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7.2 · Relate dew-point and saturation states

Learn to relate dew-point and saturation states through clear examples and targeted practice.

University of Alberta MEC E 340: Applied Thermodynamics

Psychrometry and Moist Air

Relating the water-vapour state in moist air to saturation

Begin with a closed sample of moist air: dry air and water vapour form the working mixture. State 1 is its initial condition, described by total pressure, dry-bulb temperature, and water-vapour content. Imagine cooling the sample without adding or removing water vapour. Under the idealized assumptions used here, the actual water-vapour partial pressure stays constant until condensation begins. At state 2, the vapour is just saturated; its temperature is the dew point. No equipment or cycle is being analyzed, so a mass or energy balance is not needed to determine the dew point from known vapour pressure and saturation data. All saturation values used below are supplied as problem data.

What you will learn

  • Explain dew-point temperature as a saturation condition for water vapour in moist air.
  • Relate water-vapour partial pressure to saturation pressure at the dew point.
  • Use supplied saturation data to estimate dew point or relative humidity.
  • Identify the first-condensation temperature during cooling at fixed water-vapour content.

1. Relate the dew point to a saturation state

Moist air is a mixture of dry air and water vapour. The total pressure is the sum of the component partial pressures. The water-vapour partial pressure, rather than the total pressure by itself, is the pressure to compare with water’s saturation pressure.
At a specified temperature, saturation pressure is the water-vapour pressure associated with equilibrium between vapour and liquid water. At the dew point, the actual water-vapour partial pressure equals the saturation pressure at that temperature. This relation defines the connection between the initial moist-air state and the dew-point saturation state.
As the sample cools at fixed water-vapour content, its vapour partial pressure is treated as constant before condensation. Saturation pressure decreases with decreasing temperature over the range considered. The dew point is reached when the saturation pressure has fallen to the actual vapour pressure.
pv=psat(Tdp)p_v=p_{sat}(T_{dp})
  • At the dew point, water vapour is just saturated.
  • Before condensation, the stated fixed-moisture idealization keeps the actual vapour partial pressure constant.
  • At the dew point, relative humidity is 100%.

2. Use saturation data with moist-air properties

If actual vapour pressure is known, find the dew point by locating the temperature at which supplied saturation data give that same pressure. If the target pressure lies between two tabulated saturation pressures, interpolation estimates the corresponding temperature. Interpolation should only be used when the data bracket the target; otherwise, obtain additional data or clearly identify any estimate as extrapolation.
If relative humidity and dry-bulb temperature are known, first use the saturation pressure at the dry-bulb temperature to calculate actual vapour pressure. Then use that actual pressure to find the dew point. Relative humidity is not the water-vapour content itself; it compares actual vapour pressure with saturation pressure at the current temperature.
In practical work, saturation values come from the specified property table or chart. The examples state all values needed for their calculations. Do not silently substitute remembered values for the data supplied in a problem.
ϕ=pvpsat(Tdb)\phi=\frac{p_v}{p_{sat}(T_{db})}
  • For relative humidity, use saturation pressure at the dry-bulb temperature.
  • For dew point, find where saturation pressure matches actual vapour pressure.
  • Only interpolate between supplied data that bracket the target.

3. Interpret cooling and check the result

For the cooling path considered here, state 1 is unsaturated and state 2 is the first saturated state. The sample is closed, and its water-vapour content is fixed until state 2. No condensation has occurred before that point. Further cooling can cause condensation, but calculating the amount condensed is outside this lesson.
The dew-point relation is a moist-air state relation, not a cycle-performance calculation. No cycle efficiency or coefficient of performance is needed. If analyzing actual process equipment, additional mass and energy balances would be required; they are not required just to relate a known vapour pressure to its saturation temperature.
Use physical checks. For an unsaturated state, dew point is below dry-bulb temperature. At saturation, they are equal. If a purported unsaturated state gives relative humidity above 100%, the stated data and equilibrium interpretation are inconsistent.
Tdp≤TdbT_{dp}\leq T_{db}
  • For an unsaturated state, the dew point is below the dry-bulb temperature.
  • For a saturated state, dew-point and dry-bulb temperatures are equal.
  • The first saturated state during cooling marks the onset of condensation.

Worked example

Estimate dew point from bracketed saturation data

A moist-air sample has water-vapour partial pressure 2.00 kPa. Supplied saturation data are 1.70 kPa at 15°C and 2.34 kPa at 20°C. Estimate the dew-point temperature by linear interpolation.
  1. Apply the dew-point condition
    At the dew point, actual vapour pressure equals saturation pressure. Here 2.00 kPa lies between the supplied values at 15°C and 20°C, so interpolation within that interval is appropriate.
    pv=psat(Tdp)p_v=p_{sat}(T_{dp})
  2. Interpolate between the data
    The target pressure is 0.30 kPa above the lower value, out of a 0.64 kPa pressure interval. Apply that fraction to the 5°C temperature interval.
    Tdp=15∘C+2.00−1.702.34−1.70(20−15)∘C=17.34∘CT_{dp}=15^\circ\mathrm{C}+\frac{2.00-1.70}{2.34-1.70}(20-15)^\circ\mathrm{C}=17.34^\circ\mathrm{C}
Answer: The estimated dew point is 17.34°C, or about 17.3°C.
Check: The estimate lies between 15°C and 20°C, consistent with the target pressure lying between the corresponding supplied saturation pressures.

