Well interference is the extra drawdown one pumping well causes in another. It depends on the neighbor’s rate, the distance, how long it has run, and the aquifer’s transmissivity and storativity. Every added foot of lift costs energy.
This note takes the example well from our note on transmissivity from a pumping test, puts a second well 1,000 feet away, and works out the drawdown, what distance, aquifer type and pump scheduling change, what it costs, and how to measure it on your own wells.
What is well interference?
Every pumping well draws the water level down in a cone around it. Where two cones overlap, the drawdowns add: the principle of superposition, which the U.S. Geological Survey’s treatment states as “problem solutions can be added together to obtain composite solutions” (Reilly and others, 1987). An idle well inside a neighbor’s cone shows a lower static level. A pumping well inside it shows its own drawdown plus the neighbor’s, so its pumping level sits deeper and its pump lifts farther.
Interference is mutual. Each well lowers the other in proportion to its own pumping rate, whichever was drilled first and whoever owns it.
How do you calculate the drawdown a neighboring well causes?
Use the same Cooper–Jacob straight line that gives transmissivity from a pumping test (Cooper and Jacob, 1946), written for a point at distance r from the pumping well. In district units, as the USGS gives it (Ferris and others, 1962):
s = (264 Q ÷ T) × log₁₀(0.3 T t ÷ r²S)
s is the interference drawdown, in ft, at distance r, in ft, from a well pumping Q, in gpm, for t days. T is transmissivity in gpd/ft and S is storativity, without units. The first factor, 264 Q ÷ T, is the drawdown added for every tenfold increase in time.
The straight line holds once u = r²S ÷ (4Tt), in consistent units, is small, generally less than 0.01 (Ferris and others, 1962). At a neighboring well that takes longer than at the pumping well itself, because r is the distance between the wells rather than a well radius. Earlier, or farther away, use the full Theis solution, s = Q W(u) ÷ (4πT), with W(u) the well function (Theis, 1935). This note shows both.
How much does a well 1,000 feet away take?
Take the example well from the transmissivity note, call it well A: T = 50,286 gpd/ft (6,723 ft²/day) from its constant-rate test, a confined aquifer, and an assumed storativity of 2 × 10⁻⁴, the value used for its step test. Well B, 1,000 ft away, starts pumping 1,500 gpm and keeps running. B’s factor is 264 × 1,500 ÷ 50,286 = 7.875 ft per tenfold increase in time, the same as A’s 6.3 ft scaled by 1,500 ÷ 1,200.
| Days since well B started | log₁₀(0.3 T t ÷ r²S) | Interference at well A, Cooper–Jacob (ft) | Interference at well A, full Theis (ft) |
|---|---|---|---|
| 1 | 1.878 | 14.79 | 14.81 |
| 7 | 2.723 | 21.44 | 21.44 |
| 30 | 3.355 | 26.42 | 26.41 |
| 120 | 3.957 | 31.16 | 31.15 |
After one day, A’s static level is 14.8 ft lower. By the end of a 120-day summer with B running, it is 31.2 ft lower. The growth slows with time: each tenfold increase in days adds another 7.875 ft. A’s pumping level at 1,200 gpm was 244.0 ft (a 180 ft static level plus 64.0 ft of drawdown at 24 hours); with B’s 120 days it is 275.2 ft. Check that against the pump setting and the submergence the pump needs. The effect runs both ways: over the same 120 days, A at 1,200 gpm lowers B by 24.9 ft.
These figures assume an aquifer with no boundaries and no leakage. Leakage through a confining layer flattens the curve with time; a fault or the edge of a basin steepens it, which the USGS treatment handles with image wells (Ferris and others, 1962). Treat the calculation as an indication of the size of the effect, and the measured response, below, as the record.
How much does distance help?
| Distance between the wells (ft) | Interference after 120 days (ft) |
|---|---|
| 500 | 35.90 |
| 1,000 | 31.16 |
| 2,000 | 26.42 |
| 4,000 | 21.68 |
Each doubling of distance removes the same amount: log₁₀(4) × 7.875 = 4.74 ft. Moving a new well from 1,000 to 4,000 ft away cuts the 120-day interference by 9.48 ft, about 30%. In a confined aquifer, distance helps less than most people expect.
Distance and time trade against each other, because r² and t sit in the same ratio. In the tables above, 2,000 ft at 120 days gives the same 26.42 ft as 1,000 ft at 30 days: doubling the distance does what cutting the pumping time to a quarter does.
Why does a water-table aquifer behave so differently?
Storativity sets how far and how fast the cone spreads. A confined aquifer releases water by compressing the aquifer and expanding the water, so S is tiny and the cone races outward. A water-table aquifer releases water by draining pores, and that specific yield is far larger. Here is the same pair of wells, same T, with a specific yield of 0.15, 750 times the confined value:
| Days since well B started | Confined, S = 2 × 10⁻⁴ (ft) | Water table, specific yield 0.15 (ft) |
|---|---|---|
| 1 | 14.79 | 0.002 |
| 7 | 21.44 | 1.07 |
| 30 | 26.42 | 4.39 |
| 120 | 31.16 | 8.67 |
In the water-table case, B barely reaches A in the first day, and 120 days of pumping costs A 8.67 ft instead of 31.16. Water-table aquifers also release water slowly at first as pores drain, which this simple calculation leaves out, so treat that column as an order of magnitude. Which of the two describes a district’s wells is a question for the well logs and for a test with an observation well, not for an assumed number.
