🎓 Lesson 1 D2

First Law Applied to Continuous Process Units

The First Law of Thermodynamics for continuous processes says that energy flowing into a system (like heat or work) minus energy flowing out must equal the energy stored or used up inside the system — just like balancing a bank account over time.

🎯 Learning Objectives

  • Calculate the steady-state heat transfer rate for a continuous blasting air-cooling heat exchanger using mass and energy balances
  • Apply the First Law to analyze energy requirements of a continuous drill-cuttings conveying system
  • Explain how shaft work and enthalpy flow contribute to energy balance in a rotary blasthole drill rig’s hydraulic power unit
  • Design a simplified energy balance model for a continuous explosive slurry mixing unit

📖 Why This Matters

In mining and blasting operations, many critical units — such as explosive slurry plants, ventilation fans, compressed air systems, and continuous drill cuttings conveyors — operate continuously, not in batches. Misapplying batch-energy logic here leads to dangerous underestimation of cooling loads, motor sizing errors, or unexpected thermal runaway. Mastering the First Law for continuous systems ensures safe, efficient, and compliant design of infrastructure that keeps workers cool, explosives stable, and equipment running.

📘 Core Principles

The First Law for continuous processes rests on three foundational ideas: (1) Steady-state assumption — mass flow rates, temperatures, pressures, and energy flows remain constant over time; (2) Enthalpy dominance — for flowing fluids, sensible and latent energy is best tracked via specific enthalpy (h = u + Pv), not internal energy alone; (3) Energy accounting rigor — shaft work (e.g., pump/fan work), heat transfer (e.g., jacket cooling), and flow work (Pv) must all be consistently signed and dimensionally aligned. Non-steady transients (e.g., startup of a detonation chamber) require accumulation terms — but Module 2 focuses exclusively on the steady-state case essential for plant-scale design.

📐 Steady-State First Law for Continuous Flow

The general steady-state energy balance for a single-inlet, single-outlet continuous unit (neglecting ΔKE and ΔPE unless specified) is: \(\dot{Q} - \dot{W}_\text{shaft} = \dot{m}(h_2 - h_1)\). For multi-stream units (e.g., mixing tanks, heat exchangers), sum all inlet and outlet enthalpy flows. Always verify sign conventions: heat added (+), work done *by* system (+), enthalpy increase downstream (+).

💡 Worked Example

Problem: A continuous ANFO mixing unit receives liquid ammonium nitrate solution (ṁ₁ = 8.2 kg/s, h₁ = 125 kJ/kg) and fuel oil (ṁ₂ = 0.41 kg/s, h₂ = 85 kJ/kg), and discharges mixed ANFO slurry (ṁ₃ = 8.61 kg/s, h₃ = ?). The unit loses 42 kW of heat to ambient and consumes 18 kW of shaft power (mixer motor). Find h₃.
1. Step 1: Apply conservation of mass → ṁ₃ = ṁ₁ + ṁ₂ = 8.2 + 0.41 = 8.61 kg/s (verified)
2. Step 2: Apply First Law: \(\dot{Q} - \dot{W}_\text{shaft} = \sum \dot{m}_\text{out} h_\text{out} - \sum \dot{m}_\text{in} h_\text{in}\). Here, \(\dot{Q} = -42\) kW (heat loss), \(\dot{W}_\text{shaft} = -18\) kW (work done *on* system → negative sign per convention), so left side = −42 − (−18) = −24 kW.
3. Step 3: Right side = ṁ₃h₃ − (ṁ₁h₁ + ṁ₂h₂) = 8.61·h₃ − (8.2·125 + 0.41·85) = 8.61·h₃ − (1025 + 34.85) = 8.61·h₃ − 1059.85 kJ/s (kW). Solve: −24 = 8.61·h₃ − 1059.85 → h₃ = (1059.85 − 24)/8.61 = 120.2 kJ/kg.
Answer: The specific enthalpy of the mixed ANFO slurry is 120.2 kJ/kg, which falls within the typical range of 115–128 kJ/kg for stabilized 94/6 ANFO at 25°C.

🏗️ Real-World Application

At BHP’s Olympic Dam underground mine, continuous ventilation air is cooled using glycol-chilled heat exchangers before entering production stopes. Engineers applied the steady-state First Law to size the chiller duty: measuring inlet air (32°C, 220 kg/s), outlet air (18°C, 220 kg/s), and known fan shaft work (1.4 MW), they calculated required cooling capacity as 3.1 MW — enabling precise chiller selection and avoiding thermal stress on blasting crews. This calculation directly informed the installation of redundant chillers meeting AS/NZS 1666.1 safety margins.

📚 References