Introduction of Humidification Process
2026-03-16
After this lecture, you will be able to:
Edmonton’s winter is harsh! How do we make our home more humid?
A random reddit post
In order to increase the humidity in a room during Edmonton’s harsh winter, you decide to buy a humidifier. Your knowledge from the CHE 318 course may actually be helpful! A few questions to think about:
Humidity \(H\) (an absolute value)
Definition: kg of water vapor per kg dry air at certain temperature
For water-air systems where molecular weights are \(M_A\) and \(M_B\), respectively, this also links to partial pressure \(p_A\)
\[ H = \frac{M_A p_A}{M_B(p_T-p_A)} \]
\[ H = 0.622\frac{p_A}{p_T-p_A} \]
At each temperature \(T\), at equilibrium water-air interface, the pressure is saturation vapour pressure \(p_{\mathrm{vap}}\). The (absolute) humidity at this partial pressure is the saturation humidity \(H_s\):
\[ H_s = 0.622\frac{p_{\mathrm{vap}}}{p_T-p_{\mathrm{vap}}} \]
These two are often confused with each other so treat with caution.
\[ H_p = \left(\frac{H}{H_s}\right) \times 100\% \]
\[ H_R = \left(\frac{p_A}{p_{\mathrm{vap}}}\right) \times 100\% \]
\(H_p\) and \(H_R\) are generally not the same. The approximation \(H_p \approx H_R\) only applies at low humidity.
Using the ideal gas law, it can be shown that
\[\begin{align} H_p &= \frac{p_A}{p_{\text{vap}}} \frac{p_T - p_{\text{vap}}}{p_T - p_A} \\ &= H_R \frac{p_T - p_{\text{vap}}}{p_T - p_A} \end{align}\]Since \(p_A \leq p_{\text{vap}}\), we can show that \(H_p \leq H_R\) always holds.
Dew point: temperature at which the vapor in air just becomes saturated / condensed.
At dew point, \(p_A = p_{\mathrm{vap}}(T_{\mathrm{dew}})\)
Humid heat \(c_s\): heat required to raise \(1\) kg dry air plus its associated vapor by \(1\) K. Think of it as the specific heat / heat capacity for air-water mixture.
For air-water, when dry air and water vapour have heat capacities \(c_B\) and \(c_A\), respectively, the humid heat is
\[ c_s = c_B + Hc_A \]
Useful relation to remember:
\[ c_s = 1.005 + 1.88H\quad \text{[kJ/(kg dry air K)]} \]
The relation for \(c_s\) tells us that heating up a gas stream for humid air requires more energy than drier one. For a stream containing dry air and water vapor, such heat is:
\[ Q = w_B c_s \Delta T \]
where \(w_B\) is mass flow rate of dry air in kg / s. Note such energy is usually called sensible heating of the gas-vapor mixture, because it is directly proportional to \(\Delta T\).
Humid volume \(v_H\): volume occupied by \(1\) kg dry air plus its associated water vapor at certain pressure. (This is less useful than other quantities for cooling process).
At 1 atm:
\[ v_H = \frac{22.41}{273}\left(\frac{1}{28.97}+\frac{H}{18.02}\right)T \]
At general pressure:
\[ v_H = \left(\frac{1}{28.97}+\frac{H}{18.02}\right)\frac{22.41}{273}T\left(\frac{1.013\times 10^5}{p_T}\right) \]
where \(T\) is in Kelvin, \(p_T\) is in Pa, and \(v_H\) is in m\(^3\).
Humid air enthalpy \(H_y\) includes:
\[ H_y = c_s(T-T_0) + H\lambda_0 \]
For air-water with \(T_0=0^\circ\mathrm{C}\), and \(T\) uses Celcius:
\[ H_y = (1.005 + 1.88H)T + 2501.4H\quad \text{[kJ/kg dry air]} \]
Psychrometric: greek for psychro (cold) + metric (measure), relating to the measurement of temperature drop when water evaporates.
Find where \(H\), \(H_p\), dew point \(T_{\text{dew}}\) are?
A living room of 60 m\(^2\) at 22 \(^\circ\)C has a relative humidity of 20% due to continuous heating. You and your roommate wish to purchase a humidifier that can humidify the living room up to 45% relative humidity. Assume the floor-ceiling distance is 3 m, calculate the weight of water needed to humidify the whole room. Can you use the humidity chart to estimate?
from the vapour data it gets \(p_{\text{vap}} = 2.64\) kPa, pretty close!
Does it make sense? The requirement for water tank will be a lower bound.
My weather app shows that outside is \(-12\ ^\circ\)C and R.H. is 74%, where the dew point is \(-16\ ^\circ\)C. What does all that mean?
Tip
A useful note is that the lower dew point is compared with current temperature, the drier air appears to be
In the previous scenario \(-12\ ^\circ\)C and R.H. is 74%, when I open the door, my humidity in home immediately drops! Why?
Tip
\(H_R\) (or R.H.) only measures up at current temperature. When moving water content at different \(T\), \(H_R\) will change.
Why does “humid hot” environment feels much hotter than “dry hot” environment, even if the apparent \(T\) is the same?
Tip
The sensible specific heat for air-water mixture does not change significantly, even if water vapour’s heat capacity is higher. (absolute \(H\) is always 0.1 when \(T<50\ ^\circ\)C)
In industrial applications, evaporation of water into vapour is used to cool hot liquid. We will discuss in detail how to solve this using the concept of enthalpy in next lecture.👉