Larry Bachus (a.k.a. “The Pump Guy”)

From: Jerry C.
Subject: Help!

Larry,

I just read your article, “Is Your Pump Starving to Death?”, in the July 2006 issue of Flow Control magazine (page 36) for the second time! I’m a 30 year-plus sales engineer in flow products (meters), and I’m trying to understand pumps. I’m struggling!

I understand fluid characteristics, laminar flow, etc. I even helped design a mass flowmeter, but the pump is confusing. I’m having a particularly hard time grasping how an application results in 50 PSI or 1,000 PSI! I’m told the pump does not create pressure; but if that’s the case, how does product entering a pump from a tank (60 ft. tall) get to 1,000 PSI? Help me if you can.

Thanks,

Jerry C.
Greenwood, SC

From: Larry Bachus
Re: Help!

Hello Jerry,

Well, I don’t exactly know how to answer your question. It seems you’re thinking of an application when you talk about a product entering a pump from a tank (60 ft. tall), and apparently the discharge pressure for this application is 1,000 PSI.

Pumps are somewhat difficult to understand because they combine hydraulic theory with mechanical theory, as well as some electrical theory for the motor that powers the pump. Electricians generally understand the electrical theory of a wired circuit; hydraulic technicians mostly understand the hydraulic theory of an actuated hydraulic cylinder on a piece of earth-moving equipment or industrial process; and mechanics generally understand the mechanical theory of gears, cams, pulleys, torque, chains, levers, screws, and fulcrums.

Industrial pumps mated to an independent power source (the motor), and especially centrifugal pumps, combine all three of these disciplines. Pumps are mechanical in construction, and hydraulic theory plays games in the pump when it is operating. The speed, diameters, tolerance, and the nature of the liquid will govern the horsepower consumption and amps of the motor. It’s like trying to work three unrelated Sudoku puzzles at the same time. As such, many people have only a partial understanding of pumps, which ultimately leads to some misconceptions.

Getting back to your question, you are correct, pumps don’t create pressure. Pumps convert energy into a practical and useful form. This form is called pressure. Pumps take the energy in the liquid and combine it with the energy from the motor. The pump adds these two energies and produces head, quantified in feet or meters. The centrifugal pump converts this energy first into centrifugal force and then into head or elevation.

The energy in the liquid is a function of its vapor head, temperature, the atmospheric pressure (or artificial pressure), velocity head, and static head. The energy from the motor is rated in horsepower or kilowatts. The pump combines or adds these two energies and produces head at the discharge nozzle. So, the pump doesn’t create pressure, it converts energy into head. A pressure gauge converts head into PSI.

For water, if you take the elevation in feet, and divide this by 2.31, you will get pressure in pounds per square inch or PSI. Let’s say you have a tank that is 72 ft. tall and vented to atmosphere. And, let’s say that cold water fills the tank to a height of 70 ft. A pressure gauge at the bottom of the tank would read 30 PSI (70 ft. / 2.31 = 30 PSI and 30 PSI x 2.31 = 70 ft.).

Let’s say you have a vented tank that is 24 ft. tall. And, let’s say that cold water fills the tank to a height of 23.1 ft. A pressure gauge at the bottom of the tank would read 10 PSI (23.1 ft. / 2.31 = 10 PSI and 10 PSI x 2.31 = 23.1 ft.)

With a pump, let’s say we have a submersible pump designed to develop 70 ft. of head. This is the same as saying the pump develops 30 PSI. Let’s say that the submersible pump is draining a tank. The tank holds cold water. The cold water fills the tank to a height of seven feet above the pump in the bottom of the tank. Then seven feet of energy is coming into the pump. The pump develops 70 feet of head. (The pump can add 70 ft. of energy to the liquid.) The pump can lift the cold water 77 ft. (7 + 70 = 77 ft.).

Now let’s say the same thing in pressure. Let’s say we have a pump designed to develop 30 PSI. The pressure gauge at the pump suction nozzle reads three PSI. The discharge pressure gauge will read 33 PSI (3 PSI + 30 PSI = 33 PSI).

Now let’s answer your question — I’ll use pressure. You have a vented tank of cold water. The water fills the tank to 60 ft. above your pump. Let’s convert that 60 ft. to PSI (60 / 2.31 = 26 PSI). The pressure (energy in the fluid) at the bottom of the tank is 26 PSI. This is the “GPSHa” from the article you read. This pressure is composed of the water’s vapor pressure, the temperature, the elevation, and the atmospheric pressure.

Now let’s say that one more PSI of energy is lost (as friction) in the suction pipes, elbows, and fittings leading from the tank to the pump. Therefore, the pressure gauge reading at the suction nozzle of your pump would read 25 PSI. This is the NPSHa from the article you read. If your pump develops 975 PSI (converted from the motor), then the discharge pressure gauge reading will be 1,000 PSI. That’s how a pump draining a tank 60 ft. can get 1,000 PSI out the discharge nozzle.

You work with instrumentation. Did you notice that I gathered and used certain information from the pressure gauge at the bottom of the suction vessel? I got more information from the gauge at the suction nozzle of the pump, and still other information came from a pressure gauge at the discharge nozzle of the pump. Too often, I see pumps without gauges and talk with engineers who believe their pumps are “Bad Actors.” Instrumentation is critical to understanding a pump in its fickle system. The instrumentation tech is critical to pump reliability. It’s a shame that many instrumentation techs don’t know this.

I will speak for two days on “Pumps for Instrumentation Professionals” at the upcoming ISA (Instrumentation Society of America) Conference in Houston later this year (Oct. 1-2). Maybe you can come. We’ll answer many questions and resolve many doubts regarding pumps and their fickle systems. For more details and to register for this training event, click here.

Later & Stay in Touch,

Larry Bachus, “The Pump Guy”

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 larry@bachusinc.com or 615 361-7295.

www.bachusinc.com

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 Larry Bachus (a.k.a. “Pump Guy”) demonstrates the principles of NPSHr vs. NPSHa at his Aug. 18-20 Pump Guy Seminar in the Chicago area.

Larry Bachus (a.k.a. “Pump Guy”), a regular contributor to Flow Control magazine and a widely recognized expert on pumping technology, recently presented his Pump Guy Seminar in the Chicago area to an eager crowd of pump users. Here”s what some of the attendees had to say about this training event:

• “I attended your seminar this week in Chicago, and it’s already paying for
taught me about concentric reducers on suction lines and horizontal elbows on split case pumps, which will be quite helpful on several plant system designs I’m currently working on.”
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KEY SEMINAR TOPICS INCLUDE:
• Basic Pump Principles
• NPSH
• Cavitation
• The Affinity Laws
• Work & Efficiency
• Pump Classification
• Pump Curves
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• Bearings
• Pump Packing
• Mechanical Seals
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For a sampling of Larry”s latest “PUMP GUY” columns from Flow Control magazine, see:
“Cheat Sheets: Energy, Work & Power”
“Cheat Sheets: Unwritten Pump Rules”
“Cheat Sheets: The Affinity Laws”