How to buy a Pump
In the End, It’s the Efficiency that Really Matters

In a recent pump reliability lecture, a student raised his hand and asked which computer program I recommend to specify a pump for a system. I told him I don’t recommend or use a computer program to specify and/or buy new pumps. Understand, I’m not exactly computer illiterate. I’m writing this article on a computer. I make personal purchases online. I do my company accounting with a computer, and I communicate with people all over the world by computer. But I don’t use a computer to buy a pump.
   
Buying a new pump is like buying a new TV, refrigerator or car. Would you purchase a computer program to help you buy a new lawn mower?
   
No, of course not. You’d simply consider how much money you have available to spend on the new product. You’d consider the features you need in the product. You’d consider the features you’d want in the new product. You’d consult with some friends who might already own a product similar to the one you’re considering. You’d shop and talk with a couple of dealers and salespeople. Then, and only then, you’d make your decision.
   
Purchasing a new pump is the same. It’s likely your application isn’t unique. You consider available funds, the features you need in the pump, and the accessories you want. You will probably find that numerous pump brands and models can replace your existing pumps and meet the needs of your system. Let’s practice … Let’s buy a hypothetical pump.    

Consider System Requirements
Let’s say you need to complete an operation (fill a vessel, or comply with a production schedule) at 700 GPM. The piping arrangement calls for 28 ft. of elevation differential, 20 PSI of pressure differential (20 PSI x 2.31 = 46 ft.) and six feet of resistance in the pipes and fittings. (28 ft. + 46 ft. + 6 ft. = 80 ft.) The system requires 80 ft. of head. You’ll want a pump that delivers 80 ft. of head @ 700 GPM.

Consider the Liquid(s) Pumped
What is the temperature? What is the viscosity? Does the liquid require special metallurgy?  Special gaskets and secondary seals? Maybe non-metallic? Are there abrasive solids in the flow? Let’s say your liquid contains no solids. The liquid is compatible with carbon steel. The temperature and viscosity are compatible with a centrifugal pump. You’ll find many pump manufacturers who make a carbon steel centrifugal pump that delivers 80 ft. @ 700 GPM as best efficiency coordinates. The system and liquid will lead you toward either a centrifugal pump or a PD pump.

Consider Options
Study some performance curves and talk with some pump sales reps. They can help with options: packing rings, single or double seal, in-line or end-suction, single-stage or multi-stage, surface or submersible, simple or spacer coupling, single volute or dual volute, open or closed impeller, fixed or variable speed motor.

Consider Efficiency/Economy
Most importantly, you should always carefully consider the most efficient pump that meets the system requirements, the liquid requirements, and your needs and wants. Pump efficiency is the best combination of head and flow with the least power consumption. As efficiency goes up, the monthly electric bill and cost of ownership for that pump go down.

Let’s work an example:
Let’s say you’re pumping cold water with no solids or sediment. The specific gravity of cold water is 1.0. Let’s say you have narrowed your search to three pumps that meet the head and flow requirements.
   
Pump “A” is 72 percent efficient and sells for $8,000. Pump “B” is 79 percent efficient and sells for $8,600. Pump “C” is 86 percent efficient and sells for $9,200. Pump “C” is $1,200 more to purchase than Pump “A”.
   
Let’s consider each pump within the framework of horsepower and kilowatts.

Horsepower = head x flow x sp.gr./ 3,960 x efficiency

• Hp Pump A  = 80 ft. x 700 GPM x 1.0 / 3,960 x .72 = 19.6 Hp
• Hp Pump B  = 80 ft. x 700 GPM x 1.0 / 3,960 x .79 = 17.9 Hp
• Hp Pump C  = 80 ft. x 700 GPM x 1.0 / 3,960 x .86 = 16.4 Hp

Now let’s take horsepower to kilowatts and consider the electric bill. Let’s say you pay 10-cents per kilowatt-hour for electricity and the pump runs 24 hours per day all year. There are 8,760 hours in a year.
Hp x .746 = Kw
• Pump A = 14.62 Kw x 8,760 x .10 / kwh = $12,807 / year in electricity.
• Pump B = 13.35 Kw x 8,760 x .10 / kwh = $11,694 / year in electricity.
• Pump C = 12.23 Kw x 8,760 x .10 / kwh = $10,713 / year in electricity.
    
Pump “A” is $1,200 less to purchase than Pump “C”. Yet Pump “A” costs $2,094 dollars more per year in electricity than Pump “C”.  If you keep the pump for 10 years, the cheap pump (Pump “A”) costs $20,940 more to operate if the electricity rates stay the same.  
   
Efficiency comparisons don’t have to be drastic. Even modest improvements in efficiency can justify a purchase over the life of a pump. And remember, the electric rates will rise.  
   
So, don’t concentrate on purchase price. Purchase price savings contribute very little to lifecycle cost. Concentrate on efficiency. Efficiency really reduces the total lifecycle cost of a pump.

Larry Bachus, founder of pump services firm Bachus Company Inc., is a regular contributor to Flow Control magazine. He is a pump consultant, lecturer, and inventor based in Nashville, Tenn. Mr. Bachus is a member of ASME and lectures in both English and Spanish. He can be reached at This e-mail address is being protected from spambots, you need JavaScript enabled to view it or 615 361-7295.