Thermodynamics of the Atmosphere and the Ocean

Abstract

In the previous chapters, we have used some elementary thermodynamics. Now we have the basis to go some step further and above all to apply some of our knowledge to the ocean. One of the conclusions of the previous chapter was that the atmosphere can be treated as a perfect gas. One of the gases present in the atmosphere is water vapor which can change phases in the environmental conditions of the atmosphere. That is, it can be in the gaseous form or liquid form or solidify as ice and this complicates the simple thermodynamics we used so far. These characteristics have some important consequences, for example, on the temperature profile of the atmosphere. As for the ocean the only thermodynamics we know so far is the equation of state and we will be able to obtain some interesting development concerning the stratification and stability.

Appendices

Appendix

Clausius–Clapeyron Equation

In Fig. 3.5, the isothermal for water vapor is shown when condensation takes place. The gas is compressed up to the volume $V_D$ where condensation starts. This point on the temperature remains constant until all the vapor has condensed (point $V_A$). Fluid can be considered incompressible. Consider now two isotherms at temperature T and $T-\delta T$ and connect them with two adiabats $A-B$ and $C-D$ so that we have an elementary Carnot’s cycle. The efficiency of this cycle is dT/T while the work done corresponds to the area of the cycle $de_s(V_L-V_G)$ where $V_G$ indicates the volume of the gas and $V_L$ the same for the liquid. We have then

$$\begin{aligned} de_s(V_g-V_L)=L\frac{dT}{T}\approx de_s V_G \end{aligned}$$

Because $V_G>>V_L$. Now $V_G=R_v T/e_s$ so that

$$\begin{aligned} \frac{de_s}{dT}=\frac{Le_s}{R_vT^2} \end{aligned}$$

Fig. 3.5

On the left a couple of isotherms are represented in presence of the two phases (vapor and liquid). On the right is represented the elementary Carnot cycle

The once integrated gives Eq. (3.2). The same equation can now be written in the form

$$\begin{aligned} \frac{e_s}{e_0}=\exp \left[ \frac{L}{R_v}\left( \frac{1}{T_0}-\frac{1}{T}\right) \right] =\exp \left( \frac{L}{R_v T_0}\right) \exp \left( -\frac{L}{R_v T}\right) \end{aligned}$$

where we can also write

$$\begin{aligned} \frac{L}{R_vT}=\frac{L}{R_v(T_0+C)}=\frac{L}{R_v T_0}\left( \frac{1}{1+C/T_0}\right) =\frac{L}{R_v T_0}\left( 1-\frac{C}{T_0}\right) \end{aligned}$$

where C is the temperature in degrees centigrade and we use the approximation for $C/T_0<<1$. With the substitution we have

$$\begin{aligned} e_s=e_0\exp \left( \frac{LC}{R_v T_0^2}\right) \end{aligned}$$

(3.41)

with respect to (3.4) $A=e_0$ e $b=L/R_v T_0^2$. We then a value for $b=0.0726$ which is slightly different from the value given before. This is because b is obtained as a fit to the Clausius–Clapeyron equation.

Moist Lapse Rate

We have seen how to determine the dry adiabatic lapse and at the same time we also found an equation for the moist lapse rate with some approximation. Here we illustrate a more detailed approach. We recall the (3.18) and substitute $dq_s$ from the definition of $q_s=\epsilon e_s/p$ so that

$$\begin{aligned} \frac{dq_s}{q_s}=\frac{de_s}{e_s}-\frac{dp}{p} \end{aligned}$$

And after substitution in the (3.18)

$$\begin{aligned} -L q_s\left( \frac{de_s}{e_s}-\frac{dp}{p}\right) =c_pdT+gdz \end{aligned}$$

then we get $de_s/dz=(de_s/dT)(dt/dz)$ so that

$$\begin{aligned} -\frac{dT}{dz}\left( 1+\frac{Lq_s}{c_p e_s}\frac{de_s}{dT}\right) =\frac{g}{c_p}+\frac{Lq_s}{c_p}\frac{\rho g}{p} \end{aligned}$$

And finally substituting from the C-C equation $de_s/dT=L e_s/R_vT^2$. We obtain the correct expression for the saturated gradient

$$\begin{aligned} - \frac{dT}{dz}=\Gamma _s=\Gamma _d\left( 1+\frac{Lq_s}{RT}\right) \left( 1+\frac{\epsilon L^2}{c_p R}\frac{q_s}{T^2}\right) ^{-1} \end{aligned}$$

(3.42)

That is quite different from Eq. (3.20). The ratio between the terms in parenthesis is $c_p T/\epsilon L\approx 0.2$.

