Siphon outlet velocity with head loss For a siphon of total centerline length l conveying between two reservoirs with total head difference H, the outlet velocity v (accounting for losses) is best expressed as:

Introduction / Context:
A siphon transfers liquid between elevations using atmospheric pressure and continuity. The actual discharge (and velocity) depends on the available head minus head losses along the siphon due to friction and fittings.

Given Data / Assumptions:

• Total head difference between free surfaces: H.
• Head loss in siphon: h_f (includes friction and minor losses).
• Outlet discharges to atmosphere; velocities at large reservoirs are negligible.

Concept / Approach:

Apply Bernoulli between the upstream free surface and the outlet: z_1 + p_atm/γ = z_out + p_atm/γ + v^2/(2 g) + h_f. With z_1 − z_out = H, we get v^2/(2 g) = H − h_f, hence v = √(2 g (H − h_f)).

Step-by-Step Solution:

Verification / Alternative check:

Neglecting losses (h_f ≈ 0) reduces to Torricelli form v = √(2 g H), which is an upper bound—consistent with the loss-inclusive expression.

Why Other Options Are Wrong:

(b) ignores losses and overestimates v; (c) adds losses incorrectly; (d) is dimensionally inconsistent; (e) is not a standard relation.

Common Pitfalls:

Omitting minor losses at bends and entrance; using reservoir velocities as non-negligible when they are typically small.

Final Answer:

v = √(2 g (H − h_f))