deep well

[mark aljon Matre]

can someone help or teach me how to use a deep well as a source in epanet???

here's my data

static water level 1

pumping water level 11

pump setting 24m

pump specs

flow rate 30 lps

THD 37.22m

[alphonse owusu]

Hi,

I normally use a  reservoir and pump connections for my deep well /borehole setups.

Diagram below illustrates better using your parameters

	Reservoir In Epanet

 

Node

Pump

PWL

SWL

Pumping Setting

Pumping water level

 

The reservoir in the model set to have two (2) water levels. A pattern with static water level and pumping water level. Connect your pump to the reservoir and then to your system.

Pump head should be high enough to do the TDH and static head. Ie check the resultant flow/head to have a feeling of the pump duty point on the pump curve.

 

Pump setting has minimal/no influence on the hydraulics of the pump. At 13m below pumping level, there is no issue with NPSH.

 

Alphonse

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[mark aljon Matre]

so i will use elevation of well minus the pump setting as the THD of the reservoir??

my reservoir's elevation is 118.

so 118m-24m = 94, is 94m is my THD for my reservoir??

[alphonse owusu]

If i understand you right, the levels you provided are referenced to the ground level or to a datum.

I.e static water level is 1m meaning 1m from the ground level at the borehole location.

If it so,

then head at the reservoir (representing the borehole)= ground level – pumping water level for steady state analysis.(118 – 11)

 

you are dealing with water levels in the borehole, the pump setting has no impact on the discharge head.

 

alphonse

From: epa...@googlegroups.com [mailto:epa...@googlegroups.com] On Behalf Of mark aljon Matre

Sent: 09 December 2015 11:05
To: Epanet and Development

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[mark aljon Matre]

thanks for the post and info,

now i understand this. I'm having a hard time because of their reference..

you're right my data are measured for ground level..

On Wednesday, December 9, 2015 at 2:54:42 PM UTC+8, mark aljon Matre wrote:

[Santiago Arnalich]

Raising water in water consumes almost no energy, so the depth at which the pump is installed is almost irrelevant. You can picture this by trying to lift a plastic bag full of water in a lake. It is only when you raise it up in the air that you will feel the weight

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[mark aljon Matre]

thanks sir, 

i would like to know if my thoughts are right..

my network/system is at elev. 100-120m while the TDH of my reservoir is 94m,

In order for the water to reach elev. 120m, I need to raise the head of my pump.

is that right???

sorry for these type of question, I'm a newbie in this Epanet program.

On Wednesday, December 9, 2015 at 2:54:42 PM UTC+8, mark aljon Matre wrote:

[Patrick Moore]

Mark, 

When evaluating the head of the pump you can often think of it in terms of going form water to water (or HGL to HGL) and adding some amount for headloss.  So in your case the suction water level would be 94m but you don't state what your goal system pressure is for customers at 100-120 M elevation or if the well pumps into a tank.  I would suspect each customer would need a minimum of 40-50 psi which is equivalent to 28-35 M of head.  As such the discharge HGL would be 120m + 35 m = 155M. 

Thus if you are lifting from 94 to 155m the head the static lift of the pump would need would be 155-94 = 61 m.  This is the static lift needed by the pump just to get water to barely flow.  Additional head at the pump would be necessary to overcome headloss.  If one does not know how much headloss to account for a general reasonable first guess is often 10 psi or 7.04 M  of headloss to assume.  If we add this to the static lift we would expect the pump likely needs 61 + 7 m or 68M of head to function properly and give customers a reasonable system pressure.  You stated the pump design head was already 37.22 m, which is less than this number.  You may wish to increase the head to 68 M or work out what is needed based on your specific conditions as 37 M would overcome the static lift, but would cause low pressures for customers as only 11m would be available for both friction loss and system pressure.

If you pump from one tank to another I usually estimate overflow HGL to Overflow HGL plus 10 psi as an initial guess.  If one is groundwater then use the pumping water level instead.

Hope this helps.

PS.  For your original question most people model a well as a reservoir -> pipe -> node -> pump -> discharge node to system.  The reservoir is set to the pumping water level when the pump runs.  Static water level can usually be ignored as you generally only care about the reservoir head when the pump is flowing for most cases and you don't have to worry about how to reset it in the model when the pump is off. which is not easy unless the pump runs only at specific times.

Pat Moore

[Mark Matre]

I would like to know if instead of using combination of reservoir and pump, i will use a reservoir but Its TDH will be equal to computed TDH of the pump plus the elevation of water (that is elevation of well minus PWL)??

[Patrick Moore]

Mark,

using a reservoir set at the discharge head of the well is a possibility, but this will not exactly replicate the way a pump operates.  Flow will not be governed by the pump curve which means you will have the potential to get any number of flows from the well during the simulation that may or may not be realistic.  The pump curve governs the flow of the pump  based on where the system curve crosses the pump curve.  The pump can only provide flow values based on its pump curve, while a reservoir has no such restriction.  To calculate the TDH of the well at one specific point (like the design head and flow of the pump) would be the design head plus the head of the pumping water level of the pump.  Since most pumps don't run 100% of the time at the design head and flow point, the potential for inaccuracy can be high.  It's possible you could nail the point at one time and be way off at another.  Use that type of setup at your own risk.

Pat Moore