3.1 Psychrometric Properties: DBT, WBT, DPT, RH, Humidity Ratio & Enthalpy

3.1 Psychrometric Properties: DBT, WBT, DPT, RH, Humidity Ratio & Enthalpy

Psychrometrics is the branch of thermodynamics dedicated to the evaluation of the physical, thermodynamic, and hygrometric properties of moist air (a binary mixture of dry air and water vapor). Mastery of psychrometric properties is essential for the NCEES PE Mechanical: HVAC and Refrigeration examination, serving as the quantitative backbone for heating and cooling load analysis, air-handling unit (AHU) sizing, coil selection, and indoor environmental control.

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1. Thermodynamic Model of Atmospheric Moist Air

Under standard HVAC operating conditions (temperatures between $-40^\circ\text{F}$ and $140^\circ\text{F}$ and pressures near atmospheric), dry air and water vapor behave as ideal gases with negligible error ($<0.5%$). By Dalton's Law of Partial Pressures, the total barometric pressure $P$ exerted by moist air is the sum of the partial pressure of dry air ($p_a$) and the partial pressure of water vapor ($p_v$):

$P = p_a + p_v$

Where:

• $P = \text{total atmospheric barometric pressure } (\text{psia or in. Hg})$
• $p_a = \text{partial pressure of dry air } (\text{psia})$
• $p_v = \text{partial pressure of water vapor } (\text{psia})$

Applying the ideal gas law ($p V = m R T$) to both constituents individually:

$p_a v = R_{da} T \quad \text{and} \quad p_v v = R_v T$

Where:

• $R_{da} = \frac{\bar{R}}{M_{da}} = \frac{1545.35}{28.966} = 53.352 \text{ ft}\cdot\text{lbf}/(\text{lbm}\cdot^\circ\text{R}) = 0.06855 \text{ Btu}/(\text{lbm}\cdot^\circ\text{R})$
• $R_v = \frac{\bar{R}}{M_v} = \frac{1545.35}{18.015} = 85.781 \text{ ft}\cdot\text{lbf}/(\text{lbm}\cdot^\circ\text{R}) = 0.11024 \text{ Btu}/(\text{lbm}\cdot^\circ\text{R})$
• The molecular weight ratio is $\frac{M_v}{M_{da}} = \frac{18.01528}{28.966} \approx 0.62198$

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2. Core Psychrometric State Variables

An equilibrium thermodynamic state of moist air at a fixed barometric pressure is uniquely fixed by any two independent intensive psychrometric properties (e.g., $T_{db}$ and $T_{wb}$, or $T_{db}$ and $\text{RH}$, or $T_{db}$ and $W$).

Dry-Bulb Temperature ($T_{db}$ or $t$)

The temperature of moist air indicated by an ordinary thermometer unaffected by radiation or moisture condensation/evaporation, expressed in $^\circ\text{F}$ or $^\circ\text{C}$.

Wet-Bulb Temperature ($T_{wb}$ or $t'$)

The dynamic equilibrium temperature reached by a wetted thermometer sensor exposed to a high-velocity airstream ($>900\text{ ft/min}$) under steady-state conditions where convective sensible heat transfer from the air to the wick equals the latent heat required to evaporate water from the wick into the airstream. In the NCEES handbook, thermodynamic wet-bulb temperature is rigorously treated as the adiabatic saturation temperature.

Dew-Point Temperature ($T_{dp}$ or $t_{dp}$)

The saturation temperature corresponding to the partial pressure of water vapor present in the mixture: $T_{dp} = T_{sat}(p_v)$. When moist air is cooled at constant pressure and constant humidity ratio, condensation begins exactly when $T_{db}$ reaches $T_{dp}$.

