Humidification Process II
2026-03-18
After this lecture, you will be able to:
Charts distributed in class.
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)
Before electronic sensors, humidity was measured using a two-thermometer setup

The wet bulb is a steady state evaporative cooling experiment, which can be generalized for interfacial evaporation problem:
\[ T_w < T_d \]
From Lecture 27, the specific enthalpy of moist air can be written as
\[\begin{align} H_{y} = (1.005 + 1.88H)(T - T_0) + 2501.4H \end{align}\]Consider the control volume around the wet bulb. Heat transferred from the air is used to evaporate water, linked by the heat-mass balance (need a bit prerequisite in heat transfer):
\[\begin{align} q = M_A \lambda_w A N_A \end{align}\]The evaporation rate can be written as
\[\begin{align} N_A = k_y (y_w - y) \end{align}\]We can further change \(y\) to the humidity \(H\) for dilute system. Since humidity is define as weight ratio:
\[\begin{align} H = \frac{\text{kg }H_2O}{\text{kg dry air}} \end{align}\]The mole fraction of vapor is
\[\begin{align} y = \frac{H/M_A}{1/M_B + H/M_A} \end{align}\]Since humidity is typically small
\[\begin{align} y \approx \frac{M_B}{M_A}H \end{align}\]The heat flux \(q\) in L.H.S. from the air to the wet surface is
\[\begin{align} q = h (T - T_w) A \end{align}\]where \(h\) is heat transfer coefficient in the Fourier’s law \(q = -h \Delta T\)
Combining heat transfer (\(q\)) and mass transfer (\(N_A\)) relations gives
\[\begin{align} \frac{H - H_w}{T - T_w} = - \frac{h}{M_B k_y \lambda_w} \end{align}\]Note \(\frac{H - H_w}{T - T_w}\) means the slope of a line on the psychrometric chart. The slope is almost identical to adiabatic line!
The \((T_d, H_{\text{in}})\) and \((T_w, H_{\text{out}})\) points are along the adiabatic line (no external heat exchange). For water-air system, the adiabatic line and cooling line are very close and often not distinguished.
What does the adiabatic line tell us? It is basically a process that each point has the same humid enthalpy, and no change of heat to external system:
\[ H = c_s (T - T_0) + H \lambda_0 = \text{[Const]} \]
For water-air, one handy property is that
\[ \frac{h}{M_B k_y} \approx 1.005 \approx c_s \qquad \text{[kJ / kg air]} \]
such relation allows us to use the humidity chart’s adiabatic saturation curve.
Warning
Such simplification may not be applicable for other liquid, such as benzene!
The wet-bulb experiment connects heat transfer and mass transfer, and the evaporation rate depends on the vapor pressure driving force.
Next topics we will study: