Psychrometric Chart (or Humidity Chart) A psychrometric chart is a graphical representation of the thermodynamic properties of air-water system. There are a number of different charts. They differ with respect to the barometric pressure, range of temperature, and the choice of coordinates. The most popular charts have absolute humidity plotted against dry-bulb temperature. Most published charts are for standard barometric pressure (1 atm).

Figure 8. 4-1 Psychrometric chart- SI units Figure 8.4-1 Psychrometric chart- SI units. Reference states: H2O (L, 0°C, 1 atm), dry air (0°C, 1 atm)

Figure 8. 4-2 Psychrometric chart- American Engineering units Figure 8.4-2 Psychrometric chart- American Engineering units. Reference states: H2O (L, 32°F, 1 atm), dry air (0°F, 1 atm)

The psychrometric chart has seven lines. A Absolute Humidity Axis B Specific Volume C Dry Bulb Axis D Relative Humidity Dew Point E Enthalpy F Wet Bulb & Saturation Curve G

Description of the psychometric chart The horizontal axis forms the dry-bulb temperature scale and vertical lines are lines of constant dry-bulb temperature. The dry-bulb temperature is the air temperature as measured by a thermometer, thermocouple or other conventional temperature-measuring device Dry Bulb Temperature (oC or oF) constant dry-bulb Temperature (Tdb)

[kg H2O(v)/kg Dry Air (DA)] The vertical axis is the absolute humidity (or humidity ratio, moisture content) scale and horizontal lines are lines of constant humidity. Absolute Humidity or moisture content [kg H2O(v)/kg Dry Air (DA)] constant humidity

Curves lines are used to represent relative humidity (RH or hT).

The line of saturation (100% RH) is used to provide the dew-point temperature (Tdp) and the wet-bulb temperature (Twb) scales.

The dry-bulb temperature (DBT) is the temperature of air measured by a thermometer freely exposed to the air but shielded from radiation and moisture. DBT is the temperature that is usually thought of as air temperature, and it is the true thermodynamic temperature. As a matter of fact, it indicates the amount of heat in the air and it is directly proportional to the mean kinetic energy of the air molecules. Temperature is usually measured in degrees Celsius (°C), Kelvin (K), or Fahrenheit (°F). Absolute Humidity - moist content- is the water vapor present in air, ha [kg H20(v)/kg DA]

Wet Bulb Temperature - Twb The Wet Bulb temperature is the temperature of adiabatic saturation. This is the temperature indicated by a moistened thermometer bulb exposed to the air flow. Wet Bulb temperature can be measured by using a thermometer with the bulb wrapped in wet muslin. The adiabatic evaporation of water from the thermometer and the cooling effect is indicated by a "wet bulb temperature" lower than the "dry bulb temperature" in the air. Dew Point Temperature - Tdp The Dew Point is the temperature at which water vapor starts to condense out of the air (the temperature at which air becomes completely saturated). Above this temperature the moisture will stay in the air.

Humid volume, VH (m3/kg DA). Relative humidity is the ratio of the partial pressure of water vapor in an air-water mixture to the saturated vapor pressure of water at a prescribed temperature. The relative humidity of air depends not only on temperature but also on the pressure of the system of interest. Curves lines are used to represent relative humidity (RH or hT). Relative humidity, Humid volume, VH (m3/kg DA). The humid volume is the volume occupied by 1 kg of dry air plus the water vapor that accompanies it. Lines of constant humid volume on the psychrometric chart are steep and have negative slopes.

Lines of constant dew-point are horizontal lines.

Constant wet-bulb temperatures form diagonal lines on the chart.

A different set of diagonal lines represent lines of constant specific volume.

