🎓 Lesson 4
D3
Continuity Equation in Differential and Integral Forms
The continuity equation says that fluid flowing into a space must either flow out or accumulate — nothing just disappears or appears out of nowhere.
🎯 Learning Objectives
- ✓ Derive the differential and integral forms of the continuity equation from first principles
- ✓ Apply the continuity equation to analyze transient fluid flow in mine ventilation ducts
- ✓ Calculate mass flow rates and velocity profiles in variable-area blasthole stemming columns
- ✓ Explain physical meaning of each term in both forms using mining-relevant scenarios
- ✓ Design a simplified ventilation network by enforcing continuity at junctions
📖 Why This Matters
In underground mines and open-pit operations, controlling airflow is critical for diluting diesel exhaust, managing explosive gas accumulation (e.g., methane), and ensuring worker safety during blasting. If airflow isn’t conserved — i.e., if more air enters a drift than exits — pressure imbalances develop, causing recirculation, poor fume clearance, or even backdrafting of toxic gases. The continuity equation is the foundational law that guarantees your ventilation design obeys physics — not just intuition.
📘 Core Principles
We begin with the concept of a control volume: a fixed region in space (e.g., a section of ventilation duct or a blasthole annulus). Mass conservation requires that any change in mass inside must equal net inflow minus outflow. For steady flow, accumulation vanishes (∂/∂t = 0); for incompressible fluids (e.g., water in dewatering systems), density ρ is constant, simplifying ∇·v = 0. In differential form, the equation captures local behavior — essential for CFD modeling of airflow around blast debris. In integral form, it governs system-level balancing — used daily by ventilation engineers to size fans and allocate airflow at splits and mergers.
📐 Key Calculation
The integral form is most practical for mine ventilation design: the sum of mass flow rates into a junction equals the sum exiting. For incompressible flow, volumetric flow rate Q = A·V suffices. We apply this at duct junctions, blasthole stemming interfaces, and fan inlets/outlets.
💡 Worked Example
Problem: A main intake duct (Duct A) supplies 120 m³/s to a junction. Two branch ducts exit: Duct B (diameter = 1.8 m) carries airflow to a development heading; Duct C (diameter = 1.2 m) serves a production stope. If average velocity in Duct B is 6.5 m/s, what must be the velocity in Duct C to satisfy continuity?
1.
Step 1: Compute cross-sectional area of Duct B: A_B = π × (0.9)² ≈ 2.545 m²
2.
Step 2: Calculate Q_B = A_B × V_B = 2.545 × 6.5 ≈ 16.54 m³/s
3.
Step 3: Apply continuity: Q_A = Q_B + Q_C → Q_C = 120 − 16.54 = 103.46 m³/s
4.
Step 4: Compute A_C = π × (0.6)² ≈ 1.131 m² → V_C = Q_C / A_C = 103.46 / 1.131 ≈ 91.5 m/s
Answer:
The required velocity in Duct C is 91.5 m/s — which exceeds safe limits (typical max = 15–20 m/s for mine ducts), indicating duct sizing is inadequate and redesign is needed.
🏗️ Real-World Application
At the Cadia East underground gold mine (NSW, Australia), ventilation engineers used the integral continuity equation to diagnose persistent CO accumulation in a raise-bored access shaft. Field measurements revealed 85 m³/s entering the shaft but only 62 m³/s exiting — a 23 m³/s deficit. Applying continuity, they identified unintended leakage paths into adjacent stopes via fractured ground. Sealing these paths restored balance and reduced CO exposure by >70% — validating continuity as both diagnostic and design tool.