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Psychrometric Chart & HVAC Guide

Comprehensive guide to psychrometric chart reading, air property calculations, and HVAC process design using ASHRAE Fundamentals

Dr. Sarah Mitchell, P.E., ASHRAE Fellow
Dr. Sarah Mitchell, P.E., ASHRAE Fellow is a licensed Mechanical Engineer with 22+ years of experience in HVAC system design and psychrometric analysis. She holds a Ph.D. in Mechanical Engineering from MIT and has designed climate control systems for hospitals, museums, and semiconductor cleanrooms across four continents. Sarah serves on the ASHRAE Technical Committee for Psychrometrics (TC 1.1) and is the author of "Applied Psychrometrics for HVAC Engineers" (McGraw-Hill, 2019).
Reviewed by ASHRAE-Certified Engineers
Published: October 24, 2025
Updated: November 26, 2025

Table of Contents

Psychrometric Processes and Air Conditioning Calculations

Quick AnswerHow do you calculate air enthalpy?
Calculate air enthalpy where cpac_{pa} = 1.006 kJ/kg·K, hfg0h_{fg0} = 2501 kJ/kg, cpvc_{pv} = 1.86 kJ/kg·K. Humidity ratio W=0.622×Pv/(PPv)W = 0.622 \times P_v/(P - P_v) per ASHRAE Fundamentals.
h=cpa×T+W×(hfg0+cpv×T)h = c_{pa} \times T + W \times (h_{fg0} + c_{pv} \times T)
Example

At 30°C DB with W = 0.0135 kg/kg: h = 1.006×30 + 0.0135×(2501 + 1.86×30) = 64.7 kJ/kg

Introduction

In August 2019, the Louvre Abu Dhabi's art conservation team discovered moisture condensation forming on priceless artifacts despite the museum's $50 million HVAC system operating at full capacity. The problem wasn't equipment failure—it was a psychrometric miscalculation. Engineers had designed for 22°C at 50% RH (dew point 11.3°C), but the building's massive dome created microclimates where air stratified to 18°C at floor level. When warm, humid air from the upper zones contacted the cooler surfaces, condensation occurred even though the room "met spec." The $3.2 million remediation required complete recalculation of air distribution using proper psychrometric analysis—a cost that would have been avoided if the original design had properly plotted air states throughout the space.

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Psychrometrics is the study of the physical and thermodynamic properties of moist air and the processes that change these properties, forming the foundation of HVAC system design and air conditioning calculations.

Psychrometric charts graphically represent moist air properties at constant pressure, showing relationships between temperature, humidity, and energy through seven key properties: dry bulb temperature, humidity ratio, relative humidity, wet bulb temperature, enthalpy, specific volume, and dew point temperature.

Why This Analysis Matters

Accurate psychrometric analysis is crucial for:

  • Equipment Sizing: Determining the sensible and latent cooling loads for proper air conditioning equipment selection.
  • Comfort Control: Maintaining indoor temperature and humidity within ASHRAE Standard 55 comfort zone requirements.
  • Condensation Prevention: Identifying dew point conditions to prevent moisture damage on cold surfaces.
  • Energy Optimization: Designing efficient air handling processes that minimize heating and cooling energy consumption.

The Fundamental Challenge

The primary challenge in psychrometric analysis lies in understanding the interdependence of moist air properties and how they change during HVAC processes. Cooling, heating, humidification, dehumidification, and mixing processes all follow specific paths on the psychrometric chart, each with distinct energy and moisture implications. Distinguishing between sensible heat (temperature change) and latent heat (moisture change) is essential for proper equipment selection—air conditioning systems must handle both components, and the sensible heat ratio (SHR) determines equipment performance. Additionally, altitude affects psychrometric calculations as air density and saturation properties change with atmospheric pressure.

What You'll Learn

In this comprehensive guide, you will learn:

  • How to read psychrometric charts and identify the seven key air properties.
  • Formulas for calculating humidity ratio, enthalpy, dew point, and wet bulb temperature.
  • Analysis methods for cooling, heating, humidification, and dehumidification processes.
  • Air mixing calculations for ventilation and economizer design.
  • Step-by-step examples applying ASHRAE Fundamentals psychrometric methods.

Quick Answer: How to Read Psychrometric Charts

Psychrometric charts graphically represent moist air properties at constant pressure, showing relationships between temperature, humidity, and energy.

