(dp/dv)ad = -(dp/dv)T{ p(dv/dT)p +(dU/dT)p }/cv (2.4')

From the First Law of thermodynamics, it follows that at constant pressure,

or, dividing by dT, we get

Thus, the adiabatic velocity of sound = (yp/p)1/2 (2.7')

Appendix-3 Some Selected Thermodynamic Diagrams

1. (T - S) diagram (also known as Tephigram) (Fig. 3.1) Co-ordinates:

Abscissa (x-axis) Ordinate (y-axis) Thermodynamic work:

2. Neuhoff diagram

Co-ordinates:

Abscissa (x-axis) Ordinate (y-axis) Thermodynamic work:

3. Stuve diagram

Co-ordinates:

Abscissa (x-axis) Ordinate (y-axis) where

Thermodynamic work:

4. Rossby diagram

Co-ordinates:

Abscissa (x - co-ordinate) - Specific humidity or humidity mixing-ratio Ordinate(y - co-ordinate) - Potential temperature/Equivalent potential temperature

Appendix-4 Derivation of the Equation for Saturation Vapour Pressure Curve Taking into Account the Temperature Dependence of the Specific Heats (Joos and Freeman (1967))

Let us deduce the vapour pressure curve by considering equilibrium between the liquid and the vapour phases of water at a given pressure (p)and temperature(T). At equilibrium, the temperature is, of course, the same throughout the liquid-vapour system. Because we seek the equilibrium position at a given value of T and p, we must find the minimum value of G, the Gibbs potential or enthalpy (= U — TS + p V). If we take ni mols of vapour and n2 mols of liquid, the total enthalpy of the system is

where g1 and g2 refer to the enthalpy of 1 mol of the vapour and the liquid phase respectively.

The variation in each phase is to be taken under the conditions:

T = constant, p = constant, and also, n1 + n2 = constant.

Since g1, g2 depend only upon T and p, they must remain constant, on account of the auxiliary conditions. We thus have for 8G = 0

The method of undetermined multipliers yields g1 + X = 0, g2 + X = 0 or, g1 — g2 = u1 — u2 + p(v1 — v2)+T(s2 — s1) =0

where p denotes the vapour pressure, s2 and s1 are the entropies of the vapour and the liquid phase respectively.

In computing the various terms in (4.4'), we assume that the vapour behaves like an ideal gas (this assumption may lead to considerable error if the vapour is near the point of condensation) and also note that v1 >> v2.

To compute (u1 — u2) in (4.4'), we make use of the relation (3.3.7) and write u1 = u1,0 + J cv1dT (4.5')

The third term in (4.4') may be simplified to p v1 in view of the fact that v1 >> v2. Substituting for v1 from the equation of state for vapour, we get

Putting (4.5'), (4.6') and (4.7') together and using the Mayer's relation (3.3.6), we write for the first three terms u1 -u2 + p(v1 -v2) = (u^0-u2,0) + ^ cpdT-J c2dT

where we have written c2 for cv2, since in the case of a liquid there is little external work done by change of volume, and L(0) for (u1j0 - u2,0) which measures the difference in internal energy between the liquid and the vapour phase at temperature Absolute zero and represents the latent heat of vaporization at that temperature. Now, to evaluate the term T (s2 - s1), we turn to the differential equation ds = (du + pdv)/T

and apply it first to the liquid.

In the case of the liquid, we can again neglect the volume expansion and write ds2 = c2 dT/T

Integrating between the limits 0 and T, we obtain

The term s2,0, the entropy of the liquid at temperature Absolute zero, is zero for a homogenous fluid.

For the vapour phase, we express p in terms of v in the term p dv and write the differential equation as ds1 = cv dT/T + Rdv/v Indefinite integration yields s1 = ^ cv dT/T + Rln v

We now take the lower limit of integration at temperature Absolute zero and combine all terms not dependent upon T and p to give us a single constant s1j0. This constant represents the entropy of the gas at T = 0 and p = 1. We further make the following substitutions in the second term on the right side of (4.10'):

Thus, we get

Now, the specific heat of an ideal gas does not vanish at T = 0, so a question of convergence arises in connection with the integral. To simplify matters, we can conceive of the specific heat as consisting of two parts: one part which does not vary with temperature and the other part which does and approaches zero rapidly as temperature decreases. We, therefore, write cp = cpc + cpT, where cpc is the part which remains constant and cpT is that which varies with temperature. The latter part is absent totally for a monatomic gas.

