Psychrometric chart practice problems

Nine original problems modeled on the chart questions in HVAC licensing exams. Plot them on a printed chart, read the values, then enter your answers to see whether they fall within chart-reading tolerance. Each solution follows the hand-plotting order.

All problems use standard pressure (101.325 kPa), and airflows are dry-air mass flows (kg/h). Enter a number and press Enter or Check to grade each box. Answers are computed with this site's engine (ASHRAE equations). Only the volume airflow in problem 5 is in m³/h. Tolerance (accepted range)

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1Reading a state point from two values

Air has the measured dry-bulb and wet-bulb temperatures below. Find the state point on the chart and read the other properties.

Given

Dry-bulb temperature30 °C
Wet-bulb temperature22 °C

Find

Use a point or a comma as the decimal mark, and no thousands separators (4924).

  1. %
  2. g/kg(DA)
  3. kJ/kg(DA)
  4. °C
Show solution

Plotting steps

  1. On the bottom axis, find the vertical 30 °C dry-bulb line.
  2. Find 22 °C on the saturation curve (100 % RH) and follow the wet-bulb line down to the right. It runs almost parallel to the enthalpy lines, so check that you are on the wet-bulb scale.
  3. The intersection is the state point. Split the gap between the neighboring RH curves by eye and read 50.0 %.
  4. Go horizontally right to the scale and read the humidity ratio, 13.30 g/kg(DA). Following the same line left to the saturation curve gives the dew point, 18.4 °C.
  5. Align a set square with the enthalpy lines, slide it to the state point, and read 64.2 kJ/kg(DA) on the enthalpy scale.

Intermediate values in the steps are rounded, so recomputing from them can change the last digit. Grading uses the unrounded values.

Answers and tolerance

ItemAnswerAccepted range
Relative humidity50.0 %±3 % RH
Humidity ratio13.30 g/kg(DA)±0.3 g/kg(DA)
Enthalpy64.2 kJ/kg(DA)±1.0 kJ/kg(DA)
Dew point18.4 °C±0.5 °C

Why a textbook answer may differ: Textbook answers are read by eye from the chart scales, so the last digit can differ. The values here are calculated with ASHRAE equations (Hyland-Wexler saturation pressure).

Open on the chartThe calculator opens with this problem's points and processes drawn.

2Adiabatic mixing of outdoor and return air

Outdoor air OA and return air RA from the room are mixed before the air handler. Find the state of the mixed air MA.

Given

Outdoor air OA32 °C dry-bulb, 65 % RH
Room (return) air RA27 °C dry-bulb, 50 % RH
Outdoor airflow2,500 kg/h (dry air)
Return airflow7,500 kg/h (dry air)

Find

Use a point or a comma as the decimal mark, and no thousands separators (4924).

  1. °C
  2. g/kg(DA)
  3. kJ/kg(DA)
Show solution

Plotting steps

  1. Plot OA (32 °C, 65 %) and RA (27 °C, 50 %).
  2. Join the two points with a straight line. The adiabatic mixing point always lies on it.
  3. The flow ratio is OA : RA = 1 : 3. The mix lies closer to the larger flow, so mark MA at 25 % of the line length from RA toward OA.
  4. Check with dry-bulb: (2,500 × 32 + 7,500 × 27) / 10,000 = 28.25 °C. This is the same as dividing the line by the flow ratio.
  5. From MA read the humidity ratio 13.25 g/kg(DA) horizontally and the enthalpy 62.3 kJ/kg(DA) along the enthalpy lines. The engine gives an MA dry-bulb of 28.26 °C.

Intermediate values in the steps are rounded, so recomputing from them can change the last digit. Grading uses the unrounded values.

Answers and tolerance

ItemAnswerAccepted range
Mixed air MA dry-bulb28.3 °C±0.5 °C
Mixed air MA humidity ratio13.25 g/kg(DA)±0.3 g/kg(DA)
Mixed air MA enthalpy62.3 kJ/kg(DA)±1.0 kJ/kg(DA)

Why a textbook answer may differ: Textbooks often take the mixed temperature as the weighted average 28.25 °C. Here it is solved from mass-averaged enthalpy and humidity ratio as 28.26 °C; the gap is smaller than chart-reading error.

Open on the chartThe calculator opens with this problem's points and processes drawn.

3Sensible heating and heating rate

Air is warmed by a heating coil that neither adds nor removes moisture. Find the outlet state and the heating rate.

Given

Inlet air10 °C dry-bulb, 70 % RH
Outlet dry-bulb32 °C
Airflow4,000 kg/h (dry air)

Find

Use a point or a comma as the decimal mark, and no thousands separators (4924).