Note on the Thermodynamic Diagrams

Thermodynamics diagrams are used to visualize the evolution of thermodynamic quantities that characterize the air masses. The most simple of these diagrams is the emagram that represents an air mass in a $log(p/p_0)$ (as ordinate) and T in degrees centigrade as abscissa. This means that for a dry adiabat we have to represent the curve

$$\begin{aligned} \theta =T\left( \frac{p}{p_0}\right) ^{R/c_p} \end{aligned}$$

From which we obtain the “straight lines” in logarithm coordinates

$$\begin{aligned} -\log \left( \frac{p}{p_0}\right) =\frac{c_p}{R}\log \frac{T}{\theta }\approx \frac{c_p}{R}\left( 1-\frac{T}{\theta }\right) \end{aligned}$$

(3.43)

For a given value of $\theta$ this equation represents a line of negative slope as shown in Fig. 3.6. This suggests to use a coordinate system Y, X given by

$$\begin{aligned} Y=\log \left( \frac{p_0}{p}\right) ,\,\,\,\,\,\,\,\,\,\,\,\ X=T+K\cdot Y \end{aligned}$$

(3.44)

where K is the so-called asymmetry factor (skewness) that usually is assumed $35^\circ$. An elementary way to trace the isotherms is to fix T, for example, $20^\circ$ and calculating X for different values of Y. For example, $X=20+35 \log (0.001p)$ that gives $p=1000\,\text {hPa}, X=20$ while for $p=900\,\text {hPa}, X=23.68$. Then the isotherm at $20^\circ$ is a line passing for these two points. For the dry adiabat we need to fix the potential temperature, for example, $\theta =293.15\,\text {K}$ and based on the expression $\theta =T(p/p_0)^{R/c_p}$ calculate $T=20^\circ$ for $p=1000$ hPa while $T=11.3^\circ$ for $p=900$ hPa. The other curves to represent are the moist adiabat for assigned values of $\theta _e$ and the curve of constant relative humidity. For these we need to solve correctly the moist adiabat Eq. (3.42). It may be convenient to rewrite the temperature gradient as a function of pressure dT/dp

$$\begin{aligned} \frac{dT}{dp}=\frac{1}{p}\left[ \left( \frac{R}{c_p}\right) +\left( \frac{L}{c_p}\right) q_s\right] \left( 1+\frac{L^2 q_s \epsilon }{c_pRT^2}\right) ^{-1} \end{aligned}$$

(3.45)

Fig. 3.6

On the left the skew T-pressure diagrams is shown. Solid lines with positive slope are isotherms; the solid curves are dry adiabats and those long dashed lines are moist adiabats. The pointed lines represent constant mixing ratio. On the right it is shown a detail of the same diagram. $\theta _e$ is the equivalent potential temperature while $\theta _w$ is the wet bulb potential temperature

Using this equation we found that for the saturated adiabat $\theta _e=293.15$ K going from 1000 hPa to 900 hPa the temperature goes from $20^\circ$ to $16^\circ$. As for the constant relative humidity we can proceed in the following way. We start from the definition of mixing ratio as a function of pressure

$$\begin{aligned} q=\epsilon \frac{e}{p-e} \end{aligned}$$

From which we obtain the vapor pressure

$$\begin{aligned} e_s\approx \frac{1.6 q_s p}{1+1.6q_s} \end{aligned}$$

The saturation pressure is obtained from the semi-empirical law

$$\begin{aligned} e_s=6.109 \exp \left( \frac{17.625T}{T+243.04}\right) \end{aligned}$$

(3.46)

With T in $C^\circ$ from which we obtain T once we fixed p and $q_s$. For example, for $q_s=5\,\text {g}/\,\text {kg}$ for $p=1000$ hPa we have a temperature of about $3.7^\circ$ and $2.3^\circ$ for the pressure 900 hPa. Naturally in order to represent the curves we can utilize the function “contour” in MATLAB. This is done in Fig. 3.6.

