Physics of Drip Irrigation: Pressure Calculations for Rain Barrel Design

[Cerumen]

I'm working on a drip irrigation project to present to the management at my work (a home improvement store) in hopes that my design will become a display in the store. However, even with my current knowledge of physics, I can't seem to figure out the following issue:

I'm using a 50 gallon rain barrel to collect rain water to power the irrigation system. Those not familiar with drip irrigation, it is simply a hose capped at one end with "drip locations" along its entire length, powered by some water source. The exit point for water to leave the barrel is located roughly 1.5 inches from the base of the barrel and has a diameter of 1/2 inch.

** If the rain barrel is completely full, sitting at ground level and a 10 foot x 1/2 inch irrigation hose is used (hose not perforated at this point), what pressure is exerted on the cap of the hose, ie the very end if stretched out in a line?

** Again, if the rain barrel maintains the same conditions (filled with water, hose dimensions and straight line) but is raised off the ground by, say 12 inches, what is the new pressure exerted on the cap of the hose? By what factor does this value increase as the barrel is elevated in 12 inch increments?

** How is the pressure effected across the length of the hose if 3mm perforations are made every 6 inches? Obviously, it will decrease, but by what extent?

** By what extent is this scenario changed with less water in the barrel? 1/2 full? 1/3 full?


There are more factors at play here than what I have outlined with these questions. However, with solid data on these inquiries, I can begin to calculate the other variables.
This design is clearly intended to work without the use of an electric pump.

Thank you in advance for any information in regard to this topic.

 

[Valerie Park]

The pressure exerted on the cap of the hose when the rain barrel is full and at ground level will be determined by the hydrostatic force. This is equal to the density of the water (1000 kg/m^3) multiplied by the acceleration due to gravity (9.8 m/s^2) multiplied by the height of the water in the barrel (1.5 inches). This gives you a pressure of 147.2 Pa or 0.02 psi. If the barrel is raised 12 inches off the ground, the pressure will increase by a factor of 4. The increased pressure will be 588.8 Pa or 0.085 psi. This factor will increase in multiples of 4 for each additional 12 inches of elevation. The pressure across the length of the hose with 3mm perforations every 6 inches will depend on the flow rate of water through the perforations. If the flow rate is low, the pressure will remain relatively constant. However, if the flow rate is high, then the pressure will decrease as water flows out of the holes. This can be calculated using Bernoulli's equation. The pressure will also decrease depending on how much water is in the barrel. For example, if the barrel is half full, the pressure will be half of the original value at ground level. Similarly, if the barrel is one-third full, the pressure will be one-third of the original value at ground level.

 

[sir-pinski]

First of all, congratulations on your drip irrigation project and your goal to have it displayed in the store! It sounds like you have a solid understanding of the physics behind drip irrigation, but let's go through your questions one by one to ensure we have a clear understanding of the pressure calculations.

1. If the rain barrel is completely full, sitting at ground level and a 10 foot x 1/2 inch irrigation hose is used (hose not perforated at this point), the pressure exerted on the cap of the hose would be equal to the weight of the water in the barrel, divided by the area of the cap. This can be calculated using the formula P = F/A, where P is pressure, F is force (weight of water), and A is area. The force can be calculated using the density of water (1000 kg/m3) multiplied by the volume of water in the barrel (50 gallons or 0.189 m3) and the acceleration due to gravity (9.8 m/s2). The area of the cap can be found using the formula A = πr2, where r is the radius of the cap (1/2 inch or 0.0127 m). Plugging these values into the pressure formula gives us a pressure of 68.5 kPa or 9.9 psi.

2. If the rain barrel is raised off the ground by 12 inches, the pressure exerted on the cap of the hose would increase due to the extra height of water above it. The new pressure can be calculated using the same formula as before, but with the height of water now being 12 inches less (since the barrel is raised). This would result in a pressure of 60.3 kPa or 8.7 psi. As the barrel is elevated in 12 inch increments, the pressure would decrease by the same amount each time (9.9 psi - 8.7 psi = 1.2 psi decrease for every 12 inches).

3. Adding 3mm perforations every 6 inches along the length of the hose would decrease the pressure across the hose. This is because the water is now able to flow out of the hose at these perforations, reducing the amount of water and therefore the weight of the water that is pushing down on the cap. The exact extent of this pressure decrease would depend on the number and size of the perforations, as well as the flow rate of