  1. %
  2. kJ/kg(DA)
  3. kW
Show solution

Plotting steps

  1. Plot the inlet air ① (10 °C, 70 %).
  2. Heating keeps the humidity ratio constant, so draw a horizontal line right from ①; where it meets the 32 °C dry-bulb line is the outlet ②.
  3. Read ② as 18.1 % RH and 45.82 kJ/kg(DA). The enthalpy of ① is 23.47 kJ/kg(DA).
  4. Heating rate = airflow × enthalpy difference = 4,000 / 3,600 × (45.82 − 23.47) = 24.8 kW.

Intermediate values in the steps are rounded, so recomputing from them can change the last digit. Grading uses the unrounded values.

Answers and tolerance

ItemAnswerAccepted range
Outlet RH18.1 %±3 % RH
Outlet enthalpy45.8 kJ/kg(DA)±1.0 kJ/kg(DA)
Heating rate24.8 kW±2.2 kW (includes chart-reading error: Enthalpy ±1.0 kJ/kg(DA))

Why a textbook answer may differ: The textbook shortcut, sensible heat = 1.01 × airflow × temperature difference, gives 24.7 kW, which differs by 0.6 % from the 24.8 kW here, computed from the enthalpy difference. The shortcut uses an approximate specific heat; this page uses the ASHRAE enthalpy equation.

Open on the chartThe calculator opens with this problem's points and processes drawn.

4Cooling and dehumidifying: cooling rate and condensate

The air states at a cooling coil's inlet and outlet are given. Read the enthalpies and find the cooling rate and the condensate flow.

Given

Inlet air28 °C dry-bulb, 60 % RH
Outlet air14 °C dry-bulb, 92 % RH
Airflow6,000 kg/h (dry air)

Find

Use a point or a comma as the decimal mark, and no thousands separators (4924).

  1. kJ/kg(DA)
  2. kJ/kg(DA)
  3. kW
  4. kg/h
Show solution

Plotting steps

  1. Plot the coil inlet ① (28 °C, 60 %) and outlet ② (14 °C, 92 %).
  2. Join ① and ②. Cooling with dehumidification runs down to the left: temperature and humidity ratio both fall.
  3. Enthalpy: ① 64.55, ② 37.23 kJ/kg(DA). Humidity ratio: ① 14.25, ② 9.16 g/kg(DA).
  4. Cooling rate = 6,000 / 3,600 × (64.55 − 37.23) = 45.5 kW.
  5. Condensate = 6,000 × (14.25 − 9.16) / 1,000 = 30.5 kg/h.

Intermediate values in the steps are rounded, so recomputing from them can change the last digit. Grading uses the unrounded values.

Answers and tolerance

ItemAnswerAccepted range
Inlet enthalpy64.5 kJ/kg(DA)±1.0 kJ/kg(DA)
Outlet enthalpy37.2 kJ/kg(DA)±1.0 kJ/kg(DA)
Cooling rate45.5 kW±3.3 kW (includes chart-reading error: Enthalpy ±1.0 kJ/kg(DA))
Condensate30.5 kg/h±3.6 kg/h (includes chart-reading error: Humidity ratio ±0.3 g/kg(DA))

Why a textbook answer may differ: Textbook answers are read by eye from the chart scales, so the last digit can differ. The values here are calculated with ASHRAE equations (Hyland-Wexler saturation pressure).

Open on the chartThe calculator opens with this problem's points and processes drawn.

5Supply point and airflow from the SHF line

Room sensible and latent loads and the supply temperature difference are given. Find the sensible heat ratio SHF, the humidity ratio of supply air SA, and the supply airflow.

Given

Room (return) air RA26 °C dry-bulb, 50 % RH
Room sensible load14 kW
Room latent load3.5 kW
Supply temperature difference10 K

Find

Use a point or a comma as the decimal mark, and no thousands separators (4924).

  1.  
  2. g/kg(DA)
  3. kg/h
  4. m³/h
Show solution

Plotting steps

  1. SHF = sensible / (sensible + latent) = 14 / (14 + 3.5) = 0.80.
  2. Plot the room point RA (26 °C, 50 %).
  3. Join the reference point of the SHF protractor at the chart edge to the 0.80 mark, then slide that line parallel to RA with two set squares. This is the room SHF line.
  4. With a 10 K supply difference, the supply dry-bulb is 26 − 10 = 16 °C. Where the 16 °C line meets the SHF line is the supply point SA; read 9.49 g/kg(DA) and 40.12 kJ/kg(DA).
  5. Supply airflow = total load × 3,600 / (h_RA − h_SA) = 17.5 × 3,600 / (52.91 − 40.12) = 4,924 kg/h.
  6. For volume flow, multiply by the SA specific volume 0.8316 m³/kg: 4,924 × 0.8316 = 4,095 m³/h.

