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7.3 · Read a psychrometric chart with stated pressure

Learn to read a psychrometric chart with stated pressure through clear examples and targeted practice.

University of Alberta MEC E 340: Applied Thermodynamics

Psychrometry and Moist Air

MEC E 340 Applied Thermodynamics — Study topic 7.3

A psychrometric chart is a graphical way to read properties of moist air at a specified total pressure. In this lesson, the system is moist air: dry air mixed with water vapour. A chart reading is meaningful only when the chart pressure matches the air pressure being studied. We use numbered states to keep readings and process changes organized. The examples use chart readings explicitly supplied in each problem; they do not claim to reproduce a particular published chart. Assume ordinary moist air, a steady flow where a control volume is used, and no chemical change. The purpose here is to read and interpret the chart, not to design equipment.

What you will learn

  • Identify the moist-air state and the chart pressure before reading values.
  • Locate a state from dry-bulb temperature and relative humidity or humidity ratio.
  • Read other moist-air properties from a chart without treating approximate chart readings as exact data.
  • Trace a simple air-conditioning process and interpret how its state properties change.

1. Start with pressure, state, and chart axes

Before reading anything, identify the moist-air state and the pressure stated for the chart. A chart prepared for one pressure should not be used as though it were a chart for another pressure: the plotted relationships depend on pressure. Write the pressure beside the state data and check that it matches the chart heading. If the pressure is not given, do not silently assume a standard value.
The most common horizontal coordinate is dry-bulb temperature, TdbT_{db}, the ordinary air temperature measured by a thermometer shielded from moisture effects. The vertical coordinate is humidity ratio, ω\omega, the mass of water vapour per mass of dry air, commonly reported in kilograms of water vapour per kilogram of dry air. A point on the chart represents one moist-air state at the chart pressure.
Curved and sloping lines on a chart provide other properties. Depending on the chart, these can include relative humidity, wet-bulb temperature, enthalpy, and specific volume. Read the chart legend first: line styles and scales can differ. Relative humidity, ϕ\phi, describes how close the water vapour is to saturation at the same temperature and pressure. It is not the same as humidity ratio. Enthalpy is often given per kilogram of dry air, so check its basis and units.
To locate a point, use two independent properties supplied by the problem, such as dry-bulb temperature and relative humidity, or dry-bulb temperature and humidity ratio. Follow the relevant lines to their intersection, then read other properties from the chart. Chart readings are approximate; report sensible precision rather than extra digits.
ω=mv/mda\omega = m_v/m_{da}
  • Match the stated pressure to the chart pressure before reading.
  • Locate the state from two independent properties.
  • Check each chart's legend, scales, and property units.

2. Read systematically and keep the property basis clear

A reliable reading routine prevents common mix-ups. First mark the known dry-bulb temperature on the horizontal axis. Next follow the line for the second known property until it meets that temperature. Mark the intersection as state 1, for example. From that same point, follow the chart's indicated lines or scales to read humidity ratio, relative humidity, wet-bulb temperature, enthalpy, or specific volume as needed.
Do not read a value from a nearby line just because it is easy to see. Trace the line from the state point and use the labeled scale. If two plotted lines are close, estimate between them and describe the result as approximate. A chart may display humidity ratio in grams per kilogram rather than kilograms per kilogram; convert units before using a value in a calculation.
For a steady-flow moist-air control volume, dry air is a useful mass reference because it passes through without being created or destroyed in ordinary heating or cooling. If there is no moisture addition or removal, the dry-air mass flow and humidity ratio are unchanged between inlet and outlet. If moisture is added or removed, humidity ratio changes; do not assume a constant value.
The steady-flow energy balance can connect enthalpy readings to heat transfer, but chart-reading questions may require only state properties. For a simple heater with negligible kinetic and potential energy changes and no moisture transfer, the heat added per unit dry-air mass is the outlet-minus-inlet enthalpy. Use consistent enthalpy units and sign conventions. Do not infer heat transfer or equipment performance when the problem does not provide the necessary flow or energy information.
q=h2−h1q = h_2-h_1
  • Trace from the state point to the scale for each requested property.
  • Keep track of whether humidity ratio and enthalpy are per unit dry-air mass.
  • Use a balance only when the process and supplied data justify it.

