The amazing James Watt was born in 1736 at the dawn of the industrial revolution. He was reared and educated in the coal and iron mining region of Scotland. In school, James studied math and prepared for a career as an instrumentation technician. He eventually became a mechanical engineer. His mission in life was to see machines replace human and animal labor.
Mining was, and still is, a labor intense industry. For tasks beyond the ability of a grown miner, the mining industry bred especially small horses that could go into the mines and do the pulling and hauling that a grown man could not do. It was one of man’s early attempts at genetic manipulation.
For menial labor, children were sent into the mines. Before the Child Labor Laws, children were used in the bilges of the mines because they had small lungs and didn’t consume the air/oxygen that a full-grown miner would consume.
READ ALSO:
Pump Guy Cheat Sheet No. 1:
The Relationship Between Pressure & Head
Pump Guy Cheat Sheet No. 2: 2.31 or .433?
The numbers may be different, but the result is the same
A pump’s work (output) is called Water Horsepower (WHp). It is a combination of elevation (called head rated in feet), and flow (rated in gallons per minute). Expressed as an equation:
WHp = H x Q/3960
The number 3,960 converts power into pump language. Put this into your “CHEAT SHEETS” and remember it. I’ll explain.
As stated earlier, one horsepower is 550 foot/pounds per second, and most pumps are rated in flow per minute (GPM). Multiply 550 by 60 seconds in a minute, and we have 33,000 foot/pounds per minute. Next, a gallon of water (at sea level and 70 F) weighs 8.333 pounds. Divide the 33,000 ft. pounds by 8.333 pounds per gallon and we have 3,960. The 3,960 is a horsepower expressed in pump terminology.
In James Watt’s day, water was the only liquid that needed to be moved in large quantities. This was cold water for bathing, drinking, and lifting up from the bilge of a coal mine. Beer, paint, solvents, milk, lamp oil, acid, and whiskey were made in bottles, buckets, or casks and carried from one place to another.
As technology advanced, pumps were employed to transport liquids other than water. Another element, the specific gravity was incorporated into the basic WHp formula:
WHp = H x Q x sp. gr./3960
The above is the basic formula for water horsepower and for liquids other than water. It’s a good number, but not useful. We never need to know this number.
Frequently, we need to know a variation of this number. We need to know the BHp, or brake horsepower to drive the pump. If the pump were 100 percent efficient, then the BHp would be equal to the WHp. Pumps are not 100 percent efficient. So, this brings us to another variation on the formula. Now we have:
BHp = H x Q x sp. gr./3960 x eff.
This is a relatively easy formula to understand. Lets work with it a little.
Imagine that we have a pumping system that requires 600 GPM of water, while generating 40 PSI. In pump terminology, 40 PSI is the same as 92 feet of head (We know this from a previous CHEAT SHEETS column (Feb. ’08, page 40)). Let’s say our pump is 77 percent efficient. The specific gravity of water is 1.0. To apply the formula, the H = 92, the Q is 600, and the efficiency = 77 percent. Now we have:
BHp = 92 x 600 x 1.0/3960 x 77%
BHp = 18.1 horses
We need 18.1 horses to power this pump. We must install a 20-Hp motor onto this pump because no one makes an 18.1 Hp motor. The monthly electric bill will be based on 13.5-kilowatts (Kw = Hp x .746).
Many pump specifications don’t mention or require efficiency. The specs mostly deal with the needs of the system, meaning the 92 ft. @ 600 GPM. Purchasing agents can fall into the price trap.
Suppose two different suppliers have a pump that meets the system’s requirements (92-ft. @ 600 GPM). What if one pump sells for $800 less than the other pump? What if we could save $800 by buying another comparable pump that meets the needs of the system. If the cheaper pump were 60 percent efficient, we’d have some different horsepower figures:
BHp= 92 x 600 x 1.0/3960 x 60%
BHp = 23.2 horses
The cheaper pump requires 23.2 horsepower. The monthly electric bill will be based on 17.3 Kw, and you must buy a 25 Hp motor to operate the cheaper pump with the reduced efficiency.
The difference is 5.1 horses or 3.8 kilowatts. If the pump runs for a year, this would mean an additional $3,332 dollars per year in electricity to run the inefficient pump at 10¢ per kilowatt/hour. And this $3,332 would be on top of the additional cost for the larger (25-Hp) motor. How do you feel about your $800 savings now?
An even cheaper pump might perform the same work and be only 50 percent efficient. This pump would burn 27.8 horses of electricity, and it would require a 30-Hp motor. If it ran 24-hours per day, this would be an additional $6,340 in the electric bill per year compared to the first pump.
If you’ve been following the Pump Guy in Flow Control magazine, you know I started my maintenance career as an apprentice grunt in a Birmingham steel mill. And you also know I was a pump mechanic in the Navy. I’ve stood by helpless while my bosses, chiefs, and purchasing agents made these poor decisions. And guess who was stuck with the maintenance on those cheap pumps?
Today, many government agencies send their engineers and techs to the Pump Guy Seminars. And the engineers all say, “But Larry, you don’t understand; we’re under instructions to buy the cheapest pump based on a competitive bid.” It is truly sad.
These are good people, and they understood the information, but they have their instructions. Who is responsible for this? Remember the efficiency next time you go to buy 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 [email protected] or 615 361-7295.
www.bachusinc.com