Water hammer basics: The minimum valve-closing time to qualify as “gradual closure” (so that the full Joukowsky rise does not occur) is expressed in terms of pipe length L and wave speed C as what?
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At_g = 2L / C
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Bt_g = L / C
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Ct_g = 4L / C
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Dt_g = L / (2C)
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Et_g = C / (2L)
Answer
Correct Answer: t_g = 2L / C
Explanation
Introduction / Context:
Water hammer is a transient pressure surge caused by rapid changes in flow. The characteristic time for a pressure wave to travel the pipe length and return (a round trip) governs whether a closure is considered “sudden” or “gradual” for surge calculations.
Given Data / Assumptions:
- Rigid-pipe approximation with wave speed C.
- One-dimensional transient behavior; reflection at reservoir/valve ends.
- Valve located at one end of a line of length L.
Concept / Approach:
If the valve-closing time t_c is less than 2L/C, the closure is “sudden”, and the Joukowsky equation Δp = ρ C ΔV applies for the maximum surge. If t_c ≥ 2L/C, the closure is “gradual”; the maximum head rise reduces and scales with t_c. Thus, 2L/C is the dividing line between sudden and gradual closures.
Step-by-Step Solution:
Compute the wave round-trip time: t_round = 2L/C.Define criterion: t_c ≥ t_round ⇒ gradual closure.Therefore, the minimum time for “gradual” classification is t_g = 2L/C.Verification / Alternative check:
Method of characteristics solutions and classic water-hammer charts corroborate the 2L/C threshold for end-valve systems.
Why Other Options Are Wrong:
- L/C (one-way travel) underestimates the interaction time.
- Other multiples (4L/C, L/2C) do not reflect the physics of wave reflection timing.
Common Pitfalls:
- Using local pipe segment length instead of the full length to the reflecting boundary.
Final Answer:
t_g = 2L / C