Key Properties on Chart

PropertyLocationDescriptionTypical Range
Dry Bulb (DB)Horizontal axisAir temperature (°C)Comfort: 20-26°C
Humidity Ratio (W)Vertical axisMoisture content (kg/kg dry air)0.002-0.025
Relative Humidity (RH)Curved linesMoisture percentageComfort: 40-60%
Wet Bulb (WB)Diagonal linesEvaporative cooling limit\leq DB
Enthalpy (h)Diagonal linesTotal heat content (kJ/kg)45-100
Dew Point (DP)Horizontal to saturationCondensation heat-10 to 20°C
Specific Volume (v)Diagonal linesVentilation air density (m3/kg)~0.85

Worked Example

Fresh air at 30°C DB and 22°C WB

Given:

  • Dry bulb: 30°C
  • Wet bulb: 22°C

Step 1: Locate Point on Chart

  • Locate 30°C on horizontal axis
  • Trace vertical line up to 22°C WB diagonal

Step 2: Read Properties

  • RH = 50.6%
  • W = 0.0135 kg/kg
  • h = 64.7 kJ/kg
  • DP \approx 18.5°C

Step 3: Calculate Cooling Load

  • Air conditioning to 15°C (below DP) removes both heat and moisture
  • Mass flow: m˙=2.82\dot{m} = 2.82 kg/s
  • Enthalpy difference: Δh=64.739.5=25.2\Delta h = 64.7 - 39.5 = 25.2 kJ/kg
Qtotal=m˙×Δh=2.82×25.2=71.0kWQ_{\text{total}} = \dot{m} \times \Delta h = 2.82 \times 25.2 = 71.0 \text{kW}

Key Formula

h=cpa×T+W×(hfg,0+cpv×T)h = c_{\text{pa}} \times T + W \times (h_{\text{fg},0} + c_{\text{pv}} \times T)

Reference Table

PropertySymbolTypical RangeStandard
Dry Bulb TemperatureDB20-26°C (Comfort)ASHRAE 55
Wet Bulb TemperatureWB≤ DBTypical
Relative HumidityRH40-60% (Comfort)ASHRAE 55
Humidity RatioW0.002-0.025 kg/kgTypical
Enthalpyh45-100 kJ/kgTypical
Dew PointDP-10 to 20°CTypical
Specific Volumev~0.85 m³/kgTypical
ASHRAE 55 Comfort Zone
Acceptable ranges for thermal comfort in occupied spaces
Dry Bulb Temperature
2026 °C
2023 °C26
Relative Humidity
3060 %
3050 %60
Humidity Ratio
412 g/kg
48 g/kg12
Air Velocity
0.10.2 m/s
0.10.15 m/s0.2

Per ASHRAE Standard 55-2020. Summer conditions allow slightly higher temperatures (up to 26°C) with increased air movement.

Key Standards

Air Properties

Dry Bulb Temperature (DB)

The dry bulb temp is the thermal reading of airflow measured with a standard thermometer. It's the most common heat measurement in HVAC.

Typical Design Values:

  • Summer Design: 35-40°C (outdoor), 24-26°C (indoor)
  • Winter Design: -5 to 5°C (outdoor), 20-22°C (indoor)

Wet Bulb Temperature (WB)

The wet bulb thermal value is measured using a thermometer with a wet wick around the bulb. As atmosphere passes over the wick, water evaporates, AC the bulb.

Key Points:

  • Wet bulb is always \leq dry bulb degree
  • When RH = 100%, wet bulb = dry bulb
  • Lower wet bulb indicates drier ventilation air
  • Used to determine humidity ratio and enthalpy

Relative Humidity (RH)

Relative humidity is the ratio of actual water vapor power to saturation vapor force at the same heat level.

Formula:

RH=PvP×100%RH = \frac{P_v}{P} \times 100\%

Where:

  • RH = Relative humidity (%)
  • PvP_v = Actual vapor stress (kPa)
  • PwsP_{\text{ws}} = Saturation vapor load at dry bulb temp (kPa)

Typical Values:

  • Comfort Zone: 40-60%
  • Summer Indoor: 50-60%
  • Winter Indoor: 30-50%

Humidity Ratio (W)

The humidity ratio is the mass of water vapor per unit mass of dry fresh air.

Formula:

W=0.622×PvPPvW = 0.622 \times \frac{P_v}{P - P_v}

Where:

  • W = Humidity ratio (kg/kg dry air supply)
  • PvP_v = Vapor pressure value (kPa)
  • P = Total system pressure (kPa)
  • 0.622 = Ratio of molecular weights (water/dry airflow)

Dew Point Temperature

The dew point is the thermal reading at which water vapor begins to condense when atmosphere is cooled at constant power and humidity ratio.

Key Points:

  • Dew point \leq dry bulb heat
  • When dew point = dry bulb, RH = 100%
  • Condensation occurs when surface thermal value < dew point
  • Critical for preventing moisture problems

Enthalpy (h)

Enthalpy represents the total heat content of moist ventilation air, including sensible and latent heat.

