<<

A Generalization of the Thermal Equation to Arbitrary Horizontal Flow CAPT. GEORGE E. FORSYTHE, A.C. Hq., AAF Weather Service, Asheville, N. C.

INTRODUCTION N THE COURSE of his European upper-air analysis for the Army Air Forces, Major R. C. I Bundgaard found that the shear of the observed wind field was frequently not parallel to the isotherms, even when allowance was made for errors in measuring the wind and temperature. Deviations of as much as 30 degrees were occasionally found. Since the ther- mal wind relation was found to be very useful on the occasions where it did give the direction and spacing of the isotherms, an attempt was made to give qualitative rules for correcting the observed wind-shear vector, to make it agree more closely with the shear of the (and hence to make it lie along the isotherms). These rules were only partially correct and could not be made quantitative; no general rules were available. Bellamy, in a recent paper,1 has presented a formula for the gradient wind, using for thermodynamic parameters pressure altitude and specific virtual temperature anom- aly. Bellamy's discussion is inadequate, however, in that it fails to demonstrate or account for the difference in direction between the shear of the geostrophic wind and the shear of the gradient wind. The purpose of the present note is to derive a formula for the shear of the actual wind, assuming horizontal flow of the air in the absence of frictional forces, and to show how the direction and spacing of the virtual-temperature isotherms can be obtained from this shear. A method of drawing a sketch (see accompanying figure) is given which can be used qualita- tively to examine all possible cases of gradient .

NOTATION 2 v wind velocity for cyclonic (anticyclonic) cur- v wind speed vature v0 geostrophic wind velocity RH 1 /KH — radius of horizontal curva-

Vg geostrophic wind speed ture VH horizontal acceleration « specific volume

V dv/dt T* virtual temperature t unit vector along v p pressure k unit vector pointing toward zenith — V//( ) horizontal vector gradient of ( ) n k X t, the horizontal unit vector to — ( )p horizontal isobaric vector gra- left of v dient of ( ). [The variable ( )

geopotential face and the whole plot is pro- angle between v and v0 jected onto a horizontal plane. angular rotation of earth, positive — Vff( )p is then the gradient of (negative) in northern (south- ( ) on the horizontal plane. See ern) hemisphere page 212 of Holmboe, Forsythe, ttz il sin |

KH horizontal component of curvature S* specific virtual temperature anom- of trajectory, positive (negative) aly (see Bellamy, op. tit.).

1 John C. Bellamy, "The Use of Pressure Altitude and Altimeter Corrections in Meteorology," Journal of Meteorology, vol. 2, no. 1 (1945); section 2. 2 The notation is that of Holmboe, Forsythe, and Gustin, Dynamic Meteorology, John Wiley & Sons, 1945, q. v. for detailed explanation of the symbols used here.

Unauthenticated | Downloaded 09/28/21 03:50 PM UTC 372 BULLETIN AMERICAN METEOROLOGICAL SOCIETY [Vol. 26

DERIVATION OF GENERAL EQUATION The horizontal equation of motion of air in the absence of frictional forces may be formed by equating the horizontal acceleration (v#) to the sum of the horizontal per unit mass (— 2£ls Xv) and the horizontal pressure force per unit mass3 (— VH4>P):

(1) VH = - 2xv -VH4>P.

The acceleration V// can be written as

VH = vt + IPKHU-, and the Coriolis force can be written as

- 2£lz x v = - 2ttzvn. By substitution from the last two equations into (1), it is seen that

vt + vPKnii = — 2Qzvn — VHcf>Pf or

(2) vt + (2ttzv + v2KH)n = - Vh4>v-

It is convenient to multiply equation (2) vectorially with the unit vector k, thereby rotating all vectors clockwise by 90°, as seen from above:

- vn + (2VZV + V2KH)t = - X k.

Since vt = v, the last equation can be written in its final form:

(3) - vn + (2+ VKH)V = - VHCF>P X k.

If I) is set equal to zero, the solution v of equation (3) is the gradient wind; if both V and KH are set equal to zero, v becomes the geostrophic wind. Equation (3) is now ready to be differentiated with respect to geopotential ; we obtain

(4) - + (2fi2 + vKH) g + 5) v = _ fe) xk.

