Humidity, saturation and use of the psychrometric chart

Humidity, relative and percentage saturation, dew point, wet-bulb temperature, humid heat and volume, and reading the psychrometric chart for heating, cooling, dehumidifying and mixing.

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Why it matters

Dryers, cooling towers, air-conditioning plants, compressed-air systems and humidifiers all handle mixtures of a non-condensable gas (usually air) and a condensable vapour (usually water). Their design rests on a handful of humidity measures, the dew point and wet-bulb temperature, and the psychrometric chart that ties them together.

Key ideas

Vapour–gas mixtures. Water vapour in air behaves as an ideal gas at near-atmospheric pressure. Its partial pressure p_w can never exceed the saturation (vapour) pressure p_s of water at the mixture temperature; when p_w = p_s the air is saturated.

Humidity measures.

  • Absolute (specific) humidity H, also called humidity ratio: kg water vapour per kg dry air. Dry air is used as the basis because its mass does not change when water evaporates or condenses.
  • Molal humidity H_m: kmol vapour per kmol dry air.
  • Relative humidity RH = p_w / p_s at the same temperature (×100 %).
  • Percentage (absolute) saturation = H / H_s × 100 %, where H_s is the humidity of saturated air at the same T and P. It is slightly lower than RH except at 0 % and 100 %.

Dew point. The temperature at which unsaturated air becomes saturated when cooled at constant pressure and constant humidity: p_s(T_dp) = p_w. Cooling below the dew point condenses water – the principle of dehumidification and of condensate in compressed-air lines.

Wet-bulb temperature. The steady temperature reached by a small wetted thermometer in a fast air stream, where heat transferred to the wick balances the latent heat of evaporation. It lies between the dew point and the dry-bulb temperature (all three are equal for saturated air). For the air–water system the wet-bulb temperature practically equals the adiabatic-saturation temperature (Lewis relation ≈ 1), which is why adiabatic humidification follows a constant wet-bulb line on the chart.

Humid heat and humid volume. Per kg of dry air: humid heat c_s = 1.005 + 1.88·H kJ/(kg dry air·K) and humid volume v_H = (1/29 + H/18)·R·T/P m³/kg dry air. They are used in dryer and air-conditioning energy balances.

Psychrometric chart. Plots H against dry-bulb temperature at a fixed total pressure (usually 101.325 kPa), with lines of constant RH, wet-bulb temperature, enthalpy and humid volume. Given any two independent properties you can read the rest. Common processes: sensible heating or cooling (horizontal line, H constant), cooling-and-dehumidifying (to the saturation curve, then down along it), adiabatic humidification (along a wet-bulb line), and mixing of two streams (on the straight line joining them, by lever rule on dry-air mass). Charts are for one pressure only – at other pressures use the equations.

Formulas

  • H = 0.622·p_w / (P − p_w) – kg water / kg dry air; 0.622 = 18.02 / 28.97; P total pressure, p_w vapour partial pressure (same units).
  • H_m = p_w / (P − p_w) – kmol water / kmol dry air.
  • RH = p_w / p_s(T) × 100 %
  • % saturation = H / H_s × 100 % = RH × (P − p_s) / (P − p_w)
  • p_s(T_dp) = p_w – dew point; read T_dp from steam tables.
  • c_s = 1.005 + 1.88·H – kJ/(kg dry air·K).
  • v_H = (1/28.97 + H/18.02)·R·T / P – m³/kg dry air; R = 8.314 kPa·m³/(kmol·K), T in K, P in kPa.
  • p_w ≈ p_s(T_wb) − A·P·(T − T_wb) – psychrometer equation; A ≈ 6.6 × 10⁻⁴ K⁻¹ for a well-aspirated psychrometer (empirical; check your data book).

Worked examples

Saturation pressures of water (steam tables): 15 °C 1.705 kPa; 21 °C 2.487 kPa; 22 °C 2.645 kPa; 25 °C 3.169 kPa; 30 °C 4.246 kPa; 35 °C 5.628 kPa. P = 101.325 kPa.

Example 1 (standard). Air at 30 °C has a relative humidity of 60 %. Find the humidity, percentage saturation, dew point, humid heat and humid volume.

  1. p_w = RH·p_s = 0.60 × 4.246 = 2.548 kPa.
  2. H = 0.622 × 2.548 / (101.325 − 2.548) = 0.01604 kg/kg dry air.
  3. H_s = 0.622 × 4.246 / (101.325 − 4.246) = 0.02720 kg/kg dry air; % saturation = 0.01604 / 0.02720 = 59.0 %.
  4. Dew point: p_s(T_dp) = 2.548 kPa lies between 21 °C and 22 °C; interpolating, T_dp = 21 + (2.548 − 2.487) / (2.645 − 2.487) = 21.4 °C.
  5. c_s = 1.005 + 1.88 × 0.01604 = 1.035 kJ/(kg dry air·K).
  6. v_H = (1/28.97 + 0.01604/18.02) × 8.314 × 303.15 / 101.325 = 0.881 m³/kg dry air.

Answer: H = 0.0160 kg/kg dry air; 59.0 % saturation; T_dp ≈ 21.4 °C; c_s = 1.035 kJ/(kg·K); v_H = 0.881 m³/kg dry air

Example 2 (GATE level). 1000 m³ of air at 35 °C and 70 % RH is cooled to 15 °C, leaving saturated, and the condensate is drained. The air is then reheated to 25 °C. Find the water removed and the final RH.

