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Safe Venting Limits for Chemical Refineries (Chemical Engineering) Solution

Formulas & Definitions

Problem Overview

We calculate the vertical liquid rise rate dhdt\frac{dh}{dt} inside a spherical industrial chemical dome of fixed internal radius R=12.0 mR = 12.0\text{ m} receiving liquid hydrocarbon at a constant volumetric inflow rate of dVdt=15.0 m3/min\frac{dV}{dt} = 15.0\text{ m}^3/\text{min}. The liquid volume is governed by the spherical cap relation:

V(h)=πRh2−13πh3V(h) = \pi R h^2 - \frac{1}{3}\pi h^3

Step-by-Step Mathematical Solution

Step 1: Differentiate the Spherical Cap Relation with Respect to Time

The governing formula is the Spherical Cap Volume Relation:

V=πRh2−13πh3V = \pi R h^2 - \frac{1}{3}\pi h^3
(This formula and its time derivative principles are defined in Related Rates.)

Here, the tank radius R=12.0 mR = 12.0\text{ m} is a static constant (dRdt=0\frac{dR}{dt} = 0), so we substitute R=12R = 12 before differentiation:

V=12πh2−13πh3V = 12\pi h^2 - \frac{1}{3}\pi h^3

Differentiating both sides implicitly with respect to time tt using the Power Rule and Chain Rule:

dVdt=ddt[12πh2−13πh3]=(24πh−πh2)dhdt=πh(24−h)dhdt\frac{dV}{dt} = \frac{d}{dt}\left[12\pi h^2 - \frac{1}{3}\pi h^3\right] = \left(24\pi h - \pi h^2\right)\frac{dh}{dt} = \pi h(24 - h)\frac{dh}{dt}

Isolating the vertical rise rate dhdt\frac{dh}{dt}:

dhdt=dVdtπh(24−h)\frac{dh}{dt} = \frac{\frac{dV}{dt}}{\pi h(24 - h)}

Step 2: Compute Rise Rate at Depth h=4.0 mh = 4.0\text{ m}

Substitute dVdt=15.0 m3/min\frac{dV}{dt} = 15.0\text{ m}^3/\text{min} and h=4.0 mh = 4.0\text{ m}:

dhdt=15.0π(4.0)(24−4.0)=15.0π(4.0)(20.0)=15.080.0π=316π m/min≈0.05968 m/min\frac{dh}{dt} = \frac{15.0}{\pi (4.0)(24 - 4.0)} = \frac{15.0}{\pi (4.0)(20.0)} = \frac{15.0}{80.0\pi} = \frac{3}{16\pi}\text{ m/min} \approx 0.05968\text{ m/min}

Converting to centimeters per minute:

dhdt≈5.97 cm/min\frac{dh}{dt} \approx 5.97\text{ cm/min}

Step 3: Comparative Analysis at h=2.0 mh = 2.0\text{ m} and h=12.0 mh = 12.0\text{ m}

  1. At shallow depth h=2.0 mh = 2.0\text{ m}:
    dhdt=15.0π(2.0)(24−2.0)=15.044.0π=1544π m/min≈0.1085 m/min≈10.85 cm/min\frac{dh}{dt} = \frac{15.0}{\pi(2.0)(24 - 2.0)} = \frac{15.0}{44.0\pi} = \frac{15}{44\pi}\text{ m/min} \approx 0.1085\text{ m/min} \approx 10.85\text{ cm/min}
  2. At mid-depth (equator) h=12.0 mh = 12.0\text{ m}:
    dhdt=15.0π(12.0)(24−12.0)=15.0144.0π=548π m/min≈0.03316 m/min≈3.32 cm/min\frac{dh}{dt} = \frac{15.0}{\pi(12.0)(24 - 12.0)} = \frac{15.0}{144.0\pi} = \frac{5}{48\pi}\text{ m/min} \approx 0.03316\text{ m/min} \approx 3.32\text{ cm/min}

Engineering Conclusion

In chemical process engineering and industrial storage management, the cross-sectional area of a spherical vessel varies drastically with height (Across(h)=π(2Rh−h2)=πh(24−h)A_{\text{cross}}(h) = \pi(2Rh - h^2) = \pi h(24 - h)). Notice that:

dVdt=Across(h)⋅dhdt  ⟹  dhdt=1Across(h)dVdt\frac{dV}{dt} = A_{\text{cross}}(h) \cdot \frac{dh}{dt} \implies \frac{dh}{dt} = \frac{1}{A_{\text{cross}}(h)}\frac{dV}{dt}

At the base of the tank (h=2 mh = 2\text{ m}), the horizontal surface area is small (44π m244\pi\text{ m}^2), causing the fluid level to surge rapidly upward at 10.85 cm/min10.85\text{ cm/min}. At the equator (h=12 mh = 12\text{ m}), the cross-section reaches its maximum area (144π m2144\pi\text{ m}^2), slowing the rise rate to just 3.32 cm/min3.32\text{ cm/min}.

Automated tank telemetry uses this related rates model to dynamically predict ullage compression (the compression of volatile gas headspace above the liquid). If liquid rises too quickly near the top of the vessel, the rapid reduction in vapor volume can exceed the mass flow capacity of the overhead pressure relief valves, triggering emergency flaring. By predicting dhdt\frac{dh}{dt}, digital control systems automatically modulate feed pumps to throttle charging rates as liquid depth approaches tank capacity.