Intermediate values in the steps are rounded, so recomputing from them can change the last digit. Grading uses the unrounded values.

Answers and tolerance

ItemAnswerAccepted range
Sensible heat ratio SHF0.80±0.02
Supply air SA humidity ratio9.49 g/kg(DA)±0.3 g/kg(DA)
Supply airflow (mass)4,924 kg/h±835 kg/h (includes chart-reading error: Enthalpy ±1.0 kJ/kg(DA))
Supply airflow (volume, at SA)4,095 m³/h±694 m³/h (includes chart-reading error: Enthalpy ±1.0 kJ/kg(DA))

Why a textbook answer may differ: The textbook shortcut, airflow = sensible load / (1.01 × supply temperature difference), gives 4,990 kg/h, 1.4 % away from the 4,924 kg/h here. The shortcut uses an approximate specific heat; this page solves it with the enthalpy difference (ASHRAE equations).

Open on the chartThe calculator opens with this problem's points and processes drawn.

6Full cooling cycle: mixing, coil, reheat, supply

Outdoor and return air are mixed, cooled by a coil, then reheated to supply conditions. Find the supply airflow, the mixed-air enthalpy, the cooling coil load and the reheat.

Given

Outdoor air OA33 °C dry-bulb, 60 % RH
Room (return) air RA26 °C dry-bulb, 50 % RH
Outdoor air fraction (by mass)25 %
Room sensible load20 kW
Room latent load7 kW
Supply temperature difference9 K
Coil outlet RH90 %

Find

Use a point or a comma as the decimal mark, and no thousands separators (4924).

  1. kg/h
  2. kJ/kg(DA)
  3. kW
  4. kW
Show solution

Plotting steps

  1. Plot OA (33 °C, 60 %) and RA (26 °C, 50 %), then draw the SHF = 20 / (20 + 7) = 0.741 line through RA.
  2. Supply air SA is the point on the SHF line at 26 − 9 = 17 °C (enthalpy 40.48 kJ/kg(DA), humidity ratio 9.23 g/kg(DA)).
  3. Supply airflow G = 27 × 3,600 / (52.91 − 40.48) = 7,819 kg/h.
  4. Join OA and RA and mark the mixed air MA at 25 % of the line from RA (27.8 °C, 60.23 kJ/kg(DA)).
  5. Draw a horizontal line left from SA (9.23 g/kg(DA)); where it meets the 90 % RH curve is the coil outlet CO (14.5 °C, 37.88 kJ/kg(DA)). MA → CO is the coil and the horizontal CO → SA is the reheat.
  6. Cooling coil = 7,819 × (60.23 − 37.88) / 3,600 = 48.5 kW; reheat = 7,819 × (40.48 − 37.88) / 3,600 = 5.7 kW.

Intermediate values in the steps are rounded, so recomputing from them can change the last digit. Grading uses the unrounded values.

Answers and tolerance

ItemAnswerAccepted range
Supply airflow (mass)7,819 kg/h±1,368 kg/h (includes chart-reading error: Enthalpy ±1.0 kJ/kg(DA))
Mixed air MA enthalpy60.2 kJ/kg(DA)±1.0 kJ/kg(DA)
Cooling coil load48.5 kW±12.1 kW (includes chart-reading error: Enthalpy ±1.0 kJ/kg(DA))
Reheat5.7 kW±1.4 kW (includes chart-reading error: Enthalpy ±1.0 kJ/kg(DA))

Why a textbook answer may differ: The textbook shortcut, airflow = sensible load / (1.01 × supply temperature difference), gives 7,921 kg/h, 1.3 % away from the 7,819 kg/h here. The shortcut uses an approximate specific heat; this page solves it with the enthalpy difference (ASHRAE equations).

Open on the chartThe calculator opens with this problem's points and processes drawn.

7Winter heating and spray humidification

Mixed air MA is heated by a coil and then humidified with a recirculated water spray to reach supply air SA. Find the temperature after heating, the heating rate and the humidification rate.

Given

Heating coil inlet MA10 °C dry-bulb, 50 % RH
Supply air SA27 °C dry-bulb, 30 % RH
Airflow5,000 kg/h (dry air)
HumidifierRecirculated water spray (adiabatic)

Find

Use a point or a comma as the decimal mark, and no thousands separators (4924).

