Fig. 3. A quasistatic path.
The thermodynamic equilibrium state of a quantity of dry air is defined by the volume V and pressure P of the air.
Example At 25°C 1.0kg of dry air has volume 0.847m3 and pressure 101kPa.
A quasistatic path is a succession of equilibrium states. The path may be drawn as a curve on a PV diagram, as shown in Fig. 3.
Fig. 3. A quasistatic path.
The First Law of Thermodynamics implies that each equilibrium state has an internal energy U such that in a change of equilibrium state
Ufinal - Uinitial
equals the amount of energy transferred to the air from its environment.
Quasistatic work W done by the air on its environment is defined on a quasistatic path by:
W = Integral(1, 2):P.dV,
or dW = PdV.
Quasistatic heat Q on a quasistatic path is defined by:
U2 - U1 = Q - W,
or dU = dQ - dW,
where U2 - U1 is the internal energy change, Q is the energy gained by quasistatic heat, and W is the energy lost by quasistatic work.
An adiabatic process is a change of state with the air thermally insulated from its environment. A reversible adiabatic process is called isentropic. The equilibrium states in an isentropic process are said to have the same entropy. [A change of state which is the same as a reversible adiabatic process is called "isentropic" even when the air is not thermally insulated from its environment.]
Example: In an isentropic process in air: P1V11.40 = P2V21.40.
An adiabatic process which is not reversible is not isentropic, and the final state has a greater entropy than the initial state.
Air which remains unchanged when placed in thermal contact with its environment is said to be in thermal equilibrium with its environment. An isothermal process is a change of state in which the air remains in thermal equilibrium with a constant environment. The equilibrium states in an isothermal process are said to have the same temperature. [A change of state which is the same as one which remains in thermal equilibrium with a constant environment is called "isothermal" even when the air is not in thermal contact with its environment.]
Example: In an isothermal process in air: P1V1 = P2V2. (Boyle's law)
In a process at constant volume, if the final state has a greater entropy than the initial state, then the final temperature is higher than the initial temperature.
The Second Law of Thermodynamics gives measures of absolute temperature T, and entropy phi, in terms of quasistatic heat.
Fig. 4. Isothermal and isentropic processes.
For isothermal processes in thermal equilibrium with two given environments between two isentropic sets of states, as in Fig. 4, Qab/Qcd depends only on the environments, and not on the choice of isentropic sets of states. We can therefore define the ratio of the absolute temperatures T1 and T2 by
T2 / T1 = Qab/Qcd.
The Kelvin temperature scale (K) is defined by putting T = 273.16K at the triple point of water. [The triple point is the single fixed temperature where ice, liquid and water vapour exist together in equilibrium.]
The Celsius temperature scale (°C) is defined by
Celsius temperature = Kelvin temperature - 273.15.
In an isentropic process (see Fig. 4) the quasistatic heat is zero:
Qac = 0, Qbd = 0.
The entropy difference between the two isentropic sets of states in Fig. 4 is defined to be:
phi2 - phi1 = Qab/T2 = Qcd/T1.
Entropy has units J/K.
On a quasistatic path (see Fig. 3):
d(phi) = dQ/T,
phi2 - phi1 = Integral(1, 2):dQ/T.
this gives the differential equation of energy:
dU = T.d(phi) - P.dV.
The isotherms are given by the ideal gas equation of state:
PV = mRT,
where m = mass of air (kg), and R = 287J/kgK.
Example: The pressure of 1.0kg of dry air at 25°C in a volume 1.0m3 is 85.6kPa.
The ideal gas differential equation of state is:
P.dV + V.dP = mR.dT.
The specific heat at constant volume is:
CV = (dQ/dT)dV=0 = T(d(phi)/dT)dV=0.
The specific heat at constant pressure is:
CP = (dQ/dT)dP=0 = T(d(phi)/dT)dP=0.
For dry air the specific heats are practically constant:
CV = 718J/kgK and CP = 1005J/kgK.
The properties listed below are true for ideal gases, which include air and water vapour under the conditions found in boundary layer meteorology.
