1 / 30

Psychrometrics

Understand the properties of gas-vapor mixtures, composition of dry air, humidity ratios, and wet-bulb temperatures. Learn about thermodynamic properties, including specific volume and enthalpy, and use psychrometric charts to evaluate air conditioning processes. Examples include calculating energy required for heating air and determining water removal rates during drying processes.

cbaker
Télécharger la présentation

Psychrometrics

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Psychrometrics Introduction to Food Engineering

  2. Psychrometrics • Thermodynamic properties of gas-vapor mixtures (air-water)

  3. Properties of Dry Air • Composition • N2 78 %, O2 20.9 % • Standard dry air MW = 28.9645 • Gas constant for dry air Ra = 287.055 m3.Pa/kg.K

  4. Properties of Dry Air • Specific Volume Ra = gas constant TA = absolute temperature (K) Pa = partial pressure of dry air

  5. Properties of Dry Air • Specific Heat (Cpa) • At 1 atm (101.325 kPa), T = - 40 – 60 C • Average value 1.005 kJ/kg.K • Enthalpy (heat content) • Reference 1 atm, 0 C

  6. Properties of Dry Air • Dry Bulb Temperature • Indicated by sensor

  7. Properties of Water Vapor • Moist air = dry air + water vapor • Vapor in the air is superheated steam at low pressure & temperature • Moist air is clear or foggy • MW of water = 18.01534 • Gas constant for water vapor • Rw = 461.52 m3Pa/kg.K

  8. Properties of Water Vapor • Specific Volume of Water Vapor • Below 66 C vapor follows ideal gas law

  9. Properties of Water Vapor • Specific Heat of Water Vapor • Within –71 to 184 C Cpw = 1.88 kJ/kg.K • Enthalpy of water vapor Ta = dry bulb temp

  10. Properties of Air-Vapor Mixtures • Gibbs-Dalton Law • Up to 3 atm air-water mixtures follow perfect gas laws PB = barometric (total pressure) of moist air (kPa)

  11. Dew-Point Temperature • Dew-Point Temperature Water vapors in air = steam at low pressure • Saturation temperature = dew point • Obtained from steam table at partial pressure exerted by water vapor • Below this temp, condensation of moisture

  12. Humidity Ratio • Humidity Ratio (Moisture Content) or specific humidity, W (Kg water / ky dry air)

  13. X = mole fraction

  14. Relative Humidity • Relative to maximum amount of moisture at dry bulb temperature

  15. Relative Humidity • Density Density of water vapor/density of saturated vapor at dry bulb temp of air

  16. Humid Heat • Of air-water vapor mixture • Heat required to raise temp of 1 kg dry air + water vapor by 1 K (kJ/kg dry air . K) W = humidity ratio

  17. Specific Volume • Volume of 1 kg dry air + water vapor

  18. Adiabatic Saturation of Air • Evaporation of water by sensible heat of entering air

  19. Wet-Bulb Temperature • Psychrometric wet bulb temperature • Movement of air • Thermodynamic wet bulb temperature

  20. The Psychrometric Chart

  21. Example • An air-vapor mixture is at 60 C dry bulb temp and 35 C wet bulb. Determine relative humidity, humidity ratio, specific volume, enthalpy and dew-point temp. • RH = 20 %, W = 0.026 kg/kg • Enthalpy = 129 kJ/kg dry air • Specific volume = 0.98 m3/kg dry air • Dew-point temp = 29 C

  22. Use of psychrometric chart to evaluate complex air-conditioning processes • Heating or Cooling of Air • Humidity ratio constant

  23. Example • Calculate the rate of thermal energy required to heat 10 m3/s of outside air at 30 C dry bulb temp and 80 % RH to a dry bulb temp of 80 C Mass flow rate = vol flow rate/ specific vol

  24. Mixing of Air Inverse proportion

  25. Drying • Adiabatic saturation process • Heat of evaporation is supplied only by drying air • Dry bulb temp decreases, enthalpy constant ie. constant wet bulb • Humidity ratio increases (gain moisture)

  26. Example • Heated air at 50 C, 10 % RH is used to dry rice. Air exits under saturated condition. Determine amount of water removed per kg of dry air. • W1 = 0.0078 kg/kg • Follow constant enthalpy line • W2 = 0.019 kg/kg • Moisture removed = 0.0112 kg/kg

More Related