  1. °C
  2. kW
  3. kg/h
Show solution

Plotting steps

  1. Plot MA (10 °C, 50 %) and SA (27 °C, 30 %).
  2. A recirculated spray humidifies adiabatically, so the wet-bulb temperature stays constant. Draw the wet-bulb line through SA (15.8 °C) down to the right.
  3. Heating keeps the humidity ratio, so draw a horizontal line right from MA (3.79 g/kg(DA)). Where it meets the wet-bulb line of SA is the heated air HC (34.0 °C dry-bulb).
  4. Heating rate = 5,000 / 3,600 × (43.91 − 19.61) = 33.7 kW.
  5. Humidification = 5,000 × (6.64 − 3.79) / 1,000 = 14.2 kg/h.

Intermediate values in the steps are rounded, so recomputing from them can change the last digit. Grading uses the unrounded values.

Answers and tolerance

ItemAnswerAccepted range
Dry-bulb after heating, HC34.0 °C±2.1 °C (includes chart-reading error: Temperature ±0.5 °C, Humidity ratio ±0.3 g/kg(DA))
Heating rate33.7 kW±2.8 kW (includes chart-reading error: Enthalpy ±1.0 kJ/kg(DA))
Humidification14.2 kg/h±3.0 kg/h (includes chart-reading error: Humidity ratio ±0.3 g/kg(DA))

Why a textbook answer may differ: The textbook shortcut, sensible heat = 1.01 × airflow × temperature difference, gives 33.6 kW, which differs by 0.3 % from the 33.7 kW here, computed from the enthalpy difference. The shortcut uses an approximate specific heat; this page uses the ASHRAE enthalpy equation.

Open on the chartThe calculator opens with this problem's points and processes drawn.

8Bypass factor BF and apparatus dew point ADP

The cooling coil inlet MA and outlet CO are given. Extend the coil line to find the ADP, then find BF and the cooling rate.

Given

Inlet air27 °C dry-bulb, 50 % RH
Outlet air14 °C dry-bulb, 88 % RH
Airflow8,000 kg/h (dry air)

Find

Use a point or a comma as the decimal mark, and no thousands separators (4924).

  1. °C
  2.  
  3. kW
Show solution

Plotting steps

  1. Plot the coil inlet MA (27 °C, 50 %) and outlet CO (14 °C, 88 %).
  2. Extend the line from MA through CO down to the left. It meets the saturation curve at the ADP; read 11.1 °C.
  3. BF = (t_CO − t_ADP) / (t_MA − t_ADP) = (14 − 11.1) / (27 − 11.1) = 0.18. On the chart it is the length ratio CO~ADP to MA~ADP.
  4. Cooling rate = 8,000 / 3,600 × (55.59 − 36.21) = 43.1 kW.

Intermediate values in the steps are rounded, so recomputing from them can change the last digit. Grading uses the unrounded values.

Answers and tolerance

ItemAnswerAccepted range
Apparatus dew point ADP11.1 °C±0.5 °C
Bypass factor BF0.18±0.027 (includes chart-reading error: Temperature ±0.5 °C)
Cooling rate43.1 kW±4.4 kW (includes chart-reading error: Enthalpy ±1.0 kJ/kg(DA))

Why a textbook answer may differ: Textbook answers are read by eye from the chart scales, so the last digit can differ. The values here are calculated with ASHRAE equations (Hyland-Wexler saturation pressure).

Open on the chartThe calculator opens with this problem's points and processes drawn.

9Winter heating and steam humidification

Mixed air MA is heated by a coil and then humidified by injecting steam to reach supply air SA. Find the temperature after heating, the heating rate and the humidification rate.

Given

Heating coil inlet MA14 °C dry-bulb, 40 % RH
Supply air SA30 °C dry-bulb, 28 % RH
Airflow6,000 kg/h (dry air)
HumidifierSteam humidifier (saturated steam at 100 °C, enthalpy 2,676 kJ/kg)

Find

Use a point or a comma as the decimal mark, and no thousands separators (4924).