The left-hand side of equation (4) is already in finished form; there remains only to trans- form the right-hand side. Since 8 is assumed equal to — adp, — 5(V//<£p)/5<£ is equal to

5(VH4>P) /a8p. Now the operations 8/dp and V#( )p are independent of each other, and may be interchanged. Hence 5(Vh4>p)/a8p is equal to Vh(84>/8p)p/(x, or to — (Vhocp)/

xl VhTp* x 8 T*

The general thermal wind equation takes the final form

3 The vector — VHP is the equivalent for constant-pressure analysis of the horizontal pressure-force vector — OC^HV considered in constant-level analysis. Strictly speaking, the vectors — VHP and — AXIHV are equal only when the hydrostatic equation 8 = — a8p is satisfied—that is, when the air is stationary. Since 8 must be assumed equal to — a8p in the derivation of the thermal wind equation proper, it seems reasonable to accept their equality at this point.

Unauthenticated | Downloaded 09/28/21 03:50 PM UTC Nov., 1945] ARTICLES 373

In terms of the variables used by Bellamy, op. ext., equation (5) takes the form

DISCUSSION When V = KH = 0, equation (5) assumes its usual form (9) for the geostrophic wind, stating that the geostrophic wind shear is parallel to the isotherms of virtual temperature, with a strength directly proportional to the magnitude of the horizontal isobaric virtual temperature gradient and inversely proportional to the virtual temperature itself. When the wind is not geostrophic, the terms VKH8V/8, V8(VKH)/8, and — 8(VN)/8 itself, thus causing the vector — (1 /T*)(^HTP*) Xk sometimes to, deviate markedly from the vector 21225v/5. This fact undoubtedly accounts for the lack of success frequently experienced by upper-air analysts in attempting to orient and space isotherms by

(in effect) using the actual shear vector 2ttz8v/8 as the left-hand side of equation (9). Of the three correction terms on the left-hand side of equation (5), the term — 8(vn)/8 is undoubtedly the hardest to estimate in practice. The quantity v is seen from equation (2) to be equal to the component of the pressure-force — VnP in the direction of the wind v.

Let p denote the angle between the wind v and the geostrophic wind direction vg (with p taken positive for a wind blowing toward lower pressure). Then v = \VHP\ sin 0, and the geostrophic wind law states that = 2£lzvg. Hence, from the last two equations, v = 212zVg sin p, and

(6) - vn = — 2Qevg(sin p)n.

By differentiation of equation (6) with respect to geopotential, it is seen that

(7) - = - 202 g (sin p)n - 2Q^(cos P) g n - 2l2^(sin p) g-

Now p is very difficult to determine in actual upper-air analysis, and 8p/8 is still more diffi- cult. It appears, then, that — 8(vn)/8

THE CASE OF GRADIENT W^IND

Because of the difficulty of evaluating the term — 8(vn)/84> in equation (5), the practical use of that equation is probably limited to the case of gradient wind v (1) = 0):

6V 5V b{vKll) V T (8^ ) 90 si 4+ - nvKK h a 4+ - v = ~ » ** * kV>

and to the geostrophic wind VY {V = KH = 0):

(9)

It will be noted that the quantity VKH and its variation with geopotential comprise the only difference between equation (8) and the ordinary thermal wind formula (9) for the geostrophic wind. Therefore, assuming the wind to be gradient, it is necessary to know values only of v

Unauthenticated | Downloaded 09/28/21 03:50 PM UTC and VKH on various isobaric surfaces over a given station, in order to determine the direction and spacing of the virtual-temperature isotherms above the station. As usual, nature com- pensates: The advantage of the more exact equation (8) over the usual equation (9) is paid for justly by the requirement that, in addition to the wind v, the analyst must keep track of the quantity VKH. The price does not seem high. Comparison of equations (8) and (9) reveals the properties of the gradient wind shear. Since it is presumed that the reader is familiar with the use of the geostrophic wind shear for determining the virtual temperature field, the right-hand side will be eliminated between equations (8) and (9):

In practice, v is given at various levels from wind observations, and KH can be estimated at various levels from the curvature of the contours and speed of the systems. To obtain 2ttz5Vg/8(l> at the point A (see accompanying figure), one starts by plotting 2Qz8v/8(j> (vector AB in figure). To this is added: (i) the vector BC of length VKH\8V/84>\ in the direction of the shear AB; (ii) the vector CD of length V8(VKH)/84> in the direction of the wind v at A. The vector AD is then the shear vector for the geostrophic wind. Assuming the actual wind to be gradient, the virtual-temperature isotherms at A will be parallel to AD, instead of parallel to the shear AB of the actual wind. The procedure can be modified for use with layers of appreciable thickness by such devices as using a mean wind instead of v. While qualitative rules can be drawn from equation (10), they tend to be too complicated to use. The best qualitative use of the gradient thermal wind formula appears to be to draw a sketch like the accompanying figure for each shear vector in question. This is easily done, especially if the shear vectors 20z5v/5 have already been plotted on the charts.