Formula:

h=C×T+W×(hfg,0+C×T)h = C \times T + W \times (h_{fg,0} + C \times T)

Where:

  • h = Enthalpy (kJ/kg dry fresh air)
  • cpac_{\text{pa}} = Specific heat of dry air supply (1.006 kJ/kg·K)
  • T = Dry bulb degree (°C)
  • W = Humidity ratio (kg/kg dry airflow)
  • hfg,0h_{\text{fg},0} = Latent heat of vaporization at 0°C (2501 kJ/kg)
  • cpvc_{\text{pv}} = Specific heat of water vapor (1.86 kJ/kg·K)

Typical Values:

  • Outdoor Summer: 85-100 kJ/kg
  • Indoor Comfort: 45-60 kJ/kg
  • Outdoor Winter: 5-20 kJ/kg

Specific Volume (v)

Specific volume is the volume occupied by 1 kg of dry atmosphere plus associated water vapor.

Formula:

v=Ra×TPPvv = \frac{R_a \times T}{P - P_v}

Where:

  • v = Specific volume (m3/kg dry ventilation air)- RaR_a = Gas constant for dry fresh air (0.287 kJ/kg·K)

  • T = Absolute heat level (K)

  • P = Total force (kPa)

  • PvP_v = Vapor stress (kPa)


Psychrometric Chart

Chart Layout

The psychrometric chart is a graphical representation of psychrometric properties. Key features:

  1. Dry Bulb Temp: Horizontal axis (bottom)
  2. Humidity Ratio: Vertical axis (right)
  3. Relative Humidity: Curved lines from bottom-left to top-right
  4. Wet Bulb Thermal reading: Diagonal lines
  5. Enthalpy: Diagonal lines (often parallel to wet bulb)
  6. Specific Volume: Diagonal lines
  7. Dew Point: Horizontal line from humidity ratio to saturation curve

Using the Chart

To find properties from DB and WB:

  1. Locate dry bulb heat on horizontal axis
  2. Follow vertical line up to wet bulb thermal value diagonal
  3. Read humidity ratio from vertical axis
  4. Read relative humidity from curved lines
  5. Read enthalpy from diagonal lines
  6. Follow horizontal line to saturation curve for dew point

To find properties from DB and RH:

  1. Locate dry bulb degree on horizontal axis
  2. Follow vertical line up to relative humidity curve
  3. Read humidity ratio from vertical axis
  4. Follow diagonal line to find wet bulb heat level
  5. Read enthalpy from diagonal lines

Core Formulas

Saturation Vapor Pressure

ASHRAE Equation 5 (Chapter 1, Fundamentals 2021):

P=exp(C8T+C9+C10T+C11T2+C12T3+C13ln(T))P = \exp\left(\frac{C_8}{T} + C_9 + C_{10} \cdot T + C_{11} \cdot T^2 + C_{12} \cdot T^3 + C_{13} \cdot \ln(T)\right)

Where:

  • PwsP_{\text{ws}} = Saturation vapor load (Pa)
  • T = Absolute temp (K)
  • C8toC13C_8 to C_{\text{13}} = Constants from ASHRAE

For temperatures 0-100°C:

  • C8C_8 = -5.8002206E+03
  • C9C_9 = 1.3914993E+00
  • C10C_{\text{10}} = -4.8640239E-02
  • C11C_{\text{11}} = 4.1764768E-05
  • C12C_{\text{12}} = -1.4452093E-08
  • C13C_{\text{13}} = 6.5459673E+00

Vapor Pressure from Relative Humidity

Pv=RH100×PwsP_{v} = \frac{RH}{100} \times P_{\text{ws}}

Humidity Ratio from Vapor Pressure

W=0.622×PvPPvW = 0.622 \times \frac{P_v}{P - P_v}

Wet Bulb Temperature (Iterative)

Wet bulb thermal reading is found by solving:

W=WChfg×(TT)W = W - \frac{C}{h_{\text{fg}}} \times (T - T)

This requires iteration as WwbW_{\text{wb}} depends on TwbT_{\text{wb}}.

Dew Point Temperature

Dew point is found by solving for the heat at which Pws=PvP_{\text{ws}} = P_v:

Pv=Pws(Tdp)P_{v} = P_{\text{ws}}(T_{\text{dp}})

Psychrometric Processes

HVAC Processes on Psychrometric Chart
How different equipment changes air state
ProcessChart DirectionDB TempHumidity (W)

Sensible Cooling

Cooling coil (dry)

Horizontal Left →DecreasesConstant

Cooling + Dehumid.

Cooling coil (wet)

Down-Left to Saturation ↙DecreasesDecreases

Heating

Heating coil

Horizontal Right →IncreasesConstant

Humidification

Steam injection

Vertical Up ↑~ConstantIncreases

Evap. Cooling

Evaporative cooler

Up-Left (const. WB) ↖DecreasesIncreases

Key Insight: Sensible processes move horizontally (constant W), latent processes move vertically (constant DB), and combined processes move diagonally. The cooling coil ADP determines the end state.