The internal energy is a function of temperature only: different equilibrium states with the same temperature have the same internal energy.
dU = CVdT = CPdT - PdV
CP - CV = R
Example: CP = 1005J/kgK, CV = 718J/kgK, R = 287J/kgK.
dQ = CPdT = gamma.dU, where gamma = CP/CV.
dQ = CVdT = dU.
dQ = PdV = -VdP. (Boyle's Law)
dQ = 0,
CVdT = -PdV, CPdT = VdP,
PVgamma = constant.
Example: For dry air CP/Cv = 1005(J/kgK)/718(J/kgK) = 1.40.
T/Pk = constant, where k = (CP - CV)/CP,
dT/T = k.dP/P.
Example: For dry air k = 0.286.
The potential temperature theta is the temperature the air would have with its pressure changed isentropically to 100kPa:
theta = T(100[kPa]/P)k.
Let e = vapour pressure of water, mv = mass of water vapour. Then:
eV = mvRvT, where Rv = 461.5J/kgK.
Note that Rv = R/0.622, where R is the gas constant for dry air.
The specific heats of water vapour are nearly constant from 0°C to 30°C:
CPv = 1870J/kgK, CVv = 1410J/kgK.
The specific heat of liquid water is C = 4187J/kgK.
The saturation vapour pressure es(T) in pascals over liquid water at temperature T is given by the empirical formula:
es(T) = 611.2exp[17.67T/(T + 243.5)] Pa,
where T is in degrees Celsius, and -30°C < T < 35°C.
The latent heat L of vaporization of liquid water in kJ/kg at temperature T is given by the empirical formula:
L = 2501 - 2.375T kJ/kg,
where T is in degrees Celsius, and 0°C < T < 40°C.
The total pressure P of moist air is the sum of (1) the partial pressure Pd of the dry air and (2) the vapour pressure e:
P = Pd + e.
The mixing ratio w for a quantity of air containing a mass mv of water vapour and a mass md of dry air is defined by:
w = mv/md ~ 0.622e/P.
The relative humidity is the ratio of the vapour pressure to the saturation vapour pressure:
relative humidity = e/es.
The virtual temperature Tv is the temperature of dry air which has the same density m/V as the moist air at the same pressure:
Tv ~ (1 + 0.6w)T.
The gas constant Rm for moist air satisfies the following equations:
PV = mRmT, Rm = (1 + 0.6w)R.
The specific heats of moist air, CVm and CPm, are given in terms of the specific heats of dry air by:
CVm ~ (1 + w)CV, at constant volume,
CPm ~ (1 + 0.9w)CP, at constant pressure.
In an isentropic process in moist air we have:
T/Pk = constant, where k = 0.286(1 - 0.2w).
In Thailand w is normally 0.02, or less, so the above corrections are less than 2%.
The dew point temperature Td is the temperature at which moist air becomes saturated on a path with the pressure P and the humidity ratio w constant. It is given by:
Td = (5.42×103[K])/ln(157.4×106[kPa]/wP).
The thermodynamic wet bulb temperature Tw is the temperature to which air is cooled by evaporation of water at constant pressure. It is given by:
CP(T - Tw) = L[ws(Tw) - w].
This equation must be solved iteratively to find Tw, which satisfies the inequality Td < Tw < T.
The equivalent temperature Te is the temperature reached if all the moisture in the air is condensed at constant pressure. It is given by:
Te = T + Lw/CP.
The isentropic condensation temperature Tc is the temperature at which the air becomes saturated on an isentropic path with w constant.
When saturated air is cooled by expansion, water vapour is condensed. In pseudoadiabatic processes it is assumed that the condensed water is precipitated as rain. The differential equation of the process is:
dT/T = kdP/P - (L/TCP)dws.
In reversible saturated adiabatic processes it is assumed that the condensed water remains in the air as cloud. When air contains liquid water as cloud and the air is warmed by compression, the cloud is evaporated. The theory is more complicated than the theory of pseudoadiabatic processes, but the dependence of T on P is nearly the same.
Imagine that air is lifted up to the condensation level at pressure Pc where it becomes saturated with temperature Tc. The wet bulb potential temperature thetaw is the temperature of saturated air at pressure 100kPa which reaches the same state (Tc, Pc) in a pseudoadiabatic process (see Fig. 5).
Fig. 5. Determination of the wet bulb potential temperature.
By R. H. B. Exell, 2001. King Mongkut's University of Technology Thonburi.
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