Open-channel laminar flow — For the most economical channel section (maximum discharge for a given area), vertical velocity profiles are often discussed. For laminar flow in an open channel, the vertical distribution of velocity can be reasonably assumed to be:

Introduction:
Open-channel hydraulics analyzes how velocity varies from the bed to the free surface. While most practical channels are turbulent, understanding the idealized laminar profile builds foundational insight and aids in checking numerical models. This question asks for the approximate functional form of the vertical velocity distribution in laminar open-channel flow.

Given Data / Assumptions:

• Steady, uniform, laminar flow in a wide, prismatic channel.
• No sidewall effects (2-D approximation in the vertical plane).
• Newtonian fluid; constant properties.

Concept / Approach:
Under laminar conditions, the momentum equation with constant shear stress gradient yields a quadratic (parabolic) velocity variation between the no-slip bed and the shear-free or specified shear condition at the free surface. This is analogous to plane-Poiseuille flow modified for a free surface: zero shear or specified stress at the free surface produces a parabolic profile peaking near the surface.

Step-by-Step Solution:
Apply steady, 1-D momentum balance with body force component driving flow.Integrate the constant viscosity shear relationship to obtain velocity as a quadratic in depth.Conclude that the vertical velocity profile is parabolic under laminar conditions.

Verification / Alternative check:
Analytical solutions for laminar open-channel flow (Navier–Stokes simplifications) yield quadratic velocity distributions; experiments in viscous syrup channels confirm near-parabolic profiles at very low Reynolds numbers.

Why Other Options Are Wrong:

• Hyperbolic/exponential: not supported by the laminar solution with constant viscosity.
• Straight line: resembles highly damped turbulent “log-law” approximations in certain plots, not laminar physics.
• None of these: incorrect because parabolic is the classical result.

Common Pitfalls:
Assuming turbulent-like logarithmic profiles apply in laminar regimes; ignoring that the free-surface shear boundary differs from a rigid wall but still yields a quadratic form.

Final Answer:
Parabolic