Adiabatic Lapse Rate for the Ocean

We write in a slight different way the first principle

$$\begin{aligned} T d\eta =dE+p d v_s \end{aligned}$$

(3.47)

where now $d\eta =dE/T$ is the entropy (we do not call it s to confuse it with salinity) and $v_s$ is the specific volume. Notice that all the quantities in (3.47) are referred to the unit mass. The specific heat at constant volume becomes

$$\begin{aligned} c_v=T\left( \frac{\partial \eta }{\partial T}\right) _v=\left( \frac{\partial E}{\partial T}\right) _v \end{aligned}$$

(3.48)

And the one at constant pressure

$$\begin{aligned} c_p=T\left( \frac{\partial \eta }{\partial T}\right) _p=\left( \frac{\partial E}{\partial T}\right) _p+p\left( \frac{\partial v_s}{\partial T}\right) _p \end{aligned}$$

(3.49)

From (3.47) we have

$$\begin{aligned} T\left( \frac{\partial \eta }{\partial p}\right) _T=\left( \frac{\partial E}{\partial p}\right) _T+p\left( \frac{\partial v_s}{\partial p}\right) _T \end{aligned}$$

(3.50)

At this point we derive (3.49) with respect to p and subtract it from the (3.50) derived with respect to T to obtain

$$\begin{aligned} \left( \frac{\partial \eta }{\partial p}\right) _T=-\left( \frac{\partial v_s}{\partial T}\right) _p \end{aligned}$$

(3.51)

So that

$$\begin{aligned} T d\eta =T\left( \frac{\partial \eta }{\partial T}\right) _p dT+T\left( \frac{\partial \eta }{\partial p}\right) _T dp \end{aligned}$$

And substituting from the (3.49) and (3.51) we obtain

$$\begin{aligned} Td\eta =C_p dT-T\left( \frac{\partial v_s}{\partial T}\right) _p dp \end{aligned}$$

(3.52)

For an adiabatic transformation, we have

$$\begin{aligned} dT=\frac{T}{C_p}\left( \frac{\partial v_s}{\partial T}\right) _{p,s} dp=-\frac{T}{\rho ^2 C_p}\left( \frac{\partial \rho }{\partial T}\right) _{p,s}=\frac{\alpha _T T}{\rho C_p}dp \end{aligned}$$

(3.53)

where $\alpha _T$ is the thermal expansion coefficient and the adiabatic lapse rate becomes

$$\begin{aligned} \Gamma _d=-\frac{\partial T}{\partial z}=-\frac{\partial T}{\partial p}\frac{\partial p}{\partial z}=\frac{\alpha _T T}{\rho C_p}\rho g=\frac{g \alpha _T T}{C_p} \end{aligned}$$

(3.54)

In the more general case, we consider a parcel that moves vertically with adiabatic motion. Its density will change according to

$$\begin{aligned} d\rho =\left( \frac{\partial \rho }{\partial p}\right) _{T.s}dp+\left( \frac{\partial \rho }{\partial T}\right) _{p,s}dT \end{aligned}$$

(3.55)

We have $(\partial \rho /\partial T)_{p.s}=-\rho \alpha$ while $dp=-\rho g dz$ and $dT=\Gamma _d dz$ so that the (3.55) becomes

$$\begin{aligned} d\rho =\rho \left[ -g\left( \frac{\partial \rho }{\partial p}\right) _{T.s}dp+\alpha \Gamma _d\right] dz \end{aligned}$$

(3.56)

The change in density of the environment is given by

$$\begin{aligned} \frac{d\rho }{dz}dz=\left[ \left( \frac{\partial \rho }{\partial p}\right) _{T.s}\frac{dp}{dz}+\left( \frac{\partial \rho }{\partial T}\right) _{p.s}\frac{dT}{dz}+\left( \frac{\partial \rho }{\partial s}\right) _{p.T}\frac{ds}{dz}\right] dz \end{aligned}$$

(3.57)

Stability is assured when the change of (3.56) is greater than the one given by (3.57) so that the condition for stability is

$$\begin{aligned} \alpha _T\left( \frac{dT}{dz}+\Gamma _d\right) -\beta _s \frac{ds}{dz}>0 \end{aligned}$$

(3.58)

which is our (3.32).

