Heat & Refrigeration

Deep mines get hot. Learn heat sources (virgin rock temperature, autocompression, machinery, blasting), the geothermal gradient, wet-bulb temperature and heat stress limits, air cooling power, refrigeration plants, bulk air coolers and ice plants.

At 3 km depth, rock can reach 60 °C. Air descending a shaft also warms by compression (≈ 1 °C per 100 m). Without cooling, underground workplaces become lethal. Deep mines run refrigeration plants of tens of megawatts.

An analogy

A deep mine is like working inside an oven that heats itself — the deeper you go, the hotter the walls, and even the air warms as it falls.

Heat sources

Rock heat (conduction from hot rock walls), autocompression of descending air, diesel engines (≈ 3× their rated output as heat), electric motors, blasting, fissure water, and people. In battery-electric mines, heat replaces diesel exhaust as the main ventilation driver.

Heat stress

Wet-bulb temperature combines heat and humidity — sweat cannot evaporate in saturated air. Typical limits: work restrictions above 27–28 °C wet-bulb, stop work at ≈ 32 °C. Air velocity improves cooling. Heat acclimatisation programmes protect workers.

Cooling methods

Increase airflow (limited by velocity), surface bulk air coolers on intake air, underground spot coolers, chilled water circuits, ice slurry plants (South Africa) and energy recovery turbines on chilled water returning down the shaft.

Climate simulation

Software (Ventsim, VUMA, Climsim) models heat, moisture and air psychrometrics through the network, optimising cooling placement. Thermal design is a major cost driver for deep gold and copper mines.

Formulas

• Rock temperature: VRT = T_surface + G_g × depth
• Autocompression (dry): ΔT = g z / c_p ≈ 9.8 °C per km
• Sensible cooling: Q_c = ṁ c_p ΔT = ρ Q_air c_p ΔT
• Refrigeration tons: 1 TR = 3.517 kW

Worked example: Autocompression

Given: Shaft depth 1500 m, surface 25 °C (dry)

Find: Air temperature at bottom.

• ΔT = 9.81 × 1500/1005 = 14.6 °C
• T = 25 + 14.6 = 39.6 °C

Answer: ≈ 40 °C dry-bulb before any other heat

Worked example: Cooling load

Given: 200 m³/s air (ρ 1.2 kg/m³) cooled by 8 °C (sensible)

Find: Refrigeration.

• ṁ = 240 kg/s
• Q = 240 × 1.005 × 8 = 1930 kW
• = 549 TR

Answer: ≈ 1.9 MW (≈ 550 TR)

Common mistakes

• Wrong: Using dry-bulb temperature for heat stress. — Right: Wet-bulb controls human heat stress.
• Wrong: Ignoring latent heat. — Right: Evaporation from wet rock adds large latent loads.

Applications

• Deep gold mines: Ice plants and chilled water in South Africa.
• Deep copper/nickel: Bulk air coolers in Canada and Chile.
• BEV mines: Heat-based ventilation design.

Key takeaways

• VRT rises with the geothermal gradient.
• Autocompression ≈ 9.8 °C/km.
• Wet-bulb controls heat stress.
• Q_c = ρ Q c_p ΔT; 1 TR = 3.517 kW.

Going further

• Psychrometrics: Enthalpy-based air cooling with latent loads.
• Energy recovery: Turbines on chilled water returning down shafts.
• Heat stress monitoring: Wearable sensors and work-rest regimes.