Lines of constant enthalpy at saturation are closely aligned to the lines of constant wet-bulb many charts use a single set of lines to represent constant wet-bulb temperature and constant enthalpy . An enthalpy scale is typically placed above the saturation line. Reference states for Enthalpy :- liquid water at 1 atm & 0oC, dry air at 1 atm & 0oC

Enthalpy deviation – to correct the enthalpy of humid air that is not saturated (see psychometric chart) Enthalpy deviation

Working session 11 – Problem 8.69(a) The latest weather report states that the temperature is 24°C and the relative humidity is 50%. A thermometer is mounted at the back porch of your house. What temperature would it read Using the chart determine the following values. Wet bulb temperature (oC) Dew Point (oC) Absolute Humidity (kg H2O(v)/kg DA) Specific Volume (m3/kg DA) Specific Enthalphy (kJ/kg DA) The amount of water (kg) in 1 m3 of humid air

Working session 11 – Problem 8.69(a) Wet bulb temperature = 17 oC Dew Point = 13 oC Absolute Humidity (kg H2O(v)/kg DA) = 0.0092 kg H2O(v)/kg DA Specific Volume (m3/kg DA) = 0.855 m3/kg DA Specific Enthalphy (kJ/kg DA) = (48 – 0.2) kJ/kg DA = 47.8 kJ/kg DA The amount of water (kg) in 1 m3 of humid air

What happens during heating and cooling at constant pressure? There is no change in the absolute humidity of the air-vapor mixture, as long as no condensation occurs. Cooling occurs from right to left… if superheated humid is cooled the system follows a horizontal path to the left until the saturation curve (dew pt temp & 100% RH); thereafter the gas phase follows the saturation curve Heating occurs from left to right. There is a change in the sensible heat of the air-vapor mixture. Heat must be added or subtracted to cause the temperature change. Dry Bulb Temperature Heating Cooling Condensation occurs

Example 14 – Problem 8.69(a, modified) The latest weather report states that the temperature is 24°C and the relative humidity is 50%. A sample of outside air is cooled at constant pressure. At what temperature would condensation begin? Tdp = 13 oC Calculate the rate at which heat (kJ/s) must be removed for every 1 m3/s of humid air before condensation occurs. (draw a process path on the psychometric chart) Q= ? KJ T2=13oC RH =50% 1 m3 /s HA mDA kg/s dry air 24°C RH= 50% mDA kg/s dry air 13 °C RH=100% T1 = Tdb=24oC

Ĥ1 -0.2 Ĥ2 RH =50% T2=13oC T1 = Tdb=24oC v =0.855 m3 HA/kg DA

Example 14 – Problem 8.69(a, modified) Simplified energy balance, ….. Ĥi (kJ/kg DA) Mass of dry air, Rate of heat removed,

Working session 12 – Recall Example 14 If the air is cooled to 10 °C at constant pressure, calculate the fraction of water that condenses and the rate at which heat (KJ/s) must be removed for every 1 m3/s of humid air. (draw a process path on the psychometric chart) Mass balance, Tdp= 13oC T2 =10oC RH =50% Q= ? KJ 1 m3/s HA mDA kg/s dry air m1 kg/s water 24°C RH= 50% mDA kg/s dry air m2= kg/s water 10 °C RH=100% 10°C m3= kg/s water condensed T1 = Tdb=24oC

Working session 12 – Recall Example 14 From figure 8.4.1 The fraction of water that condenses,

Working session 12 – Recall Example 14 Simplified energy balance, Reference states : liquid water at 1 atm & 0oC, dry air at 1 atm & 0oC The values of mDA (kg DA) & m3 (kg condensed water) are as previously calculated The values of Ĥ1 & Ĥ2 in kJ/kg DA are obtained from figure 8.4.2 The value of Ĥ3 in kJ/kg is relative to the specific enthalpy of liquid water at 1 atm & 0oC Substance min (kg) Ĥin(kJ/kg) @24oC mout (kg) Ĥout(kJ/kg) @10oC Humid Air mDA kg DA Ĥ1 kJ/kg DA Ĥ2 kJ/kg DA H2O (l) - m3 kg Ĥ3 kJ/kg

