an electrostatically or chemically bonded internal epoxy lining or coating to pump bowls or volute, such as Scotchkote 134, will often improve the bowl’s operating efficiency by 1%-3%.

Little known but other long-recognized simple tricks such as vane or impeller tip or cutting-edge backfilling, impeller static and dynamic balancing, and polishing of impeller inner and outer surfaces can also raise the impeller or bowl efficiency by another 1%-2%.

Try to design and select a pump to operate as close as possible to the best efficiency point (BEP) or within the best efficiency window (BEW) if possible, or to remain within plus or minus 25% of the upper and lower range of the BEW. If the pump must operate over a wide-ranging set of conditions, select a pump with a wide efficiency band and an H-Q curve shape best representing the characteristics of the system head curve, especially when a VSD (speed reduction device) or control valve is used.

When applying a VSD, make sure the pump at minimum speed can always satisfy the minimum combined system static and frictional head required to continue pumping, and to never continuously operate at any shutoff head conditions or below the minimum safe continuous flow.

For irrigation systems, seasonally checking the diameter of nozzles for erosion and replacing when indicated can pay huge dividends. Standard bore (round opening) sprinkler nozzles tend to increase and become irregular in size over time and use due to erosion—especially with abrasive conditions where uniformity and discharge can easily suffer and cause variation up to 20%-30% of catalog flow rates with well-worn nozzles. Possibly changing from a gravity, flood, furrow, or pressurized sprinkler irrigation to a drip or low-pressure system can also result in substantial water and energy savings.

Improving piping and lowering frictional losses by using larger suction and discharge piping than normal; using long radius ells and smooth piping turns; eliminating unnecessary tees, ells, restricted, or undersized piping and valves in pipelines and manifolds where feasible; and using piping materials with the lowest internal friction factor, such as PVC, PE, cement-lined ductile iron, or internally coated or lined steel or iron pipe over conventional steel can also positively benefit a system and lower friction losses.

For systems with inline suction or discharge screens or filtration equipment, be sure the backwash or cleaning cycles and intervals are performed according to the manufacturer’s recommendations and timers are properly set and adjusted. Verify any screen or filtration element or media is sized for the applicable system flow rate and operated at no more than the recommended or maximum pressure differential before backwash or cleaning and for the intended service and backwash flow rate and duration.

For motors, increasing conductor sizes, optimizing voltages, improving the power factor using online capacitors, using high efficiency or VFD-rated motors, using and maintaining the proper grade and weight of lubricating oil and grease, and adding or changing at the recommended frequencies are all critical. Also, operating motors at 75%-100% of full load duty; providing or enhancing an existing air supply to a motor using a clean, cool, uniform, and fully ventilated operating environment; and derating a motor’s horsepower for a given load or when used on an open-delta power system can raise the efficiency of a pumping plant by as much as 5% through electrical changes alone.

 A Combination of Efficiencies

Never forget the efficiency of a pumping plant is derived from the combined efficiencies of the pump, motor, and any transmission means such as drop and offset cable for submersible units or lineshaft and bearings for vertical turbine pumps (VTPs). Changing out old lineshaft sleeves and rubber bearings on a VTP during a routine or emergency repair (particularly old, stiff, tight, inflexible, and cracked rubber bearings); eliminating or changing from a standard multiple vee-belt drive to a single grooved or serpentine belt arrangement for a belt-drive pump; adjusting and maintaining the proper and uniform tension of drive belts; increasing the size of the drop cable; or providing enhanced or supplemental water cooling over a submersible motor or air cooling over a vertical hollow shaft/vertical solid shaft motor may only raise the efficiency by a half point or so, but it all adds up.

Although it is always important to maximize the overall efficiency of a pumping plant, for most clients it is just as important to lower or stabilize the present operating costs and extend reliability, particularly the electrical power costs when applicable, and extend or maintain the service life of the existing pumping plant (pump and motor) or the individual components (pump or motor) as much as possible.

