Pumping Well Drawdown Thiem Theis Calculator
Our hydrology & water resources calculator computes pumping well drawdown thiem theis accurately. Includes formulas and worked examples.
Reviewed for accuracy by Daniel Agrici, Founder & Lead Developer
Pumping Well Drawdown Thiem Theis Calculator
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Formula: s = (Q / 4piT) x W(u)
Worked example โ Drawdown s = 0.912 m at r = 100 m after 1 day | u = 5.79e-4 | W(u) = 6.878
Formula
s = (Q / 4piT) x W(u)
The Theis equation s = (Q / 4piT) x W(u) calculates drawdown in a confined aquifer due to pumping. s is the water-level decline (m) at a given point, Q is the pumping rate (mยณ/s), T is aquifer transmissivity (mยฒ/s) representing the rate of horizontal water flow, and W(u) is the Theis well function of u = rยฒS / (4Tt), where r is radial distance from the well (m), S is storativity (dimensionless), and t is pumping duration. The result predicts how far and how deeply the cone of depression extends around a production well.
Worked Examples
Example 1: Productive Sand and Gravel Aquifer, One Day of Pumping
Problem:A confined sand and gravel aquifer is pumped at Q = 500 L/min. Transmissivity T = 0.005 m2/s (432 m2/day), storativity S = 0.0001. Find the drawdown in an observation well r = 100 m from the pumping well after t = 1440 min (1 day).
Solution:Convert to SI: Q = 500 / 60000 = 0.008333 m3/s; t = 1440 x 60 = 86400 s u = r2 S / (4 T t) = (100^2 x 0.0001) / (4 x 0.005 x 86400) = 1.0 / 1728 = 5.787e-4 W(u) = -0.5772 - ln(5.787e-4) + u - ... = -0.5772 + 7.4547 + 0.0006 = 6.878 Q / (4 pi T) = 0.008333 / (4 x 3.14159 x 0.005) = 0.008333 / 0.062832 = 0.13263 m s = 0.13263 x 6.878 = 0.912 m
Result:Drawdown s = 0.912 m at r = 100 m after 1 day | u = 5.79e-4 | W(u) = 6.878
Example 2: Low-Transmissivity Aquifer, High Abstraction Rate
Problem:A municipal well draws Q = 2000 L/min from a tight confined aquifer with T = 0.0008 m2/s (69.1 m2/day) and S = 0.0002. What is the drawdown at r = 50 m after t = 600 min (10 hours)?
Solution:Convert to SI: Q = 2000 / 60000 = 0.033333 m3/s; t = 600 x 60 = 36000 s u = (50^2 x 0.0002) / (4 x 0.0008 x 36000) = 0.5 / 115.2 = 4.340e-3 W(u) = -0.5772 - ln(4.340e-3) + u - ... = -0.5772 + 5.4398 + 0.0043 = 4.867 Q / (4 pi T) = 0.033333 / (4 x 3.14159 x 0.0008) = 0.033333 / 0.010053 = 3.3157 m s = 3.3157 x 4.867 = 16.14 m
Result:Drawdown s = 16.14 m at r = 50 m after 10 hours | u = 4.34e-3 | W(u) = 4.867
Example 3: Two Observation Wells: Theis Difference Reproduces Thiem
Problem:Q = 1200 L/min, T = 0.002 m2/s, S = 0.0005, after t = 4320 min (3 days). Run the calculator twice, at r1 = 20 m and r2 = 200 m, then compare the drawdown difference with the Thiem steady-state expression s1 - s2 = Q / (2 pi T) x ln(r2 / r1).
Solution:Q = 1200 / 60000 = 0.02 m3/s; t = 4320 x 60 = 259200 s; 4 T t = 4 x 0.002 x 259200 = 2073.6 m2 Q / (4 pi T) = 0.02 / 0.0251327 = 0.79577 m At r1 = 20 m: u1 = (400 x 0.0005) / 2073.6 = 9.645e-5, W(u1) = 8.6694, s1 = 0.79577 x 8.6694 = 6.899 m At r2 = 200 m: u2 = (40000 x 0.0005) / 2073.6 = 9.645e-3, W(u2) = 4.0737, s2 = 0.79577 x 4.0737 = 3.242 m Theis difference: s1 - s2 = 6.899 - 3.242 = 3.657 m Thiem check: Q / (2 pi T) x ln(200/20) = 0.02 / 0.0125664 x ln(10) = 1.59155 x 2.30259 = 3.665 m
Result:s1 = 6.899 m, s2 = 3.242 m, difference 3.657 m โ within about 0.2% of the Thiem value 3.665 m
Frequently Asked Questions
What is Pumping Well Drawdown (Thiem/Theis)?
Drawdown is the decline in hydraulic head produced by pumping - the difference between the pre-pumping static water level and the level measured while the well is running. Thiem and Theis are the two classical analytical solutions for that decline: Thiem (1906) is the steady-state case, which compares drawdown at two observation radii and needs neither time nor storativity, while Theis (1935) is the transient case and predicts drawdown at any radius after any elapsed pumping time. Both underpin the same practical work - deriving transmissivity and storativity from a pumping test, spacing production wells so their cones of depression do not interfere, and delineating capture zones for wellhead protection.
How is Pumping Well Drawdown (Thiem/Theis) calculated?
