where in the last we used k = R/cP. Note that the formula fails when z >6 cp/g ~ 30 km.
The isentropic (6 = constant) profile is often observed in the daytime boundary layer and up to the LCL where the air is well mixed vertically. Above this level the lapse rate is smaller because of warming due to condensation of water vapor in the parcel into droplets. □
When does the hydrostatic approximation not work? This can happen in unusual circumstances, but first consider typical conditions. If the force of gravity is not balanced by the vertical gradient of the pressure field, the parcel must be accelerating vertically. Suppose the imbalance is 1% or an acceleration of ~ 0.1ms-2. After only 10 s a parcel starting at rest would have a vertical velocity of 1ms-1, a very rare occurrence except in a thunderstorm. Vertical velocities are typically of the order of 0.01 ms-1, which suggests that large imbalances are very rare. Arguments can also be constructed from three-dimensional considerations that at synoptic scales (scales that match the typical observing stations on a weather map, a few hundred kilometers) one finds that horizontal motions are typically on the order of 1 to 10 ms-1 and vertical motions are of the order of centimeters per second. These arguments can be found in the first few chapters of most dynamics books.
The mechanical potential energy per unit mass of a parcel (called the gravitational geopotential) is ^ (z) = gz where z is its height above some reference level (typically sea level).3 We can write for the change in potential from one level to another:
This is the amount of work performed in lifting a 1 kg parcel from zbeiow to zabove (not counting buoyancy forces, just gravity). The geopotential can be turned around slightly to be considered a function of the pressure level of the parcel. So instead of ^(z) we can think of V(p). This is just the gravitational potential energy per unit mass of a parcel at pressure level p. Now the change in gravitational potential energy in going from one pressure level to another is
where zPabove and zPbelow are the elevations above the reference level for which the pressures are pabove and pbelow. The geopotential height, Zp, is defined to be the height in meters of the pressure level for a given value of the potential energy per unit mass:
The geopotential height, Zp1 (x, y), is the elevation of the surface for a given pressure p = p1. The height of this constant pressure surface is a function of x and y (longitude and latitude) over the Earth's surface. All meteorologists are familiar with the 500 hPa height field, since it is so important in weather forecasting.
Example 6.6: height field of an isothermal atmosphere Suppose the temperature is everywhere T0. What is the 500 hPa height field?
3 If z were large enough we would have to take into account the z dependence of g = g(z) (due to the weakening of the gravitational force with distance from the Earth's center) and use fz g(z) dz, but this is seldom important in studies of weather and climate of the troposphere.
Answer: For an isothermal atmosphere we have p _ poe-z/H (6.32)
or after taking natural logs of each side, z = H ln —. (6.33)
For a temperature of 3oo K, the scale height H _ Rd To/g is 8786 m. If the reference pressure is fixed at 1ooohPa, using ln 2 = o.693, we have Z5oo _ 6o9om.
Note that because it is much colder at the poles, the 5oohPa height field is lower at the poles (T ~ 25o K) than at the Equator (TEquator ~ 3oo K), or about 5ooo m versus 6ooo m. Roughly speaking, the height field scales inversely to T, but remember this is for an isothermal atmosphere.
In our solution we also had to specify the surface pressure. If there is more atmospheric mass above one position on the Earth (as measured by surface pressure), this will lift the geopotential height field. So the height field is determined by the amount of mass above the reference surface (this sets the surface pressure) at a point and the thermal structure of the air above that point. Keep in mind that the surface pressure (proportional to the mass above the site) varies no more than a few percent from time to time. It falls as much as 8% or 9% in the eye of the most intense (hurricane strength) tropical storms. □
The vertically averaged temperature in a layer can be defined as rp above t dlnp
T _ " pbelow rpabove dlnp ^pbelow r
where d ln p _ dp/p. (Note that dz a -dp/p at least locally according to the hydrostatic equation.) This justifies the use of d ln p as our integration increment.
After applying the hydrostatic equation we find:
g zbelow dz g z z
R ln(pabove/pbelow) R ln(pabove/pbelow)
zabove - zbelow _ — In-_ H ln- [thickness] (6.36)
g pabove pabove and for dry air H _ 29.3T is the (local) scale height. The quantity Az _ zabove -zbelow is called the thickness of the layer lying between the two pressure surfaces.
Clearly the thickness is a measure of the local vertically averaged temperature of the layer.
Example 6.7 How thick is the 500 to 600hPa layer when the average temperature
Example 6.8 The 1000 hPa to 500 hPa thickness is often used in weather discussions to describe the average temperature in the lower part of the troposphere. What is the average temperature for a sea level temperature of 295 K and a lapse rate of 6Kkm—11?
Answer: Use T = (1/Az) f^z (T0 — Tz) dz and (6.36). Then the temperature T = 278 K. The thickness is « 5647 m. □
It is important to notice that thickness is a measure of temperature. Cold layers are thin, warm layers thick, in direct proportion to the average Kelvin temperature in the layer.