Worked example

Find dew point from relative humidity

Moist air has dry-bulb temperature 30°C and relative humidity 40%. The supplied saturation pressure at 30°C is 4.24 kPa. Additional supplied saturation data are 1.60 kPa at 14°C and 1.70 kPa at 15°C. Find actual vapour pressure and estimate dew point by interpolation.
  1. Calculate actual vapour pressure
    Relative humidity is actual vapour pressure divided by saturation pressure at the dry-bulb temperature. Multiply the stated relative humidity by the supplied saturation pressure at 30°C.
    pv=ϕpsat(Tdb)=0.40(4.24 kPa)=1.696 kPap_v=\phi p_{sat}(T_{db})=0.40(4.24\,\mathrm{kPa})=1.696\,\mathrm{kPa}
  2. Interpolate for the dew point
    The target pressure, 1.696 kPa, is bracketed by the supplied values at 14°C and 15°C. Interpolate within this 1°C interval; do not extrapolate from data that fail to bracket the target.
    Tdp=14∘C+1.696−1.601.70−1.60(15−14)∘C=14.96∘CT_{dp}=14^\circ\mathrm{C}+\frac{1.696-1.60}{1.70-1.60}(15-14)^\circ\mathrm{C}=14.96^\circ\mathrm{C}
Answer: The actual vapour pressure is 1.696 kPa, and the interpolated dew point is approximately 15.0°C.
Check: The dew point is below the 30°C dry-bulb temperature, as expected for a state with relative humidity below 100%.

Worked example

Locate first condensation during cooling

A moist-air sample at total pressure 100 kPa has water-vapour partial pressure 1.70 kPa and initial dry-bulb temperature 20°C. It is cooled at fixed total pressure without adding or removing water vapour. Supplied saturation pressures are 1.70 kPa at 15°C and 2.34 kPa at 20°C. Determine the first-condensation temperature and whether the initial sample is saturated.
Cooling to the dew point
Cooling to the dew pointDry-bulb temperature, THumidity ratio, ωcool at fixed vapour content120°C, unsaturated215°C, saturatedSchematic · not to scale

Schematic, not to scale. Closed moist-air sample at fixed total pressure and fixed vapour content before condensation; state 1 cools to first saturated state 2. No saturation curve is shown.

  1. Check the initial state
    At 20°C, the supplied saturation pressure is 2.34 kPa, which exceeds the actual vapour pressure. The initial sample is therefore unsaturated.
    1.70 kPa<2.34 kPa1.70\,\mathrm{kPa}<2.34\,\mathrm{kPa}
  2. Find the first saturated state
    During the stated cooling, actual vapour pressure stays at 1.70 kPa until condensation begins. The supplied data show saturation pressure reaches 1.70 kPa at 15°C, so this is the dew point and first-condensation temperature.
    pv=psat(15∘C)=1.70 kPap_v=p_{sat}(15^\circ\mathrm{C})=1.70\,\mathrm{kPa}
Answer: The initial sample is unsaturated. First condensation begins at 15°C.
Check: The dew point is 5°C below the initial dry-bulb temperature, consistent with cooling an unsaturated sample to saturation.

Common mistakes and how to avoid them

Using total pressure in place of water-vapour partial pressure.
Correction: Compare the water-vapour partial pressure with saturation pressure.
Using saturation pressure at the dew point to calculate relative humidity at the initial state.
Correction: Use saturation pressure at the initial dry-bulb temperature for relative humidity.
Assuming dew point always equals dry-bulb temperature.
Correction: They are equal at saturation; for an unsaturated state, dew point is lower.
Interpolating when the target pressure falls outside the supplied saturation data.
Correction: Interpolation requires bracketing values. Obtain additional data or explicitly identify any estimate outside the range as extrapolation.

Lesson summary

  • Dew point is the temperature at which actual water-vapour partial pressure equals saturation pressure.
  • Relative humidity compares actual vapour pressure with saturation pressure at the current dry-bulb temperature.
  • At fixed vapour content during cooling, the first saturated state marks the onset of condensation.
  • Use supplied saturation data and interpolate only when those data bracket the target.

Check your understanding

Question 1

For an unsaturated moist-air state, how does dew-point temperature compare with dry-bulb temperature?
  1. Dew point is below dry-bulb temperature.
  2. Dew point is above dry-bulb temperature.
  3. They must be equal.
  4. The temperatures cannot be compared using saturation.
Show answer and explanation
Dew point is below dry-bulb temperature.
An unsaturated state must be cooled to reach saturation, so its dew point is below its current dry-bulb temperature.

Question 2

A sample has actual vapour pressure 1.50 kPa. Supplied saturation data give 1.50 kPa at 13°C. At what temperature does it reach saturation while cooling at fixed vapour content?
  1. At 13°C.
  2. At the total pressure expressed in °C.
  3. Only when its relative humidity is zero.
  4. The dew point cannot be related to saturation pressure.
Show answer and explanation
At 13°C.
The dew point is where saturation pressure equals actual vapour pressure. The supplied datum identifies that temperature as 13°C.

Key terms

Dew-point temperature
The temperature at which water vapour in a moist-air state becomes just saturated when cooled at fixed vapour content.
Saturation pressure
The pressure of water vapour in equilibrium with liquid water at a specified temperature.
Water-vapour partial pressure
The portion of a gas mixture’s total pressure contributed by water vapour.
Relative humidity
The ratio of actual water-vapour partial pressure to saturation pressure at the same temperature.

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