The Theis route is two steps: form the dimensionless group u = r2 S / (4 T t), then evaluate s = (Q / 4piT) x W(u), where W(u) is the exponential integral E1(u). Using the defaults on this page - Q = 500 L/min converted to 0.008333 m3/s, T = 0.005 m2/s, S = 0.0001, r = 100 m and t = 1440 min converted to 86400 s - u = 5.787e-4, W(u) = 6.878, Q / (4 pi T) = 0.1326 m and s = 0.912 m. The Thiem route skips both S and t and returns only the difference between two radii, s1 - s2 = Q / (2 pi T) x ln(r2 / r1), so it is valid only after the cone of depression has stopped expanding.
When does the Thiem steady-state equation replace the Theis solution for well drawdown?
Thiem applies only after the cone of depression stops expanding, which needs leakage, induced river infiltration or a constant-head boundary to balance Q; many confined aquifers never get there and drawdown keeps creeping up with the logarithm of time. The drawdown difference between two radii converges far sooner, because W(u1) - W(u2) approaches 2 ln(r2/r1) and Theis therefore tends to Q / (2 pi T) x ln(r2/r1). Example 3 above gives 3.657 m against the Thiem value 3.665 m, a 0.2 percent gap.
How is transmissivity obtained from the straight-line slope of a pumping well drawdown plot?
Cooper and Jacob (1946) rewrote Theis for u < 0.01 as s = (2.303 Q / 4 pi T) x log10(2.25 T t / r2 S), so drawdown against log time plots as a straight line. Transmissivity comes from its slope: T = 0.183 Q / (drawdown per log cycle), because 2.303 / 4 pi = 0.1833. Storativity follows from the time intercept t0 where the fitted line crosses zero drawdown, S = 2.25 T t0 / r2. On a distance-drawdown plot the constant doubles to 0.366, since 2.303 / 2 pi = 0.3665.
Why does the Theis equation underestimate drawdown in an unconfined pumped aquifer?
Theis holds T constant, but unconfined pumping dewaters the section, so saturated thickness b and T = Kb both fall and real drawdown exceeds the value predicted from the initial thickness. The Jacob correction subtracts s2 / (2b) from the observed drawdown to give its confined equivalent: when s reaches 10 percent of b that correction is 5 percent of s, and at 25 percent of b it is 12.5 percent. Delayed gravity drainage also flattens the mid-time curve, which is why Neuman (1972) is used for unconfined tests.
How far does the drawdown cone around a pumping well extend?
Theis gives no true finite radius of influence - drawdown is non-zero at every distance and simply decays with W(u). The practical figure comes from setting the Cooper-Jacob drawdown to zero, 2.25 T t / (r2 S) = 1, which gives R = 1.5 sqrt(T t / S). With the defaults on this page (T = 0.005 m2/s, S = 0.0001, t = 86400 s) that is 1.5 x sqrt(4.32 x 10^6) = about 3.1 km after one day. R grows as the square root of time, so pumping 100 times longer widens it only tenfold.
How do wellbore storage and partial penetration distort drawdown near the pumping well?
Both are near-field effects Theis ignores. Early on, part of the discharge comes from water stored in the casing, so measured drawdown lags theory and the early log-log data lie on a unit slope; Kruseman and de Ridder treat the effect as spent after t = 25 rc2 / T, just 50 s for a 0.1 m casing radius at T = 0.005 m2/s. A partially penetrating well adds vertical flow; by the Hantush criterion that can be ignored beyond r = 1.5 b sqrt(Kh/Kv), about 142 m for b = 30 m and Kh/Kv = 10.
How do I combine drawdown from several pumping wells in a well field?
The Theis solution is linear in Q, so drawdowns superpose: total decline at a point is the sum of Q_i / (4 pi T) x W(u_i), each u_i built from that well and its own radius r_i. The same superposition covers rate changes over time and aquifer boundaries through image wells - a recharge boundary is matched by an injection image at the mirrored position, a no-flow barrier by an identical pumping image (Ferris and others, USGS Water-Supply Paper 1536-E, 1962). It breaks down once unconfined dewatering makes the problem non-linear.
What does a step-drawdown test add to a constant-rate pumping well test?
A constant-rate test returns the aquifer properties T and S. A step test - typically three to five equal steps of 60 to 120 minutes at rising rates - splits the drawdown measured in the pumped well itself into aquifer loss and well loss, s = BQ + CQ^n (Jacob 1947; Rorabaugh 1953 allowed n from 1.5 to 3.5, with n = 2 in routine use). Well efficiency is BQ / s. Walton (1962) read C under 0.5 min2/m5 as a properly developed well, 0.5 to 1.0 as mild deterioration and 1 to 4 as severe clogging.
Which unit mismatches break the Thiem and Theis drawdown equations?
The group u = r2 S / (4 T t) is dimensionless only when r, T and t share one length and time base. This page converts internally - L/min divided by 60000 gives m3/s, minutes times 60 give seconds - but transmissivity must be typed in m2/s: entering 432, the m2/day equivalent of 0.005 m2/s, makes T 86400 times too large and shrinks the Q / (4 pi T) coefficient by the same factor. Storativity is dimensionless too: a specific yield of 0.15 where a confined 0.0001 belongs lifts u 1500-fold and cuts W(u) from 6.88 to 0.28.
References
- Theis (1935) - The relation between the lowering of the piezometric surface and the rate and duration of discharge of a well using ground-water storage
- USGS TWRI 3-B3 - Type Curves for Selected Problems of Flow to Wells in Confined Aquifers
- Lohman (1972) - Ground-Water Hydraulics, USGS Professional Paper 708 (Theis and Thiem solutions, W(u) tables)
- Kruseman & de Ridder - Analysis and Evaluation of Pumping Test Data (ILRI Publication 47)
Background & Theory
History
Reviewed for accuracy by Daniel Agrici, Founder & Lead Developer ยท Editorial policy
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