Psychrometric chart exam-style problems: five original practice problems

How this article works

Exam problems are solved by reading points on the chart, or by applying shortcut formulas to enthalpy and humidity ratio. Shortcut formulas use convenient values such as an air density of 1.2 kg/m³ and a specific heat of 1.005 kJ/kg·K, so the answers differ from the exact calculation by a few percent. All problems below were written by this site; the conditions are set as a problem and the solution is shown step by step. The answers are engine values, with the shortcut-formula values in brackets or beside them. In an exam, follow the approximation the problem specifies and pick the answer closest to the chart reading.

Problem 1: mixed state

Problem. Outdoor air at 33 °C / 60 % (25 %) is mixed with return air at 26 °C / 50 % (75 %) by dry air mass. Find the dry-bulb temperature, humidity ratio, enthalpy and relative humidity of the mixture.

Solution. Outdoor: W = 19.11 g/kg, h = 82.17 kJ/kg. Return: W = 10.50 g/kg, h = 52.91 kJ/kg. Mixing is a mass-weighted average of enthalpy and humidity ratio.

W_M = 0.25 × 19.11 + 0.75 × 10.50 = 12.65 g/kg
h_M = 0.25 × 82.17 + 0.75 × 52.91 = 60.23 kJ/kg
From h = 1.006 t + W (2501 + 1.86 t), t_M = 27.77 °C and the relative humidity is 54.1 %.

Shortcut. Mixing dry-bulb linearly gives 27.75 °C, 0.02 K from the engine value. Mixing enthalpy and humidity ratio first and then reading the dry-bulb off the chart is the accurate way. The mixed point lies on the straight line joining outdoor and return points, 75 % of the way from the outdoor point.

Problem 2: cooling coil load

Problem. The mixed air of problem 1, 6,000 CMH at the coil inlet state, is cooled by a cooling coil to 14 °C / 90 %. Find the cooling load, its sensible and latent parts, and the condensate.

Solution. The mixed air has a specific volume of 0.8698 m³/kg, so the dry air mass flow is ṁ = 6,000 / 3600 / 0.8698 = 1.916 kg/s. At the coil outlet h = 36.72 kJ/kg and W = 8.96 g/kg.

Total load Q = ṁ (h_M − h_C) = 1.916 × (60.23 − 36.72) = 45.0 kW
Sensible Q_s = ṁ (1.006 + 1.86 W_M) (t_M − t_C) = 27.2 kW and latent Q_l = Q − Q_s = 17.9 kW
Condensate ṁ (W_M − W_C) = 1.916 × 0.00369 × 3600 = 25.5 kg/h.

The coil's sensible ratio is Q_s / Q = 0.60, so about 40 % of the heat goes into dehumidification. Split sensible and latent the way the problem specifies.

Problem 3: sensible heat ratio and supply airflow

Problem. A room at 26 °C / 50 % has a sensible load of 12 kW and a latent load of 3 kW. With a supply temperature difference of 10 K, find the sensible heat ratio, the supply temperature, the supply airflow (CMH) and the supply humidity ratio.

Solution. SHF = 12 / (12 + 3) = 0.80. Supply temperature 26 − 10 = 16 °C. With the room's W = 10.50 g/kg the moist air specific heat is c_p = 1.0255 kJ/kg·K and ṁ = 12 / (1.0255 × 10) = 1.170 kg/s. With a supply specific volume of 0.8316 m³/kg the airflow is 3,503 CMH. The supply humidity ratio is W_S = W_R − 3 / (ṁ × (2501 + 1.86 × 26)) = 9.49 g/kg.

Shortcut. With density 1.2 kg/m³ and specific heat 1.005 kJ/kg·K, Q_s = 1.2 × V × 1.005 × ΔT gives V = 12 / (1.2 × 1.005 × 10) × 3600 = 3,582 CMH, which is 2.3 % from the engine value. On the chart, draw the sensible-heat-ratio line of 0.80 through the room point and read the supply point at the supply temperature.

Problem 4: bypass factor

Problem. The air enters a cooling coil at 28 °C and leaves at 16 °C, and the mean coil surface temperature (apparatus dew point, ADP) is 10 °C. Find the bypass factor and the contact factor.

