Psychrometrics & Air Processes

14% of exam

Moist-air properties, the psychrometric chart, and sensible, latent, mixing, humidification, and dehumidification processes at sea level and altitude.

6 concepts

Exam areas

B. Heating/Cooling Processes

Moist-Air Properties and the Psychrometric Chart

The six interlocking properties of moist air — dry-bulb, wet-bulb, dew point, humidity ratio, enthalpy, specific volume — and how to fix a state and read processes on the chart.

Almost every HVAC and refrigeration problem on this exam begins by fixing the state of moist air, and almost every wrong answer traces back to a state read incorrectly. Moist air is a two-component mixture — dry air plus a little water vapor — so it takes exactly two independent intensive properties to pin the state at a fixed total pressure. Give the exam any two independent ones — dry-bulb with wet-bulb, dew point, humidity ratio, relative humidity or enthalpy, for instance (never dew point with humidity ratio, which both measure only the moisture) — and the psychrometric chart hands you all the rest in seconds. The NCEES PE Mechanical Reference Handbook collects the governing relations in §7 Psychrometrics, but the chart is what you will actually use under time pressure, so this concept builds fluency in both the equations behind the chart and the act of reading it.

Humidity ratio is the conserved bookkeeping variable

Humidity ratio

W

is the mass of water vapor carried by each pound of dry air,

W = M_w/M_{da}

, in

\text{lb}_w/\text{lb}_{da}

(or grains, with

7000\,\text{gr} = 1\,\text{lb}_w

). It is the natural conserved quantity for air processes because the dry-air mass flow never changes as air is heated, cooled, humidified, or mixed — only the water riding along with it does. In terms of partial pressures it follows directly from the ideal-gas mole ratio scaled by the molar-mass ratio

18.01528/28.9645 = 0.62198

(§7.1; the handbook rounds it to

0.622

in the partial-pressure forms):

W = 0.62198\,\dfrac{p_w}{p - p_w}

Relative humidity, dew point, and saturation

Relative humidity

\phi

is the ratio of the actual vapor mole fraction to its saturated value at the same temperature and pressure, very nearly

\phi = p_w/p_{ws}(t)

, where

p_{ws}(t)

is the saturation pressure of water at the dry-bulb temperature. Dew point

t_d

is the temperature to which you must cool the air at constant pressure and constant

W

until it first saturates — that is, until

p_{ws}(t_d) = p_w

. Saturation humidity

W_s(t,p)

is the largest

W

the air can hold at that temperature; degree of saturation is

\mu = W/W_s

. Dew point is set by

W

alone and is therefore the cleanest indicator of absolute moisture content.

\phi = \dfrac{p_w}{p_{ws}(t)}, \qquad t_d : \; p_{ws}(t_d) = p_w

Wet-bulb temperature: the chart's diagonal

The thermodynamic wet-bulb temperature

t^{*}

is the temperature reached when water evaporates adiabatically into the airstream until it saturates at constant total pressure. It is depressed below the dry-bulb whenever the air is unsaturated, and the dry-bulb, wet-bulb, and dew point all collapse to one value at saturation. The ASHRAE working relation (ASHRAE Handbook—Fundamentals, Ch. 1) for

W

from a measured dry-bulb

t

and wet-bulb

t^{*}

(with

W_s^{*}

the saturation humidity at

t^{*}

) is the equation you use whenever a sling-psychrometer reading is given; the NCEES handbook itself gives

W

only from the partial pressures, so reach for this correlation when only the two temperatures are reported:

W = \dfrac{(1093 - 0.556\,t^{*})\,W_s^{*} - 0.240\,(t - t^{*})}{1093 + 0.444\,t - t^{*}}

Enthalpy and specific volume

Specific enthalpy is referenced to a pound of dry air, not a pound of mixture:

h = 0.240\,t + W(1061 + 0.444\,t)

\text{Btu/lb}_{da}

, where the first term is dry-air sensible heat and the parenthesis is the vapor's latent-plus-sensible content. Lines of constant enthalpy on the chart run almost parallel to constant wet-bulb lines — close, but not identical, which matters in precise work. Specific volume is also per pound of dry air,

v = 0.370\,(t + 459.67)(1 + 1.6078\,W)/p

\text{ft}^3/\text{lb}_{da}

; you need it to convert between volumetric airflow (CFM) and the mass flow that actually carries the energy.

h = 0.240\,t + W(1061 + 0.444\,t)

Locating a state and reading a process

Plot a state by walking in from any two property scales: dry-bulb along the bottom, humidity ratio up the right edge, wet-bulb and enthalpy along the upper-left diagonal scale, relative-humidity as the family of curves, with saturation as the leftmost boundary. Once the point is fixed, a process is just an arrow. A horizontal arrow (constant

W

) is pure sensible heating or cooling; a vertical arrow (constant dry-bulb) is pure latent change; an arrow up-and-left at constant wet-bulb is adiabatic humidification. Reading the endpoints off the chart and differencing the properties is the entire method.