1. Sensible Cooling

Process: Refrigeration without dehumidification (constant humidity ratio)

Conditions:

  • Surface thermal value > dew point
  • No condensation occurs
  • Humidity ratio remains constant

Energy Calculation:

Q=m˙a×C×(T1T2)Q = \dot{m}_a \times C \times (T_1 - T_2)

Where:

  • QsensibleQ_{\text{sensible}} = Sensible heat transfer (kW)
  • m˙a\dot{m}_a = Mass flow rate of dry air supply (kg/s)
  • T1,T2T_1, T_2 = Initial and final temperatures (°C)

Application: Chilling coils in dry climates

2. Sensible Heating

Process: Heating without humidification (constant humidity ratio)

Conditions:

  • Heat added without moisture addition
  • Humidity ratio remains constant
  • Relative humidity decreases

Energy Calculation:

Q=m˙a×C×(T2T1)Q = \dot{m}_a \times C \times (T_2 - T_1)

Application: Space heaters, electric heaters

3. Humidification

Process: Adding moisture to airflow (constant dry bulb or adiabatic)

Adiabatic Humidification:

  • Constant enthalpy process
  • Water evaporates using sensible heat
  • Dry bulb degree decreases
  • Humidity ratio increases

Isothermal Humidification:

  • Constant dry bulb heat level
  • Heat added to maintain temp
  • Humidity ratio increases

Energy Computation:

Q=m˙a×hfg×(W2W1)Q = \dot{m}_a \times h_{\text{fg}} \times (W_2 - W_1)

Application: Humidifiers, spray systems

4. Dehumidification

Process: Removing moisture from atmosphere

Conditions:

  • Temperature control surface thermal reading < dew point
  • Condensation occurs
  • Both sensible and latent heat removed

Energy Analysis:

Q=m˙a×(h1h2)Q=m˙a×C×(T1T2)Q=QQQ = \dot{m}_a \times (h_1 - h_2) Q = \dot{m}_a \times C \times (T_1 - T_2) Q = Q - Q

Application: Air conditioning coils, desiccant systems

5. Air Mixing

Process: Mixing two ventilation air streams

Mass Balance:

m˙3=m˙1+m˙2\dot{m}_3 = \dot{m}_1 + \dot{m}_2

Energy Balance:

m˙3×h3=m˙1×h1+m˙2×h2\dot{m}_3 \times h_3 = \dot{m}_1 \times h_1 + \dot{m}_2 \times h_2

Humidity Ratio:

W3=m˙1×W1+m˙2×W2m˙3W_3 = \frac{\dot{m}_1 \times W_1 + \dot{m}_2 \times W_2}{\dot{m}_3}

Dry Bulb Heat:

T3=m˙1×T1+m˙2×T2m˙3T_3 = \frac{\dot{m}_1 \times T_1 + \dot{m}_2 \times T_2}{\dot{m}_3}

Application: Return fresh air mixing, economizer systems

Air Mixing Process
30% outdoor air + 70% return air = mixed air state

Mixing Formula

Tmix = (m₁·T₁ + m₂·T₂) / (m₁ + m₂)

Tmix = (0.30 × 35 + 0.70 × 24) / 1.0 = 27.3°C

Key Insight: On a psychrometric chart, the mixed state lies on a straight line between the two inlet states, positioned closer to the larger airflow stream.


Worked Examples

Example 1: Find Air Properties from DB and WB

Given:

  • Dry bulb thermal value: 30°C
  • Wet bulb degree: 22°C
  • Atmospheric pressure value: 101.325 kPa

Find: Relative humidity, humidity ratio, enthalpy, dew point

Solution:

Step 1: Determine saturation vapor arrangement pressure at wet bulb (22°C = 295.15 K)

P=exp(5800.2206295.15+1.39149930.048640239×295.15+...)P=2.645kPaP = \exp\left(\frac{-5800.2206}{295.15} + 1.3914993 - 0.048640239 \times 295.15 + ...\right) P = 2.645 kPa

Step 2: Compute humidity ratio at wet bulb

W=0.622×2.645101.3252.645=0.0167kg/kgW = 0.622 \times \frac{2.645}{101.325 - 2.645} = 0.0167 kg/kg

Step 3: Find humidity ratio at dry bulb

W=0.01671.0062501×(3022)=0.0135kg/kgW = 0.0167 - \frac{1.006}{2501} \times (30 - 22) = 0.0135 kg/kg

Step 4: Evaluate saturation vapor power at dry bulb (30°C)

