Subtopic: Thermodynamic principles of I.C. engines
Question Statement:
A Cipolletti weir has the crest length of 1.2 m and a crest water level of 0.5 m. The average approach velocity of water in m s⁻¹ on the crest is
Options:
(A) 0.49
(B) 1.18
(C) 1.19
(D) 1.32
Official Answer Key:
C
Step-by-Step Solution & Explanation
Discharge through a Cipolletti (trapezoidal, side slopes 1H:4V) weir: Q = 1.859×L×H^1.5 (SI units).
Q = 1.859 × 1.2 × (0.5)^1.5 = 1.859 × 1.2 × 0.3536 = 0.789 m³/s.
The Cipolletti weir's flow section on the crest is trapezoidal (side slope 1:4, i.e. 0.25 m horizontal per 1 m vertical), so the wetted flow area is: A = (L + 0.25H)×H = (1.2 + 0.25×0.5)×0.5 = 1.325×0.5 = 0.6625 m² (NOT simply L×H, which would treat the section as rectangular — a common error that leads to option (D)).
Average approach velocity = Q/A = 0.789/0.6625 = 1.19 m/s. Matches option (C) exactly; using the rectangular (uncorrected) area instead gives Va ≈ 1.31–1.32 m/s, corresponding to the distractor (D).
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