Energy Loss in Pipe Flow – Entrance Loss for Sharp-Edged Inlet What is the head loss at the entrance of a pipe with a sharp-edged inlet (v is the average velocity in the pipe)?
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A0.5 * v^2 / (2g)
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Bv^2 / (2g)
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C1.5 * v^2 / (2g)
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Dapproximately zero
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E2 * v^2 / (2g)
Answer
Correct Answer: 0.5 * v^2 / (2g)
Explanation
Introduction:Entrance loss is one of the standard minor losses in internal flows. Its magnitude depends on how the fluid is introduced into the pipe, especially the inlet geometry.
Given Data / Assumptions:
- Entrance is sharp-edged (square-edged).
- v is the uniform mean velocity inside the pipe.
- Discharge into the pipe from a large reservoir or plenum.
Concept / Approach:
The head loss due to entrance is expressed as h_e = K_e * v^2 / (2g). For a sharp-edged entrance, empirical data give K_e approximately 0.5. Streamline contraction and subsequent expansion to fill the pipe core create energy dissipation.
Step-by-Step Solution:
Step 1: Write the generic relation: h_e = K_e * v^2 / (2g).Step 2: For a sharp-edged entrance, take K_e ≈ 0.5.Step 3: Therefore, h_e ≈ 0.5 * v^2 / (2g).Verification / Alternative check:
Smoother bell-mouth inlets have much smaller K_e (approaching 0.04 to 0.1), while square-edged inlets have K_e close to 0.5. This range corroborates the selected value.
Why Other Options Are Wrong:
v^2 / (2g) and 2 * v^2 / (2g): Too large for a plain sharp edge; typical only for combinations of fittings or exits.1.5 * v^2 / (2g): No standard entrance configuration yields such a high coefficient.Approximately zero: Only for well-rounded bell-mouth entrances, not sharp-edged.
Common Pitfalls:
Memorizing a single coefficient for all entrances, or mixing entrance and exit losses. Always link the coefficient to the geometry.
Final Answer:
0.5 * v^2 / (2g)