A friend contacted me for help in sizing a pump for a drip irrigation system for an exterior garden cultivation project. The idea was to store water in Rain Barrels to provide some head elevation, and use a pump to provide the additional head needed to adequately sprinkle the water to the specific portions throughout garden.
My friend provided me an aerial view of the planned construction of the distribution system along with some references to the diameter of pipes he was expecting to use.
One of the first things I would need to accurately size everything, is first knowing the final drip rate expected at the end. If an expected amount is really needed a recommended value could be 1/4 inch of water per day found through this Link.
The first thing I was going to do was calculate the major head losses across the system
Calculating the major head loss across the system, can be found using the formula below
$H_L = F * (L/D) * \frac{V^2}{2g}$
Where
- F: Friction factor due to roughness on the interior of the pipe
- L: Length of pipe
- D: Diameter of pipe
- V: Average liquid velocity within the pipe
- g: Acceleration of gravity
Looking off the provided reference guide, I made a table to start calculating the major head losses across all the reported stretches of pipe.
Color Ref. | Pipe Length [ft] | Diameter [inch] | Major Head Loss |
Red | 15 | 1 | 1.40625E-05 |
Red | 18 | 1 | 0.000016875 |
Red | 50 | 1 | 0.000046875 |
Red | 90 | 1 | 0.000084375 |
Red | 50 | 1 | 0.000046875 |
Blue | 12 | .75 | 0.000015 |
Blue | 15 | .75 | 0.00001875 |
Blue | 4 | .75 | 0.000005 |
I would want to know if the final exit height will be higher than the water surface, like if the rain barrels were located on a low elevation relative to the final destination of the plants. |
If the final drip line height is lower than the rain barrel surface water level height (This would result in needing a small pump if at all needed) |
Does the water have to be ejected from the opening out a certain distance to the plants? |
Then from the tees on the mainline a fitting.for poly-pipe would be attached then from.that line I would poke holes.for 1/4 inch transfer barbs and attach 1/4 inch drip.line to that.
$ \frac{ P_1}{ \rho g } + \frac{ ( V_1 )^2 } {2 g } + Z_1 - H_L + H_P = \frac{ P_2 } { \rho g } \frac{ (V_2)^2 }{ 2 g } + Z_2 $
$H_P = \frac{ (1 ft/s)^2}{ 2 * 32 ft/s^2 } + (1/2 ft) + 2.47 x 10^-4 ft $
No comments:
Post a Comment