14.2 Psychrometric charts and processes

Air-conditioning processes analysis

Psychrometric charts let you read the thermodynamic properties of moist air at a glance. Instead of solving equations every time, you plot a state point on the chart and immediately find temperature, humidity, enthalpy, and more. They're the primary tool for analyzing how air changes as it moves through HVAC equipment.

Interpreting psychrometric charts

A psychrometric chart packs several properties onto one diagram:

• Dry-bulb temperature ($T_{db}$) runs along the x-axis.
• Humidity ratio ($\omega$, kg water per kg dry air) runs along the y-axis.
• Relative humidity ($\phi$) appears as a family of curved lines, with the saturation curve ($\phi = 100\%$) forming the upper-left boundary of the chart.
• Wet-bulb temperature ($T_{wb}$) lines slope downward to the right from the saturation curve.
• Enthalpy ($h$) lines run roughly parallel to the wet-bulb lines (and are sometimes printed on a separate oblique scale to avoid clutter).
• Specific volume ($v$) lines slope steeply from lower-left to upper-right.

To fix a state point you need any two independent properties. The most common pair is $T_{db}$ and $\phi$, but $T_{db}$ and $\omega$, or $T_{db}$ and $T_{wb}$, work just as well. Once the point is located, every other property can be read directly.

Analyzing air-conditioning processes on the chart

Every air-conditioning process is a path between two state points. The direction of that path tells you what's happening to the air:

For adiabatic mixing of two air streams, the mixed state falls on a straight line connecting the two inlet states, divided by the mass-flow-rate ratio.

Psychrometric processes

Sensible heating and cooling

Sensible heating raises the dry-bulb temperature without adding or removing moisture. On the chart this is a horizontal line moving to the right at constant $\omega$.

• Example: Air enters a heating coil at 20 °C and leaves at 30 °C. The humidity ratio stays the same; only $T_{db}$, $h$, and $v$ change.

Sensible cooling is the mirror image: a horizontal line moving to the left. This only works as long as the air is not cooled below its dew-point temperature ($T_{dp}$). If it is, moisture condenses and you no longer have a purely sensible process.

Humidification and dehumidification

Pure humidification at constant $T_{db}$ is a vertical line moving upward on the chart. You're adding water vapor without changing the air's dry-bulb temperature.

• Example: Steam injection into air at 25 °C raises $\omega$ from 0.010 to 0.015 kg water/kg dry air while $T_{db}$ stays at 25 °C.

Pure dehumidification at constant $T_{db}$ is a vertical line moving downward. In practice, true isothermal dehumidification requires a desiccant or similar device, because cooling-based dehumidification always lowers $T_{db}$ as well.

Combined processes

Cooling and dehumidification is the most common air-conditioning process. Air passes over a cooling coil whose surface temperature is below the air's dew point. Both $T_{db}$ and $\omega$ decrease, so the process line slopes downward and to the left.

• Example: Air at 30 °C and 50 % RH is cooled to 15 °C and roughly 90 % RH. On the chart, the actual path curves toward the apparatus dew point (the effective coil surface temperature) along the saturation curve.

Heating and humidification moves the state point upward and to the right. This happens, for instance, when air passes through a warm-water spray chamber.

• Example: Air at 20 °C and 30 % RH is conditioned to 30 °C and 50 % RH.

Evaporative (adiabatic) cooling follows a line of approximately constant wet-bulb temperature. The air gives up sensible heat to evaporate water, so $T_{db}$ drops while $\omega$ rises, with enthalpy nearly constant.

Adiabatic mixing of two streams is represented by a straight line connecting the two inlet states. The mixed-air state divides that line in inverse proportion to the mass flow rates:

$T_{db,3} = \frac{\dot{m}_1 \, T_{db,1} + \dot{m}_2 \, T_{db,2}}{\dot{m}_1 + \dot{m}_2}$

$\omega_3 = \frac{\dot{m}_1 \, \omega_1 + \dot{m}_2 \, \omega_2}{\dot{m}_1 + \dot{m}_2}$

• Example: Mixing 1 kg/s of air at 20 °C with 2 kg/s at 30 °C gives $T_{db,3} \approx 26.7\,°C$. The mixed point sits closer to the stream with the larger mass flow rate.

Air properties calculation

Locating initial and final states

1. Identify the two known properties for the initial state (e.g., $T_{db}$ and $\phi$).
2. Find the intersection of those two property lines on the chart. That's your state point.
3. Read off all other properties at that point: $\omega$, $h$, $T_{wb}$, $T_{dp}$, $v$.

