Most people believe the major cost of their water feature is the initial installation, but with features in the range of 1,500 gallons and up, chances are that the operating costs of the pump will exceed construction costs over the life of the feature. Hard to believe? Consider a 10´ x 16´ pond about 2´ deep. The pond’s volume should be recirculated at least once per hour, around 2,400 GPH, but these days most of my clients want to see about double that flow, so let’s figure the actual flow of 4,800 GPH over the life of the feature.
A typical direct-drive, solids-handling pump might be rated at 1,200 Watts, so the only other information needed is the average cost per kilowatt per hour from the electric bill. Let’s assume it’s $0.10/kW/hr; that is, 10¢ per thousand Watts per hour:
Hourly Cost to Run = Pump Wattage x $/kW/hour
1,200 Watts x .10/1,000Watts/hr = $0.12/hr, or 12¢ per hour to run.
From there the rest is easy. The cost per month is the hourly cost times 24 hours times x 30 days = 720 hr/mo; so Monthly Cost to Run = Pump Wattage x $/kW/hour x 720 hr/mo
The cost per month is therefore 12¢ times 720, or $86.40 per month to run the pump continuously. See where this is going? Add in a new replacement pump at a modest $500 every 5 years and that’s almost $11,000 over the 15-year life of the feature, just in pumping costs. At that rate, it’s easy to see the importance of the twin challenges of proper pump selection. First, you’ll want to choose the pump that’s the least costly to run and still does the job. We’ll talk about address this in Choosing the Right Pump next issue. Today I’m going to talk about getting the most out of whatever pump you’ve got, because what most people overlook is the right plumbing to get the most GPH possible.
Why pay the dough and not get the flow?
Can you imagine paying close to $100 per month and getting only half the flow the pump is capable of? It happens every day, everywhere. Before I actually measured what was going on, I always thought that the outlet size of the pump was a good indicator of what tubing size to use. I’d see the 1.5˝ or 2˝ outlet and set up the tubing or piping accordingly. Totally coincidentally, I was often unhappy with the output of the pump, and returned a lot of “bad” pumps. Turns out it wasn’t the pumps that were bad, it was my understanding of the plumbing required to deliver full flow. The loss of GPH that I was complaining about was ALWAYS a result of friction in my plumbing system. Enter Deb Spencer of Water’s Edge, in the nick of time. In her marvelous ‘Pump It Up!’ lectures Deb has been teaching water feature contractors like me how to use Friction Loss Charts to calculate actual flows through any size pipe for years. The idea is simple: use the right size pipe to minimize friction losses, and you’ll get all the flow you’re paying for. Don’t, and you’ll have to upsize pumps and operating costs, sometimes burning out pumps prematurely to boot.
Guidelines to Minimize Friction Loss
Using the charts, I came up with these simple guidelines. (See Figure 1) As you can see, some of these numbers look a little off. Everybody knows you can push a lot more than 1950 GPH through 1½˝ pipe, right? Absolutely. The key is, pumps have to work much harder to push through smaller pipe, and you lose flow to friction. If you want to maximize the flow of just lifting water from one level to the next while minimizing stress on pumps and plumbing, these limits, corresponding to 5 feet per second of water velocity, are what every pipe and irrigation manufacturer recommends.
Seeing is Believing…
Predictably, shoving another set of “Guidelines” under people’s noses and demanding that they follow them ‘for their own good’ wasn’t very successful, so I set up this simple rig to visually demonstrate what is going on. It’s just 4 equal lengths of 1¼˝, 1½˝, 2˝ and 3˝ corrugated tubing sequentially attached to the same pump and used to fill a bucket. Using a pump that can (optimally) deliver 6,600 GPH at 1´ of head, 1¼˝ tubing fills a ten gallon bucket in about 30 seconds. Switching to 1½˝ cuts the time IN HALF, to about 15 seconds, and through 2˝ pipe it fills in a little over 8 seconds! Most folks don’t want to bother, but switching up to the recommended 3˝ pipe increases flow by a further 25%, filling the bucket in 6 seconds flat!
But don’t take my word for it – try it yourselves! (But use a big bucket – all that flow into a 5-gallon pail splashes like a Shamu show!)
Afterword – Circumference = 2πr versus Volume = πr2
For those of you who have to know why, I offer these graphic explanations. (For the rest of you, skip it, there’s no quiz this week.) Remember the formulas above from Geometry? (I didn’t either.) The dramatic results of the Bucket Test show the relationship between the inner circumference of the pipe, where friction happens, and the pipe’s area, which determines the flow it can handle. Take a look at this diagram. (Figure 2) Friction at the inner surface of the pipe is represented by arrows. With water flowing at the same speed in each pipe, the friction generated should be in direct proportion to the circumference. Looking at the proportions, when the small circle has a radius of 1, the circumference is 2π. The large circle will have a circumference twice as large at 4π, so we would expect about double the friction in the larger circle. However, the total flow a pipe can handle is proportional to the area of the pipe, which grows exponentially, while tubing size and circumference increase only arithmetically. With r = 1, the area of the small circle is π; the larger is 4π. With four times the area we can expect quadruple the flow through the larger pipe with only double the friction.
The proportion of circumference to area changes radically as the pipe diameter goes from small to large. In 1˝ tubing the radius of ½˝ gives a circumference four times greater than the area (C = π, A = ¼π), so friction is exaggerated and flow minimized. As tubing size increases the area increases exponentially, so the larger the pipe, the proportionately smaller the friction losses. 2˝ pipe has a circumference only double the area (C =2π, A = π). 4˝ pipe has circumference and area equal at 4π, and larger diameter pipe has greater area than circumference, maximizing volume and minimizing friction losses. This is why small pipes are so prone to friction losses, and why larger pipe increases so much in carrying capacity as friction drops.
Next issue we’ll talk about Choosing the Perfect Pump, from the types of pumps to choose from, to what size to use, to how to cut electric costs in half. See you then!