Difficulty: Medium
Correct Answer: 4.0 m
Explanation:
Introduction / Context:
Discharge through short tubes and mouthpieces differs from that through thin-plate orifices because the jet expands to run full, altering the discharge coefficient. For an external mouthpiece on a large tank, we estimate the head from the measured discharge using the appropriate coefficient of discharge.
Given Data / Assumptions:
Concept / Approach:
For a mouthpiece: Q = C_d * a * sqrt(2 * g * H). Rearranging gives H = (Q / (C_d * a))^2 / (2 * g). We insert the measured Q and standard C_d to compute the head over the mouthpiece centerline.
Step-by-Step Calculation:
a = π * (0.10)^2 / 4 ≈ 7.854 × 10^-3 m^2.Velocity V = Q / a ≈ 0.05948 / 7.854e-3 ≈ 7.58 m/s.Ideal jet speed sqrt(2gH) = V / C_d ≈ 7.58 / 0.82 ≈ 9.24 m/s.Compute head H = (9.24^2) / (2 * 9.81) ≈ 85.4 / 19.62 ≈ 4.35 m.Closest listed value ≈ 4.0 m.
Verification / Alternative Check:
Using C_d in the 0.80–0.83 range gives H between roughly 4.2 and 4.5 m; 4.0 m is the closest discrete option and consistent with typical textbook rounding and experimental variability.
Why Other Options Are Wrong:
2.0–3.0 m: too small for the measured discharge and area with realistic coefficients.6.0 m: overestimates head unless coefficients are unreasonably low.
Common Pitfalls:
Final Answer:
4.0 m
Discussion & Comments