What does interference cost in energy?
Lifting one acre-foot one foot takes 1.024 kWh at 100% efficiency (UC Agriculture and Natural Resources rounds it to 1.02). Divide by the pumping plant’s wire-to-water efficiency, the measure explained in our note on pump efficiency tests:
Added kWh per acre-foot = 1.024 × added lift ÷ wire-to-water efficiency
Added lift in ft; efficiency as a decimal. Multiply by the energy rate for dollars per acre-foot.
At the pump note’s example values, 71.4% wire-to-water and $0.18 a kWh, every foot of interference adds 1.434 kWh, about 26 cents, to every acre-foot well A pumps. B’s 31.16 ft after 120 days adds 44.69 kWh per acre-foot, $8.04. On 1,000 acre-feet pumped at that lift, that is about $8,044 a year.
Energy is only half of it. A fixed-speed pump facing more head moves back along its curve and delivers less water, by an amount only its own curve can tell; our note on reading a pump curve shows where to look.
Does staggering run times reduce interference?
Yes, while the wells run, though not the long-term trend. Suppose B runs 12 hours a day at 1,500 gpm for 30 days. Superposition treats each start and stop as adding or subtracting a well (Reilly and others, 1987). On day 30, the interference at A swings between 7.75 ft at the end of B’s rest and 18.66 ft at the end of B’s run.
- If A runs during the same 12 hours as B, A averages 16.11 ft of interference while it pumps.
- If A runs while B rests, A averages 10.27 ft: 5.84 ft less, or 36%. At the example’s energy figures, that is about $1,508 a year on 1,000 acre-feet pumped in those hours.
- The middle of the swing tracks the volume pumped. B pumping the same daily volume around the clock, at 750 gpm, would hold A 13.20 ft lower.
Time-of-use tariffs push every well into the same off-peak hours, which is the overlapping case. Where the tariff windows allow, alternate the wells that sit closest together. Our note on scheduling well pumps around time-of-use rates covers the windows.
How do you measure interference on your own wells?
An idle neighbor is an observation well. Rest well A until its level is steady, start B at a measured rate, and record A’s level every minute from a transducer or the SCADA historian. Plot A’s drawdown against log time. Once u is small, about 17.9 hours in for the example wells, the points fall on a straight line, and the USGS treatment gives two numbers from it (Ferris and others, 1962):
T = 264 Q ÷ Δs and S = 0.3 T t₀ ÷ r²
Q is B’s rate in gpm; Δs is the drawdown at A over one log cycle of time, in ft; t₀ is the time, in days, where the straight line extended back crosses zero drawdown; r is the distance between the wells in ft.
For the example wells, the line crosses zero at 19.1 minutes, 0.01326 days, so S = 0.3 × 50,286 × 0.01326 ÷ 1,000² = 2.0 × 10⁻⁴. That is the storativity a test on the pumping well alone cannot give, as the transmissivity note explains. Start from steady conditions, log the run times of every other well nearby, and note the barometric pressure, because all of them move the level too.
Waterwell IQ keeps each well’s tests, and the transmissivity read from them, in one record, read the same way every time; the Waterwell IQ methodology page explains how.
Quick answers
Does a neighboring well reduce my well’s yield?
It lowers the pumping level, so the pump lifts farther. A fixed-speed pump then delivers less water, by an amount its own curve sets, and uses more energy for each acre-foot. If the pumping level nears the pump setting, submergence becomes the limit.
How far apart should district wells be?
There is no single distance; compute it from the aquifer’s own T and S. In a confined aquifer each doubling of distance removes only log₁₀(4), about 0.6, times the neighbor’s drawdown per log cycle. In a water-table aquifer distance and time count for much more.
Is interference the same as a falling water table?
No. Interference is local and recovers after the neighbor stops. A regional decline shows in the static levels of wells beyond any neighbor’s reach, and it does not recover when one well rests.
See the tests on your own wells read the same way every time. Request a walkthrough →
Sources
- Theis, C.V. The relation between the lowering of the piezometric surface and the rate and duration of discharge of a well using ground-water storage. Transactions, American Geophysical Union 16(2): 519–524. 1935. The nonequilibrium equation and the well function W(u).
- Cooper, H.H., Jr., and C.E. Jacob. A generalized graphical method for evaluating formation constants and summarizing well-field history. Transactions, American Geophysical Union 27(4): 526–534. 1946. The straight-line (Cooper–Jacob) method.
- Ferris, J.G., D.B. Knowles, R.H. Brown and R.W. Stallman. Theory of aquifer tests. U.S. Geological Survey Water-Supply Paper 1536-E. 1962. s = (264Q ÷ T) log(0.3Tt ÷ r²S) in gpm, gpd/ft, days and ft; T = 264Q ÷ Δs and S = 0.3Tt₀ ÷ r² from an observation well; u generally less than 0.01; image wells for boundaries.
- Reilly, T.E., O.L. Franke and G.D. Bennett. The principle of superposition and its application in ground-water hydraulics. U.S. Geological Survey Techniques of Water-Resources Investigations 03-B6. 1987. Adding the drawdowns of several wells, and of a well’s starts and stops.
- University of California Agriculture and Natural Resources. Energy and cost required to lift or pressurize water. Publication IG6-96. 1.02 kWh to lift one acre-foot one foot at 100% efficiency.