In a fluid such as the atmosphere a low density parcel embedded in a denser environment will rise. Consider a fluid of uniform density. Figure 6.6 shows an arbitrary volume isolated inside a box of fluid of uniform density (picture a thin film enclosing a portion of the same uniform fluid). The fluid including the enclosed parcel is at rest. Therefore, all components of the forces acting on the enclosed interior portion (parcel) must sum vectorially to zero. We need only consider
Figure 6.6 Diagram of an irregular volume in a fluid. The weight of the mass in the volume is peVg, where pe is the density of the fluid, V is the volume and g is the acceleration due to gravity. The pressure forces of surrounding fluid indicated by the arrows pointing inward exactly balance the downward pointing force due to gravity. But if a fluid of lower density, say p, is substituted in the same irregular volume, there will be a net upwards force on this mass.
6.4 Archimedes' Principle
the vertical components, since the horizontal components balance without any gravitational contribution. The weight of the interior volume indicated by the downward pointing vector (see Figure 6.6) has magnitude Mg = peVg. The subscript e indicates the density of the uniform fluid (the environment). The pressure forces exerted by the surrounding fluid must add up vectorially to a vector pointing upwards with magnitude Mg. Now suppose the isolated volume is carved out and refilled with matter of a different density, say p. The force exerted on this volume (parcel) is upwards and of magnitude peVg, but the weight of the parcel is only p Vg. There is thus a net upward force of
If the density of the parcel p is less than the environmental density, the force will be upwards. This is the buoyancy force on the parcel and the formula is called Archimedes' Principle. Often we want to know the force per unit mass or acceleration. We can use Newton's Second Law (F = Ma where a is the vertical acceleration):
We may use p = pe and for an ideal gas p = p/RT; hence a=< T,-11 Tg
[acceleration by buoyancy]. (6.41)
Note that if T > Te, the force is upward; and if T < Te, the force is downward. (Proof that warm air rises as we learned in elementary school!)
Example 6.9 Consider a thunderstorm shaft. Let the temperature inside the shaft near the surface be 5 K warmer than that outside ~ 300 K. Then the vertical acceleration of a parcel in the shaft is approximately 0.163 m s-2. If the acceleration is constant the velocity after time t is v = at and the altitude is z = 2at2. The parcel reaches 4 km in about 3.7 min at which time it has a vertical velocity of 36 m s-1.
Of course, this analysis is very simplified because the effects of latent heat release are ignored. In addition it was assumed that the differential temperature between the interior of the parcel and the environment remained constant during the ascent. In ensuing sections we will see how these effects are taken into account. □
A slab of air (parcel) sits at rest in an environment with a certain temperature profile. The forces are balanced on the parcel. What happens if the parcel is nudged upwards or downwards by a tiny amount? Do the forces in the new position balance or do they impart a restoring force or perhaps a repelling force? First consider a parcel at point A in Figure 6.7. Note that a straight line segment slanted upwards into the upper left quadrant direction passes through the point at A. This straight line segment is a dry adiabat passing through A. If a parcel is lifted adiabatically at A, it moves along this adiabat and it finds itself warmer than the environmental curve which lies to its left. In this case the parcel, being warmer than the environment, will experience a buoyant force upwards. Hence, a parcel at point A under a small perturbation upwards will experience a net buoyant force to continue going upward. If the same parcel is nudged downwards from A it will experience a downward buoyant force. We say the point A is unstable with respect to dry convection.
Next consider the point C. An analysis similar to the above shows that the point C is stable with respect to infinitesimal perturbations. The point B is neutral. In fact, we can quickly see that all points between the surface and B along the environmental curve are unstable. We say this is an unstable layer. Similarly, the points above B on the curve as shown form a stable layer.
Obviously, the key to these analyses is the local slope of the environmental profile compared to the slope of a dry adiabat. Clearly dr dZ
The vertical derivative of the potential temperature (see Section 3.3.3), d6/dz, provides an even simpler rule. The derivative of 6 with respect to z can be calculated from its definition, 6 = T (p/p0)—K (with k = Rd/cp):
T dz V p J p dz ' The hydrostatic equation can be used to simplify the second term:
This gives d6 6
— = — (Td — Te) [derivative of potential temperature]. dz T
In the last equation we introduced the local environmental lapse rate: Te This last expression immediately tells us that d6
z neutral neutral
Figure 6.8 Schematic of three soundings on a 0-z diagram. The one with dO /dz > o is stable; the neutral sounding has dO /dz _ o; the unstable one has dO/dz < o.
In the 0 versus z diagram, the dry adiabats are vertical lines (Figure 6.8). If 0 is increasing with altitude, the layer is stable. If 0 is decreasing with altitude, the layer is unstable. We will talk more about stability in Chapter 7, when we work with thermodynamic diagrams.