Exam strategy

Identify the two independent properties the problem gives, fix the state, and only then compute. When given a sling psychrometer's dry-bulb and wet-bulb, use the ASHRAE wet-bulb relation or the chart; do not confuse wet-bulb with dew point unless the air is saturated. Carry humidity ratio in

\text{lb}_w/\text{lb}_{da}

for energy balances and convert to grains only for the per-grain latent factor,

4{,}840/7{,}000 = 0.691

. Remember that

h

and

v

are per pound of dry air, so a CFM-to-mass conversion always divides volumetric flow by the specific volume at the state in question, never by a generic

13.33\,\text{ft}^3/\text{lb}

unless the state truly is standard air.

Key equations

Humidity ratio from partial pressures

W = 0.62198\,\dfrac{p_w}{p - p_w}

$W$ in $\text{lb}_w/\text{lb}_{da}$ ; $p_w$ = vapor partial pressure, $p$ = total pressure (psia). $7000\,\text{gr}=1\,\text{lb}_w$ .

Saturation humidity ratio

W_s = 0.62198\,\dfrac{p_{ws}}{p - p_{ws}}

Maximum $W$ at temperature $t$ ; $p_{ws}$ = saturation pressure of water at $t$ (psia).

Relative humidity

\phi = \dfrac{p_w}{p_{ws}(t)}

Ratio of actual to saturated vapor partial pressure at the same dry-bulb $t$ and $p$ .

Dew point condition

p_{ws}(t_d) = p_w

$t_d$ = temperature where saturation pressure equals the actual vapor pressure; set by $W$ alone.

Humidity ratio from wet-bulb

W = \dfrac{(1093 - 0.556\,t^{*})\,W_s^{*} - 0.240\,(t - t^{*})}{1093 + 0.444\,t - t^{*}}

$t$ = dry-bulb, $t^{*}$ = wet-bulb (°F), $W_s^{*}$ = saturation humidity at $t^{*}$ . ASHRAE USCS correlation (not in the NCEES handbook, which gives $W$ only from partial pressures).

Moist-air specific enthalpy

h = 0.240\,t + W(1061 + 0.444\,t)

$\text{Btu/lb}_{da}$ ; $t$ in °F. First term dry-air sensible, parenthesis vapor latent + sensible.

Moist-air specific volume

v = \dfrac{0.370\,(t + 459.67)(1 + 1.6078\,W)}{p}

$\text{ft}^3/\text{lb}_{da}$ ; $t$ in °F, $p$ in psia. Per pound of DRY air, not mixture.

Degree of saturation

\mu = \dfrac{W}{W_s(t,p)}

Dimensionless; close to but not equal to relative humidity $\phi$ .

Worked examples

Fix a state from dry-bulb and wet-bulb

Problem. Air at sea level ( $p = 14.696\,\text{psia}$ ) reads $80\,^\circ\text{F}$ dry-bulb and $67\,^\circ\text{F}$ wet-bulb. Find the humidity ratio, relative humidity, enthalpy, specific volume, and dew point.

Solution. Saturation humidity at the wet-bulb:

p_{ws}(67) = 0.3277\,\text{psia}

, so

W_s^{*} = 0.62198(0.3277)/(14.696-0.3277) = 0.01419

. Humidity ratio:

W = \dfrac{(1093 - 0.556\cdot 67)(0.01419) - 0.240(80-67)}{1093 + 0.444\cdot 80 - 67} = \dfrac{(1055.7)(0.01419) - 3.12}{1061.5} = \dfrac{11.86}{1061.5} = 0.01117\,\text{lb}_w/\text{lb}_{da}

(

78\,\text{gr/lb}

). Vapor pressure:

p_w = Wp/(0.62198 + W) = (0.01117)(14.696)/(0.63315) = 0.2593\,\text{psia}

. With

p_{ws}(80)=0.5074

\phi = 0.2593/0.5074 = 0.511 = 51.1\%

. Enthalpy:

h = 0.240(80) + 0.01117(1061 + 0.444\cdot 80) = 19.20 + 0.01117(1096.5) = 31.4\,\text{Btu/lb}_{da}

