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• Thermodynamics is the science of energy conversion involving heat and other forms of energy, most notably mechanical work. It studies and interrelates the macroscopic variables, such as temperature, volume and pressure, which describe physical, thermodynamic systems.

Chapter 4 Chapter 4

Meteorology \ Dr. Mazin sherzad

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The Ideal Gas Law

•An equation of state describes the relationship among pressure,

temperature, and density of any material.

•All gases are found to follow approximately the same equation of state,

which is referred to as the “ideal gas law (equation)”.

•Atmospheric gases, whether considered individually or as a mixture,

obey the following ideal gas equation: gas equation:

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Question: Calculate the density of water vapor which exerts a pressure of 9 mb at 20°C. Answer:

Use the ideal gas law: Pv= ρRvT

Pv = 9 mb = 900 Pa (a SI unit)

Rv = R* / Mv = 461 J deg-1 kg-1

T = 273 + 20 (°C) = 293 K.

So we know the density of water vapor is:

ρ = Pv/ (RvT) = 900 / (461*293) = 6.67 x 10-3 kg m-3

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VIRTUAL TEMPERATUREMoist air has a lower apparent molecular weight that dry air. The gas constant for 1 kg of moist air is larger than that for 1 kg of dry air. But the exact value of the gas constant of moist air would depend on the amount of water vapor contained in the air. It is inconvenient to calculate the gas constant for moist air. It is more convenient to retain the gas constant of dry air and use a fictitious temperature in the ideal gas equation. This fictitious temperature is called “virtual temperature”. This is the temperature that dry air must have in order to has the same density as the moist air at the same pressure. Since moist air is less dense that dry air, the virtual temperature is always greater than the actual temperature. actual temperature.

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Laws of thermodynamics

The four principles (referred to as "laws"):

The zeroth law of thermodynamics, which underlies the basic definition of temperature

The second law of thermodynamics, which states that the entropy of an isolated macroscopic system never decreases or (equivalently) that perpetual motion machines are impossible

The third law of thermodynamics, which concerns the entropy of a perfect crystal at absolute zero temperature, and which implies that it is impossible to cool a system all the way to exactly absolute zero.

The first law of thermodynamics, which mandates conservation of energy, and states in particular that the flow of heat is a form of energy transfer.

Meteorology \ Dr. Mazin sherzad

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Laws of thermodynamics

Meteorology \ Dr. Mazin sherzad

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pdVTdSdU

TdS = Q thermal energy and pdV = WTherefore we can say dQ=dU+dW, where:

U Internal energy (is the total energy contained by a thermodynamic system)

(S) entropy (is a thermodynamic property that is a measure of the energy not available for work in a thermodynamic process)

The first law of thermodynamics, which mandates conservation of energy, and states in particular that the flow of heat is a form of energy transfer.

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Enthalpy• Enthalpy is a measure of the total energy of a thermodynamic system. It includes

the internal energy

The enthalpy of a system is defined as:

H is the enthalpy of the system (in joules),

U is the internal energy of the system (in joules),

p is the pressure at the boundary of the system and its environment, (in pascals )

V is the volume of the system, (in cubic meters).

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Consider a gas contained in a cylinder that is fitted with a piston:

When V1 > V2 (the gas is compressed) work is done on the gas and W < 0

When V2 > V1 (the gas expands) work is done by the gas and W > 0 The work done in going from

volume V1 to volume V2 depends on the path of integration and as such is not an exact differential.

dW = Fdx

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Specific heat: The ratio of the heat added to a system to the change in temperature of the system dq/dT The units for specific heat are J kg-1 K-1

Specific heat at constant volume (cv)

At constant volume a gas does no work and the first law of thermodynamics reduces to dq = duSpecific heat at constant pressure (cp)

Defined as:

In this case work is done by the gas, since as heat is added to the gas the gas expands (dW = pdV).

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Adiabatic process: dq = 0

For both processes V and α decreaseFor the isothermal process (shown bycurve AB) this implies that p mustincrease For the adiabatic process (shown bycurve AC) the internal energy, and thustemperature, increases.

For the same mass of gas at the samevolume (points B and C) the samplewith the higher temperature (C) willalso have a higher pressure, hence theadiabat (AC) on the p-V diagram issteeper than the isotherm (AB)

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Specific Heat of Dry Air

• For ordinary calculations value of cp = 1.0 kJ/kg.K (equal to kJ/kg.oC)

-is normally accurate enough• For higher accuracy cp = 1.006 kJ/kg.K (equal to kJ/kg.oC) - is better

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Dry Adiabatic Lapse Rate

• Consider an air parcel undergoing an adiabatic change in pressure, with no phase change of any water substance in the air parcel. For this air parcel the first law of thermodynamics can be written as:

dPdTc p

dPP

RTdTc p

dz

dP

P

RT

dz

dTC p

gdz

dP

Since this change in pressure implies a change in elevation:

From the hydrostatic equation :

Combining these equations give:

Combining these equations give:

Pc

gRT

dz

dT

p

pc

g

11

2

1004

81.9

KkgJ

sm

The value of cp and cv for dry air are:cp = 1004 J kg-1 K-1cv = 717 J kg-1 K-1

The value of cp and cv for dry air are:cp = 1004 J kg-1 K-1cv = 717 J kg-1 K-1

where p is the pressure, the density, g the acceleration of gravity, and Z the geometric height.

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KmCkmC /10/8.9

Dry and Moist Adiabatic Lapse RatesDry adiabatic lapse rate is constant = 10ºC/km.Moist adiabatic lapse rate is NOT a constant. It depends on the temperature of saturated air parcel.The higher the air temperature, the smaller the moist adiabatic lapse rate.

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Example 3: 2 kg of ice at -10 oC and 3 kg of water at 70 oC are mixed in an insulated container. Find

a) Equilibrium temperature of the system b) Entropy produced.

Homework???