Basin Storage Change by Trapezoid Rule
Mediumstoragemass-balancetrapezoid
A basin starts with 1,000 m³ of storage at t = 0. Measured inflows at t = 0, 300 and 600 s are 1, 3 and 2 m³/s; measured outflows at the same times are 0.5, 1.0 and 1.5 m³/s. Using trapezoidal integration of net flow, what is the maximum storage among the three tabulated times?
- A2,350 m³
- B1,750 m³
- C750 m³
- D2,500 m³
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Answer: B. 1,750 m³
ΔS = ∫(I − O)dt, trapezoid per interval. Net flow at the three times is 0.5, 2.0 and 0.5 m³/s. Interval 1: (0.5 + 2.0)/2 · 300 = 375 m³, so S₁ = 1,375. Interval 2: (2.0 + 0.5)/2 · 300 = 375 m³, so S₂ = 1,750 m³, the largest tabulated value. 2,350 m³ adds inflow volume (1,350) and ignores outflow. 750 m³ leaves out the initial 1,000 m³ and gets 750 from the net volume of both intervals. 2,500 m³ uses 600 s for each interval instead of 300 s.
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