  1. Inlet: p_w1 = 0.70 × 5.628 = 3.940 kPa; H₁ = 0.622 × 3.940 / (101.325 − 3.940) = 0.02516 kg/kg dry air.
  2. Outlet of cooler (saturated at 15 °C): H₂ = 0.622 × 1.705 / (101.325 − 1.705) = 0.01065 kg/kg dry air.
  3. Humid volume of inlet air: v_H1 = (1/28.97 + 0.02516/18.02) × 8.314 × 308.15 / 101.325 = 0.908 m³/kg dry air.
  4. Dry air: 1000 / 0.908 = 1101 kg.
  5. Water condensed = 1101 × (0.02516 − 0.01065) = 16.0 kg.
  6. Reheating does not change H, so p_w stays 1.705 kPa; RH at 25 °C = 1.705 / 3.169 = 53.8 %.

Answer: ≈ 16.0 kg water removed; final RH ≈ 54 %

Common mistakes

  • Basing humidity on kg of wet air instead of kg of dry air.
  • Using RH and percentage saturation interchangeably in precise work.
  • Taking p_s at the dew point or wet-bulb temperature when the mixture temperature is meant (RH uses p_s at the dry-bulb temperature).
  • Expecting RH to stay constant during heating – H stays constant; RH falls.
  • Reading a 101.325 kPa chart for air at a different total pressure.
  • Converting a volume of humid air to dry-air mass with the dry-air density instead of the humid volume.

For GATE CH

Expect NAT questions on humidity from partial pressure or RH, dew point from steam-table data, water condensed in a cooler, mixing of two air streams, and dryer balances using humid heat. Practise working on a dry-air basis and checking whether the air reaches saturation in each step.

Quick check

  1. Air at 101.325 kPa has p_w = 2.0 kPa. What is its humidity?
  2. For saturated air, how do the dry-bulb, wet-bulb and dew-point temperatures compare?
  3. Air at 20 °C and 60 % RH has p_w = 1.403 kPa. What is its dew point (p_s at 12 °C = 1.403 kPa)?
  4. When unsaturated air is heated at constant pressure, what happens to H and RH?
  5. Which basis is used for humidity, and why?

Answers: 1. 0.0125 kg/kg dry air; 2. they are equal; 3. 12 °C; 4. H constant, RH decreases; 5. kg of dry air, because its mass is unchanged by evaporation or condensation.

Interview questions

All Process Calculations interview questions

Try answering each one aloud before you open it.

  1. 1.What is humidity and how is it measured?Concept

    Humidity is the amount of water vapor present in the air. It is typically measured in terms of relative humidity, which is the ratio of the current absolute humidity to the highest possible absolute humidity at that temperature, expressed as a percentage. Instruments like hygrometers or psychrometers are used to measure humidity.

  2. 2.Explain the concept of saturation in the context of humidity.Concept

    Saturation occurs when the air contains the maximum amount of water vapor it can hold at a given temperature and pressure. At this point, the relative humidity is 100%, and any additional water vapor will condense into liquid water. This is an important concept in processes like drying and air conditioning.

  3. 3.What is a psychrometric chart and what information can it provide?Concept

    A psychrometric chart is a graphical representation of the physical and thermal properties of moist air. It provides information such as dry bulb temperature, wet bulb temperature, relative humidity, dew point, specific humidity, and enthalpy. Engineers use it to analyze air conditioning processes and to design HVAC systems.

  4. 4.Why is the wet bulb temperature important in industrial processes?Application

    The wet bulb temperature is important because it reflects the lowest temperature that can be achieved by evaporative cooling. It is used in designing cooling towers and air conditioning systems. It also helps in determining the efficiency of evaporative cooling processes.

  5. 5.What happens if air is cooled below its dew point?Application

    If air is cooled below its dew point, the water vapor in the air will begin to condense into liquid water. This is because the air can no longer hold the same amount of moisture at the lower temperature. This principle is used in dehumidification and in processes like refrigeration.

  6. 6.How does relative humidity affect human comfort and industrial processes?Application

    Relative humidity affects human comfort as it influences the body's ability to cool itself through evaporation of sweat. High humidity can make temperatures feel hotter, while low humidity can cause dryness and irritation. In industrial processes, relative humidity can affect drying rates, product quality, and equipment efficiency.

  7. 7.Explain how a cooling tower uses the principles of humidity and saturation.Application

    A cooling tower uses the principles of humidity and saturation by allowing water to evaporate into the air, which cools the remaining water. As the air becomes saturated with water vapor, it carries away heat from the water, effectively cooling it. This process relies on the wet bulb temperature and the air's capacity to absorb moisture.

  8. 8.Estimate the relative humidity of air at 101.325 kPa with a dry-bulb temperature of 30 °C and a wet-bulb temperature of 25 °C.Numerical

    Use the psychrometer equation p_w ≈ p_s(T_wb) − A·P·(T − T_wb) with A ≈ 6.6 × 10⁻⁴ K⁻¹: p_w = 3.169 − 6.6 × 10⁻⁴ × 101.325 × 5 = 2.83 kPa. Then RH = p_w / p_s(30 °C) = 2.83 / 4.246 ≈ 67 %. A psychrometric chart gives about the same value; the constant A is empirical, so quote it from your data book.

  9. 9.Determine the dew point of air at 20 °C and 60 % relative humidity.Numerical

    The vapour partial pressure is p_w = RH × p_s(20 °C) = 0.60 × 2.339 = 1.403 kPa. The dew point is the temperature at which this equals the saturation pressure; steam tables give p_s = 1.403 kPa at about 12 °C. So the dew point is ≈ 12 °C – any surface colder than that will collect condensate.

  10. 10.What is the significance of the dew point in weather forecasting and industrial applications?Application

    The dew point is significant in weather forecasting as it indicates the temperature at which dew, frost, or fog may form. In industrial applications, it helps in controlling humidity levels in processes like drying, storage, and air conditioning. It is a critical parameter for ensuring product quality and process efficiency.

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