3. Compare states along a process

A process is a change from one moist-air state to another. Label the inlet as state 1 and the outlet as state 2. On a temperature–humidity-ratio chart, mark both points and connect them with a simple directed segment to show the change. This segment indicates the stated end states; it is not a claim that the actual path has a particular shape unless the process information supports it.
For sensible heating with no moisture transfer, dry-bulb temperature and enthalpy increase while humidity ratio remains constant. On the chart, the two points therefore have the same humidity ratio and different dry-bulb temperatures. For cooling without moisture removal, temperature falls while humidity ratio remains constant. If the problem says moisture is removed, the outlet humidity ratio is lower; use the stated outlet information or read it from the chart rather than assuming a numerical change.
A chart is a graphical property source, not a substitute for identifying the physical process. A single state point does not tell you whether air was heated, cooled, humidified, or dehumidified. The process description and the two state points together give that information. Likewise, a line that looks like a familiar process on one chart should not be assigned that process unless its end states and assumptions agree.
For any calculated change, check the direction against the process. Heating should not be reported as a temperature decrease; moisture removal should not be reported as an increase in humidity ratio. If an energy calculation is included, verify its units and whether heat is positive into or out of the air.
Δh=h2−h1\Delta h = h_2-h_1
  • Use numbered endpoints to describe a process clearly.
  • Constant humidity ratio is justified only when there is no moisture transfer.
  • Treat a chart path as schematic unless the process path is specified.

Worked example

Locate a state from dry-bulb temperature and relative humidity

A moist-air state is at 25 °C and 50% relative humidity. The stated pressure is 100 kPa, and the supplied chart is also for 100 kPa. A chart-reading exercise supplies the following readings at the intersection: humidity ratio 0.010 kg water vapour/kg dry air, enthalpy 50 kJ/kg dry air, and wet-bulb temperature 18 °C. Identify the state properties requested and explain the chart-reading sequence.
  1. Check pressure and known properties
    The stated pressure matches the supplied chart. Use dry-bulb temperature and relative humidity as the two known properties that locate state 1.
    P=100 kPa,Tdb,1=25 ∘C,ϕ1=50%P=100\ \mathrm{kPa},\quad T_{db,1}=25\ ^\circ\mathrm{C},\quad \phi_1=50\%
  2. Locate the intersection
    Find 25 °C on the dry-bulb axis and follow the 50% relative-humidity line to its intersection. The problem supplies the readings at that point, so no other chart values need to be assumed.
    ω1=0.010 kg/kgda,h1=50 kJ/kgda,Twb,1=18 ∘C\omega_1=0.010\ \mathrm{kg/kg_{da}},\quad h_1=50\ \mathrm{kJ/kg_{da}},\quad T_{wb,1}=18\ ^\circ\mathrm{C}
  3. Report with the correct basis
    The humidity ratio and enthalpy are stated per kilogram of dry air. Keep those bases when using the values later; these are chart readings, not extra-precision measurements.
Answer: At 100 kPa, state 1 has Tdb,1=25 ∘CT_{db,1}=25\ ^\circ\mathrm{C}, ϕ1=50%\phi_1=50\%, ω1=0.010 kg/kgda\omega_1=0.010\ \mathrm{kg/kg_{da}}, h1=50 kJ/kgdah_1=50\ \mathrm{kJ/kg_{da}}, and Twb,1=18 ∘CT_{wb,1}=18\ ^\circ\mathrm{C}.
Check: The pressure used to locate the state is the same as the chart pressure. Humidity ratio and enthalpy retain their dry-air basis.

