# 1.11: Process Terminology

Processes associated with constant temperature are isothermal. For example, eqs. (1.9a) and (1.13a) apply for an isothermal atmosphere. Those occurring with constant pressure are isobaric. A line on a weather map connecting points of equal temperature is called an isotherm, while one connecting points of equal pressure is an isobar. Table 1-6 summarizes many of the process terms.

Table 1-6.Process names. (tendency = change with time)
Name Constant or equal

contour

isallobar

isallohypse

isallotherm

isanabat

isanomal

isentrope

isobar

isobath

isobathytherm

isoceraunic

isochrone

isodop

isodrosotherm

isoecho

isogon

isogram

isohel

isohume

isohyet

isohypse

isoline

isoneph

isopleth

isopycnic

isoshear

isostere

isotach

isotherm

entropy (no heat exchange)

height

pressure tendency

height tendency

temperature tendency

vertical wind speed

weather anomaly

entropy or potential temp.

pressure

water depth

depth of constant temperature

thunderstorm activity or freq.

time

dew-point temperature

wind direction

(generic, for any quantity)

sunshine

humidity

precipitation accumulation

height (similar to contour)

(generic, for any quantity)

cloudiness

(generic, for any quantity)

density

wind shear

specific volume (1/ρ)

speed

temperature

Sample Application

Name the process for constant density.

From Table 1-6: It is an isopycnal process.

Exposition: Isopycnics are used in oceanography, where both temperature and salinity affect density.

HIGHER MATH • Hypsometric Eq.

To derive eq. (1.26) from the ideal gas law and the hydrostatic equation, one must use calculus. It cannot be done using algebra alone. However, once the equation is derived, the answer is in algebraic form.

The derivation is shown here only to illustrate the need for calculus. Derivations will NOT be given for most of the other equations in this book. Students can take advanced meteorology courses, or read advanced textbooks, to find such derivations.

Derivation of the hypsometric equation:

Given: the hydrostatic eq:

\ \begin{align}\frac{dP}{dz}=-\rho\cdot |g|\tag{1.25c}\end{align}

and the ideal gas law:

\ \begin{align}P=\rho \cdot \Re_{d} \cdot T_{v}\tag{1.23}\end{align}

First, rearrange eq. (1.23) to solve for density:

$$\rho=P /\left(\Re_{d} \cdot T_{v}\right)$$

Then substitute this into (1.25c):

$$\frac{\mathrm{d} P}{\mathrm{d} z}=-\frac{P \cdot|g|}{\mathfrak{R}_{d} \cdot T_{v}}$$

One trick for integrating equations is to separate variables. Move all the pressure factors to one side, and all height factors to the other. Therefore, multiply both sides of the above equation by dz, and divide both sides by P.

$$\frac{\mathrm{d} P}{P}=-\frac{|g|}{\Re_{d} \cdot T_{v}} \mathrm{d} z$$

Compared to the other variables, g and ℜd are relatively constant, so we will assume that they are constant and separate them from the other variables. However, usually temperature varies with height: T(z). Thus:

$$\frac{\mathrm{d} P}{P}=-\frac{|g|}{\Re_{d}} \cdot \frac{\mathrm{d} z}{T_{v}(z)}$$

Next, integrate the whole eq. from some lower altitude z1 where the pressure is P1, to some higher altitude z2 where the pressure is P2:

$$\int_{P_{1}}^{P_{2}} \frac{\mathrm{d} P}{P}=-\frac{|g|}{\Re_{d}} \cdot \int_{z_{1}}^{z_{2}} \frac{\mathrm{d} z}{T_{v}(z)}$$

where |g|/ℜd is pulled out of the integral on the RHS because it is constant.

The left side of that equation integrates to become a natural logarithm (consult tables of integrals).

The right side of that equation is more difficult, because we don’t know the functional form of the vertical temperature profile. On any given day, the profile has a complex shape that is not conveniently described by an equation that can be integrated.

Instead, we will invoke the mean-value theorem of calculus to bring Tv out of the integral. The overbar denotes an average (over height, in this context).

That leaves only dz on the right side. After integrating, we get:

$$\left.\ln (P)\right|_{P_{1}} ^{P_{2}}=-\left.\frac{|g|}{\Re_{d}} \cdot \overline{\left(\frac{1}{T_{v}}\right)} \cdot z\right|_{z_{1}} ^{z_{2}}$$

Plugging in the upper and lower limits gives:

$$\ln \left(P_{2}\right)-\ln \left(P_{1}\right)=-\frac{|g|}{\Re_{d}} \cdot \overline{\left(\frac{1}{T_{v}}\right)} \cdot\left(z_{2}-z_{1}\right)$$

But the difference between two logarithms can be written as the ln of the ratio of their arguments:

$$\ln \left(\frac{P_{2}}{P_{1}}\right)=-\frac{|g|}{\Re_{d}} \cdot \overline{\left(\frac{1}{T_{v}}\right)} \cdot\left(z_{2}-z_{1}\right)$$

Recalling that ln(x) = –ln(1/x), then:

$$\ln \left(\frac{P_{1}}{P_{2}}\right)=\frac{|g|}{\Re_{d}} \cdot \overline{\left(\frac{1}{T_{v}}\right)} \cdot\left(z_{2}-z_{1}\right)$$

Rearranging and approximating $$\overline{1 / T_{v}} \approx 1 / \overline{T_{v}}$$ (which is NOT an identity), then one finally gets the hypsometric eq:

\ \begin{align} \left(z_{2}-z_{1}\right) \approx \frac{\Re_{d}}{|g|} \cdot \overline{T_{v}} \cdot \ln \left(\frac{P_{1}}{P_{2}}\right)\tag{1.26}\end{align}