14.7 Psychrometrics: Relative Humidity, Wet-Bulb Temperature, and the Psychrometric Chart

14.7 Psychrometrics: Relative Humidity, Wet-Bulb Temperature, and the Psychrometric Chart

NCEES lists "Psychrometrics (e.g., relative humidity, wet bulb)" as the final sub-topic of the Thermodynamics and Heat Transfer area. Psychrometrics treats moist air as a mixture of dry air and water vapor, and the whole subject rests on the fact that the water vapor is present at such low partial pressure that it behaves as an ideal gas.

The Three Temperatures

The universal ordering:

$T_{db} \ge T_{wb} \ge T_{dp}$

At saturation all three are equal, and the gap between them measures how dry the air is. The wet-bulb depression $(T_{db} - T_{wb})$ is therefore a direct index of drying potential — which is why a sling psychrometer determines relative humidity from two temperature readings, and why evaporative cooling works well in a desert and poorly in a swamp.

Relative Humidity

$\phi = \frac{P_v}{P_{\text{sat}}(T_{db})}\times100\%$

where $P_v$ is the actual partial pressure of water vapor and $P_{\text{sat}}$ is the saturation pressure at the dry-bulb temperature.

The critical implication: because $P_{\text{sat}}$ rises steeply with temperature, heating air without adding moisture lowers its relative humidity even though the absolute moisture content is unchanged. Air at 10 °C and 80% RH heated to 25 °C falls to about 30% RH — the reason winter indoor air feels dry and requires humidification. Conversely, cooling raises RH until saturation is reached, at which point condensation begins.

Humidity Ratio (Specific Humidity)

$\omega = \frac{m_v}{m_a} = 0.622\,\frac{P_v}{P - P_v} \qquad [\text{kg water/kg dry air}]$

The 0.622 is the molar-mass ratio $M_v/M_a = 18.02/28.97$.

Why everything is per kilogram of dry air: during humidification, dehumidification, and mixing, the mass of water changes but the mass of dry air is conserved. Using dry air as the basis makes it the invariant reference, so $\omega$ changes only when moisture is actually added or removed.

Enthalpy of Moist Air

$h = 1.006\,T + \omega(2{,}501 + 1.86\,T) \qquad [\text{kJ/kg dry air, } T \text{ in }°\text{C}]$

Latent heat dominates. At $\omega = 0.010$ the latent term is $25.0$ kJ/kg against a sensible term of about $20$ kJ/kg at 20 °C. This is why removing moisture is thermodynamically expensive and why air-conditioning loads are reported as separate sensible and latent components.

Processes on the Psychrometric Chart

The chart plots $\omega$ (vertical) against $T_{db}$ (horizontal), with the saturation curve ($\phi = 100%$) as the upper-left boundary. Lines of constant RH, wet-bulb, and enthalpy overlay it.

Worked Example 1: Full State Determination

Air at 30 °C dry-bulb and 50% relative humidity, at 101.3 kPa. Find $P_v$, $\omega$, $h$, and the dew point.

Vapor pressure, with $P_{\text{sat}}(30°\text{C}) = 4.246$ kPa:

$P_v = \phi P_{\text{sat}} = 0.50(4.246) = 2.123\ \text{kPa}$

Humidity ratio:

$\omega = 0.622\frac{2.123}{101.3 - 2.123} = 0.622\frac{2.123}{99.18} = 0.622(0.02141) = 0.01332\ \text{kg/kg dry air}$

Enthalpy:

$h = 1.006(30) + 0.01332[2{,}501 + 1.86(30)] = 30.18 + 0.01332(2{,}556.8) = 30.18 + 34.06 = \boxed{64.2\ \text{kJ/kg dry air}}$

Note that the moisture carries more energy (34.1 kJ/kg) than the dry air itself (30.2 kJ/kg), at only 1.3% moisture by mass.

Dew point is the saturation temperature corresponding to $P_v = 2.123$ kPa. From the table, 2.123 kPa lies between 15 °C (1.706) and 20 °C (2.339). Interpolating:

$T_{dp} = 15 + 5\left(\frac{2.123-1.706}{2.339-1.706}\right) = 15 + 5(0.659) = \boxed{18.3\ °\text{C}}$

Practical consequence: any surface in this air below 18.3 °C will sweat. A chilled-water pipe at 7 °C or a single-glazed window on a cold day condenses, which is why insulation and vapor barriers are specified against the design dew point rather than the air temperature.

Worked Example 2: Cooling Coil with Sensible and Latent Loads

2.0 kg/s of dry air enters a coil at 30 °C, $\omega_1 = 0.0133$ (the state above), and leaves at 14 °C, 95% RH. Find the total, sensible, and latent loads.

Exit state, with $P_{\text{sat}}(14°\text{C}) \approx 1.598$ kPa:

$P_v = 0.95(1.598) = 1.518\ \text{kPa}$ $\omega_2 = 0.622\frac{1.518}{101.3-1.518} = 0.622\frac{1.518}{99.78} = 0.00946\ \text{kg/kg}$ $h_2 = 1.006(14) + 0.00946[2{,}501+1.86(14)] = 14.08 + 0.00946(2{,}527.0) = 14.08 + 23.90 = 37.98\ \text{kJ/kg}$

Total load: $\dot{Q}_{\text{total}} = \dot{m}_a(h_1-h_2) = 2.0(64.2 - 37.98) = 2.0(26.2) = \boxed{52.5\ \text{kW}}$

Sensible load (temperature change at the exit humidity, using moist-air specific heat ≈ 1.02 kJ/kg·K): $\dot{Q}_s = \dot{m}_a c_p(T_1-T_2) = 2.0(1.02)(30-14) = 32.6\ \text{kW}$

Latent load (moisture removal): $\dot{Q}_L = \dot{m}_a(\omega_1-\omega_2)(2{,}501) = 2.0(0.0133-0.00946)(2{,}501) = 2.0(0.00384)(2{,}501) = 19.2\ \text{kW}$

Check: $32.6 + 19.2 = 51.8$ kW, within rounding of 52.5 kW ✓

Sensible heat ratio: $SHR = 32.6/52.5 = 0.62$ — so 38% of the cooling capacity goes to removing water, not to lowering temperature. Condensate rate: $2.0(0.00384) = 0.0077$ kg/s $= 27.7$ kg/h.

The design lesson: a coil sized on temperature change alone would be undersized by about 60%. And because dehumidification requires the coil surface to be below the dew point, a chilled-water system running at too high a supply temperature can meet the sensible load while removing no moisture at all — the classic cause of a space that is cool but clammy.

Worked Example 3: Evaporative Cooling

Air at 38 °C and 20% RH passes through an evaporative cooler with 85% effectiveness. The wet-bulb temperature is 22 °C. Find the exit dry-bulb temperature.

$T_{\text{out}} = T_{db} - \varepsilon(T_{db}-T_{wb}) = 38 - 0.85(38-22) = 38 - 0.85(16) = 38 - 13.6 = \boxed{24.4\ °\text{C}}$

A 13.6 °C drop with no refrigeration — the process is adiabatic, converting sensible heat into latent heat as water evaporates, so it moves up along a constant wet-bulb line.

The hard limit: the exit temperature can never fall below the wet-bulb temperature, which is what makes evaporative cooling effective in dry climates (large wet-bulb depression) and nearly useless in humid ones. At 30 °C and 80% RH the wet-bulb is about 27 °C, so even a perfect cooler could drop the air only 3 °C — and it would leave the space saturated.