Working session 12 – Recall Example 14 Reference states : liquid water at 1 atm & 0oC, dry air at 1 atm & 0oC Rate of heat removed, Substance min (kg) Ĥin(kJ/kg) @24oC mout (kg) Ĥout(kJ/kg) @10oC Humid Air 1.1696 kg DA 47.8 kJ/kg DA 29.5 kJ/kg DA H2O (l) - 0.0019 kg 41.89 kJ/kg

Working session 13 Enthalpy 13 oC 32 40% Calculate the amount of sensible heat that must be added to 10 kg of air at 13°C dry-bulb and 40% relative humidity to raise the temperature of the air to 32°C dry bulb. What is the corresponding relative humidity? H2 Enthalpy H1 40% 13 oC 32

Working session 13 Enthalpy 40% 13 32 oC Ĥ2 Ĥ1 Locate the 13°C dry bulb and 40% RH point. Follow the enthalpies line to Ĥ1. Move horizontally to intercept the vertical line of 32°C dry bulb Follow enthalpies line Ĥ2. Ĥ2 = 42-0.45 = 41.55 kJ/kg DA. Ĥ1 = 22.75-0.075 = 22.675 kJ/kg DA. Ĥ2 Enthalpy Ĥ1 40% 13 32 oC The relative humidity is now ~ 12%.

Test Your Self Wet solids pass through a continuous dryer. Hot dry air enters the dryer at a rate of 400 kg/min and picks up the water that evaporates from the solids. Humid air leaves the dryer at 50°C containing 2.44 wt% water vapor and passes through a condenser in which it is cooled to 10°C. The pressure is constant at 1 atm throughout the system. Draw and label a process flowchart of the above process. At what rate (kg/min) is water evaporating in the dryer? Use the psychrometric chart to estimate the wet-bulb temperature, relative humidity, dew point, and specific enthalpy of the air leaving the dryer. Use the psychrometric chart to estimate the absolute humidity and specific enthalpy of the air leaving the condenser. Use the results of parts (b) and (c) to calculate the rate of condensation of water (kg/min) and the rate at which heat must be transferred from the condenser (kW).

Test Your Self (Problem 8.73) At what rate (kg/min) is water evaporating in the dryer? ~ 10 kg/min Use the psychrometric chart to estimate the absolute humidity, wet-bulb temperature, relative humidity, dew point, and specific enthalpy of the air leaving the dryer. AH=0.025kg H2O/kg DA, Twb~32.8oC, RH~32%, Tdp~28.6oC, Ĥ=~115 kJ/kg DA Use the psychrometric chart to estimate the absolute humidity and specific enthalpy of the air leaving the condenser. AH=0.0078kg H2O/kg DA, Ĥ=29.5 kJ/kg DA Use the results of parts (b) and (c) to calculate the rate of condensation of water (kg/min) and the rate at which heat must be transferred from the condenser (kW). rate of water condensed = 6.88 kg/min, Q=-565 kJ/s

What is humidification and dehumidification? A. Humidification adds moisture to the air which increase the absolute humidity. B. Dehumidificaiton remove moisture from the air which decrease the absolute humidity. Dehumidification Humidification

How is humidity increased? A. Water is added in vapor form. B. Water is converted from liquid to gas. C. There is an increase in the energy level. Humidification

How is humidity decreased? A. There is a change from gas to liquid. B. There is a decrease in the energy level. C. With the loss of energy, condensation occurs. Dehumidification

Example 15 How much moisture is needed to raise the relative humidity of an air-vapor mixture from 12% to 50% and the temperature stays at 75°F. Water needed, = final absolute humidity - initial absolute humidity = 0.0094 - 0.0022 = 0.0072 lbs water / lb dry air lbs water/ lb dry air 0.0094 50% 0.0022 12% 75°

Working session 14 How much heat is needed to evaporate the water that’s needed to increase the relative humidity from 12% to 50% and the temperature stays at 75°F. lbs water/ lb dry air 50% 0.0094 0.0022 12% 75°db