DACUM Codes
To help meet your professional needs, this column covers skills and competencies found in DACUM charts for drillers, pump installer, and geothermal contractors. DO refers to the drilling chart and GO refers to the geothermal chart. The letter and number immediately following is the skill on the chart covered by the column. This column covers: PIF-1, 2, 3, 4, 5, 6, 7, 8, 9. More information on DACUM and the charts are available at www.NGWA.org/Certification and click on “Exam Information.”

This simply indicates even though you may have the means and ability to increase the efficiency of the pumping plant today, the resultant cost savings may be so incremental and minor as not to be worth the current investment and cash outlay given the current power costs and age and condition of the plant. Many clients, particularly those such as water purveyors who receive revenue based on the direct volume of water they produce and sell, will often examine this bottom-line factor to be sure their pumping plants are delivering the expected water flow at the lowest unit cost possible—but at the same time may feel the extra cost to improve a specific pumping plant is not worth the disruption, downtime, or added investment.

In most cases the rate and volume of lost and “unaccounted-for” water within a water system is generally a much greater concern to the system’s decision makers and operators than gaining one to three points of pumping plant efficiency. This is especially true since this type of condition (system leakage) often tends to increase in volume over a short period of time, and given all unaccountable water has already incurred the cost for treatment and pumping, the cost of this lost water strikes directly at the water system’s bottom line, lowers the available water for new growth, and can never be recovered as revenue.

This is one of the primary reasons water utilities are so concerned with leakage and lost water, and is the prime motivator behind investing in a leak detection survey followed by the indicated level of leak repairs. This in itself can often be one of the most important and useful methods of energy and water conservation.

Finally, implementing a preventive maintenance program can pay huge dividends by servicing and maintaining the plant’s components, thus maintaining efficiency and providing advance notice of approaching or impending failure.

 Formulas and Examples

Beyond the simple evaluation of conducting upgrades or changes to pumping plants, some method needs to typically be used to verify if the proposed improvements are cost effective with a reasonable period of payback.

The formulas (1) through (4) and examples with answers (a) through (p) have been included to assist with these evaluations. For all examples, assume a straight-line cost recovery without factoring for utility base fees, connection, power factor, or demand charges, inflation, or depreciation.

  • Operating cost per hour =

(.000189) × GPM × TDH × Incremental power cost (in $/kW-hr)

Ƞ1 (in %)

  • Cost per 1000 gallons of water2 =

.00315 × Incremental power cost (in $/kW-hr) × TDH

Ƞ (in %)

 1 Ƞ = overall pumping plant or individual component efficiency (in %)

2 For other volumes, divide the desired volume by 1000 and multiply by formula (2)

Example 1

Determine the (a) operating cost per hour for a pump with an efficiency of 75% and motor with an efficiency of 90% that delivers 500 GPM against a TDH of 200 feet, plus (b) the cost for this unit to produce a volume of 1 million gallons of water. Power cost = $0.12/kW-hr.

  • Operating cost per hour =

.000189 × 500 GPM × 200′ TDH × 0.12/kW-hr = 2.268 = $3.36/hour

.75 (Pump eff.) × .90 (Motor eff.)               .675

  • Cost per 1 million gallons =

.00315 × 0.12 ×200′ TDH (× 1000) = .0756 × 1000 = $112.00/million gallons

.75 × .90                                    .675

In addition to the previous equations that can be used to determine the delivered cost of water on an existing or new pumping plant, there are various other methods used to determine the potential power and cost savings from conducting improvements to a pumping plant. A few are shown below:

(3) Potential savings of power (kW-hrs/year) = IKW × T × (1 – NA/NO)

(4) Potential cost savings (in $/year) = kW-hrs/year × Incremental energy cost (in $/kW-hr)

Where:

IKW = instantaneous input power to pumping plant or pump or motor, in kW = (HP × .746)

T = operating period (time) per year, in total operating hours

NA = active or current system, plant, or component efficiency (in %, expressed as a decimal)

NO = optimal or desired system, plant, or component efficiency (in %, expressed as a decimal)

Cost = expressed in U.S. dollars ($)

Example 2

A fully loaded 100 HP submersible pump has a field tested and verified bowl efficiency of 63% at 1000 GPM, but the manufacturer’s pump curve indicates a new pump should exhibit an efficiency of 78% at the same flow and head (condition of service). If the unit operates at 3500 hours per year, what is the (c) potential energy savings per year and (d) cost reduction from energy savings per year, both assuming a new pump is installed? Power costs = $0.10 per kW-hr.

