Water hammer basics: How does the peak pressure rise due to a sudden valve closure vary with the fluid density (all else equal)?

Introduction / Context:Water hammer (pressure surge) occurs when flow is rapidly decelerated, launching a compression wave that propagates along the pipe. The classic Joukowsky relation captures the instantaneous pressure rise for a sudden change in velocity. Understanding how density affects this rise is critical for surge protection, valve operation strategies, and pipe material selection.

Given Data / Assumptions:

• Elastic pipe–fluid system characterized by wave speed c.
• Sudden valve closure causing velocity change ΔV.
• Fluid is slightly compressible; pipe is slightly elastic.

Concept / Approach:The Joukowsky equation states Δp = ρ * c * ΔV for a rapid closure without significant reflection effects during the initial instant. For fixed ΔV and fixed c, the pressure rise is directly proportional to fluid density ρ. Even though c itself depends on fluid bulk modulus and pipe elasticity, the primary proportionality to ρ remains explicit in the formula.

Step-by-Step Solution:

Verification / Alternative check:Wave speed c ≈ sqrt(K/ρ) adjusted for pipe elasticity. If ρ increases modestly while K and pipe properties remain fixed, c decreases slightly, but the product ρ * c still grows with ρ for common ranges, aligning with the direct proportionality in the Joukowsky relation when c is treated as system-given.

Why Other Options Are Wrong:

• Inverse or square-root relations: Do not match the fundamental Δp = ρ * c * ΔV dependence.

Common Pitfalls:Ignoring the role of pipe elasticity; assuming incompressible rigid-pipe conditions where c → ∞ (nonphysical); confusing mass density with specific weight when comparing fluids.

Final Answer:directly proportional to density of fluid