Hydrometry – Cup/propeller current meter calibration and use Given two calibration points (50 s counts): 12 rev → 0.25 m/s and 30 rev → 0.46 m/s. What velocity will 50 revolutions in 1 minute indicate?

Difficulty: Medium

Correct Answer: 0.60

Explanation:


Introduction / Context:
Current meters are calibrated by relating flow velocity v to rotation rate N (revolutions per second). A linear relation v = a + b * N is commonly used over practical ranges. With two calibration points, we can determine a and b and then compute velocity for any observed count rate.


Given Data / Assumptions:

  • Calibration point 1: 12 revolutions in 50 s → N1 = 12/50 rps, v1 = 0.25 m/s.
  • Calibration point 2: 30 revolutions in 50 s → N2 = 30/50 rps, v2 = 0.46 m/s.
  • Observation: 50 revolutions in 60 s → N = 50/60 rps.
  • Assume linear v–N relation: v = a + b * N within this range.


Concept / Approach:

Use the two given calibration pairs to solve for a and b. Then substitute the measured N to get the indicated velocity. This is the standard field practice for converting meter counts to flow velocity in open-channel gauging.


Step-by-Step Solution:

1) Compute N1 = 12/50 = 0.24 rps; N2 = 30/50 = 0.60 rps.2) Solve a + b * 0.24 = 0.25 and a + b * 0.60 = 0.46 → subtracting gives b * (0.60 − 0.24) = 0.21 → b = 0.21 / 0.36 = 0.58333.3) Back-substitute for a: a = 0.25 − 0.58333 * 0.24 = 0.11.4) Observation N = 50/60 = 0.83333 rps → v = a + b * N = 0.11 + 0.58333 * 0.83333 ≈ 0.596 ≈ 0.60 m/s.


Verification / Alternative check:

Check reasonableness: 0.60 m/s lies above 0.46 m/s, consistent with the higher rotation rate (50 in 60 s is faster than 30 in 50 s). Linear interpolation/extrapolation is mild and within calibration range.


Why Other Options Are Wrong:

0.42 and 0.50 are too low for the higher rotation rate; 0.73 is too high relative to the linear calibration.


Common Pitfalls:

Using revolutions per minute directly without converting to rps in the calibrated formula, or mixing up the 50 s calibration period with the 60 s observation period.


Final Answer:

0.60

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