Substituting for cp and setting for p = 1 a very small lower limit in (4.11'), we get si (T, 1) - i1 (To, 1 )= jT cpxdT/T + cpc ln T - cpc ln To (4.12')

In (4.12'), the lower limit of the integral has been put to zero because of the very rapid decline in the value of cpT with temperature. Since the contribution of cpT to the integral is negligible even at T = 1 °A, we may interpret s1,0 approximately as the entropy at T = 1 and p = 1. Strictly, it is the limiting value of the difference s1 (T, 1) - cpc ln T0 as T0 approaches zero.

Combining (4.8'), (4.9') and (4.11') and substituting in (4.4'), and solving for ln p, we get the vapour pressure equation ln p = -L(0)/RT +(1 /RT^T c2 dT - (1 /RT) cpT dT +(11/R) jfT cpT dT/T + (cpc/R)ln T - (1/R) jf^dT/T +(s1,0 - i2,0 - cpc)/R (4.13')

It is easy to see from (4.13') that if the variation of the specific heat with temperature be ignored, it will reduce to the Clausius-Clapeyron equation (4.8.5) with only a few changes of symbols. In (4.13'), we have used p for es, cp for cp', c2 for cw, R for Rv, and a different constant for A.

Appendix-5 Theoretical Derivation of Kelvin's Vapour Pressure Relation for er/es

We derive the formula following a procedure adopted by Wallace and Hobbs (1977): Let us suppose that a volume of air resting on a plane surface of water in a closed space is supersaturated with water vapour and that a certain number of water vapour molecules combine or condense unto themselves spontaneously at constant pressure and temperature without the aid of an aerosol nucleus to form a viable water droplet of volume V and surface area A. If cl and cv be the chemical potentials of the liquid and the vapour phases of water respectively, and n the number of molecules per unit volume of the liquid, the condensation leads to a decrease in the Gibbs free energy of the system by n V (cv - cl), where we define chemical potential for a particular phase of the material as a measure of Gibbs free energy per molecule of that phase at constant pressure and temperature. Now, the gain in Gibbs free energy in creating the surface area of the droplet is equal to A o, where o is the work required to create a unit area of the vapour-liquid interface and is usually called the surface energy of water. So, the net increase in Gibbs free energy of the system due to formation of the droplet is given by

where G(= U - TS + pV) denotes the Gibbs free energy of the system (3.7.6).

Now, the chemical potential of a particular phase of water changes when the vapour pressure of that phase changes reversibly at constant temperature. The changes are related by the following expression dc = v de (5.2')

where e denotes vapour pressure and v the volume of one molecule of that phase.

Applying (5.2') to the case of a molecule each of the vapour and liquid phases of water, we obtain for the difference in chemical potential between the two phases, the expression d(cv - cl) = (vv - vl)de (5.3')

where vv and vl denote the volume occupied by a molecule of the vapour and the liquid phase respectively.

Since vv >> vl, (5.3') may be simplified to d(cv - cl)=vvde (5.4')

Substituting for vv from the ideal gas law for one mol of water vapour, evv = kT, where k is Boltzmann's constant, in (5.4'), we get d(cv - cl)=KT(de/e) (5.5')

But it can be shown that when the vapour pressure e reaches the saturation value es which is the equilibrium vapour pressure over a plane surface of water, cv = cl. Integrating (5.5'), we, therefore, get

For a droplet of radius R, (5.7') becomes

Fig. 5.1' Variation of aG with the radius R of the droplet in the case when e/es < 0, and when e/es > 0 (After Wallace and Hobbs, 1977, published by Academic Press, Inc)

As long as e < es, ln (e/es) is negative and the right side of (5.8') is positive for all values of R, as shown in Fig. 5.1''.

