Transcript Laws of thermodynamics

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1
Chapter 4
• 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.
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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 TEMPERATURE
Moist 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 first law of thermodynamics, which mandates conservation of energy,
and states in particular that the flow of heat is a form of energy transfer.
 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.
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Laws of
thermodynamics
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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.
dU  TdS  pdV
TdS = Q thermal energy and pdV = W
Therefore 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)
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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:
dW = Fdx
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.
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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
In this case work is done by the gas, since as heat is added to the gas the
gas expands (dW = pdV).
Specific heat at constant volume (cv)
At constant volume a gas does no work and the first
law of thermodynamics reduces to dq = du
Specific heat at constant pressure (cp)
Defined as:
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Adiabatic process:
dq = 0
For both processes V and α decrease
For the isothermal process (shown by
curve AB) this implies that p must
increase For the adiabatic process (shown by
curve AC) the internal energy, and thus
temperature, increases.
For the same mass of gas at the same
volume (points B and C) the sample
with the higher temperature (C) will
also have a higher pressure, hence the
adiabat (AC) on the p-V diagram is
steeper 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
•
The value of cp and cv for dry air are:
cp = 1004 J kg-1 K-1
cv = 717 J kg-1 K-1
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:
c p dT  dP
Combining these equations give:
RT
c p dT 
dP
P
Since this change in pressure implies a change in elevation:
From the hydrostatic equation :
Combining these equations
give:
dT
gRT

dz
cp P
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dP
 g
dz
 
g
cp
Cp
dT RT dP

dz
P dz
where p is the pressure, the density, g the acceleration of gravity, and
Z the geometric height.
 9.81 m

1004
J
kg 1
s 2
K 1
 9.8 C / km  10 C / Km


Dry and Moist Adiabatic Lapse Rates
 Dry 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???
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