Help with water pump calculations

In summary, the pump Jack found has a flow rate of 6 LPM and a head pressure of 80 psi. It would take 1-1½ minutes to fill a 2 gallon container with this pump.
  • #1
DannySmythe
18
2
I have a rain barrel 20 feet below a deck. I want to pump water from the rain barrel up to a small container (2 gallons) once a day. I'm thinking it should take less than 15 minutes to fill.
I've found a pump I like that has the following specs, but it may be too powerful, but it's cheap ($15)
Head Pressure 80 psi ( I think this converts to about 180 feet for water?)
Flow rate 6 LPM. (I assume that this is the flow rate with 0 head pressure through it's 1/2" port?)
How do I use the 6 LPM to calculate the flow rate through say a 1/2" hose?
I don't want it dribbling out and taking an hour to fill, but I also don't want it splashing out with force.
 
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  • #2
Flow rate stays constant, no matter the size of the hose, so it will always be 6 LPM.

Since 2 gallons is roughly 8 L, it will take about 1-1½ minute to do the job.

If you want to know the outlet velocity, then you have to divide the flow rate by the area of the hose, using the proper units. 1 L is 0.001 m³, so the flow rate is 0.006 m³/min. ½" is 0.0127 m, so the area is 0.7854 * 0.0127² = 0.000127 m². Thus the outlet velocity is 0.006 / 0.000127 = 47 m/min or 155 ft/min. That is less than 2 mph. If I didn't make any mistake, it means that if you let the hose flow on the ground, you could walk faster than the water front. I doubt one would consider this «splashing out with force».
 
  • #3
I'm going to disagree with Jack and ask that you send additional specs about the pump. I have a hard time believing the $15 pump is a positive displacement (constant flow) pump.
 
  • #4
I found a pump (Rule IL200) that had a chart of lift vs flow so I got it.
Just for fun here are the numbers:
Max Flow 3.3 GPM
Max lift 25'
Flow at 15' (my application) = 1.3 GPM
So it will take a couple of minutes to fill my 2 gallon container.
Thanks to all for info.
 
  • #5
I take it you found the performance curve? Yes, pump manufacturers will often quote max flow and max head, but that doesn't tell you what the flow is for your application: you need to find your operating point on the curve. Glad you figrued it out.
 
  • #6
russ_watters said:
I take it you found the performance curve? Yes, pump manufacturers will often quote max flow and max head, but that doesn't tell you what the flow is for your application: you need to find your operating point on the curve. Glad you figrued it out.
The chart only had a few entries. Lucky for me one was 15' head.
But in thinking about it, assuming the hose is the same diameter as the pump port, isn't the flow just proportional to (Max head - App head)? Where App head is the head needed for the application.
For example at an 'App head' of 0' the flow will be the rated flow F and at an 'App head' of 'Max head' the flow will be 0. So can't I just draw a straight line between these 2 points and pick my operating point? Or is it really that non-linear?
 
  • #7
1) It's non-linear, very much so.
2) Be aware that 15' is the total dynamic head. This is the static head (difference between the elevation of the pipe discharge and the surface elevation of the supply water) plus the dynamic head ("pressure" loss due to the friction between the fluid and the pipe as it flows through plus all losses due to direction changes and fittings). At 1-2 gpm in a 0.5" pipe that shouldn't be all that much, but depending on the length of pipe, could start to add up. For 25 ft. of hose I'd estimate an additional 1 ft. of dynamic head at 1.3 gpm. Since fluid systems tend to balance out, I'd take a thumbsuck guess and say you'll likely see something like 1.2 gpm. Not that that makes a huge difference for your application, but you should be aware of how pumps are actually sized for next time.
 
  • #8
Good to know.
Thank you.
 

Related to Help with water pump calculations

1. How do I calculate the required flow rate for my water pump?

The required flow rate for a water pump can be calculated by dividing the total volume of water needed by the time it needs to be pumped. This will give you the flow rate in gallons per minute (GPM). For example, if you need to pump 1000 gallons of water in 5 minutes, the required flow rate would be 200 GPM.

2. What factors should be considered when selecting a water pump for a specific application?

There are several factors to consider when selecting a water pump, including the required flow rate, head (vertical distance between the pump and the highest point of discharge), horsepower, and the type of pump (centrifugal, submersible, etc.). It is also important to consider the type of liquid being pumped and any special requirements, such as the need for a corrosion-resistant pump or one with specific pressure capabilities.

3. How do I calculate the total dynamic head (TDH) for my water pump?

The total dynamic head (TDH) is the total resistance that the water pump must overcome to move the desired amount of water. It is calculated by adding the static head (vertical distance between the pump and the highest point of discharge), the friction loss (due to pipes, fittings, etc.), and the pressure head (any additional pressure needed). This calculation will give you the TDH in feet.

4. Is it necessary to account for the pump's efficiency when calculating water pump requirements?

Yes, it is important to account for the pump's efficiency when calculating water pump requirements. The efficiency of a pump is the percentage of the input power that is converted into useful work. A more efficient pump will require less power to achieve the same flow rate and head, resulting in lower operating costs.

5. What is the best way to determine the power requirements for a water pump?

The power requirements for a water pump can be determined by multiplying the flow rate (in GPM) by the total dynamic head (in feet) and dividing the result by the pump's efficiency. This will give you the required power in horsepower (HP). It is also important to factor in any additional power needed for features such as variable speed drives or automatic controls.

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