Flow in a flowing system always goes from higher pressure to lower pressure in that the pressure provides the driving force to move the fluid or gas. Pressure usually drops uniformly as a function of the square of net flow increases. Therefore, pressure at the flow source must quadruple to double flow in most flow piping systems. On this basis, and the wealth of physical data for many piping and tubing systems, you can calculate water flow from net pressure changes in a defined system.

- Skill level:
- Easy

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## Instructions

- 1
Define the flow-pressure relationship in the application. In this example, clear filtered water in a 4-foot-high swimming pool is being drained to fill an in-ground pool that will replace it. The new pool is 100 feet away, and the ground there is about 5 feet lower, so a 100-foot-long garden hose with a smooth inner diameter of 5/8 inch is being used to let gravity do its work of draining the old pool. From this information, you can calculate the water flow rate as the first pool empties from static head pressure alone.

- 2
Calculate the pressure head heights at the beginning and end of the emptying process. With the first pool filled, the water level is 4 feet high. The water flows through the hose from a drain at the bottom of the first pool, and from there drops 5 feet over the 100 feet of ground to the top of the in-ground pool, where it pours in from the open hose end at the top. Static pressure height therefore equals 4 feet + 5 feet = 9 feet when draining begins, and 5-feet when the pool is empty.

- 3
Calculate the head pressure at each of the head heights. Since each 2.31 feet of water height converts to 1 pound per square inch (psi), the available static pressure at a sealed hose outlet will be 9/2.31 = 3.9 psi when full, and 2.16 psi when empty.

- 4
Calculate the flow as a result of pressure at both pressure values from Step 3. Since the 5/8 inch inner diameter (ID) inner hose liner is smooth PVC plastic, its flow performance is virtually identical to PVC plastic pipe of the same ID, which happens to be nominal ½ inch Schedule 40 PVC pipe with a 0.622-inch ID. Consulting the pressure loss chart for this PVC pipe, it has a loss of 35.5 psi at a flow of 10 gpm. Dividing the 3.9 psi above by 35.5 psi = 0.1098, and taking the square root results in 0.331 X 10 gpm = 3.31 gpm for the starting flow. Dividing 2.16 psi by 35.5 psi and taking the square root results in 0.2467 X 10 gpm = 2.466 gpm.

- 5
Validate the flow results for this equivalent-pipe pressure-to-flow calculation method against an online calculator for garden hose. Dividing the 3.9 psi pressure loss value by 40 psi (normal water spigot pressure, which would drive 11 gpm with this hose) and extracting the square root yields 0.3122 X 11 gpm = 3.43 gpm. This shows a difference of (3.43-3.31 gpm)/3.31 gpm of 3.7 per cent, which is reasonable for flow in similar plastic pipes. Dividing 2.16 psi/40 psi and extracting the square root yields 0.232 X 11 gpm = 2.55 gpm versus 2.466 gpm using the chart, again a reasonable 3.4 per cent difference.

#### Tips and warnings

- Hoses provide a great way to transfer water from a higher place to a lower place by gravity if you have the time to allow them to work.
- Always wear safety eyewear when working with pipes and flow systems.

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#### References

- Arian Industrial Control and Instrumentation: ISO-5167 standard and its mass flow rate formula
- Aubuchon Hardware/Learning Center: All about garden hose/flows
- US Department of Energy/Ask a Scientist: Garden Hose Pressure
- Washington State University Extension/Irrigation: Garden Hose Flow Rate and Time