Static Stability of the Ocean

The previous conclusions can be used to relate the static stability to buoyancy frequency N. Consider a water parcel at pressure p and depth $-z$ with salinity S that we move vertically without exchanging salt or heat with the environment up to the level $-z+\delta z$ and pressure $p-\delta p$. At this level the properties of the parcel are $\rho ', s, T+\delta T$. The change in temperature $\delta T$ corresponds to the change in pressure $\delta T=(dT/dp)_{ad}\delta p$. Substituting for the hydrostatic equilibrium we have $\delta T=-(dT/dp)_{ad}\rho g\delta z=-\Gamma _d\delta z$ where $\Gamma _d$ is the adiabatic lapse rate. The force acting on the parcel at level $-z+\delta z$ is

$$\begin{aligned} F=\delta V g(\rho _2-\rho ') \end{aligned}$$

(3.59)

Dove $\delta V$ is the volume of the parcel and $\rho _2$ is the density of the environment. The acceleration is then

$$\begin{aligned} a=\frac{F}{m}=\frac{\delta V g(\rho _2-\rho ')}{\delta V \rho '} \end{aligned}$$

(3.60)

The densities appearing in the (3.60) can be found in such way that

$$\begin{aligned} a=\frac{g}{\rho }\left[ \rho +\left( \frac{\partial \rho }{\partial z}\delta z \right) _e-\rho -\left( \frac{\partial \rho }{\partial z}\delta z \right) _p\right] \left[ 1+\left( \frac{1}{\rho }\frac{\partial \rho }{\partial z}\delta z\right) _p\right] ^{-1} \end{aligned}$$

(3.61)

where e indicates the environment and p the parcel. The environment change can be written as

$$\begin{aligned} \left[ \frac{\partial \rho }{\partial z}\delta z \right] _e=\left[ \frac{\partial \rho }{\partial S}\frac{\partial S}{\partial z}+\frac{\partial \rho }{\partial T}\frac{\partial T}{\partial z}+\frac{\partial \rho }{\partial p}\frac{\partial p}{\partial z}\right] _e\delta z \end{aligned}$$

While for the particle

$$\begin{aligned} \left[ \frac{\partial \rho }{\partial z}\delta z \right] _e=\left[ -\frac{\partial \rho }{\partial T}\Gamma _d+\frac{\partial \rho }{\partial p}\frac{\partial p}{\partial z}\right] _p\delta z \end{aligned}$$

At this point we can make a few simplifications. In (3.61) for $\delta z \rightarrow 0$ the denominator becomes 1. Also $(\partial p/\partial z)_w=(\partial p/\partial z)_e$ and also $(\partial \rho /\partial p)_w=(\partial \rho /\partial p)_e$ so that the (3.61) becomes

$$\begin{aligned} \frac{a}{g}=\frac{1}{\rho }\left[ \frac{\partial \rho }{\partial S}\frac{\partial S}{\partial z}+\frac{\partial \rho }{\partial T}\left( \frac{\partial T}{\partial z}+\Gamma _d\right) \right] \delta z \end{aligned}$$

(3.62)

It is clear that if $\delta z$ is negative, that is, the particles moves downward and the acceleration a is positive we have a stable situation and this means that the term in square parenthesis is negative. As in the case of the atmosphere in this case the parcel will oscillate around its initial position. We can define as stability the ratio

$$\begin{aligned} E=-\frac{1}{\delta z}\left( \frac{a}{g}\right) \end{aligned}$$

(3.63)

So that we will have stability for $E>0$ and instability for $E<0$. To find the Brunt–Väisällä frequency we observe that in (3.62) the term involving salinity is an order of magnitude less than the other terms and it can be neglected. So that we remain with

$$\begin{aligned} \frac{a}{g}=\frac{1}{\rho }\left[ \frac{\partial \rho }{\partial z}+\left( \frac{\partial \rho }{\partial T}\right) _{ad}\left( \frac{\partial T}{\partial z}\right) _{ad}\right] \delta z=\frac{1}{\rho }\left[ \frac{\partial \rho }{\partial z}+\left( \frac{\partial \rho }{\partial z}\right) _{ad}\right] \delta z \end{aligned}$$

where

$$\begin{aligned} \left( \frac{\partial \rho }{\partial z }\right) _{ad}=\left( \frac{\partial \rho }{\partial p }\frac{\partial p}{\partial z}\right) _{ad}=-\frac{g}{c^2} \end{aligned}$$

(3.64)

where we have used the hydrostatic equilibrium and c is the velocity of sound (a little bit different from (3.26)). The buoyancy frequency is then

$$\begin{aligned} N^2=gE=\left[ -\frac{1}{\rho }\frac{\partial \rho }{\partial z}-\frac{g}{c^2}\right] \end{aligned}$$

(3.65)

Data show that the oscillation periods change from a few minutes to a few hours in the deep ocean.