Hydraulic Reservoir Surge Control (Civil Engineering) Solution

Problem Overview

We model the water accumulation rate inside a flood detention basin modeled as a horizontal triangular prism of fixed length $L = 100.0\text{ m}$ and cross-sectional width-to-depth constraint $w = h$ . Inflow during a storm is $\frac{dV}{dt} = 25.0\text{ m}^3/\text{min}$ . We calculate the vertical rise rate $\frac{dh}{dt}$ at $h = 4.0\text{ m}$ and evaluate emergency spillway bypass criteria.

Step-by-Step Mathematical Solution

Step 1: Establish the Volume Formula via Geometric Constraint

The volume of a triangular prism is the cross-sectional triangular area multiplied by the prism length:

V = \left(\frac{1}{2} \cdot w \cdot h\right) \cdot L

(This formula and its geometric reduction principles are defined in Geometric Similarity in Related Rates and Related Rates.)

The basin length $L = 100.0\text{ m}$ is a static constant ( $\frac{dL}{dt} = 0$ ). The cross-sectional geometry enforces the proportional constraint $w = h$ . Substituting both $L = 100$ and $w = h$ into the volume formula before differentiating:

V = \frac{1}{2}(h)(h)(100.0) = 50.0 h^2

Step 2: Differentiate with Respect to Time

Differentiate both sides implicitly with respect to time $t$ using the Power Rule and Chain Rule:

\frac{dV}{dt} = \frac{d}{dt}[50.0 h^2] = 100.0 h \frac{dh}{dt}

Isolating the water level rise rate $\frac{dh}{dt}$ :

\frac{dh}{dt} = \frac{1}{100.0 h} \frac{dV}{dt}

Step 3: Compute Rise Rate at Depth $h = 4.0\text{ m}$

Substitute $\frac{dV}{dt} = 25.0\text{ m}^3/\text{min}$ and $h = 4.0\text{ m}$ :

\frac{dh}{dt} = \frac{25.0}{100.0 (4.0)} = \frac{25.0}{400.0} = \frac{1}{16}\text{ m/min} = 0.0625\text{ m/min}

Converting to centimeters per minute:

\frac{dh}{dt} = 0.0625\text{ m/min} \times 100\text{ cm/m} = 6.25\text{ cm/min}

Step 4: Emergency Bypass Protocol Evaluation

Remaining Freeboard Depth to Spillway: $\Delta h = H_{\text{spillway}} - h = 6.0\text{ m} - 4.0\text{ m} = 2.0\text{ m}$
Additional Volume Required to Fill Basin from $h = 4.0\text{ m}$ to $h = 6.0\text{ m}$ : $V(6.0) = 50.0(6.0)^2 = 50.0(36.0) = 1,800.0\text{ m}^3$ $V(4.0) = 50.0(4.0)^2 = 50.0(16.0) = 800.0\text{ m}^3$ $\Delta V = 1,800.0 - 800.0 = 1,000.0\text{ m}^3$
Exact Time to Spillway Overtopping at $\frac{dV}{dt} = 25.0\text{ m}^3/\text{min}$ : $t_{\text{overtop}} = \frac{\Delta V}{\frac{dV}{dt}} = \frac{1,000.0\text{ m}^3}{25.0\text{ m}^3/\text{min}} = 40.0\text{ minutes}$
Safety Verdict: Because the true time to overtop is $40.0\text{ minutes}$ (which exceeds the $25.0\text{ minute}$ mandatory bypass threshold), emergency discharge gates do not need to be opened immediately. However, supervisory control systems must remain on high alert because if inflow surges above $40.0\text{ m}^3/\text{min}$ , the safety margin collapses.

Engineering Conclusion

In civil hydraulic engineering, retention basin cross-sections expand with depth: the horizontal water surface area at depth $h$ is $A_{\text{surface}}(h) = w \cdot L = h \cdot 100 = 100h$ . Because $\frac{dV}{dt} = A_{\text{surface}}(h) \cdot \frac{dh}{dt}$ , the rate of vertical rise $\frac{dh}{dt} = \frac{1}{100h}\frac{dV}{dt}$ decreases as the reservoir fills.

When water is shallow ( $h = 1\text{ m}$ ), the level rises at $25\text{ cm/min}$ . At $h = 4\text{ m}$ , the rise rate decelerates to $6.25\text{ cm/min}$ . Civil engineers use this related rates relationship in automated supervisory control and data acquisition (SCADA) networks. Acoustic depth sensors continuously measure $h$ and compute $\frac{dh}{dt}$ , allowing algorithms to dynamically estimate time-to-crest and modulate automated sluice gates to prevent earthen embankment breach without unnecessarily flooding downstream wash channels.