This means that in this case when e/es < 0, AG is always positive and the system never attains thermodynamic equilibrium. In the case when e > es, the second term on the right side of (5.8' is negative, and AG first increases with R but decreases after R reaches a particular value, r.

Since 5(AG)/dR = 0 at R = r, we get from (5.8'), ln(er/es ) = (2a/r)(1/nkT) (5.9')

This is Kelvin's formula (5.1).

Appendix-6 Values of Thermal Conductivity Constants for a Few Materials, Drawn from Sources, Including 'International Critical Tables'(1927), 'Smithsonian Physical Tables' (1934), 'Landholt-Bornstein' (1923-1936), 'McAdams' (1942) and others

Material |
Density |
Specific heat |
Thermal |
Thermal | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||

P (gm/cc) |
(Cal/gm/C) |
Conductivity k(Cal/s/cm/C) |
Diffusivity (k/Pc) | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||

Soil, wetted (43% |
1.67 |
0.53 |
0.0017 |
0.0019 | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||

water) | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||

Quartz sand |
1.65 |
0.19 |
0.00063 |
0.0020 | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||

(medium, fine and | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||

dry) | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||

Quartz sand (8.3% |
1.75 |
0.24 |
0.0014 |
0.0033 | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||

moisture) | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||

Sandy clay (15% |
1.78 |
0.33 |
0.0022 |
0.0037 | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||

moisture) |
(continued) Material Density Specific heat Thermal Thermal p (gm/cc) (Cal/gm/C) Conductivity Diffusivity k(Cal/s/cm/C) (k/pc) Soil, very dry Some wet soils Wet mud Rocks and building materials: Brick masonry (at20C) Concrete, av. Stone Concrete, dams Granite (at0C) Limestone (at 0C) Marble Sandstone Traprock Rock material (av for earth) 1.50 2.47 0.60 0.20 0.20 0.0004-8 0.003-8 0.0020 0.0015 0.0022 0.0058 0.0065 0. 0048 0.0055 0.0062 0.002-3 0.004-.010 0.0022 0.0044 0.0048 0.0107 0.0127 0.0081 0.0097 0.0113 0.0075 0.010 (Room temperature (°C) assumed in all cases, except where specifically given). Appendix-7 Physical Units and Dimensions The following three Tables give the basic and derived S.I. units used in the text.
Table 7.2 Derived units Variable Name Symbol (unit) Acceleration - m s-2 Pressure Pascal Pa (N m-2) Energy Joule J(Nm)
Appendix-8 Some Useful Physical Constants and Parameters Angular velocity of the earth (Q) Mean radius of the earth Acceleration due to gravity (g) at msl at latitude 45° Mass of the earth Mean density of the earth Gravitation constant (G) Universal Gas constant (R*) Mean molecular weight of dry air Mean molecular weight of water vapour Gas constant for dry air (R) Mechanical equivalent of heat Specific heat of dry air at constant pressure (cp) 7.292 x 10-5 radian s-1 6.37 x 106m 9.80616ms-2 5.988 x 1024kg 5.50 x 103kg m-3 6.673 x 10-11Nm-2kg- 8.314 x 103JK-1 28.9kg (kmol)-1 18.016kg (kmol)-287J K-1 kg-1 4.18 J (Cal)-1 1004 J K-1 kg-1 Specific heat of dry air at constant volume (cv) Ratio of the specific heats (y) Latent heat of condensation at 0 °C Freezing point of water Standard sea level pressure Standard sea level temperature Standard sea level density Planck's constant (h) Boltzmann's constant (k) Velocity of light (c) Avogadro number (N) Stefan-Boltzmann constant (a) Wien's constant 717JK-1 kg-1 2.5 x 106 J kg-1 273.16K 101.325 kPa 288.15K 1.225 kg m-3 6.625 x 10-34J s 1.380 x 10-23JK-1 2.997 x 108ms-1 6.0247 x 1023 (mol)-1 5.6687 x 10-8Wm-2 K-4 289.78 x 10-5m K |

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