Question 3 of 6: Mine Ventilation — Cooling/Dehumidification, Psychrometrics and a Cooling-Coil Process

Question 3: Mine Ventilation — Cooling/Dehumidification, Psychrometrics and a Cooling-Coil Process (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

3.1 — Refrigeration and cooling towers

Refrigeration. A mechanical vapour-compression (occasionally absorption) chiller circulates chilled water or brine to air-handling cooling coils, either at a bulk air cooler (BAC) plant on surface conditioning the entire intake airstream, or as spot coolers placed underground close to hot working faces for localised relief. Air drawn across the chilled coil by a fan is cooled below its dew point, so it is both sensibly cooled (dry-bulb temperature drops) and dehumidified (moisture condenses out on the coil and drains away) in the same pass. Components: compressor, condenser, expansion valve, evaporator/cooling coil, chilled-water circulating pumps and pipework, and the air-handling unit/fan that draws mine air across the coil. Physically, heat is removed from the air at the evaporator/cooling coil by the refrigerant absorbing it as it evaporates (latent heat of the refrigerant, not the air), then rejected at the condenser to a secondary heat sink (surface ambient air or a cooling-tower water loop) as the refrigerant is compressed and condensed back to liquid, closing the cycle.

Cooling towers. A cooling tower rejects the heat picked up by the condenser-water loop to the atmosphere by direct-contact evaporative cooling: warm condenser water is distributed over a fill/packing material inside the tower while ambient air is drawn or forced through it (counter-flow or cross-flow); a small fraction of the water evaporates into the passing airstream, absorbing latent heat from the remaining bulk water and cooling it before it is recirculated to the condenser. Components: distribution deck/spray nozzles, fill/packing, induced- or forced-draft fan, cold-water collection basin, and the pump/pipework loop back to the condenser. Because the dominant heat-rejection mechanism is evaporation (latent heat carried away in the water vapour added to the air), not sensible heating of the air passing through, cooling-tower performance is governed by the ambient WET-bulb temperature, not dry-bulb — the tower can, in principle, cool the water to within a few degrees of the entering air's wet-bulb temperature, but never below it.

3.2 — Enthalpy and the psychrometric chart

Enthalpy. The total heat content of moist air per unit mass of DRY air — the sensible heat of the dry air and its water vapour, plus the latent heat of vaporisation carried by the water vapour present — expressed in kJ/kg dry air (or Btu/lb dry air). It is the quantity actually conserved (together with the dry-air mass flow G) in a fan/coil heat balance such as $q=G(h_1-h_2)$, which is why enthalpy, not dry-bulb temperature alone, is the correct basis for sizing a cooling/dehumidifying coil.

Psychrometric chart. A graphical plot of the state properties of moist air — dry-bulb temperature on the x-axis, humidity ratio (moisture content) on the y-axis, with superimposed curves of constant relative humidity, wet-bulb temperature and enthalpy — that lets any two independent properties (e.g. dry-bulb and wet-bulb) fix the complete state of the air and every other property be read off directly, avoiding the underlying psychrometric equations for routine estimating.

3.3 — Cooling-coil process: heat and moisture removed

Given.

Find. The specific heat removed (3.3.2, q/G) and the specific moisture removed (3.3.3, Gw/G) between states 1 and 2.

Approach. Find the humidity ratio at state 1 from the dry-bulb/wet-bulb pair via the Carrier (ASHRAE) psychrometric equation, and at state 2 directly from saturation (dry-bulb = wet-bulb = dew point); compute the moist-air enthalpy at each state; then $q/G=h_1-h_2$ and $G_w/G=W_1-W_2$.

1. Part 3.3.1 — sketch the process. On the psychrometric chart the process runs from state 1 (32.2°C DB, 26.7°C WB, unsaturated) down and to the left to state 2 (21.1°C, saturated) — both dry-bulb temperature and humidity ratio fall together as the coil removes both sensible and latent heat, with the end state landing exactly on the saturation curve.
   Cooling-coil process on the psychrometric chart: state 1 (entering, unsaturated) moves down-left to state 2 (leaving, saturated at 21.1°C) as the coil removes both sensible and latent heat.
2. Part 3.3.2 — specific enthalpy change (change in heat). Using the Carrier equation to get the humidity ratio at state 1 from the 26.7°C wet-bulb reading, then the standard moist-air enthalpy relation $h=1.006T_{db}+W(2501+1.86T_{db})$ at each state: $$W_1 = 0.01988\ \text{kg/kg dry air}, \qquad h_1 = 1.006(32.2)+0.01988(2501+1.86\times32.2) = 83.31\ \text{kJ/kg}$$ $$W_2 = 0.622\dfrac{p_{ws}(21.1)}{P-p_{ws}(21.1)} = 0.01572\ \text{kg/kg dry air}, \qquad h_2 = 1.006(21.1)+0.01572(2501+1.86\times21.1) = 61.15\ \text{kJ/kg}$$ $$\boxed{\dfrac{q}{G} = h_1-h_2 = 83.31-61.15 = 22.15\ \text{kJ/kg dry air}\ \ (9.52\ \text{Btu/lb dry air})}$$ (Multiply by the actual dry-air mass flow rate G, kg/s, to get q in kW = kJ/s = 1000 W per kJ/kg·kg/s.)
3. Part 3.3.3 — specific moisture removed. $$\boxed{\dfrac{G_w}{G} = W_1-W_2 = 0.01988-0.01572 = 0.004164\ \text{kg water/kg dry air}\ \ (29.15\ \text{grains/lb dry air})}$$ (Multiply by G, kg/s dry air, to get the moisture condensed out at the coil in kg/s, or convert G to lb/hr for lb/hr.)