Valve head-loss basics: the dimensionless loss coefficient K for a FULLY OPEN globe valve is approximately (order of magnitude) _____ under standard fittings data.

Difficulty: Easy

Correct Answer: 10

Explanation:


Introduction / Context:
When designing piping networks, engineers often estimate pressure drops across valves and fittings using a dimensionless loss coefficient, K. The globe valve is a high-loss valve because the flow makes multiple turns through the body and seat. This question checks the typical order-of-magnitude K value for a fully open globe valve.


Given Data / Assumptions:

  • Globe valve fully open.
  • Single-phase incompressible flow regime.
  • Head loss estimated by h_f = K * v^2/(2g), where v is mean velocity in the adjacent straight pipe of the same nominal size.


Concept / Approach:

Standard handbooks tabulate K for fittings. Typical values (order of magnitude) are: long-radius elbow K ≈ 0.2–0.3, standard elbow ≈ 0.9, gate valve (fully open) ≈ 0.15–0.2, swing check ≈ 2, angle valve ≈ 2, and globe valve ≈ 10. The higher K for globe valves reflects substantial turbulence and separation inside the body due to abrupt directional changes.


Step-by-Step Solution:

Recall head-loss relation: h_f = K * v^2/(2g).Identify fitting type: fully open globe valve → high K.Select tabulated order-of-magnitude value: K ≈ 10.


Verification / Alternative check:

Piping design charts and Crane-style correlations consistently put fully open globe valves near K ≈ 10 (varies with size and style, but order of magnitude remains about 10).


Why Other Options Are Wrong:

0.2 is typical of a gate valve fully open, not a globe valve. 25 and 75 are too large for standard globe patterns (would grossly overpredict head loss). 300 is unrealistic for a single valve and more typical of many fittings in series.


Common Pitfalls:

Confusing K with valve Cv; mixing mean pipe velocity with vena contracta velocity; using partially open data for fully open conditions.


Final Answer:

10

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