## Wednesday, June 27, 2012

### Designing Irrigation Systems

A college mentor—an engineer by the way—used to always say “two plus two has to equal four”. He and an economics professor would conduct programs together from an engineering and economics perspective. The economist would begin by saying that the engineer carries calculations out to the fourth decimal then adds 10 % for error. Then the engineer would say if you laid all economists end to end the world would be a whole lot better off.

The point is that with an irrigation system, the output will not exceed the input. The input in such a case would be the flow rate or capacity of the pump and/or water supply. Without getting too technical there are other parameters that will interfere with the output as well. Friction loss in the pipe or tubing, elevation differences, equipment efficiencies, pressure losses through sprinklers and/or emitters and evaporation are aspects that interfere with design.

It makes little difference whether it is a 160 acre center pivot or a one acre drip irrigation system. These things have to be accounted for. However, the pump flow rate is probably the starting point. As an example, let’s look at two different scenarios. Here we’ll assume all the pressure losses, sprinkler pressures, elevations, etc. have been accounted for.

First example: One acre of blueberries

Blueberry spacing = 4 feet

Row spacing and length = 12 feet apart & 100 feet long

Water flow rate = 2.3 gpm

Drip emitter spacing and capacity = 2 feet, ½ gallon per hour (gph)

How many rows will the pump flow rate handle?

Again assuming all the pressure losses, etc. has been accounted for;

2.3 gpm x 60 min/hr = 276 emitters

½ gph per emitter

276 emitters x 2 feet (spacing) = 552 feet of row

So if the rows are 100 feet long, that pumping capacity will only water 5 ½ rows or only cover about 15 % of the patch. The one acre patch then would have to be watered in seven zones. To reduce the number of zones, the pump capacity has to be increased. Theoretically, to operate as one zone, pump capacity would have to increase from 2.3 gpm to 16 gpm assuming there are no additional efficiency losses.

Second example: 160 acre center pivot on corn

Water flow rate = 1000 gallons per minute (gpm)

There are 27,154 gallons in one acre-inch (A-In) which is one inch of water covering one acre. So if one inch of water was to be applied to the 160 acres, how long would it take?

160 acres x 27,154 gallons = 72 hours or 3 days

1000 gpm x 60 min/hr

So if the acres are doubled, either the time doubles or the flow rate has to double. If the pump has the capability of increasing the flow rate, pressure losses in the pipe increase. The bottom line is that changing one parameter changes everything else.

Equipment and material suppliers generally have a good handle on irrigation system design. Just remember 2 + 2 has to = 4.

(By Ed Browning, MU Extension Natural Resource Engineering Specialist)