Worked example

Read an outlet state after sensible heating

Air at a stated pressure of 100 kPa enters a simple heater at state 1 and leaves at state 2. There is no moisture transfer. A supplied 100 kPa chart gives state 1 as Tdb,1=20 ∘CT_{db,1}=20\ ^\circ\mathrm{C}, ω1=0.008 kg/kgda\omega_1=0.008\ \mathrm{kg/kg_{da}}, and h1=40 kJ/kgdah_1=40\ \mathrm{kJ/kg_{da}}. The outlet dry-bulb temperature is 35 °C. The chart reading at 35 °C and the same humidity ratio gives h2=55 kJ/kgdah_2=55\ \mathrm{kJ/kg_{da}}. Find the enthalpy increase per kilogram of dry air.
Sensible heating at 100 kPa
Sensible heating at 100 kPaDry-bulb temperature, THumidity ratio, ωSensible heating120 °C235 °CSchematic · not to scale

Moist air at the stated 100 kPa pressure; schematic temperature–humidity-ratio plot, not to scale. No moisture transfer is assumed.

  1. Set the process assumptions
    The control volume is the heater, with moist air flowing from state 1 to state 2. No moisture transfer means the humidity ratio is unchanged. The stated pressure matches the supplied chart.
    ω2=ω1=0.008 kg/kgda\omega_2=\omega_1=0.008\ \mathrm{kg/kg_{da}}
  2. Read the outlet enthalpy
    At the outlet dry-bulb temperature, follow the constant-humidity-ratio line from state 1 to state 2 and use the supplied chart reading. The value is given, so it is not necessary to estimate it independently.
    h2=55 kJ/kgdah_2=55\ \mathrm{kJ/kg_{da}}
  3. Calculate the enthalpy change
    Subtract inlet enthalpy from outlet enthalpy. The result is positive, consistent with air being heated. It is an enthalpy increase, not a heat-transfer rate; a rate would also require dry-air mass flow and applicable balance assumptions.
    Δh=55−40=15 kJ/kgda\Delta h=55-40=15\ \mathrm{kJ/kg_{da}}
Answer: The enthalpy increase is 15 kJ/kgda15\ \mathrm{kJ/kg_{da}}. With negligible kinetic and potential energy changes and no shaft work, this also equals heat added per unit dry-air mass for the stated heater model.
Check: The humidity ratio stays fixed because there is no moisture transfer, and both temperature and enthalpy increase during heating.

Worked example

Interpret a stated dehumidification change

A moist-air stream is cooled and dehumidified from state 1 to state 2 at the stated chart pressure of 95 kPa. The supplied 95 kPa chart readings are: state 1, Tdb,1=28 ∘CT_{db,1}=28\ ^\circ\mathrm{C}, ω1=0.014 kg/kgda\omega_1=0.014\ \mathrm{kg/kg_{da}}, h1=64 kJ/kgdah_1=64\ \mathrm{kJ/kg_{da}}; state 2, Tdb,2=16 ∘CT_{db,2}=16\ ^\circ\mathrm{C}, ω2=0.009 kg/kgda\omega_2=0.009\ \mathrm{kg/kg_{da}}, h2=39 kJ/kgdah_2=39\ \mathrm{kJ/kg_{da}}. State the changes in humidity ratio and enthalpy per kilogram of dry air. Do not infer equipment performance.
Cooling and dehumidification
Cooling and dehumidificationDry-bulb temperature, THumidity ratio, ωCooling and dehumidification1Inlet2OutletSchematic · not to scale

Moist air at the stated 95 kPa pressure; schematic temperature–humidity-ratio plot, not to scale. The endpoint readings are supplied; the connecting segment does not assert a detailed path.

