Dillon Nichols

pollinator

Posts: 597

Location: Victoria BC

27

posted 1 year ago

So my parents are tired of watering by hand. I offered to help set up some irrigation before I take off for parts unknown.

I've measured the area they need irrigated, figured out the maximum required inches per week, and converted the resulting acre-inches into gallons.

Where I'm having trouble with the math is figuring out how long the irrigation can run without runoff. I realize this is a bit of an exercise in aggravation; it won't be particularly hard to observe this once the irrigation is installed, and act accordingly from there. In any case, the soil is such a mixed bag it will probably vary substantially between beds. However, now that it has annoyed me, I want to sort out the math even if it is of dubious utility.

I found this page with an example of what I need... sort of. http://extension.psu.edu/plants/vegetable-fruit/news/2013/determining-how-long-to-run-drip-irrigation-systems-for-vegetables

So, according to their example and chart, an available water holding capacity of 0.12 inches of water/inch depth of soil and a flow-rate of 0.45GPM per 100' means a max time per application of 110 minutes. They specify a 10-inch deep rootzone, and that irrigation is happening at 50% soil depletion.

0.45GPM*110 minutes=49.5 gallons for a 100' row 30" wide, aka 250sf.

A cubic foot of water would cover 12 feet with one inch of water. With 7.48 Gallons per cubic foot, this gives us 6.62 cubic feet of water, or ~79.4 square feet at 1" depth. Spread over the whole 250sf row, this is 0.3176" of water.

Using the holding capacity provided for the example soil of 0.12 inches of water per inch of soil depth, multiplied by the stated depth of 10", multiplied by the area of the row, multiplied by the available water holding capacity:

0.12*10*250*0.5=150 inches, which spread over the entire row would become 0.6" of water.

Obviously something is wrong here, since these numbers should match, no? Anyone see where I've gone wrong, presuming it is I who have gone wrong? Or, have a better example?

I've measured the area they need irrigated, figured out the maximum required inches per week, and converted the resulting acre-inches into gallons.

Where I'm having trouble with the math is figuring out how long the irrigation can run without runoff. I realize this is a bit of an exercise in aggravation; it won't be particularly hard to observe this once the irrigation is installed, and act accordingly from there. In any case, the soil is such a mixed bag it will probably vary substantially between beds. However, now that it has annoyed me, I want to sort out the math even if it is of dubious utility.

I found this page with an example of what I need... sort of. http://extension.psu.edu/plants/vegetable-fruit/news/2013/determining-how-long-to-run-drip-irrigation-systems-for-vegetables

So, according to their example and chart, an available water holding capacity of 0.12 inches of water/inch depth of soil and a flow-rate of 0.45GPM per 100' means a max time per application of 110 minutes. They specify a 10-inch deep rootzone, and that irrigation is happening at 50% soil depletion.

0.45GPM*110 minutes=49.5 gallons for a 100' row 30" wide, aka 250sf.

A cubic foot of water would cover 12 feet with one inch of water. With 7.48 Gallons per cubic foot, this gives us 6.62 cubic feet of water, or ~79.4 square feet at 1" depth. Spread over the whole 250sf row, this is 0.3176" of water.

Using the holding capacity provided for the example soil of 0.12 inches of water per inch of soil depth, multiplied by the stated depth of 10", multiplied by the area of the row, multiplied by the available water holding capacity:

0.12*10*250*0.5=150 inches, which spread over the entire row would become 0.6" of water.

Obviously something is wrong here, since these numbers should match, no? Anyone see where I've gone wrong, presuming it is I who have gone wrong? Or, have a better example?

Bryant RedHawk

gardener

Posts: 2289

Location: Vilonia, Arkansas - Zone 7B/8A stoney, sandy loam soil pH 6.5

183

posted 1 year ago

1) width of the bed, which is generally 30 inches or 2.5 feet;

2) number of gallons in an acre inch of water: 27,154;

3) number of square feet in an acre: 43,560 and; 4) drip tube flow rate: this varies, for the example we’ll use 0.45 gpm/100 ft.

First, determine how much drip tape is needed by dividing the area in an acre by the row spacing. 43,560 ft2 ÷ 6 ft = 7,260 ft of drip tape needed.

Next, determine the area of the acre to which the water will be applied. 7,260 ft drip tape × 2.5 ft wide beds = 18,150 ft2. (0.42 acre.)

Determine the number of gallons of water needed to apply. 27,154 gal/acre-in × 0.42 acre = 11,405 gal.

Determine the number of 100 feet units of drip tape used. 7,260 ft ÷ 100 ft = 72.6 units.

Determine the number of gallons per minute needed using the drip tube flow rate. 72.6 units × 0.45 gpm = 32.67 gpm. If your well does not have this capacity, you will need to water in zones.

Lastly, determine the amount of time to run the system. 11,405 gal needed/32.67 gpm = 349 minutes or 5.8 hours.

It appears that your 0.6 is a G.P.H. adsorbed calculation.

I think you should first get actual figures for the space you are going to irrigate along with the type of irrigation you are going to use. I also don't see your time computation in this which is a necessary part of the calculations.Using the holding capacity provided for the example soil of 0.12 inches of water per inch of soil depth, multiplied by the stated depth of 10", multiplied by the area of the row, multiplied by the available water holding capacity:

0.12*10*250*0.5=150 inches, which spread over the entire row would become 0.6" of water.

1) width of the bed, which is generally 30 inches or 2.5 feet;

2) number of gallons in an acre inch of water: 27,154;

3) number of square feet in an acre: 43,560 and; 4) drip tube flow rate: this varies, for the example we’ll use 0.45 gpm/100 ft.

First, determine how much drip tape is needed by dividing the area in an acre by the row spacing. 43,560 ft2 ÷ 6 ft = 7,260 ft of drip tape needed.

Next, determine the area of the acre to which the water will be applied. 7,260 ft drip tape × 2.5 ft wide beds = 18,150 ft2. (0.42 acre.)

Determine the number of gallons of water needed to apply. 27,154 gal/acre-in × 0.42 acre = 11,405 gal.

Determine the number of 100 feet units of drip tape used. 7,260 ft ÷ 100 ft = 72.6 units.

Determine the number of gallons per minute needed using the drip tube flow rate. 72.6 units × 0.45 gpm = 32.67 gpm. If your well does not have this capacity, you will need to water in zones.

Lastly, determine the amount of time to run the system. 11,405 gal needed/32.67 gpm = 349 minutes or 5.8 hours.

It appears that your 0.6 is a G.P.H. adsorbed calculation.

We love visitors, that's why we live in a secluded cabin deep in the woods. "Buzzard's Roost (Asnikiye Heca) Farm." Promoting permaculture to save our planet Tuwa tokiya kola

Watch the full PDC and ATC from home. As much or as little as you want: http://kck.st/2q6Ycay. |