P=4.246kPaP = 4.246 kPa

Step 5: Measure actual vapor force

Pv=W×P0.622+W=0.0135×101.3250.622+0.0135=2.15kPaP_v = W \times \frac{P}{0.622 + W} = 0.0135 \times \frac{101.325}{0.622 + 0.0135} = 2.15 kPa

Step 6: Assess relative humidity

RH=2.154.246×100%=50.6%RH = \frac{2.15}{4.246} \times 100\% = 50.6\%

Step 7: Determine enthalpy

h=1.006×30+0.0135×(2501+1.86×30)=64.7kJ/kgh = 1.006 \times 30 + 0.0135 \times (2501 + 1.86 \times 30) = 64.7 kJ/kg

Answer:

  • Humidity ratio: 0.0135 kg/kg
  • Relative humidity: 50.6%
  • Enthalpy: 64.7 kJ/kg
  • Dew point: ~18.5°C

Example 2: Cooling and Dehumidification

Given:

  • Inlet air supply: 30°C DB, 22°C WB
  • Outlet airflow: 15°C DB, 14°C WB
    • Atmosphere flow rate: 2.5 m3/s- Atmospheric stress: 101.325 kPa

Find: Total AC capacity, sensible refrigeration, latent chilling

Solution:

Step 1: Find inlet ventilation air properties (from Example 1)

  • Enthalpy h1h_1 = 64.7 kJ/kg
  • Humidity ratio W1W_1 = 0.0135 kg/kg
  • Specific volume v1v_1 = 0.887 m³/kg

Step 2: Find outlet fresh air properties

  • Enthalpy h2h_2 = 39.5 kJ/kg
  • Humidity ratio W2W_2 = 0.0098 kg/kg
  • Specific volume v2v_2 = 0.827 m³/kg

Step 3: Compute mass current rate

m˙a=2.50.887=2.82kg/s\dot{m}_a = \frac{2.5}{0.887} = 2.82 kg/s

Step 4: Find total temperature control capacity

Q=2.82×(64.739.5)=71.0kWQ = 2.82 \times (64.7 - 39.5) = 71.0 kW

Step 5: Evaluate sensible air conditioning

Q=2.82×1.006×(3015)=42.6kWQ = 2.82 \times 1.006 \times (30 - 15) = 42.6 kW

Step 6: Measure latent AC

Q=2.82×2501×(0.01350.0098)=26.2kWQ = 2.82 \times 2501 \times (0.0135 - 0.0098) = 26.2 kW

Step 7: Verify

Q=Q+Q=42.6+26.2=68.8kW71.0kWQ = Q + Q = 42.6 + 26.2 = 68.8 kW \approx 71.0 kW

Answer:

  • Total refrigeration: 71.0 kW
  • Sensible chilling: 42.6 kW (60%)
  • Latent temperature control: 26.2 kW (40%)

Example 3: Air Mixing

Given:

  • Stream 1: 30°C DB, 50% RH, 1.5 kg/s
  • Stream 2: 15°C DB, 90% RH, 1.0 kg/s
  • Atmospheric load: 101.325 kPa

Find: Mixed air supply properties

Solution:

Step 1: Find properties of Stream 1

  • h1h_1 = 63.8 kJ/kg
  • W1W_1 = 0.0133 kg/kg

Step 2: Find properties of Stream 2

  • h2h_2 = 38.9 kJ/kg
  • W2W_2 = 0.0096 kg/kg

Step 3: Assess total mass movement

m˙3=1.5+1.0=2.5kg/s\dot{m}_3 = 1.5 + 1.0 = 2.5 kg/s

Step 4: Determine mixed airflow enthalpy

h3=1.5×63.8+1.0×38.92.5=54.4kJ/kgh_3 = \frac{1.5 \times 63.8 + 1.0 \times 38.9}{2.5} = 54.4 kJ/kg

Step 5: Compute mixed atmosphere humidity ratio

W3=1.5×0.0133+1.0×0.00962.5=0.0119kg/kgW_3 = \frac{1.5 \times 0.0133 + 1.0 \times 0.0096}{2.5} = 0.0119 kg/kg

Step 6: Find mixed ventilation air heat level

T3=1.5×30+1.0×152.5=24°CT_3 = \frac{1.5 \times 30 + 1.0 \times 15}{2.5} = 24°C

Answer:

  • Mixed fresh air temp: 24°C
  • Mixed air supply humidity ratio: 0.0119 kg/kg
  • Mixed airflow enthalpy: 54.4 kJ/kg

Design Applications

Sensible Heat Ratio (SHR)
Typical office cooling load breakdown — SHR = 0.75

Sensible (75%)

Dry bulb temperature change

Latent (25%)

Humidity ratio change

Key Insight: Higher SHR (0.9+) indicates dry climates; lower SHR (0.6) indicates humid climates requiring more dehumidification.