. Specific volume:

v = 0.370(539.67)(1 + 1.6078\cdot 0.01117)/14.696 = 0.370(539.67)(1.01796)/14.696 = 13.8\,\text{ft}^3/\text{lb}_{da}

. Dew point: solve

p_{ws}(t_d) = 0.2593\,\text{psia}\Rightarrow t_d = 60.3\,^\circ\text{F}

. Sanity:

t_d = 60.3 < t^{*} = 67 < t = 80

, exactly the required ordering for unsaturated air. Answers:

W = 0.0112\,\text{lb}_w/\text{lb}_{da}

\phi = 51.1\%

h = 31.4\,\text{Btu/lb}_{da}

v = 13.8\,\text{ft}^3/\text{lb}_{da}

t_d = 60.3\,^\circ\text{F}

W = \dfrac{(1055.7)(0.01419) - 3.12}{1061.5} = 0.0112\ \text{lb}_w/\text{lb}_{da}

Convert CFM to dry-air mass flow

Problem. An air handler delivers $4000\,\text{CFM}$ of the $80\,^\circ\text{F}/67\,^\circ\text{F}$ air above ( $v = 13.8\,\text{ft}^3/\text{lb}_{da}$ ). What dry-air mass flow does it move, and how much water vapor (lb/hr) rides along with it?

Solution. Dry-air mass flow:

\dot m_{da} = \dfrac{\dot V}{v} = \dfrac{4000\,\text{ft}^3/\text{min}}{13.8\,\text{ft}^3/\text{lb}_{da}} = 289.9\,\text{lb}_{da}/\text{min}

, i.e.

\dot m_{da}\times 60 = 17{,}390\,\text{lb}_{da}/\text{hr}

. Water carried:

\dot m_w = W\,\dot m_{da} = 0.01117(17{,}390) = 194\,\text{lb}_w/\text{hr}

. Sanity check on the magnitude: standard air is

0.075\,\text{lb}/\text{ft}^3

, giving roughly

4000\times 0.075\times 60 = 18{,}000\,\text{lb/hr}

of mixture — our dry-air value of

17{,}400\,\text{lb/hr}

is just below that because this air is warmer and lighter than

70\,^\circ\text{F}

standard air. Answers:

\dot m_{da} = 1.74\times 10^{4}\,\text{lb/hr}

\dot m_w = 194\,\text{lb/hr}

\dot m_{da} = \dfrac{\dot V}{v}, \qquad \dot m_w = W\,\dot m_{da}

Common pitfalls

•Confusing wet-bulb with dew point. They coincide only at saturation; for the $80/67$ air above, $t^{*}=67\,^\circ\text{F}$ but $t_d=60\,^\circ\text{F}$ . Dew point tracks $W$ ; wet-bulb tracks enthalpy.
•Treating $h$ and $v$ as per pound of MIXTURE. Both are referenced to a pound of DRY air, so CFM-to-mass conversions divide by the moist-air specific volume $v$ in $\text{ft}^3/\text{lb}_{da}$ .
•Using $\phi = W/W_s$ as exact. That ratio is the degree of saturation $\mu$ ; relative humidity is the partial-pressure (mole-fraction) ratio and differs by a few percent.
•Reading the chart at the wrong pressure. A sea-level chart is only valid near sea level; at altitude the saturation curve and every property line shift (separate concept).
•Carrying humidity ratio in grains into an enthalpy or mass balance. Energy balances need $\text{lb}_w/\text{lb}_{da}$ ; grains are for the per-grain latent factor ( $4{,}840/7{,}000 = 0.691$ ) only.
•Assuming standard air ( $v\approx 13.33\,\text{ft}^3/\text{lb}$ , $\rho = 0.075\,\text{lb}/\text{ft}^3$ ) for every state. Use the actual specific volume at the actual temperature and humidity when precision matters.

References

NCEES PE Mechanical Reference Handbook — §7 Psychrometrics
NCEES PE Mechanical Reference Handbook — §7.4 Thermodynamic Properties of Moist Air
ASHRAE Handbook—Fundamentals, Chapter 1, Psychrometrics

Sensible, Latent, Mixing and the 1.10 / 4,840 / 4.5 Coefficients

The standard-air airflow equations for sensible, latent, and total load, the sensible heat ratio, and how to mix two airstreams on the chart.