Working session 14 :- The amount of heat …. BTU needed = final enthalpy - initial enthalpy BTU needed = 28.0 BTU/lb dry air - 21.0 BTU’s/lb dry air BTU needed = 7.0 BTU per lb dry air H2 50% H1 12% 75°db

Adiabatic Cooling Process Air undergoing adiabatic cooling through contact with liquid water moves along a constant wet-bulb temperature line on the psychrometric chart from its initial condition towards the 100% relative humidity curve

Adiabatic Cooling Adiabatic cooling is said to have taken place when a warm gas is brought into contact with a cold liquid, causing the gas to cool and some of the liquid to evaporate and with an assumption that no heat is lost to its surrounding during the gas-to-liquid heat transfer process. Examples of common adiabatic cooling process Spray cooling , spray humidification Spray dehumidification Drying Spray drying

Adiabatic Saturation Temperature (Tas) Assuming; (Cp)air, (Cp)H2O, and (DHvap)H2O, are independent of temperature at the prevailing process conditions. The enthalpy changes undergone by the unevaporated liquid water and the solid (if there is one) in going from T2 to T4 are negligible compared to the changes undergone by the entering wet air and the evaporated water. The heat required to raise liquid water from T2 to T3 is negligible compared to the heat of vaporization of water.

~0 ~0

and the process is adiabatic (Q=0), then the simplified energy equation becomes; If the outlet temperature T3 is low enough, the air leaves saturated with water (i.e. RH = 100%) then the temperature corresponding to this condition is called the adiabatic saturation temperature (Tas) & happens to coincide with the wet bulb temperature (i.e. Tas = Twb)

What is evaporative cooling? Through evaporation, moisture in the air accumulates and the air temperature decreases. Sensible heat from the air vaporizes water from its liquid to gaseous phase. There is no loss or gain of heat within the system because the amount of sensible heat removed equals latent heat added to the water. The process follows a constant enthalpy line and hence the process is also known as adiabatic cooling The maximum temperature reduction is the difference between starting dry bulb temperature and its wet bulb temperature. Addition of moisture Lowering of dry bulb temp Evaporative cooling lowers the dry bulb temperature. It is used where a lower temperature is desired.

Example 16 A chicken producer is considering installing evaporative cooling in a breeding herd building. The outside air has a 32°C dry bulb temperature and 35 percent relative humidity, What is the lowest temperature that can theoretically be obtained from the air coming off the cooling pads. (~ 21oC ) However, due to the inefficiency of evaporative coolers, the temperature of the air coming off the cooling pads will be 2oC above the lowest attainable temperature in part (a). Estimate the relative humidity under these conditions. (~ 23oC , RH~85%)

Working session 15 A chicken producer is considering installing an evaporative cooling system in a breeding herd building. Air enters the cooler at 35°C dry bulb and 30% relative humidity and leaves at 27°C Determine the amount of water (kg/hr) needed for an evaporative cooler when 1000 m3/hr humid air passes through a livestock building per hour. Calculate also the flowrate (m3/hr) of air entering the cooler and the corresponding relative humidity of the air leaving the cooler.

Working session 15 kg water/ kg dry air 0.0143 30% 0.0106 Dry Bulb Temp. 27° 35° Vhumid ~ 0.8694 m3/kg DA 0.888 m3/kg DA

Working session 15 Read absolute humidity value for 35°C and 30% relative humidity  y1=0.0106 kg water/kg DA Extend enthalphy line toward saturation curve. Read absolute humidity value for 27°C  y3=0.0143 kg water/kg DA. Water added/kg dry air: 0.0143 - 0.0106 = 0.0037 kg water/kg DA Kg of dry air required = (1000 m3 air /hr) x (kg DA /0.8694 m3 air) = 1150kg DA /hr Total water added = 0.0037 kg water/kg DA x 1150 kg DA/hr = 4.256 kg/hr water Air flowrate entering a cooler = 1150 kg DA /hr x 0.888 m3/kg DA = 1021 m3/hr Relative humidity of air leaving a cooler = ~ 65%