7.1 Calculate humidity ratio and relative humidity | University of Alberta MEC E 340: Applied Thermodynamics

University of Alberta MEC E 340: Applied Thermodynamics

Psychrometry and Moist Air

MEC E 340 Applied Thermodynamics — Study topic 7.1

Consider a small control volume containing moist air at a numbered state, state 1. Its working mixture is dry air plus water vapour; for these calculations, assume the mixture behaves as an ideal-gas mixture and the water vapour is below saturation. The total pressure is the sum of the dry-air and water-vapour partial pressures. This topic uses that mixture model and water saturation-pressure data to calculate two different measures of moisture. No device energy balance is needed unless a separate problem asks about heating or cooling: the focus here is the state of the moist air.

What you will learn

Distinguish humidity ratio from relative humidity.
Calculate humidity ratio from total pressure and water-vapour partial pressure.
Calculate relative humidity from water-vapour partial pressure and supplied saturation-pressure data.
Find one humidity measure from the other when temperature, pressure, and suitable property data are known.

1. Define the two measures and their basis

Humidity ratio, also called the humidity ratio of moist air, compares the mass of water vapour with the mass of dry air in the same sample. It is not the mass of vapour divided by the total mass of moist air. Its units are kilograms of water vapour per kilogram of dry air, commonly written as kg water/kg dry air.

Relative humidity compares the actual water-vapour partial pressure with the saturation pressure of water at the same temperature. It is a fraction, often reported as a percentage. The saturation pressure is a property of water at the specified temperature; use a supplied value or an identified property table rather than guessing it.

The two measures answer different questions. Humidity ratio gives a mass ratio and, at fixed total pressure, is linked to vapour partial pressure. Relative humidity shows how close the vapour pressure is to the saturation pressure at the current temperature. Therefore, relative humidity can change when temperature changes even if no water vapour is added or removed.

\omega=m_v/m_{da}

Humidity ratio is on a dry-air mass basis.
Relative humidity uses saturation pressure at the same temperature as the air.
Always state pressure units and identify the source of saturation-pressure data.

2. Governing relations and calculation sequence

For the ideal-gas-mixture model, each constituent has a partial pressure. Let

P

be the total moist-air pressure,

p_v

the water-vapour partial pressure, and

p_{da}

the dry-air partial pressure. Dalton’s relation gives

P=p_{da}+p_v

. The mass ratio follows by applying the ideal-gas relation to each constituent at the same temperature and volume. The molar-mass ratio of water vapour to dry air is represented here by 0.622.

To calculate humidity ratio, first obtain

p_v

and

P

in the same pressure units. The dry-air partial pressure is

P-p_v

. Then use the pressure ratio in the humidity-ratio equation. The result is dimensionless as a ratio, but conventionally reported as kg water/kg dry air.

To calculate relative humidity, divide the actual vapour partial pressure by the saturation pressure at the air temperature. Multiply the fraction by 100 when expressing it as a percentage. If humidity ratio is given instead of

p_v

, rearrange the humidity-ratio equation to find

p_v

before calculating relative humidity.

These relations assume an ideal-gas mixture and use the water saturation pressure corresponding to the stated temperature. They do not require a heat or work balance. In a larger applied problem, first identify the moist-air state and any process assumptions; calculate these state properties using the data provided.

\omega=0.622\frac{p_v}{P-p_v},\qquad \phi=\frac{p_v}{p_{sat}(T)}

Use absolute total pressure and partial pressure in matching units.
Obtain saturation pressure from stated data or a named property source.
Under the stated model, relative humidity at saturation is 1, or 100%.

3. Check data, units, and physical meaning

A reliable solution begins by listing the known state information: total pressure, temperature where needed, and either vapour partial pressure, humidity ratio, or relative humidity. Mark which values are supplied and which must be calculated. When converting a percentage to a fraction, divide by 100 before substituting.

For a physically ordinary unsaturated moist-air state,

p_v

is less than or equal to

p_{sat}(T)

, so relative humidity is no greater than 100%. If a calculation gives more than 100%, check for a percentage/fraction error, mismatched temperatures, inconsistent pressure units, or an incompatible set of supplied data. Do not silently alter the data.

A calculated humidity ratio must be nonnegative and should be interpreted on the dry-air basis. Avoid calling it a percentage. Conversely, relative humidity is usually reported as a percentage, not as kg/kg. Keep these different units and meanings visible in the final answer.

p_{da}=P-p_v

Check that $p_v\leq P$ and, for an unsaturated state, $p_v\leq p_{sat}(T)$ .
A ratio reported as a percentage must be multiplied by 100.
State the ideal-mixture assumption and the source of any saturation-pressure value.

Worked example

Humidity ratio from partial pressure

Moist air is at state 1 with total pressure

P=100.0\ \mathrm{kPa}

and water-vapour partial pressure

p_v=2.00\ \mathrm{kPa}

. Assuming an ideal-gas mixture, calculate its humidity ratio.

Identify the state data
The control volume contains moist air at state 1. Total pressure and water-vapour partial pressure are given in the same units, so their difference is the dry-air partial pressure.
$p_{da}=100.0-2.00=98.0\ \mathrm{kPa}$
Apply the mixture relation
The humidity ratio is water-vapour mass divided by dry-air mass. For the ideal-gas-mixture model, use the given molar-mass ratio 0.622 with the partial-pressure ratio.
$\omega=0.622\frac{2.00}{98.0}=0.0127\ \mathrm{kg\ water/kg\ dry\ air}$

Answer: The humidity ratio is

0.0127\ \mathrm{kg\ water/kg\ dry\ air}

.