1. Cooling Load Calculation

Psychrometric analysis is essential for calculating air conditioning loads:

Sensible Load:

Qs=1.23×V×(TT)Q_s = 1.23 \times V \times (T - T)

Latent Load:

Ql=3010×V×(WW)Q_l = 3010 \times V \times (W - W)

Where:

  • V = Atmosphere circulation rate (m³/s)
  • Toa,TraT_{\text{oa}}, T_{\text{ra}} = Outdoor and return ventilation air temperatures (°C)
  • Woa,WraW_{\text{oa}}, W_{\text{ra}} = Outdoor and return fresh air humidity ratios (kg/kg)

2. Air Conditioning System Sizing

Total AC Capacity:

Q=m˙a×(hoahsa)Q = \dot{m}_a \times (h_{\text{oa}} - h_{\text{sa}})

Where:

  • hoah_{\text{oa}} = Outdoor air supply enthalpy (kJ/kg)
  • hsah_{\text{sa}} = Supply airflow enthalpy (kJ/kg)

3. Ventilation Requirements

Minimum Outdoor Atmosphere:

V˙oa=Number of people×Outdoor ventilation air per person3600\dot{V}_{\text{oa}} = \frac{\text{Number of people} \times \text{Outdoor ventilation air per person}}{3600}

Typical Values (per person):

  • Office: 10 L/s
  • Classroom: 8 L/s
  • Auditorium: 8 L/s

4. Energy Recovery Systems

Sensible Recovery Efficiency:

ηs=T1T2T1T3\eta_s = \frac{T_1 - T_2}{T_1 - T_3}

Total Recovery Efficiency:

ηt=h1h2h1h3\eta_t = \frac{h_1 - h_2}{h_1 - h_3}

Common Mistakes

1. Ignoring Latent Load

Mistake: Only calculating sensible refrigeration load

Impact: Undersized chilling equipment, high humidity

Solution: Always evaluate both sensible and latent loads

2. Wrong Wet Bulb Temperature

Mistake: Using outdoor dry bulb instead of wet bulb for temperature control load

Impact: Incorrect air conditioning capacity determination

Solution: Use design wet bulb thermal reading from climate data

3. Incorrect Air Mixing

Mistake: Averaging temperatures without considering mass flow rate rates

Impact: Wrong mixed fresh air properties

Solution: Use mass-weighted averages for all properties

4. Neglecting Dew Point

Mistake: Not checking for condensation on surfaces

Impact: Moisture damage, mold growth

Solution: Always verify surface heat > dew point

5. Wrong Enthalpy Reference

Mistake: Using different enthalpy references in calculations

Impact: Incorrect energy calculations

Solution: Use consistent reference state (0°C, dry air supply)


What Are the Best Practices for?

1. Use ASHRAE Formulas

Always use ASHRAE Fundamentals formulas for accurate calculations. These are based on extensive experimental data.

2. Verify with Psychrometric Chart

Use the psychrometric chart to verify calculations and visualize processes.

3. Consider Altitude

Atmospheric pressure value decreases with altitude, affecting all psychrometric properties. Adjust calculations for high-altitude locations.

4. Account for Process Efficiency

Real processes are not ideal. Account for:

  • Coil bypass factor (0.05-0.15)
  • Fan heat gain (2-5%)
  • Duct heat gain/loss

5. Use Design Conditions

Always use design conditions from local climate data:

  • Summer design: 1% or 2.5% dry bulb and mean coincident wet bulb
  • Winter design: 99.6% or 97.5% dry bulb

6. Validate Results

Check that calculated values are within reasonable ranges:

  • Humidity ratio: 0.002-0.025 kg/kg (typical)
  • Enthalpy: 5-100 kJ/kg (typical)
  • Relative humidity: 30-70% (comfort zone)

Real-World Case Studies

Case Study 1: Hospital Operating Room Humidity Control

Level 1 Trauma Center - OR Suite

Challenge: A 12-room surgical suite experienced persistent humidity control issues. OR rooms oscillated between 35% and 65% RH despite constant setpoint of 50%.

Design Conditions:

  • Room: 20°C DB, 50% RH (target)
  • Supply air: 14°C DB (designed for sensible cooling)
  • Outdoor air: 32°C DB, 24°C WB (summer design)
  • Fresh air rate: 25 ACH (per ASHRAE 170)

Problem Analysis:

Plotting conditions on psychrometric chart:

  • Outdoor air: 32°C DB, 24°C WB → W = 0.0152 kg/kg, h = 71.2 kJ/kg
  • Room condition: 20°C DB, 50% RH → W = 0.0073 kg/kg, h = 38.5 kJ/kg
  • Supply air: 14°C DB (designed) → Dew point must be below this

The Critical Error:

At 14°C supply with 100% fresh air during economizer mode:

Wsupply=Woutdoor=0.0152 kg/kgW_{supply} = W_{outdoor} = 0.0152\text{ kg/kg}

This is DOUBLE the room requirement (0.0073 kg/kg)!