  1. °C
  2. kW
  3. kg/h
Show solution

Plotting steps

  1. Plot MA (14 °C, 40 %) and SA (30 °C, 28 %).
  2. The slope of a steam humidification line is the heat-moisture ratio u = Δh / ΔW, equal to the enthalpy of the supplied steam (u = 2,676 kJ/kg). On the outer u scale of the protractor, join the reference point to the 2,676 mark and slide this line to SA with two set squares. It is almost vertical and leans very slightly to the right as it rises.
  3. Heating keeps the humidity ratio, so draw a horizontal line right from MA (3.95 g/kg(DA)). Where it meets the u line through SA is the heated air HC (29.6 °C dry-bulb).
  4. Heating rate = 6,000 / 3,600 × (39.87 − 24.07) = 26.3 kW.
  5. Humidification = 6,000 × (7.38 − 3.95) / 1,000 = 20.6 kg/h.
  6. Check: u = (h_SA − h_HC) / (W_SA − W_HC) = (49.06 − 39.87) / (7.38 − 3.95) × 1,000 = 2,676 kJ/kg; this equals the steam enthalpy.

Intermediate values in the steps are rounded, so recomputing from them can change the last digit. Grading uses the unrounded values.

Answers and tolerance

ItemAnswerAccepted range
Dry-bulb after heating, HC29.6 °C±0.5 °C
Heating rate26.3 kW±3.3 kW (includes chart-reading error: Enthalpy ±1.0 kJ/kg(DA))
Humidification20.6 kg/h±3.6 kg/h (includes chart-reading error: Humidity ratio ±0.3 g/kg(DA))

Why a textbook answer may differ: Textbooks often treat steam humidification as a vertical line at constant dry-bulb, so the air after heating is taken at the supply temperature 30.0 °C and the heating rate comes to 27.0 kW. Here the supplied steam is saturated at 100 °C (enthalpy 2,676 kJ/kg) and the heat it carries is included, so the humidification line leans slightly along u = 2,676 kJ/kg: the air after heating is 29.6 °C and the heating rate is 26.3 kW. This is the same definition as the calculator's winter heating and humidification.

Open on the chartThe calculator opens with this problem's points and processes drawn.

Tolerance (accepted range)

Readings between scale lines vary a little from person to person, even on the same chart. An answer counts as correct when its difference from the reference is within the range below, boundary included.

QuantityToleranceReason
Temperature (dry-bulb, dew point, ADP)±0.5 °CHalf of a 1 °C division
Humidity ratio±0.3 g/kg(DA)About a third of a 1 g/kg division
Enthalpy±1.0 kJ/kg(DA)Error from sliding a set square to the slanted enthalpy scale
Relative humidity±3 % RHError from splitting the gap between 10 % curves by eye
Ratios (SHF, BF)±0.02 · calculated from readings, so widened to the propagated reading error if larger (up to ±25 % of the answer)Misreading the ADP by 0.5 °C shifts BF by about 0.02
Heat rate (kW)±3 % of the answer · calculated from readings, so widened to the propagated reading error if larger (up to ±25 % of the answer)Enthalpy-difference reading error carries straight into the result
Airflow (kg/h, m³/h)±3 % of the answer · calculated from readings, so widened to the propagated reading error if larger (up to ±25 % of the answer)Divided by an enthalpy difference, so the same as heat rate
Condensate and humidification (kg/h)±5 % of the answer · calculated from readings, so widened to the propagated reading error if larger (up to ±25 % of the answer)Based on a humidity-ratio difference of a few g/kg, so the relative error is larger

Frequently asked questions

My chart reading is slightly off from the answer. Is it wrong?

Usually not. Readings between scale lines differ a little from person to person. This page accepts values read directly from the chart within ±0.5 °C for temperature, ±0.3 g/kg for humidity ratio, ±1.0 kJ/kg for enthalpy and ±3 % RH (percentage points) for relative humidity. For values calculated from readings, such as heat rate, airflow and condensate, each problem uses how much those reading errors grow through the formula (at least ±3 % and at most ±25 % of the answer). Real exam grading rules are not published, so show your plotting steps and equations.

Why do the answers here differ from my textbook?

Textbooks use values read by eye from the chart, or shortcuts such as sensible heat = 1.01 × airflow × temperature difference. This page calculates with ASHRAE Handbook equations, so decimals can differ. Each problem states the size of that difference under "Why a textbook answer may differ".

How do I draw the SHF line on the chart?

On the SHF protractor at the chart edge, join the reference point to the SHF value, then slide that line parallel to the room point using two set squares. The supply point is where this line reaches the room temperature minus the supply temperature difference. Use "Open on the chart" in problems 5 and 6 to see the SHF line drawn by the calculator.

Where can I get a blank chart to practice on?

The blank chart page (https://psychart.kr/en/download/) has free A4 and A3 printable charts. "Print problems only" on this page prints the problems without solutions.

Last updated 2026-10-18.