Humidification operations and cooling towers

Why it matters

Cooling towers reject the waste heat of almost every refinery, power plant and chemical plant, and humidification and dehumidification are used in air conditioning, drying and textile plants. These are simultaneous heat and mass transfer operations between air and water, so you need psychrometry (humidity, wet-bulb temperature, enthalpy) and the Merkel enthalpy-driving-force method to size and troubleshoot them.

Key ideas

Psychrometric quantities (air–water at total pressure P).

• Absolute (molal or mass) humidity Y: kg water vapour per kg dry air. Dry-air basis is used because dry air flow is constant through the equipment.
• Saturation humidity Y_s: humidity when the partial pressure of water equals its vapour pressure at the gas temperature.
• Relative humidity = p_A/p_A^sat (as %); percentage humidity = Y/Y_s. They are close but not equal.
• Dew point: temperature at which the air, cooled at constant humidity, becomes saturated.
• Humid heat c_s and humid volume: heat capacity and volume of moist air per kg dry air.
• Enthalpy of moist air per kg dry air, referenced to 0 °C liquid water and dry air.

Adiabatic saturation and wet-bulb temperature. Air contacted with recirculated water in an insulated chamber cools and humidifies until it is saturated at the adiabatic saturation temperature. The wet-bulb temperature is the steady temperature of a small wetted thermometer in moving air, set by a balance between convective heat in and latent heat out. For air–water the Lewis relation h_G/(k_Y·c_s) ≈ 1 holds, so the wet-bulb and adiabatic-saturation temperatures are practically equal – a coincidence of air–water (Lewis number close to 1), not true for other systems.

Cooling towers. Warm water flows down over fill (splash bars or film packing); air flows up (counter-flow) or across (cross-flow), driven by a tall natural-draft shell or by fans (forced or induced draft). A small fraction (about 1–2 % per 10 K of cooling) evaporates and carries away most of the heat as latent heat; sensible heat transfer adds the rest.

• Range = water inlet − water outlet temperature.
• Approach = water outlet − inlet air wet-bulb temperature. The wet-bulb, not the dry-bulb, is the theoretical limit, and approaches below about 3–5 K become very expensive.
• Effectiveness = range/(range + approach).
• Make-up water = evaporation + drift (windage) + blowdown (to limit dissolved-solids build-up).

Merkel method. Combining heat and mass transfer with the Lewis relation gives a single driving force: the difference between the enthalpy of saturated air at the local water temperature, H*, and the actual air enthalpy, H. On an H–T diagram:

• the equilibrium curve is the saturated-air enthalpy H* versus water temperature (curved upwards);
• the operating line is straight with slope L·c_L/G_s, from an energy balance;
• tower height Z = H_tOG × N_tOG, where N_tOG = ∫ dH/(H* − H) (evaluate numerically).
• Minimum air rate: the operating line touches the equilibrium curve (often tangentially); practical G_s is 1.2–1.5 times the minimum.

Effect of conditions. A higher ambient wet-bulb (hot humid weather) raises the cold-water temperature; more air flow lowers the approach but costs fan power and increases drift; more fill area reduces height of a transfer unit.

Formulas

Absolute humidity (air–water): Y = 0.622·p_A/(P − p_A); general Y = (p_A/(P − p_A))·(M_A/M_B) Relative humidity: RH = p_A/p_A^sat Humid heat: c_s = 1.005 + 1.88·Y kJ/(kg dry air·K) Humid volume: v_H = (1/29 + Y/18)·(22.414)·(T/273.15)·(101.325/P) m³/kg dry air Enthalpy: H = (1.005 + 1.88·Y)·t + 2501·Y kJ/kg dry air (t in °C) Wet-bulb balance: h_G (T − T_w) = k_Y λ_w (Y_w − Y); air–water Lewis relation h_G/(k_Y·c_s) ≈ 1 Cooling-tower energy balance: L·c_L (T_L2 − T_L1) = G_s (H₂ − H₁) Merkel: Z = (G_s/(K_Y·a))·∫ dH/(H* − H) Range = T_in − T_out; approach = T_out − T_wb; effectiveness = range/(range + approach)

• p_A – partial pressure of water, kPa; P – total pressure, kPa; L – water flow, kg/s; G_s – dry-air flow, kg/s; c_L – 4.186 kJ/(kg·K); H – kJ/kg dry air; K_Y·a – kg/(m³·s); λ_w – latent heat at T_w, kJ/kg. Take vapour pressures from steam tables.