Saturation Pressure ($p_{ws}$)

The maximum equilibrium vapor pressure of water at a given dry-bulb temperature. In ASHRAE / NCEES reference tables, $p_{ws}$ is tabulated as a function of temperature. For manual estimation between $32^\circ\text{F}$ and $140^\circ\text{F}$, Antoine-type equations or table lookups are used:

• At $T = 32^\circ\text{F}$: $p_{ws} = 0.08866 \text{ psia} = 0.1805 \text{ in. Hg}$
• At $T = 60^\circ\text{F}$: $p_{ws} = 0.2563 \text{ psia} = 0.5218 \text{ in. Hg}$
• At $T = 70^\circ\text{F}$: $p_{ws} = 0.3632 \text{ psia} = 0.7392 \text{ in. Hg}$
• At $T = 80^\circ\text{F}$: $p_{ws} = 0.5073 \text{ psia} = 1.0326 \text{ in. Hg}$

Relative Humidity ($\phi$ or $\text{RH}$)

The ratio of the actual water vapor partial pressure $p_v$ to the saturation vapor pressure $p_{ws}$ at the same dry-bulb temperature:

$\text{RH} = \phi = \frac{p_v}{p_{ws}(T_{db})} \times 100\%$

Humidity Ratio ($W$ or $\omega$)

The ratio of the mass of water vapor to the mass of dry air in the mixture. Also referred to as moisture content or specific humidity:

$W = \frac{m_v}{m_{da}} = \frac{M_v p_v}{M_{da} p_a} = 0.62198 \frac{p_v}{P - p_v} \quad \left[\frac{\text{lb}_{w}}{\text{lb}_{da}}\right]$

To express $W$ in grains of moisture per pound of dry air ($1\text{ lb} = 7,000\text{ grains}$):

$W_{\text{grains}} = 7000 \times W = 7000 \times 0.62198 \frac{p_v}{P - p_v} = 4353.9 \frac{p_v}{P - p_v} \quad \left[\frac{\text{grains}}{\text{lb}_{da}}\right]$

Degree of Saturation ($\mu$)

The ratio of the actual humidity ratio $W$ to the saturation humidity ratio $W_s$ at the same dry-bulb temperature and total pressure:

$\mu = \left. \frac{W}{W_s} \right|_{T, P} = \frac{0.62198 \frac{p_v}{P - p_v}}{0.62198 \frac{p_{ws}}{P - p_{ws}}} = \phi \left[ \frac{P - p_{ws}}{P - p_v} \right] = \frac{\phi}{1 + (1 - \phi)\frac{p_{ws}}{P - p_{ws}}}$

Notice that $\mu < \phi$ whenever $\phi < 100%$, though for typical HVAC temperatures ($p_{ws} \ll P$), $\mu \approx \phi$.

Specific Volume of Moist Air ($v$)

The volume of mixture per unit mass of dry air, expressed in $\text{ft}^3/\text{lb}{da}$ (or $\text{m}^3/\text{kg}{da}$):

$v = \frac{R_{da} T}{p_a} = \frac{R_{da} T}{P - p_v} = \frac{R_{da} T (1 + 1.6078 W)}{P}$

Where $T$ is in absolute temperature ($^\circ\text{R} = ^\circ\text{F} + 459.67$).

Specific Enthalpy ($h$)

The total energy per unit mass of dry air, defined as the sum of dry air sensible enthalpy and water vapor enthalpy referenced to $0^\circ\text{F}$ dry air and $32^\circ\text{F}$ saturated liquid water:

$h = h_{da} + W h_v = c_{p,da} T_{db} + W (h_{g0} + c_{p,v} T_{db})$

Using standard USCS coefficients ($c_{p,da} = 0.240\text{ Btu}/\text{lb}\cdot^\circ\text{F}$, $h_{g0} = 1061.2\text{ Btu}/\text{lb}$, $c_{p,v} = 0.444\text{ Btu}/\text{lb}\cdot^\circ\text{F}$):

$h = 0.240 T_{db} + W (1061 + 0.444 T_{db}) \quad \left[\frac{\text{Btu}}{\text{lb}_{da}}\right]$

In SI units ($c_{p,da} = 1.006\text{ kJ}/\text{kg}\cdot^\circ\text{C}$, $h_{g0} = 2501\text{ kJ}/\text{kg}$, $c_{p,v} = 1.86\text{ kJ}/\text{kg}\cdot^\circ\text{C}$):

$h = 1.006 T_{db} + W (2501 + 1.86 T_{db}) \quad \left[\frac{\text{kJ}}{\text{kg}_{da}}\right]$

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3. Property Relationships & Thermodynamic Inequalities

