# What is the formula to work out time to remove water from an open vessel with vacuum?

We have an open container sitting at 1000mm off the ground and a receiver tank sitting one metre above it. Container holds 400l of water and is connected to the receiver by 80mm id pipe. I was planning on using extraction fans sitting in the open end of the receiver to create a vacuum to remove the water.

What is a formula I could use to work out how inHg needed to be produced by the fans to remove water in x amount of time?

So if your fan doesn't produce a minimum of 1m of water (which is about 2.9 inHg), there would be no flow. Afterwards, it depends upon your fan curve. You would find that the system curve would likely equal 2.9 inHg + K*Q^2, where Q is the flowrate and K is a factor that depends upon how you move the flow - note with smooth curves K is small, and with sharp turns and disruptive contours, K is large.

Wherever your pump curve intersects this system curve is the flow rate. Plain and simple. Once you know the flowrate, divide the volume to move by the flowrate to gain the time estimate.

Edit:

With a wet/dry vacuum style compressor, you're most like to see a curve similar to this one:

At the 40" of water (approx. 1 meter), the vacuum would remove 300 cfm of air (and therefore water) a minute. At less than 100 cfm removal, the vacuum would stall. So, you should expect removal between 100-300 cfm, or between 3-9 seconds.

• Thanks Mark, id bump it but i dont have enough points. i dont have any data on the fan, but i rigged it up and tested and it pulls 2.6m of water. Is there any way to work it out off that. The piping consists of only 2 130c/l rad elbows and 1m straight flex. what would be a guestimate on the K factor. Cheers Jun 5, 2018 at 6:38
• What kind of fan?
– Mark
Jun 5, 2018 at 7:00
• lets say im starting a bit back yardish with the tops off of 2 1200w wet and dry vacuums Jun 5, 2018 at 7:07
• your a legend, thanks for taking the time Jun 6, 2018 at 10:27