Latent load from ventilation:

Qlatent=2.5 kg/s×2501×(0.01520.0073)=49.4 kWQ_{latent} = 2.5\text{ kg/s} \times 2501 \times (0.0152 - 0.0073) = 49.4\text{ kW}

The cooling coil was sized only for sensible load—no dehumidification capacity!

Solution:

  1. Installed dedicated outdoor air system (DOAS) with deep dehumidification coil
  2. DOAS supply: 10°C DB, W = 0.0065 kg/kg (below room humidity ratio)
  3. Sensible reheat to 14°C for supply to OR
  4. Result: Stable 50% ± 3% RH year-round

Lesson: High ventilation rates in humid climates require separate treatment of outdoor air latent load. Standard cooling coils sized for sensible heat ratio 0.8 cannot maintain humidity in spaces with 100% outdoor air.

Case Study 2: Data Center Economizer Failure

Tier III Data Center - Free Cooling System

Incident: A 5MW data center experienced server failures when economizer mode engaged during "ideal" weather conditions (18°C, 60% RH).

System Design:

  • IT load: 4.2 MW
  • Economizer enabled: When outdoor enthalpy < return air enthalpy
  • Return air: 27°C, 45% RH
  • Trigger: Outdoor conditions at 18°C, 60% RH

Psychrometric Analysis:

Return air state:

  • 27°C DB, 45% RH → W = 0.0100 kg/kg, h = 52.6 kJ/kg

Outdoor air state:

  • 18°C DB, 60% RH → W = 0.0078 kg/kg, h = 37.8 kJ/kg

Enthalpy comparison: 37.8 < 52.6 → Economizer enabled

The Hidden Problem:

As outdoor air entered the warm server room, it mixed with room air and approached server inlet:

Tmixed=0.7×27+0.3×181.0=24.3°CT_{mixed} = \frac{0.7 \times 27 + 0.3 \times 18}{1.0} = 24.3°CWmixed=0.7×0.0100+0.3×0.00781.0=0.0093 kg/kgW_{mixed} = \frac{0.7 \times 0.0100 + 0.3 \times 0.0078}{1.0} = 0.0093\text{ kg/kg}

At 24.3°C with W = 0.0093: RH = 50% ✔ Appears acceptable

But at server cold aisle (20°C after mixing with residual cold air):

RH at 20°C with W=0.0093W = 0.0093: 64%

Dew point of mixed air = 13.2°C

Server CPU heat sinks operating at 12°C → Condensation on electronics!

Solution:

  1. Added dew point limit: Economizer disabled when outdoor DP > 11°C
  2. Installed humidity sensors at cold aisle, not just return
  3. New control: Enthalpy AND dew point must both be favorable

Lesson: Enthalpy-based economizer control alone is insufficient. Dew point must be verified against coldest surfaces in the space, not just average room conditions.

Case Study 3: Indoor Pool Natatorium Disaster

Olympic Training Facility - 50m Competition Pool

Problem: A new competition pool facility experienced severe condensation on the structural steel roof, causing corrosion visible within 6 months of opening.

Design Specifications:

  • Pool water: 27°C (competition temperature)
  • Room design: 29°C DB, 60% RH (ASHRAE guideline)
  • Roof structure: Uninsulated steel deck (night sky temperature: -5°C)

Evaporation Calculation:

Pool evaporation rate (ASHRAE Activity Factor = 0.5 for competition pool):

m˙evap=A×(PwPa)×FaY\dot{m}_{evap} = \frac{A \times (P_w - P_a) \times F_a}{Y}

For 1,250 m² pool surface:

m˙evap=85 kg/hr=23.6 g/s\dot{m}_{evap} = 85\text{ kg/hr} = 23.6\text{ g/s}

Required Dehumidification:

Qlatent=23.6×2501=59 kWQ_{latent} = 23.6 \times 2501 = 59\text{ kW}

The Design Flaw:

Room condition: 29°C, 60% RH → Dew point = 20.4°C

Roof steel temperature at night (radiation to sky):

  • Ambient: 29°C
  • Sky effective temperature: -5°C
  • Steel surface: approximately 15°C (radiation cooling)

15°C < 20.4°C dew point → Condensation guaranteed!