Worked examples

Example 1 – psychrometric properties (standard). Given: air at 30 °C, 101.325 kPa, relative humidity 60 %; p^sat of water at 30 °C = 4.246 kPa (steam tables).

1. p_A = 0.60 × 4.246 = 2.548 kPa.
2. Y = 0.622·p_A/(P − p_A) = 0.622 × 2.548/98.78 = 0.01604 kg/kg dry air.
3. c_s = 1.005 + 1.88 × 0.01604 = 1.035 kJ/(kg dry air·K).
4. H = 1.035 × 30 + 2501 × 0.01604 = 31.06 + 40.12 = 71.2 kJ/kg dry air.
5. Dew point: temperature where p^sat = 2.548 kPa; between 21 °C (2.487 kPa) and 22 °C (2.645 kPa) → about 21.4 °C.

Y = 0.0160 kg/kg dry air, H = 71.2 kJ/kg dry air, dew point ≈ 21.4 °C.

Example 2 – cooling tower balance (GATE level). Given: 10 kg/s of water cooled from 40 °C to 30 °C; inlet air wet-bulb 25 °C; dry-air flow 10 kg/s; take the inlet air enthalpy as that of saturated air at 25 °C (p^sat = 3.169 kPa); c_L = 4.186 kJ/(kg·K).

1. Range = 10 K, approach = 30 − 25 = 5 K; effectiveness = 10/15 = 0.667.
2. Inlet air: Y = 0.622 × 3.169/(101.325 − 3.169) = 0.02008; H₁ = (1.005 + 1.88 × 0.02008) × 25 + 2501 × 0.02008 = 76.3 kJ/kg dry air.
3. Heat duty Q = L·c_L·ΔT = 10 × 4.186 × 10 = 418.6 kW.
4. H₂ = H₁ + Q/G_s = 76.3 + 41.9 = 118.2 kJ/kg dry air.
5. Evaporation ≈ Q/λ ≈ 418.6/2400 = 0.174 kg/s (about 1.7 % of the water circulated).

Q = 418.6 kW; exit-air enthalpy 118.2 kJ/kg dry air; effectiveness 66.7 %; evaporation ≈ 0.17 kg/s.

Common mistakes

• Basing humidity on total moist air instead of dry air.
• Using dry-bulb temperature as the cooling limit – the limit is the wet-bulb.
• Treating relative and percentage humidity as identical at high humidity.
• Using the Lewis relation (T_wb = T_as) for non-water systems such as toluene–air.
• Writing heat removed in kJ when flows are in kg/s – the answer is kW.
• Forgetting drift and blowdown in make-up water.

For GATE CH

Expect calculations of humidity, relative humidity, dew point, humid heat and enthalpy from vapour-pressure data; cooling-tower range, approach, effectiveness and energy balances; exit air enthalpy or minimum air rate on the H–T diagram; and conceptual questions on wet-bulb versus adiabatic saturation temperature and the Lewis relation.

Quick check

1. Why is humidity expressed per kg of dry air?
2. Define approach in a cooling tower.
3. When does the wet-bulb temperature equal the adiabatic saturation temperature?
4. A tower cools water from 42 °C to 32 °C with a wet-bulb of 27 °C. What is its effectiveness?

Answers: 1. dry-air flow stays constant while water vapour changes; 2. cold-water temperature minus inlet-air wet-bulb temperature; 3. when the Lewis relation h_G/(k_Y·c_s) ≈ 1 holds, as for air–water; 4. 10/(10 + 5) = 66.7 %.