Understanding the mathematical and physical relationship among the three key temperatures ($T_{db}, T_{wb}, T_{dp}$) is a high-yield concept on the PE exam:

NCEES CBT Handbook Tip — Wet-Bulb Equation: The relation connecting wet-bulb temperature $T_{wb}$ to partial vapor pressure $p_v$ at standard pressure is Carrier's equation: $p_v = p_{ws}' - \frac{(P - p_{ws}')(T_{db} - T_{wb})}{2800 - 1.3 T_{wb}}$ Where $p_{ws}'$ is the saturation pressure evaluated at the wet-bulb temperature $T_{wb}$ (in $^\circ\text{F}$). On the exam, when reading psychrometric charts, you can locate state points directly via chart grid lines without solving Carrier's equation manually.

4. Comprehensive Property Lookup & Conversion Reference

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5. Step-by-Step Worked Example: Exact Psychrometric State Determination

Problem Statement

Moist air at standard sea level atmospheric pressure ($P = 14.696\text{ psia}$) has a dry-bulb temperature of $T_{db} = 80.0^\circ\text{F}$ and a wet-bulb temperature of $T_{wb} = 67.0^\circ\text{F}$. From steam saturation tables:

• Saturation pressure at $80^\circ\text{F}$: $p_{ws}(80^\circ\text{F}) = 0.5073\text{ psia}$
• Saturation pressure at $67^\circ\text{F}$: $p_{ws}'(67^\circ\text{F}) = 0.3288\text{ psia}$

Calculate:

1. Partial vapor pressure $p_v$ (psia)
2. Humidity ratio $W$ (in $\text{lb}w/\text{lb}{da}$ and $\text{grains}/\text{lb}_{da}$)
3. Relative humidity $\text{RH}$ (%)
4. Specific enthalpy $h$ ($\text{Btu}/\text{lb}_{da}$)
5. Specific volume $v$ ($\text{ft}^3/\text{lb}_{da}$)

Solution Steps

Step 1: Calculate actual partial vapor pressure $p_v$ via Carrier's equation $p_v = p_{ws}' - \frac{(P - p_{ws}')(T_{db} - T_{wb})}{2800 - 1.3 T_{wb}}$ $p_v = 0.3288 - \frac{(14.696 - 0.3288)(80.0 - 67.0)}{2800 - 1.3(67.0)}$ $p_v = 0.3288 - \frac{(14.3672)(13.0)}{2800 - 87.1} = 0.3288 - \frac{186.774}{2712.9} = 0.3288 - 0.06885 = 0.25995\text{ psia} \approx 0.2600\text{ psia}$

Step 2: Calculate humidity ratio $W$ $W = 0.62198 \frac{p_v}{P - p_v} = 0.62198 \times \frac{0.2600}{14.696 - 0.2600} = 0.62198 \times \frac{0.2600}{14.4360} = 0.011195\text{ lb}_w/\text{lb}_{da}$ $W_{\text{grains}} = 0.011195 \times 7000 = 78.37\text{ grains}/\text{lb}_{da}$

Step 3: Calculate relative humidity $\text{RH}$ $\text{RH} = \frac{p_v}{p_{ws}(80^\circ\text{F})} \times 100\% = \frac{0.2600}{0.5073} \times 100\% = 51.25\%$

Step 4: Calculate mixture specific enthalpy $h$ $h = 0.240 T_{db} + W (1061 + 0.444 T_{db})$ $h = 0.240(80.0) + 0.011195(1061 + 0.444 \times 80.0)$ $h = 19.20 + 0.011195(1061 + 35.52) = 19.20 + 12.28 = 31.48\text{ Btu}/\text{lb}_{da}$

Step 5: Calculate specific volume $v$ $v = \frac{R_{da} T}{P - p_v} = \frac{53.352 \times (80.0 + 459.67)}{144 \times (14.696 - 0.2600)} = \frac{53.352 \times 539.67}{144 \times 14.4360} = \frac{28792.1}{2078.78} = 13.85\text{ ft}^3/\text{lb}_{da}$

Total moist air density $\rho = \frac{1 + W}{v} = \frac{1 + 0.011195}{13.85} = 0.0730\text{ lb}_m/\text{ft}^3$.