Consider a level z0 in a stable layer of the atmosphere such as point C in Figure 6.7. Since the layer is stable, a parcel displaced upwards will experience a restoring force tending to push it back to the point z0. Similarly a downwards displacement below zo will result in an upwards restoring force. The acceleration a of a parcel slightly displaced along a dry adiabat at this point is given by Archimedes' Principle:
where Te(z) is the environmental temperature profile or sounding; Ta(z) is the local adiabat passing through the curve Te(z) at height z0 (this is point C in Figure 6.7); T0 is the value of the environmental temperature at the point of intersection, z0, T0 = Te(z0). Both environmental and adiabatic curves cross at this point. Recall that dTe/dz = —Te and dTa/dz = —Td. Then the environmental and adiabatic temperatures near the point z0 can be written as:
where we have used the approximate sign (~) to indicate that we are using only the tangent to the environmental curve; of course, the adiabatic lapse rate curve is a straight line, so the approximation is not necessary in the second equation. For notational convenience let x = z — z0. Then we can equate the acceleration to the second time derivative of x:
This is the familiar harmonic oscillator equation, whose solution is x(t) = A sin ut + B cos ut
where A and B are constants depending on the initial conditions,4 and (in units rad s—!)
This frequency is called the Brunt-Vaisala frequency. The frequency in cycles per second (Hz), f = u/2n, is
Physics refresher: oscillator notation The angular frequency of a linear oscillator is denoted by u. Its units are radians per second. The corresponding frequency f is given by u/2n in units of cycles per second or hertz. The period of the oscillation is P = 1/f = 2n/u.
When the atmosphere is stable (Td > re) a parcel will oscillate; if the atmosphere is unstable, then u2 < 0 and the trigonometric functions become a mixture of exponentials at least one of which is growing in time. To see this go back to the differential equation for the oscillator (6.54) and instead of —u2 insert X2 > 0. Then the solutions become AeXt + Be—Xt. (You can insert this in the differential equation to satisfy yourself.) This means that instead of oscillating, the parcel "runs away." Another important observation is that for small amplitude oscillations (when the linear formula is valid) the frequency is independent of the amplitude of the displacement.
4 We usually are not interested in the initial conditions of the oscillation, but rather the (angular) frequency, rn.
An alternative expression for m in terms of the potential temperature is often useful. Referring to the last subsection we find:
This particular representation shows explicitly that if d0/dz is positive (stable atmospheric layer), then the quantity m is a real number and the oscillations will occur. If the slope is negative, then the frequency becomes an imaginary number, which means that the parcel's displacement will grow exponentially in time, either up or down depending on the initial perturbation. Of course in the real atmosphere the parcel does not accelerate all the way to infinity, but instead its motion is limited by the failure of our linear analysis which assumed small deviations. Unstable layers lead to overturning and mixing of the parcels within the layer.
Example 6.10 Suppose the environmental lapse rate is 0Kkm-1 (isothermal atmosphere) and T0 = 300 K, then what is the oscillation frequency?
which corresponds to a period of 5.7 min. □
It is typical of atmospheric vertical oscillations that the characteristic time is of the order of a few minutes.
6.7 Where is the LCL?
In this section we derive an approximate formula that shows that the LCL (lifting condensation level, see Section 5.9) is determined as a function of the temperature and mixing ratio, T0 and w0, of a parcel at the surface. Consider what happens to the saturation vapor pressure for this parcel. As the parcel rises its temperature falls from its surface value,
where Ta (z) is used to denote the temperature of the parcel as it rises adiabatically to altitude z above the surface. The saturation vapor pressure only depends on the temperature inside the parcel. We can use the integrated form of the Clausius-Clapeyron equation (5.15) to evaluate the saturation vapor pressure es(Ta(z)) in the parcel as a function of altitude (along the dry adiabat):
where Rw = 461.5 Jkg-1 K-1 is the gas constant for water vapor, and L is the enthalpy of vaporization (latent heat) for water (L = 2.500 x 106 J kg-1). We can simplify the last equation by noting that
which leads to:
As the parcel is lifted adiabatically (by some external mechanism) its mixing ratio, w = w0, will remain fixed since it is a conservative quantity. The external air pressure can be written as an exponential function, p0e-z/H, to a good approximation, where H is the atmospheric scale height. As a parcel rises adiabatically its internal pressure will adjust to that of the external pressure at each altitude z. Since w0 = ee(z)/p(z) we can obtain a formula for the vertical dependence of the actual vapor pressure e(z) in the parcel e(z) = = 1.608 W0 p0e-z/H (6.65)
e where the atmospheric scale height, H, can be taken to be RdT0/g with T0 the temperature near the surface. The value of Hw 1.7 km) is much smaller than typical values of H .An example is shown in Figure 6.9 where H was taken to be 8.3 km. At the surface e(z) ^ es(z), but the gap narrows as the parcel is lifted adiabatically. The value of e(z) will catch up with es (z) as z increases. The saturation mixing ratio changes as the parcel rises, as shown in Figure 6.10.
Consider the T-z diagram shown in Figure 6.11 to see how the temperature of the parcel decreases linearly with altitude as it rises. At the same time the temperature versus height for ws = constant also decreases but more slowly. The intersection of the two curves is the LCL. This latest view is the one usually shown in thermodynamic diagrams which are the subject of the next chapter.
By equating the forms for es(T(z)) to that of e(z) we find a formula for zlcl ln (es(T0)e/w0P0) ,, , zLCL = 1/Hw- 1/H ■ (6"66)
We can simplify this further by noting that the argument of the logarithm reduces to ws/w0 = 1/r where r is the relative humidity at the reference level (surface or sea level):
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