Difficulty: Easy
Correct Answer: v^2 / (2g)
Explanation:
Introduction:
Minor losses in pipe flow include losses at entrances, exits, bends, valves, and fittings. The exit loss is a standard case when a pipe discharges into a large tank or the atmosphere.
Given Data / Assumptions:
Concept / Approach:
The kinetic energy per unit weight carried by the jet inside the pipe is v^2 / (2g). On discharge to a large reservoir, this kinetic energy is dissipated as turbulence, so the corresponding head is lost. Hence the exit loss coefficient K_exit = 1.
Step-by-Step Solution:
Verification / Alternative check:
Energy equation between a section just inside the pipe and a point in the reservoir at the same elevation shows the velocity head is not recovered in pressure but is dissipated, confirming the result.
Why Other Options Are Wrong:
0.5 * v^2 / (2g): Entrance, not exit, typically uses approximately 0.5 for sharp edges.1.5 * v^2 / (2g) and 2 * v^2 / (2g): No standard basis for such high coefficients at a plain exit.Negligible: The velocity head is not recovered; the loss is significant.
Common Pitfalls:
Confusing entrance and exit coefficients or assuming kinetic energy is recuperated in pressure at the outlet to a large tank.
Final Answer:
v^2 / (2g)
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