Larry Bachus
(a.k.a. “Pump Guy”)

From: G.M.
To: Larry Bachus
Subject: Efficiency

Larry,

I work for a chemical company in Europe. I have your excellent book, Everything You Need to Know About Pumps, and I read the Pump Guy in Flow Control magazine every month. I am charged with improving pump efficiency at our facility.

Due to a couple of mysterious failures, we have a number of vertical pumps in the workshop. After checking them, we’ve found the wear rings have nearly double the design clearances.

That being the case, I am using your calculation as an aid to show the process engineers the true cost of inefficiency. However, being of that ilk, they are sure to ask a number of pertinent questions when I present my case.

Looking at the formula you used in the January ‘09 Pump Guy Column in Flow Control, (“A Different Article”, page 42), could you give me some idea as to where the conversion factor (0.000189) is calculated from? I don’t need or expect chapter and verse, but I know I will be asked. I want to be able to reply rather than have a vacant expression wander over my face.

Again, thanks for your superb book. I have given the book to my maintenance techs, and they’re quite comfortable with it, as opposed to those weightier textbooks and manuals we’re all too familiar with.

G.M.

From: Larry Bachus
To: G.M.
Subject: RE: Efficiency

Hello G.M.,

Thanks for the kind words on the book and the column. I applaud your persistence. Many good presentations fall flat when someone asks the unexpected, and the presenter doesn’t know the answer.

These days, many people in industry are intrigued with the buzzword “Lifecycle Cost.” (I hate buzzwords.) Every day, the lifecycle cost of a pump becomes increasingly illusive, as more and more elements (procurement cost, disposal cost, “J” factor cost, storage cost, etc.) are added to the supposed, proposed total lifecycle cost. (Can you see why I hate buzzwords?)

However, one element of the “lifecycle cost” is easily quantified and practical to understand. This is the energy cost of operation. How much energy, in kilowatts, at a cost per kilowatt-hour, does a pump consume in operation? And, 0.000189 is the conversion factor.

I admit, 0.000189 is a strange conversion factor. I would venture to say it isn’t in any university textbook. It converts mechanical and hydraulic terminology into pump terminology, and then prepares the terminology for conversion into the cost of energy.

In the United States, we are still stuck in imperial measurements (pressure in PSI, horsepower, flow in GPM, and feet of head). We rate pumps in horsepower, although operating cost (power consumption) is rated in kilowatt-hours. The formula for yearly operating costs is:

0.000189 x gpm x TDH x $kwh x sp.gr. x 8,760 / PumpEff. x MotorEff.

Your question deals with the conversion factor (0.000189). The factor derives from hydraulic and mechanical terminology in imperial units, and then converts the units into metric kilowatts.

You need these few tidbits of information:
• James Watt (the electricity unit ‘watt’ is named for James) experimented with a workhorse and determined one horsepower = 550 ft. pounds per second.
• Next, there are 60-seconds in a minute.
• One gallon of ambient water weighs 8.333 pounds.
• And finally, kilowatts = horsepower x .746.

Now, let’s apply these tidbits of information. Here we go:

0.000189 = .746 / 550ft.lbs/sec. x 60sec./min. / 8.333lbs./gal.

The formula can be reduced to:

0.000189 = .746 / 3,960

The conversion factor (0.000189) is based on water. In the “operation cost” formula: The specific gravity incorporates liquids other than water; the motor efficiency incorporates losses in the electric motor.

The “$kwh” can personalize the formula to your actual cost and currency (Dollars, Euros, Pounds, Rupees, Pesos, Rand) of electricity at your location.

The “8,760” is the total hours in a year (365 days x 24-hrs. per day). This can be adjusted to the actual operating hours. For example, if the pump operates eight hours per day, five days per week, in a 50-week work year, then you could use 2,000 operating hours per year.

And, that’s how you manipulate the formula. This formula will slowly vanish as older engineers retire and as imperial units disappear in favor of metric units.

I run into Brits, Aussies, Canadians and Gringos all over the world. I was recently giving pump lectures in India. The older guys like the formula. The younger guys say, “What?”

A number of years ago, I operated my own pump rebuild shop in the Caribbean. I had to learn and use the metric system of measurements. Bouncing decimals is so easy.

Only the United States, Myanmar (Burma) and Liberia cling to imperial units. I only wish the progenitors of the metric system had considered metric time. There’ll be more on that later. Stay in touch.

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 a recent Pump Guy Seminar.

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