Following process lines

Once you have the initial state, the type of process tells you which line to follow:

• Sensible heating/cooling: Move horizontally to the target $T_{db}$. Read the new $h$ (and confirm $\omega$ hasn't changed).
• Humidification/dehumidification at constant $T_{db}$: Move vertically to the target $\omega$. Read the new $h$ and $\phi$.
• Cooling and dehumidification: The idealized path follows a straight line toward the apparatus dew point (ADP) on the saturation curve. In practice, the air leaves somewhere between the ADP and the inlet state, depending on the coil's bypass factor.
• Evaporative cooling: Follow the constant $T_{wb}$ line toward the saturation curve.

At the final state, read all properties the same way you did for the initial state.

Adiabatic mixing calculations

1. Locate both inlet states on the chart.
2. Draw a straight line between them.
3. Divide the line segment using the inverse mass-flow-rate ratio. If $\dot{m}_1 = 1$ kg/s and $\dot{m}_2 = 2$ kg/s, the mixed point sits $\frac{2}{3}$ of the way from state 1 toward state 2.
4. Read $T_{db,3}$, $\omega_3$, and $h_3$ at the mixed point.

Because the process is adiabatic and no moisture is added or removed externally, the energy balance closes automatically: $\dot{m}_1 h_1 + \dot{m}_2 h_2 = (\dot{m}_1 + \dot{m}_2) h_3$.

Energy requirements for air-conditioning

Sensible heating and cooling

For a steady-flow process with no moisture change, the rate of heat transfer is:

$\dot{Q} = \dot{m}_a \, c_p \, (T_2 - T_1)$

where $\dot{m}_a$ is the mass flow rate of dry air and $c_p \approx 1.005$ kJ/(kg·°C) for dry air at typical HVAC conditions.

• Example: Heating 1 kg/s of air from 20 °C to 30 °C:

$\dot{Q} = 1 \times 1.005 \times (30 - 20) = 10.05 \text{ kW}$

A more general (and more accurate) approach uses enthalpy directly from the chart: $\dot{Q} = \dot{m}_a (h_2 - h_1)$. This accounts for the small contribution of water vapor to the specific heat.

Humidification and dehumidification

When moisture content changes, you must use the enthalpy-based equation because latent energy is involved:

$\dot{Q} = \dot{m}_a \, (h_2 - h_1)$

Read $h_1$ and $h_2$ from the psychrometric chart at the initial and final states.

• Example (humidification): 1 kg/s of air at 25 °C and 50 % RH ($h_1 \approx 50.4$ kJ/kg) is humidified to 25 °C and 70 % RH ($h_2 \approx 56.3$ kJ/kg):

$\dot{Q} = 1 \times (56.3 - 50.4) = 5.9 \text{ kW}$

For dehumidification, $h_2 < h_1$, so heat is removed from the air ($\dot{Q}$ is negative, meaning cooling is required).

Combined processes and adiabatic mixing

Cooling and dehumidification involves both sensible and latent loads. The total cooling rate is still:

$\dot{Q}_{total} = \dot{m}_a \, (h_1 - h_2)$

You can split this into sensible and latent components if needed:

• Sensible load: $\dot{Q}_s = \dot{m}_a \, c_p \, (T_1 - T_2)$
• Latent load: $\dot{Q}_L = \dot{m}_a \, h_{fg} \, (\omega_1 - \omega_2)$

where $h_{fg} \approx 2501$ kJ/kg is the enthalpy of vaporization of water at typical conditions.

The sensible heat ratio (SHR) is $\dot{Q}_s / \dot{Q}_{total}$. Many psychrometric charts include a protractor (SHR scale) on the side that lets you draw the correct process-line slope for a given SHR.

Heating and humidification works the same way in reverse: $\dot{Q} = \dot{m}_a (h_2 - h_1)$, where $h_2 > h_1$.

Adiabatic mixing requires no external energy input. The enthalpy of the mixed stream satisfies:

$h_3 = \frac{\dot{m}_1 h_1 + \dot{m}_2 h_2}{\dot{m}_1 + \dot{m}_2}$

If the mixed state falls above the saturation curve on the chart, some moisture will condense out (fog formation), and the actual exit state will lie on the saturation curve with a lower enthalpy than the simple mixing calculation predicts.