Solution. The bypass factor is the fraction of air that passes the coil without touching the surface, BF = (t_out − t_ADP) / (t_in − t_ADP) = (16 − 10) / (28 − 10) = 0.333. The contact factor is 1 − BF = 0.667. Inlet, outlet and ADP lie on one straight line on the chart, and the outlet is BF of the way from the ADP toward the inlet. Problems that find the outlet temperature from BF use the same equation.

Problem 5: heating and steam humidification

Problem. A winter room is held at 22 °C / 40 %. Outdoor air at 0 °C / 50 % is mixed at 25 %, and 5,000 CMH (at room state) is heated and then humidified with steam to be supplied at 28 °C with the room's humidity ratio. Find the mixed state, the heating load and the humidification rate.

Solution. Room W = 6.56 g/kg, specific volume 0.8449 m³/kg, ṁ = 5,000 / 3600 / 0.8449 = 1.644 kg/s. Outdoor W = 1.88 g/kg, so mixed W_M = 5.39 g/kg, mixed dry-bulb 16.54 °C and h_M = 30.29 kJ/kg.

Humidification = ṁ (W_S − W_M) = 1.644 × 0.00117 × 3600 = 6.9 kg/h. Allowing for the heat the steam brings, the air before humidification has h = h_S − ΔW × 2676 = 44.92 − 0.00117 × 2676 = 41.79 kJ/kg and dry-bulb 27.86 °C. Heating Q = ṁ (h_H − h_M) = 1.644 × (41.79 − 30.29) = 18.9 kW.

With adiabatic spray instead of steam the dry-bulb falls along a constant wet-bulb line, so the air must be heated higher. Textbooks that ignore the steam's heat take the heated temperature as the supply temperature, so their heating load differs a little; follow the method the problem specifies. A fuller comparison is in Winter office humidification load.

Easy things to miss in an exam

• Mixing is by mass. The ratio by airflow (CMH) differs from the ratio by mass when densities differ. Check the basis the problem gives.
• Load is an enthalpy difference. Coil load is airflow times the enthalpy difference, not a temperature difference. A temperature difference alone gives only the sensible part.
• Sensible heat ratio is sensible over total. Not latent over sensible. It is 0.80 in problem 3.
• BF uses the inlet and the ADP. It is (t_out − t_ADP) / (t_in − t_ADP), and the denominator is the difference between inlet and ADP.
• The state behind the airflow. The problem says which state the CMH refers to. Problem 2 here uses the coil inlet; problem 5 uses the room state.
• Units. Humidity ratio appears in g/kg and kg/kg; divide by 1000 when converting g/kg to kg/kg.

Try this calculation on the site

Open problem 1 mixing on the chartOpen the problem 3 ratio line on the chart

The problem 1 link plots outdoor, return and the 25 % mixed point so you can see the mixed dry-bulb and humidity. The problem 3 link draws the sensible heat ratio line of 0.80 through the room point so you can read the supply point at the supply temperature. For coil calculation and reheat see Summer cooling coil selection and reheat, for outdoor-air decisions see Economizer changeover, for flow units see unit conversion, for a set of practice problems see practice problems, and for plotting see How to read a psychrometric chart.

References

• ASHRAE Handbook Fundamentals 2017 (SI), chapter 1 Psychrometrics: humidity ratio, enthalpy, specific volume, saturation vapour pressure (Hyland-Wexler), and the mixing, cooling and humidifying processes. The numbers here were computed with this site's engine, which implements those equations at 101.325 kPa.
• Bypass factor, contact factor, sensible heat ratio and apparatus dew point (ADP) are the definitions common to air conditioning textbooks. The density of 1.2 kg/m³ and specific heat of 1.005 kJ/kg·K in exam shortcuts vary a little between textbooks and problems.
• The steam humidification balance follows the Humidifiers chapter of ASHRAE Handbook HVAC Systems and Equipment; the steam enthalpy of 2676 kJ/kg is the 100 °C saturated steam value (IAPWS-IF97 2675.6).
• All problems are original practice problems written by this site; no past exam question is copied. The situations and numbers are example values.