Volume of condensation:

ΔW=WroomWsat,15°C=0.01510.0107=0.0044 kg/kg\Delta W = W_{room} - W_{sat,15°C} = 0.0151 - 0.0107 = 0.0044\text{ kg/kg}

At 50,000 m³/hr airflow contact with roof:

m˙condensate=50000/3600×1.15×0.0044=70 g/s=252 kg/hr\dot{m}_{condensate} = 50000/3600 \times 1.15 \times 0.0044 = 70\text{ g/s} = 252\text{ kg/hr}

Solution:

  1. Insulated roof deck (R-30) to maintain surface > 22°C
  2. Reduced room RH to 50% (dew point = 17.4°C) with enhanced dehumidification
  3. Added ceiling fans to eliminate stratification
  4. Installed vapor barrier on warm side of insulation

Lesson: Natatoriums require dew point analysis at ALL surfaces, including those with radiation losses. Standard comfort design (60% RH) is often too humid for buildings with high thermal mass or radiation exposure.

Quick Reference Card

Psychrometric Process Summary

ProcessChart MovementW Changes?h Changes?Application
Sensible CoolingHorizontal leftNoDecreasesDry climate AC
Sensible HeatingHorizontal rightNoIncreasesWinter heating
Cooling + DehumidDiagonal down-leftDecreasesDecreasesTypical summer AC
Adiabatic HumidAlong constant hIncreasesConstantEvaporative coolers
Isothermal HumidVertical upIncreasesIncreasesSteam humidifiers
Air MixingStraight lineWeighted avgWeighted avgEconomizer systems

Critical Dew Points to Remember

ApplicationTarget RHTypical DPColdest Surface Limit
Office50% at 24°C13°CWindows (>14°C)
Hospital OR50% at 20°C9°CSurgical instruments
Data Center45% at 24°C11°CCPU heat sinks (>12°C)
Museum50% at 21°C10°CExterior walls (>11°C)
Natatorium50% at 29°C17°CRoof structure (>18°C)

Design Checklist

Our cooling load calculations reflect real-world conditions and safety factors.

Our cooling load calculations reflect real-world conditions and safety factors.

Our engineering team developed this methodology based on internal testing and real-world validation.

Conclusion

Psychrometric calculations are fundamental to HVAC system design. By understanding the relationships between temperature, humidity, and energy, engineers can design efficient heating and cooling systems, optimize energy consumption, ensure occupant comfort, and prevent moisture problems.

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Key Takeaways

  • Calculate enthalpy using h=cpa×T+W×(hfg,0+cpv×T)h = c_{pa} \times T + W \times (h_{fg,0} + c_{pv} \times T)—enthalpy represents total heat content including sensible and latent heat
  • Determine humidity ratio from vapor pressure using W=0.622×(Pv/(PPv))W = 0.622 \times (P_v/(P - P_v))—humidity ratio is mass of water vapor per unit mass of dry air
  • Use psychrometric chart to visualize HVAC processes—chart shows relationships between all air properties enabling process design
  • Calculate total cooling load from enthalpy difference—Qtotal=m˙a×(h1h2)Q_{total} = \dot{m}_a \times (h_1 - h_2) where enthalpy difference includes both sensible and latent components
  • Maintain comfort conditions per ASHRAE 55—24°C/50% RH summer, 20-22°C/30-50% RH winter for optimal occupant comfort
  • Prevent condensation by keeping surface temperatures above dew point—dew point is temperature where condensation begins, critical for moisture control

Further Learning

References & Standards

Primary Standards

ASHRAE Fundamentals Handbook Chapter 1: Psychrometrics. Provides comprehensive psychrometric formulas, property calculations, and chart reading methods. Includes ASHRAE Equation 5 for saturation vapor pressure and all psychrometric property relationships.

ASHRAE Standard 55 Thermal Environmental Conditions for Human Occupancy. Specifies comfort conditions (24°C/50% RH summer, 20-22°C/30-50% RH winter) that guide psychrometric design targets.

ASHRAE Standard 62.1 Ventilation for Acceptable Indoor Air Quality. Specifies minimum ventilation rates (10 L/s per person for offices) that affect psychrometric calculations through ventilation heat gain.

Supporting Standards & Guidelines

ASHRAE Handbook - HVAC Systems and Equipment Chapter 4: Air Handling and Distribution. Provides guidance on air handling system design and psychrometric processes.

Further Reading

  • ASHRAE Technical Resources - American Society of Heating, Refrigerating and Air-Conditioning Engineers resources
  • McQuiston, Parker, Spitler: Heating, Ventilating, and Air Conditioning Analysis and Design - Comprehensive HVAC textbook
  • Stoecker, Jones: Refrigeration and Air Conditioning - Reference for refrigeration and air conditioning principles

Note: Standards and codes are regularly updated. Always verify you're using the current adopted edition applicable to your project's location. Consult with local authorities having jurisdiction (AHJ) for specific requirements.


Disclaimer: This guide provides general technical information based on international HVAC standards. Always verify calculations with applicable local codes and consult licensed professionals for actual installations. HVAC system design should only be performed by qualified professionals. Component ratings and specifications may vary by manufacturer.

Frequently Asked Questions

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