(c) Potential savings (kW-hrs/year) = (100 HP × .746) × 3500 hrs/year × (1 – .63/.78)

Potential savings (kW-hrs/year) = (74.6 kW) × 3500 hrs/year × (.1923) = 50,211.54 kW-hrs/year

(d) If energy costs $0.10/kW-hr, potential cost savings = $50,211.54 × .10 = $5021.15/year

Example 3 

A 368-input horsepower pumping plant has an overall efficiency of 62% and operates at 2500 hours per year. If the estimated repair costs are $64,000 with a desire to recover these costs over five years of operation through energy savings alone with a unit electrical cost of 16 cents per kilowatt-hour ($0.16/kW-hr), (e) how much energy in kilowatt-hours must be conserved each year to meet the target goal? and (f) what will the new unit efficiency have to be?

(e) Determine the number of kilowatt-hours needed in 5 years: $64,000/$0.16/5 years = 80,000 kW-hrs/year

(f) Use Formula (3) to ascertain the required new or optimum unit efficiency (NO):

(f1) 80,000 kW-hrs/yr = (368 IHP × .746) × 2500 hrs/yr × (1 – .62/NO)

(f2) 80,000 kW-hrs/yr= 274.53 IKW × 2500 hrs/yr × (1 – .62/NO) = [.1166 = (1 – .62/NO)] = .7018 (70.2%)

The above results demonstrate the final outcome from the proposed repairs must result in a combined minimum unit efficiency of 70.2%. If feasible, this increase in the overall combined unit efficiency could be the sum from separate improvements made in just the pump—by changing the motor to an energy efficient or premium efficiency style, lowering the power transmission losses, or a combination from all available modifications.

In reality, improving this plant’s efficiency will also likely lower the input horsepower slightly; therefore, the actual length of payback would generally be somewhat longer than shown above.

However, this example remains a testament as to how much energy can be conserved with just a few minor improvements in component or total efficiency. The important factor to consider is the separate efficiencies of the pump and motor and power transmission method result in the overall efficiency of the pumping plant. Therefore, any modification up or down to one component will negatively or positively impact the final overall efficiency.

Example 4

Determine: (h) the years of cost recovery for retrofitting a pumping plant installation from a fully loaded standard efficiency 50 HP motor (Ƞ =90%) to a fully loaded premium efficiency 50 HP motor (Ƞ =94.5%) using the existing pump (Ƞ =82%). Power costs are $0.15 per kW-hr at 6000 hours per year of operation if the new motor plus installation costs $9000?

(g) Potential cost savings ($/yr) = (50 HP × .746) × 6000 hrs/yr × (1- .90/.945) × $0.15/kW-hr

Potential cost savings ($/yr) = (37.3 kW) × 6000 hrs/yr × (.04762) × $0.15/kW-hr = $1598.60/yr

(h) Cost recovery = $9000/$1598.60/yr = 5.63 years = full cost recovery in six years or less

Example 5

After conducting a pumping plant performance test, a client would like to evaluate if changing an entire 250 HP vertical turbine pump and electric motor from an existing plant (tested pump efficiency = 71% and motor efficiency = 85% = 60.35% (.6035) plant efficiency) to a higher efficiency plant will recover the initial investment in (A) 10 years or (B) 5 years, plus conserve 15% or 20% of the present energy costs per year using power savings only. What new pump and motor (unit) efficiency would be required for both conditions A or B? Tested COS: 2500 GPM @ 300 feet TDH

Power costs: $0.13/kW-hr, assumed hours/year of operation = 4000, cost of new pump and motor = $46,000

(i) Current input power: 2500 GPM × 300′ TDH = 266.752 BHP × .746 = 199.00 kW/.85 (M.E.) = 234.11 IKW

3960 × .71 (P.E.)

