Discharge through an external mouthpiece — practical formula Let a be the cross-sectional area of the external mouthpiece and H the head above its centerline. The discharge Q is approximately:

Difficulty: Easy

Correct Answer: Q = 0.855 * a * sqrt(2 * g * H)

Explanation:


Introduction / Context:
Mouthpieces are short tubes fitted at tank orifice openings. For a well-designed external mouthpiece running full, empirical studies give a practical coefficient close to 0.855 for the discharge relation.



Given Data / Assumptions:

  • External (protruding) mouthpiece of short length.
  • Steady incompressible flow; fully aerated jet; negligible approach velocity.
  • Coefficient of discharge C_d ≈ 0.855 (typical value).


Concept / Approach:
Ideal discharge from Bernoulli/continuity is Q_ideal = a * sqrt(2 * g * H). Real discharge is Q = C_d * Q_ideal with C_d < 1 to account for contraction and losses through the mouthpiece.



Step-by-Step Solution:

Write ideal flow: Q_ideal = a * sqrt(2 g H).Apply coefficient: Q = C_d * a * sqrt(2 g H).Use C_d ≈ 0.855 for external mouthpiece → Q ≈ 0.855 a sqrt(2 g H).


Verification / Alternative check:
Compare to sharp-edged orifice (C_d ≈ 0.62): the mouthpiece tends to have a higher C_d when running full because it controls contraction and reduces separation.



Why Other Options Are Wrong:

  • 1.855: too high; implies C_d > 1.
  • 0.585: too low for a full, well-formed mouthpiece.
  • 0.855 * a * (2 g H): missing the square root, wrong dimensions.
  • Coefficient = 1: assumes ideal flow without losses.


Common Pitfalls:
Mixing orifice and mouthpiece coefficients; forgetting the square-root dependence on head.



Final Answer:
Q = 0.855 * a * sqrt(2 * g * H)


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