# Lecture 8 Saturated Adiabatic Processes Phase Changes Liquid Gas (Vapor) Energy absorbed Energy released Solid (Ice) melting evaporation deposition freezing.

## Presentation on theme: "Lecture 8 Saturated Adiabatic Processes Phase Changes Liquid Gas (Vapor) Energy absorbed Energy released Solid (Ice) melting evaporation deposition freezing."— Presentation transcript:

Phase Changes Liquid Gas (Vapor) Energy absorbed Energy released Solid (Ice) melting evaporation deposition freezing sublimation condensation

Latent Heat Heat released or absorbed during a phase change of water AMS Glossary AMS Glossary

Latent Heats at 0  C Latent heats of vaporization (evaporation) and condensation (L v ): ~ 600 cal  g -1 Latent heats of fusion (freezing) and melting (L f ): ~ 80 cal  g -1 Latent heats of sublimation and deposition (L s ): ~ 680 cal  g -1 Conversion: 1 cal  g -1 = 4.1855 x 10 3 J  kg -1

Exercise Convert L v to J  kg -1 L v = (600 cal  g -1 ) x 4.1855x 10 3 J  kg -1 = 2.5 x 10 6 J  kg -1 = 2.5 x 10 6 J  kg -1

Exercise Suppose that the mixing ratio (w) of a parcel is 2.0 x 10 -2 (20g water vapor per kg of dry air) (20g water vapor per kg of dry air) Suppose that 10% of the vapor condenses How much does the parcel warm? (Assume constant pressure.) (Assume constant pressure.)

Solution, Part 1  Q = -L v  m v, where m v is the mass of water vapor (Why the negative sign?) (Why the negative sign?)  q  -L v  w  q  -L v  w = - (2.5 x 10 6 J  kg -1 )  (-2.0 x 10 -3 ) = - (2.5 x 10 6 J  kg -1 )  (-2.0 x 10 -3 ) = 5.0 x 10 3 J  kg -1 = 5.0 x 10 3 J  kg -1

Solution, Part 2 Temperature change at constant pressure: But, c p  1000 J  kg -1  kg -1 

A Saturated Adiabatic System Consider a parcel consisting of dry air, water vapor, and liquid water Closed system, saturated Closed system, saturated Consider adiabatic transitions i.e., no heat enters or leaves system i.e., no heat enters or leaves system

Saturated Adiabatic Transitions Expansion  cooling  condensation i.e., water vapor decreases, liquid water increases i.e., water vapor decreases, liquid water increases Condensational heating partially offsets cooling due to expansion Condensational heating partially offsets cooling due to expansion Result: Cooling rate less than dry adiabatic rate Result: Cooling rate less than dry adiabatic rate Compression  warming  evaporation i.e., water vapor increases, liquid water decreases i.e., water vapor increases, liquid water decreases Evaporative cooling partially offsets heating due to compression Evaporative cooling partially offsets heating due to compression Result: Heating rate less than dry adiabatic rate Result: Heating rate less than dry adiabatic rate

Comparison z Temperature Unsaturated parcel Saturated parcel

Dry and Saturated Adiabats Temperature Pressure Dry Saturated

Temperature of Lifted Parcel Consider a parcel that is initially unsaturated Parcel is lifted to LCL and beyond

Temperature z LCL

Temperature z LCL

Temperature z LCL

Temperature z LCL

Temperature z LCL

Temperature z LCL

Temperature z LCL

Temperature z LCL

Temperature z LCL

Temperature z LCL

Temperature z LCL

Temperature z LCL

Temperature z LCL

Temperature z LCL

Wet-Bulb Temperature, T w The wet-bulb temperature is the temperature associated with a wick-covered thermometer on a psychrometer.

Measuring Humidity: Sling psychrometer A psychrometer consists of two glass thermometers with one covered with a wick (cloth) that is wet. This measures the ‘wet-bulb’ temperature.

Physical Basis of T w Evaporation will occur on a wick at a steady rate (as sling is slung). Heat must be continually supplied in an amount = the latent heat of vaporization of the water. Heat is taken from the air passing over the wick of the thermometer – resulting in a drop in temperature!

Physical Basis of T w If air is saturated, no evaporation takes place. Temperature of the wet bulb is same as dry bulb. If air is very dry, the wet-bulb depression may be substantial. Wet-bulb depressionis the T-T w

Dew Point vs. Wet Bulb Temperature Dew point is when we cool a volume of air until saturation is reached while keeping moisture content constant. Wet-bulb temperature we obtain if we cool air to saturation by evaporating water into it. Therefore, saturation is reached at a higher temperature than dew point. T d ≤ T w ≤ T

Normand’s Rule States that the wet-bulb temperature may be determined by lifting a parcel of air adiabatically to its LCL and then following a moist adiabat from that temperature back down to the parcel’s actual temperature.

Relationship between dew point, wet- bulb and dry bulb temperatures

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