10/28/97A F Emery1 PSYCHROMETRICS and ELEMENTARY PROCESSES (SI Units) Ashley F. Emery University of Washington

10/28/97A F Emery2 Psychrometrics The study of a mixture of dry air and water vapor Although precise thermodynamic relations are available for moist air, we will treat moist air as a mixture of ideal gases

10/28/97A F Emery3 Why study psychrometrics? The degree of moisture has a strong effect on 1) heating, cooling, and comfort 2) insulation, roofing, stability and deformation of building materials 3) sound absorption, odor levels, ventilation 4) industry and agriculture

10/28/97A F Emery4 Dry Air and Water Vapor Nitrogen Oxygen Argon Carbon Dioxide Dry Air Component% by volMW Effective MW Water Vapor

10/28/97A F Emery5 IDEAL GAS PV=mRT P = pressure Pa. V= volume cu. meter. m=mass kg R=gas constant T=temperature K=C+273

10/28/97A F Emery6 Dalton’s Law partial pressures

10/28/97A F Emery7 Mixture of Gases letting and remembering that we obtain

10/28/97A F Emery8 IMPORTANT PROPERTIES Humidity ratio, W Humidity ratio is the mass of water vapor per unit mass of dry air. Units are kg/kg

10/28/97A F Emery9 IMPORTANT PROPERTIES Saturated Humidity ratio, Saturation is when the air contains the maximum amount of water vapor at its current temperature. The saturation pressure is taken from the steam tables at the moist air temperature.

10/28/97A F Emery10 IMPORTANT PROPERTIES Relative humidity, Relative humidity is defined as the ratio of the partial pressure of the water vapor to the saturation pressure at the same temperature

10/28/97A F Emery11 IMPORTANT PROPERTIES Degree of saturation, Degree of saturation is the ratio of the amount of water contained in the moist air to that which would be contained if the air were saturated

10/28/97A F Emery12 IMPORTANT TEMPERATURES Dry Bulb Dew Point = temperature of moist air at rest = temperature at which the water vapor will condense out of the moist air. It is the temperature for which W is the saturated humidity ratio

10/28/97A F Emery13 IMPORTANT TEMPERATURES Adiabatic Saturation Temperature, it is the temperature at which liquid water would evaporate into the moist air without any heat addition to the system satisfies

10/28/97A F Emery14 IMPORTANT TEMPERATURES Wet Bulb Temperature, Is the temperature reached by evaporative cooling. A cotton sock is wrapped around a thermometer, saturated with distilled water. The water evaporates and the resulting temperature is called the wet bulb temperature. It is a close approximation to

10/28/97A F Emery15 Thermodynamic Properties enthalpy, kJ/kg of dry air =1.0 T +W(1.805 T ) specific volume, cu. m /kg of dry air

10/28/97A F Emery16 Example moist air at 26C dry bulb, 18C dew point, 101 kPa a) dp P =saturation pressure at the dew point Temperature. From the steam tables dp P = kPa    W b) if the 26C moist air were saturated, ws P = kPa and    s W

10/28/97A F Emery17 Example (continued)Example a)degree of saturation  s W W  d)relative humidity  Pws w P  e) enthalpy h= *( *26) = kJ/kg f)volume ) *1000. (101 )27326(*   da v = cu. m./kg

10/28/97A F Emery18 W h rh Psychrometric Chart saturation line

10/28/97A F Emery19 Simple Heating W 12 CTTT20 2,4C * 1,10C 1 

10/28/97A F Emery20 Simple Heating, solution  WW *(1.805* )=16.55 kJ/kg = 10.05kJ/kg-da CTTT20 2,4C * 1,10C 1  *(1.805* )=26.60 kJ/kg

10/28/97A F Emery21 Simple Heating and Humidification W 1 w T 17C T 20C TTT10C, * 3, 3,4C * 1,10C 1  3

10/28/97A F Emery22 Simple Heating and Humidification, Solution W  W  W *(1.805* )=47.91 kJ/kg-da = = kg/kg-da w T 17C T 20C TTT10C, * 3, 3,4C * 1,10C 1 

10/28/97A F Emery23 Simple Heating and Humidification, Solution W  W  W = *42.01 =31.01 kJ/kg-da w T 17C T 20C TTT10C, * 3, 3,4C * 1,10C 1  *(1.805* )=47.91 kJ/kg-da

10/28/97A F Emery24 Dehumidification and Cooling W  W  W CTCTCTT4 * 2,10 2,17 * 1,20C 1  The answer is the same as for the previous problem since the end points are the same BUT how can we actually go from point 1 to point 2?? ALSO T w =?

10/28/97A F Emery25 Dehumidification and Cooling, solution W 2 1 1’ 2’ 1 to 1’ by cooling 1’ to 2’ by cooling and dehumidification 2’ to 2 by heating

10/28/97A F Emery26 Dehumidification and Cooling, solution W to 1’ by cooling 15.4C *(1.805* )= ( ) = kJ/kg-da 1’ 2’

10/28/97A F Emery27 Dehumidification and Cooling, solution W 2 1 1’ to 2’ by cooling and dehumidification 1’ 2’  W  W ' 2  T assume that the water leaves at C

10/28/97A F Emery28 Dehumidification and Cooling, solution W 2 1 1’ to 2’ by cooling and dehumidification 1’ 2’ '1  h '2,  w h 1.98 '2  h -4.5 '2  T '2'1   da m q    da m w m  

10/28/97A F Emery29 Dehumidification and Cooling, solution W 2 1 2’ to 2 by heating 1’ 2’  h 1.98 '2  h 2'2   da m q  

10/28/97A F Emery30 Dehumidification and Cooling, solution W 2 1 1’ 2’ '2   da m q   '2'1   da m q   '11   da m q     da m q  

10/28/97A F Emery31 Difference between Humidification and Dehumidification W 1 2 W 2 1 1’ 2’ Water is injected at 10C Water is rejected at -4.5C   da m q   21   da m q  

10/28/97A F Emery32 Adiabatic Mixing of 2 Streams 1 2 3

10/28/97A F Emery33 Adiabatic Mixing of 2 Streams W 1 2 3

10/28/97A F Emery34 Adiabatic Mixing of 2 Streams, example m 3 /min at 16C dry bulb and rh=50% is mixed with 50 m 3 /min at 26C dry bulb and 15C wet bulb. min kg/0.826/100 1 / 1, v da m   /50 2 / 2,  v da m 

10/28/97A F Emery35 Adiabatic Mixing of 2 Streams, example C db, 42% rh 5.8C dp w 1 = w 2 = w 3 = h 1 = h 2 = h 3 = rh 2 = 29%