Obtaining the geostrophic shear vector from the shear of the gradient wind.

It is seen from equation (8) that for the gradient wind shear 8v/8(f> (assumed different from 0) to be parallel to the isotherms, it is both necessary and sufficient that the term V8(VKH) /8 be parallel to 5v/5<£. And the latter is equivalent to saying that at least one of the following must be true: (i) v is zero; (ii) v is parallel to 8v/8] (iii) vKH does not change with height.

NUMERICAL EXAMPLE To illustrate the relative magnitudes of the three terms on the left-hand side of equation (10) for one particular case, let us consider an example at latitude 40°N, where 2ilz — 10~4. Mts units will be used. At one level, say 30,000 dynamic decimeters (near 10,000 ft), assume the wind v is gradient and is from 310° with a speed of 30 m s_1 (approx 67 mph). At the 33,000- dynamic-decimeter level (near 11,000 ft), also assume the wind is gradient but is from 302° with a speed of 34 m s_1 (approx 76 mph). Then 5v is from 258° with a magnitude of 6 m s-1. Assume that RH is equal to 6 degrees of latitude at both levels, or 0.67 X 106 m. Then, assuming cyclonic curvature, KH = 1 /RH = 1.5 X 10~6 m-1. The quantity vKH is 4.5 X 10"5 s_1 at the lower level and 5.1 X 10~5 s_1 at the upper level, whence 8(vKH) = 0.6 X 10-5 s-1.

Unauthenticated | Downloaded 09/28/21 03:50 PM UTC We take 8(f) as 3,000 dynamic decimeters (about 1,000 ft), and use the following mean values of v and vKH: v = 32 m s"1 from 306°, vKH = 4.8 X lO"5 s"1. We find that:

2Q* ^ = 2.0 X 10~7 m"1 from 258°; 8(f> 8v VKH r=1.0X 10-7 m"1 from 258°; 8(f)

8(VKH) v = 0.64 X 10"7 m"1 from 306° 8(f)

These three vectors are drawn to scale in the accompanying figure. The difference between the shear vector 5v/5<£ and the geostrophic shear vector 8vg/8(f) in this case is 9° in direction and 75 per cent in magnitude. No single example can be regarded as typical, but the above is certainly a possible one. Although KH and 5v/5<£ are fairly large, no change of curvature with height has been assumed. If the shear 5v/5<£ had been less in magnitude, other things being equal, the angle between 5v/50 and 8vg/8(f) would have been larger than 9°. For comparison, the magnitude of the term — 8(vn)/8(f> on the left-hand side of the complete equation (5) will be estimated. The example will be the same, except, of course, that the wind can no longer be assumed gradient. Assume that in equation (7) is 0° at the lower level (30,000 dyn dm) and 3° at the upper level (33,000 dyn dm). Then at the lower level equation (7) reduces to the following:

8(vn) 8(3

In this case the geostrophic wind speed turns out to be 39 m s-1 at the lower level, and at the same level it is found that

(11) - ^^ = 0.7 X 10~7 m"1 from 40°. 8(f)

The direction is known to be 40° at this level, because at a level where /3 is zero the wind must be gradient. For comparison, the term (11) has been plotted as a dotted vector in the accom- panying figure; the term is seen in this case to be an appreciable correction to 2ttz8v/8(f). How- ever, as stated previously, the impossibility of determining accurate to 2 or 3 degrees makes the term — 8(i)n)/8(f) usually impossible to consider in practice. The assumed change in 0 of 3 degrees per thousand feet is by no means extreme, and occasional observations show changes in (3 up to 9 degrees through the same thickness.

Melting Snow from Airfield by Underground Pipes The writer estimates that to eliminate all snow from a runway 1000 yd long by 100 ft wide under the worst weather conditions, would require approximately 72 tons of coal /24 hr, or an equivalent in fuel oil. Most efficient method of operation is to maintain a moder- ate heat when snow is predicted and then boost the temperature immediately as snow starts to fall. The heated-surface method of snow removal also eliminates the need of runways being temporarily out of use while being cleared by mechanical means. Added safety is another factor as there will be no ice-coated slipper surfaces. Cost compares well with that of mechanical snow removal. During an average winter it will be found necessary to take care of snowfalls for not more than 10-15 days at full power, and for no more than 60-70 days at 50% full power during periods of predicted snowfalls.—T. N. Addam, in Heat. $ Vent., June 1945, pp. 87-89.

Unauthenticated | Downloaded 09/28/21 03:50 PM UTC