Estimate ideal groundwater transit and its limits

Turn a volume flux into an average motion

The default column has Darcy flux q = 0.08 m/day. That value divides volumetric flow by the bulk cross-section, including solid material. To estimate the average velocity in the connected effective pore space, divide by effective porosity n_e. Here n_e = 0.25, a dimensionless fraction.

Thus v = q/n_e = 0.08/0.25 = 0.32 m/day. The sign follows q; positive means left to right. This is an average linear pore velocity in the simplified column, not a claim that every particle follows a straight identical microscopic path. USGS distinguishes specific discharge, or Darcy velocity, from average water velocity normalized by effective porosity. USGS explanation of groundwater velocity terminology.

Effective porosity concerns the pore volume participating in the modeled flow. It is not the same quantity as specific yield, which concerns water released by drainage. The browser keeps saturation and porosity fixed; it does not calculate how much an aquifer would release when its water level falls.

Estimate a conditional crossing time

For a constant nonzero average pore velocity, the ideal advective transit time is length divided by speed magnitude: t = 100/|v|. The vertical bars mean use the nonnegative magnitude of v, so reversing direction does not produce negative elapsed time.

At the default, t = 100 m / 0.32 m/day = 312.5 days. The start is the higher-head left endpoint. Head boundaries are maintained, so the calculation assumes the same flow persists while the modeled parcel crosses. It is not a simulation of the boundaries draining down or the gradient relaxing over time.

Make a calculation ladder on paper: head difference, gradient, Darcy flux, pore velocity, crossing time. Keep the units on each line. Substituting K directly for v would give 50 days here, while substituting q would give 1,250 days. Both substitutions confuse quantities that happen to share velocity units.

Reverse the flow and identify the starting endpoint

Change left head from 14 to 6 m while K stays 2 m/day. The flux becomes −0.08 m/day and pore velocity becomes −0.32 m/day. The higher-head endpoint is now on the right. The ideal crossing time remains 312.5 days because the speed magnitude has not changed.

Capture both settings and label the start and finish for each. A fixed left-to-right travel story would be wrong in the reversed case even though the displayed positive duration is unchanged. Direction belongs in the explanation alongside time.

At left head 20 m and K = 10 m/day, the driving gradient is 0.1, q is 1 m/day and v is 4 m/day. The ideal crossing takes 25 days. Check the full chain before interpreting that shorter result; it reflects two changed inputs, not evidence that one change alone caused the entire difference.

Handle zero flow without inventing a time

Set left head equal to the fixed right head of 10 m. Both q and v become zero. There is no finite advective crossing time: the chosen flow mechanism does not carry water across the column. Zero days would mean instantaneous travel and would be the opposite interpretation.

This result does not claim that diffusion or other microscopic processes cease. Those processes are outside the model. The browser’s absence of a finite advective time is a statement about its defined transport mechanism, not a universal statement that nothing can move.

Keep transport predictions within their evidence

The transit estimate omits dispersion, diffusion, sorption, chemical reactions, heterogeneous pathways and changing boundary conditions. It is therefore not a predicted first arrival or concentration curve for a contaminant. Different solutes can interact differently with the medium, and an average advective time does not describe the full spread of arrival times.

Write a claim-and-evidence ledger: the supplied parameters support the ideal calculation; a real transport forecast would need information about pathways, transport properties and boundary histories. This model also provides no pumping or safe-yield assessment. More arithmetic digits would not supply those missing mechanisms.

Transfer to a different ideal column

A paper-only column is 60 m long, with q = −0.03 m/day and effective porosity 0.20. Find average pore velocity, crossing time and starting side under the same positive-right convention.

Check the groundwater-travel transfer

v = −0.03/0.20 = −0.15 m/day. The ideal crossing time is 60/0.15 = 400 days, starting on the higher-head right side and moving left. The positive duration uses speed magnitude; the negative velocity retains direction. This remains an ideal advective estimate, not a contaminant-arrival forecast.