Reading velocities from a flow net On a flow net, the spacing between two consecutive streamlines at section 1 is 1 cm, and at the next downstream section it is 0.5 cm. If the local velocity at section 1 is 1.0 m/s, what is the local velocity at section 2 (assume equal potential drop between corresponding sections)?

Difficulty: Medium

Correct Answer: 2.0 m/sec

Explanation:


Introduction / Context:
Flow nets are graphical solutions of Laplace’s equation used in seepage and potential flow. The spacing of streamlines and equipotentials encodes discharge per flow tube, enabling velocity comparisons between sections without direct computation of the velocity field.


Given Data / Assumptions:

  • Two adjacent streamlines form a stream tube of width s.
  • Equal potential drop between compared sections (same number of potential squares).
  • Incompressible, steady flow; constant thickness normal to the net.


Concept / Approach:

In a given stream tube, discharge q is conserved: q = v * A_tube. With constant thickness, A_tube ∝ s (streamline spacing). Therefore, v is inversely proportional to s. Halving the spacing doubles the velocity.


Step-by-Step Solution:

Conservation: q_1 = v_1 * s_1 = q_2 = v_2 * s_2.Solve for v_2: v_2 = v_1 * (s_1 / s_2).Given v_1 = 1.0 m/s, s_1 = 1 cm, s_2 = 0.5 cm → v_2 = 1.0 * (1 / 0.5) = 2.0 m/s.


Verification / Alternative check:

Uniform net thickness and equal potential drop ensure the stream tubes carry the same discharge; the inverse proportion confirms the result.


Why Other Options Are Wrong:

(a) and (b) ignore the reduced spacing; (d) and (e) exaggerate the change beyond the inverse-spacing rule for the given ratios.


Common Pitfalls:

Comparing sections across different potential drops; reading spacing obliquely rather than perpendicular to streamlines; neglecting varying thickness in 3D scenarios.


Final Answer:

2.0 m/sec

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