  1. Confirm chart pressure and states
    Use the 95 kPa chart for both endpoints. The problem supplies each endpoint's chart readings, so the task is to compare state 1 and state 2 rather than invent a process curve.
    P=95 kPaP=95\ \mathrm{kPa}
  2. Find the humidity-ratio change
    Subtract the inlet humidity ratio from the outlet humidity ratio. A negative result indicates less water vapour per kilogram of dry air at the outlet, consistent with the stated dehumidification.
    Δω=0.009−0.014=−0.005 kg/kgda\Delta\omega=0.009-0.014=-0.005\ \mathrm{kg/kg_{da}}
  3. Find the enthalpy change
    Use the same outlet-minus-inlet convention for enthalpy. The decrease is consistent with the lower outlet temperature and the supplied lower outlet enthalpy. This comparison alone does not determine a heat-transfer rate or device performance.
    Δh=39−64=−25 kJ/kgda\Delta h=39-64=-25\ \mathrm{kJ/kg_{da}}
Answer: The humidity ratio decreases by 0.005 kg/kgda0.005\ \mathrm{kg/kg_{da}}, and enthalpy decreases by 25 kJ/kgda25\ \mathrm{kJ/kg_{da}}.
Check: Both changes are negative from inlet to outlet, matching the stated cooling and dehumidification. No mass flow or energy-transfer rate was supplied, so neither is calculated.

Common mistakes and how to avoid them

Reading a chart without checking its pressure.
Correction: Match the chart pressure to the pressure stated for the moist air before locating or reading a state.
Treating relative humidity and humidity ratio as interchangeable.
Correction: They are different properties. Use the chart's lines and labels to identify each one.
Assuming humidity ratio stays constant in every heating or cooling process.
Correction: It stays constant only when there is no moisture transfer. Follow the process description.
Reporting chart readings with unjustified precision or the wrong mass basis.
Correction: Use sensible approximate precision and preserve units such as kilograms per kilogram of dry air.

Lesson summary

  • A psychrometric chart applies at a specified pressure; check that pressure first.
  • Use two independent known properties to locate a moist-air state, then read requested properties from the chart legend and scales.
  • Keep dry-air mass bases and units explicit. Chart values are approximate.
  • Compare numbered inlet and outlet states to identify process changes; use constant humidity ratio only when moisture transfer is absent.

Check your understanding

Question 1

A problem states 90 kPa, but the only available chart is labelled 100 kPa. What is the sound first step?
  1. Use the 100 kPa chart because pressure does not affect the chart.
  2. Find a chart or supplied data for 90 kPa before making the reading.
  3. Change the stated air temperature until it matches the chart.
  4. Assume the humidity ratio is zero.
Show answer and explanation
Find a chart or supplied data for 90 kPa before making the reading.
Psychrometric chart relationships depend on pressure, so the pressure must match the stated condition.

Question 2

In a heater with no moisture transfer, which property is unchanged between inlet and outlet?
  1. Dry-bulb temperature
  2. Enthalpy
  3. Humidity ratio
  4. Relative humidity
Show answer and explanation
Humidity ratio
With no moisture transfer, the water-vapour mass per dry-air mass remains unchanged. Temperature and enthalpy can rise.

Question 3

A chart gives inlet enthalpy 47 kJ/kg dry air and outlet enthalpy 59 kJ/kg dry air. What is the enthalpy change using outlet minus inlet?
  1. −12 kJ/kgda-12\ \mathrm{kJ/kg_{da}}
  2. 12 kJ/kgda12\ \mathrm{kJ/kg_{da}}
  3. 106 kJ/kgda106\ \mathrm{kJ/kg_{da}}
  4. 12 kJ12\ \mathrm{kJ}
Show answer and explanation
12 kJ/kgda12\ \mathrm{kJ/kg_{da}}
The change is 59−47=12 kJ/kgda59-47=12\ \mathrm{kJ/kg_{da}}. The units remain per kilogram of dry air.

Key terms

Psychrometric chart
A graphical set of moist-air property relationships for a stated pressure.
Dry-bulb temperature
The ordinary air temperature represented on the chart's temperature axis.
Humidity ratio
Water-vapour mass divided by dry-air mass in a moist-air mixture.
Relative humidity
A measure of how close the water vapour is to saturation at the same temperature and pressure.
Chart reading
An approximate property value obtained by locating a state and following the chart's labeled lines or scales.

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