Check: The result is positive, and the vapour partial pressure is below the total pressure. The value is a mass ratio, not a percentage.

Worked example

Relative humidity from humidity ratio

At state 1, moist air has total pressure

P=98.0\ \mathrm{kPa}

and humidity ratio

\omega=0.0100\ \mathrm{kg\ water/kg\ dry\ air}

. Its temperature is

25^\circ\mathrm{C}

. For this example, use the supplied saturation pressure

p_{sat}=3.17\ \mathrm{kPa}

at that temperature. Find relative humidity.

Find the vapour partial pressure
Rearrange the humidity-ratio relation to isolate $p_v$ . The total pressure and humidity ratio are supplied; the saturation-pressure datum is also explicitly supplied for the stated temperature.
$p_v=\frac{P\omega}{0.622+\omega}=\frac{(98.0)(0.0100)}{0.632}=1.55\ \mathrm{kPa}$
Calculate relative humidity
Compare the actual vapour partial pressure with saturation pressure at the same temperature. Convert the resulting fraction to a percentage.
$\phi=\frac{1.55}{3.17}=0.489\approx48.9\%$

Answer: The relative humidity is approximately 48.9%.

Check: The vapour pressure is less than the supplied saturation pressure, so the result is below 100% and is consistent with an unsaturated state.

Worked example

Humidity ratio from relative humidity

At state 1, moist air is at

P=100.0\ \mathrm{kPa}

and

20^\circ\mathrm{C}

. The stated saturation-pressure datum is

p_{sat}=2.339\ \mathrm{kPa}

at this temperature, and measured relative humidity is 50.0%. Assuming an ideal-gas mixture, calculate humidity ratio.

Convert relative humidity to vapour pressure
Relative humidity is supplied as a percentage, so use 0.500 as the fraction. Multiply this by the saturation pressure at state 1 to obtain the actual water-vapour partial pressure.
$p_v=0.500(2.339)=1.1695\ \mathrm{kPa}$
Calculate the mass ratio
Use total pressure minus vapour partial pressure as the dry-air partial pressure. Substitution gives the humidity ratio on a dry-air mass basis.
$\omega=0.622\frac{1.1695}{100.0-1.1695}=0.00736\ \mathrm{kg\ water/kg\ dry\ air}$

Answer: The humidity ratio is approximately

0.00736\ \mathrm{kg\ water/kg\ dry\ air}

.

Check: The partial pressure is less than both total pressure and the supplied saturation pressure. The computed humidity ratio is positive and appropriately reported as a mass ratio.

Common mistakes and how to avoid them

Using total moist-air mass in the denominator of humidity ratio.

Correction: Humidity ratio is the mass of water vapour divided by the mass of dry air.

Using temperature in place of saturation pressure in the relative-humidity relation.

Correction: Obtain

p_{sat}

at the stated temperature from supplied data or an identified property source, then divide

p_v

by that pressure.

Putting relative humidity into an equation as 50 instead of 0.50.

Correction: Convert the percentage to a fraction before multiplying by saturation pressure.

Reporting humidity ratio as a percentage or relative humidity in kg/kg.

Correction: Report humidity ratio in kg water/kg dry air and relative humidity as a fraction or percentage.

Lesson summary

Humidity ratio is water-vapour mass per dry-air mass.
For the ideal-gas-mixture model, calculate humidity ratio from vapour and dry-air partial pressures.
Relative humidity is actual vapour partial pressure divided by saturation pressure at the same temperature.
When converting between the measures, keep pressure units consistent and use saturation-pressure data supplied for the state.

Check your understanding

Question 1

At a stated temperature,

p_v=1.20\ \mathrm{kPa}

and supplied

p_{sat}=2.40\ \mathrm{kPa}

. What is relative humidity?

50%
2%
100%
The humidity ratio is $0.50\ \mathrm{kg/kg}$

Show answer and explanation

50%

Relative humidity is

p_v/p_{sat}=1.20/2.40=0.50

, or 50%. It is not a humidity ratio.

Question 2

Which mass basis defines humidity ratio?

Water-vapour mass divided by dry-air mass
Dry-air mass divided by water-vapour mass
Water-vapour mass divided by total moist-air mass
Total moist-air mass divided by dry-air mass

Show answer and explanation

Water-vapour mass divided by dry-air mass

By definition, humidity ratio compares water-vapour mass with dry-air mass.

Question 3

For an unsaturated moist-air state, which comparison should normally hold?

$p_v<p_{sat}(T)$
$p_v>P$
$p_{sat}(T)<0$
$p_v=P+p_{sat}(T)$

Show answer and explanation

p_v<p_{sat}(T)

An unsaturated state has actual vapour partial pressure below the saturation pressure at the same temperature.

Key terms

Humidity ratio: Mass of water vapour divided by mass of dry air in a moist-air sample.
Relative humidity: Ratio of actual water-vapour partial pressure to saturation pressure at the same temperature.
Partial pressure: The pressure contribution assigned to one constituent of a gas mixture.
Saturation pressure: Pressure of water vapour at saturation for a specified temperature, obtained from property data.

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7.1 · Calculate humidity ratio and relative humidity

University of Alberta MEC E 340: Applied Thermodynamics

MEC E 340 Applied Thermodynamics — Study topic 7.1

What you will learn

1. Define the two measures and their basis

2. Governing relations and calculation sequence

3. Check data, units, and physical meaning

Humidity ratio from partial pressure

Relative humidity from humidity ratio

Humidity ratio from relative humidity

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through MEC E 340

About this lesson and its review