Hydraulic jump energy loss: If D1 and D2 are the upstream and downstream conjugate depths of a hydraulic jump, the loss of specific head at the jump equals

Difficulty: Medium

Correct Answer: (D2 - D1)^3 / (4 * D1 * D2)

Explanation:


Introduction / Context:
A hydraulic jump in a rectangular channel converts supercritical flow to subcritical flow, dissipating energy. Designers must quantify the head loss across the jump to size stilling basins and appurtenances.


Given Data / Assumptions:

  • D1 = supercritical depth upstream of the jump.
  • D2 = subcritical (sequent) depth downstream.
  • Rectangular prismatic channel, steady flow.


Concept / Approach:
From the specific energy equation and momentum (sequent depth relation), the loss of specific head h_L across the jump can be expressed purely in terms of D1 and D2.


Step-by-Step Solution:
Specific energy upstream: E1 = D1 + V1^2/(2g).Specific energy downstream: E2 = D2 + V2^2/(2g).Using continuity and momentum to eliminate velocities yields the standard head loss:h_L = E1 - E2 = (D2 - D1)^3 / (4 * D1 * D2).


Verification / Alternative check:
Check limiting behavior: If D2 ≈ D1, numerator tends to zero faster than denominator, so h_L → 0, as expected for a vanishing jump.


Why Other Options Are Wrong:

  • Squares in numerator (option b) underestimates loss.
  • (D2 + D1)^3 (option c) has wrong dependence and gives nonzero loss even when D2 = D1.
  • Option d lacks the cubic dependence characteristic of jump loss.


Common Pitfalls:
Using depth difference squared instead of cubed; forgetting the product 4 * D1 * D2 in the denominator.


Final Answer:
(D2 - D1)^3 / (4 * D1 * D2)

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