Measuring static suction head from pump discharge pressure 1

[beps]

Hi all,

I have a pump pulling up water from a large well through a suction pipe (1 inch PVC pipe) with a foot valve. Suction pipe travels horizontally from pump for 15 feet before descending into well.

Access to the well is very difficult and therefore, measuring the amount of water in the well is not feasible on a regular basis.

However, if I knew the static suction head I would have a very good idea of how full the well is.

I have a measurement of the pumps discharge pressure which currently is 6 bar (200 feet head) during operation.

The value of static discharge head is nil (0), since the pump provides distribution of water at it's same level.

Your help would be much appreciated!

Cheers

 

[rotw]

Hs: Static suction head
Pd: Pressure at pump discharge = 6 bar
P_loss: Pressure losses through suction
Patm: Atmospheric pressure
Delta_H: Total pump head (read from pump curve)
rho: Fluid (water) density

Hs = Delta_H - (Pd - Patm + P_loss) / rho x gravity

P_loss is to be estimated using a basic hydraulic loss calculation for the losses through pipe and components (valve etc.) at your operating flowrate (the flowrate that gets you 6 bar on the discharge).

 

[LittleInch]

What's the question?

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[beps]

Question was: How can one measure static suction head from pump discharge pressure?

thanks rotw for your answer. However I'm a bit confused, since all measurements are in meters / feet of head, except rho and gravity, which have different units.

These are the numbers I have in meters of head:

Delta H = 55m
Pd = 61.2m
P_loss: 2m
Patm = 10.3m
Delta H = 55
Rho = ?
Gravity = ?

Am I missing something?

 

[rotw]

P and H are symbols not for the same physical quantities...should not mix ...
In SI unit -> P (Pa) = rho (kg/m3) x g (9.81 m/s2) x H (m)

Pd = 61.2m 6 x 10^5 Pa
P_loss= 2m 0.2 x 10^5 Pa
Patm = 10.3m 1.013 x 10^5 Pa
Delta H = 55 m
Rho = ~1000 kg/m3
Gravity = ~9.81 m/s2

Hs = 55 - (6 - 1.013 + 0.2) x 10^5 / (1000 x 9.81) = 2.12 m

By the way, is this an academic exercise?

 

[beps]

Fantastic! Thank you so much!

Actually this is a real life situation have a 150 year old well under the living room of my house. Access is via large heavy trap door which has cement tiles above it to fit in seamlessly with the rest of the floor and sealed at the edges... So not at all convenient to open often. The well receives rainwater but has no overflow. So need to make sure I have an approx idea of how full it is.

Thanks again

 

[rotw]

no problem.
small correction on acceleration unit

 

[beps]

Ok excuse my ignorance but when I calculate:

55 - (6 - 1.013 + 0.2) x 10^5 / (1000 x 9.81)

I am getting 507.95 as the answer and not 2.12

What am I doing wrong?

 

[rotw]

you did not do the calculation correctly...

PS: By the way, out of simple curiosity, how did you calculate the losses on suction (to arrive to 0.2 x 10^5 Pa)?

 

[beps]

Ok used a scientific calculator and got to your answer. And 2m is just about where the water level should be as I only recently checked.

For the losses on suction I used an Online Hazens-Williams Calculator...

 

[LittleInch]

Sorry guys, but for me this doesn't add up.

Rotw, As soon as you introduce Patm into your equation, it means the Pd needs to be in absolute units. Hence your Pd is 7 bara or in metres is now ~72m. Also Hs should really be a negative number it should be head liquid - head of inlet in relative terms??

If you look at it in gauge pressure terms then you have a negative pressure at the inlet as you're lifting water and have a friction loss, so inlet pressure is ~ -0.4 barg

You say the differential head of the pump is 55m so ~ 5.4 barg, makes the discharge 5 barg, not 6.

so something isn't correct here.

The difficulty of doing this method is essentially the accuracy of your differential head. Depending on the pump curve, a small error in flow calculation can make a big difference in differential head. when you're only dealing in a few metres it will make a big difference. Your maximum practical lift is probably no more than 5 to 6 metres.

Much better IMO, to simply install a pressure gauge on your inlet which measures in absolute or negative gauge pressure and then just subtract the friction losses and convert to metres head.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.