(j) Current yearly power costs: 234.11 IKW × .13/kW-hr × 4000 hrs/yr = $121,739.34/yr (936,440 kW-hrs)

(k) (A) Required power savings/yr ($): $121,739.34 × .15 (15%) = $18,261/yr + $46,000/10 yrs = $23,061/yr

(l) (A) Required power (energy) savings/yr (kW-hrs): $23,061 per yr./.13 kW-hr = 177,392 kW-hrs/yr

(m) (A) Required plant eff. (NO): 177,392 kW-hrs/yr = 234.11 IKW × 4000 hrs/yr × (1 – .6035/NO) = .7445 (74.45%)

(n) (A) Required pump efficiency: Given a typical efficiency of 95% (.95) for a premium efficient 250 HP, 1800 RPM motor, the required minimum efficiency for a new bowl assembly would be .7445/.95 = .7837 (78.37%)

(B) What if the conditions were changed to a desired unit payback of 5 years plus a 20%/year decrease of power?

(o) (B) The required yearly cost reduction of power rises to $33,548.00 per year/$0.13 = 258,060 kW-hrs/yr

(p) (B) Required plant eff. (NO): 258,060 kW-hrs/yr = 234.11 IKW × 4000 hrs/yr × (1 – .6035/NO) = .833 (83.3%)

Goal A would be entirely feasible by using standard components and construction, but the higher level of plant efficiency (83.3%) in scenario B with a motor efficiency of 95% would now require a bowl assembly minimum efficiency of 87.7% (.833/.95), which represents a value close to the upper practical limit for most VTPs.

However, goal B is still possible with a large diameter bowl (typically around a 12-inch to 16-inch) along with other specific efficiency changes, each providing a minor and incremental improvement, but when added together making a substantial difference in the final pump efficiency. Efficiencies include polishing, backfilling and dynamic balancing of impellers (or even consider the use of higher efficiency semi-open impellers); using low-friction materials for impellers; using a smaller diameter (12-inch or 14-inch) bowl with more stages and trimming of impellers to increase individual impeller efficiency; grinding out and smoothing flow passages in impellers and bowls; designing and operating close to the BEP; and applying internal bowl coatings. In totality, all of these could raise the bowl efficiency from a starting point of 84%-86% to the desired level of 87.8% or greater through these incremental improvements of 0.50%-1.5% each.

In this example, assuming the plant efficiency and power costs both remain reasonably stable after the initial five years of operation, the resulting energy and cost savings would increase following the initial cost recovery of the unit from 15% to approximately 18%-19% for scenario A and from 20% to 27%-28% for scenario B, resulting in much lower energy usage than the current consumption. In many other cases, the improvement rise in the pump’s efficiency can be enough by itself to lower the brake horsepower sufficiently to permit the use of a smaller HP motor during a unit revision or change-out.

This type of example illustrates the practical upper limit for energy savings and a reasonable payback interval for most VTP or submersible pump applications, as well as the many variables in a typical calculation and how much making a minor change to a single value of one parameter can significantly impact the other values. However, always remember each application is different and must be evaluated on a case by case basis to determine the precise hours of operation, if variable flow and head conditions occur during unit operation, plus the desired and minimum acceptable level of cost recovery and energy conservation.

The last four examples underscore how rapidly energy reduction along with an associated cost recovery for a relatively small increase in motor or pump efficiency alone or together can occur over a fairly short time without even touching the other component, or conversely, the potential savings in operating costs by improving the efficiency of a single element or both the pump and motor.

_______________

This concludes this seven-part series on well and pump rehabilitation and efficiency. I hope it has been informational and beneficial and will be helpful to you in future years.

Next month, we will embark on a new and different topic.

Until then, as always, work safe and smart.


Ed Butts, PE, CPI, is the chief engineer at 4B Engineering & Consulting, Salem, Oregon. He has more than 40 years of experience in the water well business, specializing in engineering and business management. He can be reached at epbpe@juno.com.

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