Using Slichter’s velocity relation for groundwater flow, compute the seepage velocity given: viscosity factor = 1.00, Slichter’s constant C = 400, effective size d = 0.5 mm, and hydraulic gradient i = 1/80. Choose the closest value (m/day).

Difficulty: Medium

Correct Answer: 1.25 m/day

Explanation:


Introduction / Context:
Empirical groundwater-flow relations like Slichter’s are useful for quick estimates of seepage velocity in granular aquifers when effective size and gradient are known. This problem applies the standard form to obtain a daily velocity.


Given Data / Assumptions:

  • Viscosity factor (temperature corrected) = 1.00 (typical reference condition).
  • Slichter’s constant C = 400.
  • Effective grain size d = 0.5 mm.
  • Hydraulic gradient i = 1/80 = 0.0125.
  • Use the common form v = C * i * d^2 with d in mm and v in m/day.


Concept / Approach:
Slichter’s empirical velocity relation is expressed as:
v = C * i * d^2where v is the apparent seepage velocity (m/day), i is hydraulic gradient, d is effective size (mm), and C is a temperature-dependent constant (dimension-adjusted). With the viscosity factor set to 1.00, we use the supplied C directly.


Step-by-Step Solution:

Compute d^2: d = 0.5 mm → d^2 = 0.5^2 = 0.25.Compute i: i = 1/80 = 0.0125.Compute v: v = 400 * 0.0125 * 0.25.First multiply 400 * 0.0125 = 5.0, then 5.0 * 0.25 = 1.25 m/day.


Verification / Alternative check:
Order of magnitude is reasonable for coarse sand under a mild gradient; daily seepage of about a meter is credible in such materials.


Why Other Options Are Wrong:

  • Values < 1 m/day (0.25, 0.50, 0.75) underestimate the computed result.
  • 1.00 m/day is close but still below the exact calculation.


Common Pitfalls:

  • Using d in meters rather than millimeters in the Slichter form provided with C.
  • Forgetting to square the grain size.


Final Answer:
1.25 m/day

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