Flow net discharge — if Nf is the number of flow channels, Nd is the number of potential drops, and H is the total head loss, the total discharge q through the section can be expressed (for constant K) as:

Introduction / Context:
Flow nets provide a graphical solution of Laplace’s equation for steady seepage in two dimensions. By counting equal-flow channels and equal-potential drops in a properly drawn net, engineers can estimate seepage discharge without solving differential equations, provided permeability and thickness are known or absorbed into a constant K.

Given Data / Assumptions:

• Nf = number of flow channels between boundaries.
• Nd = number of equipotential drops between head boundaries.
• H = total head difference across the domain.
• K is a constant that includes permeability k and, if applicable, section thickness b (so K may be k * b).

Concept / Approach:

For a good-quality flow net with approximate curvilinear squares, discharge per unit width is proportional to k * (Nf / Nd) * H. If the section has constant thickness b, then q = k * b * (Nf / Nd) * H. Grouping the constants into K yields the compact relation used for quick estimates and checks.

Step-by-Step Solution:

Verification / Alternative check:

Compare with analytical solutions for simple geometries (e.g., confined flow between parallel sheets) to validate counts and scaling. Sensitivity to small miscounts highlights the value of neat, proportional squares.

Why Other Options Are Wrong:

Nd/Nf inverts the proportionality and under/overestimates discharge.

Multiplying Nf and Nd or squaring H lacks theoretical basis from Laplacian potential theory.

Dividing by Nf * Nd contradicts the additive behavior of channels and potential drops.

Common Pitfalls:

Poorly drawn nets that are not near-squares; forgetting to include section thickness b in K; inconsistent units for H and k.

Final Answer:

q = K * (Nf / Nd) * H