Thermal Wind Balance Questions and Answers

Answer: A

Differentiating the geostrophic relation with respect to pressure and substituting the hydrostatic relation eliminates the geopotential and leaves the vertical shear expressed through the horizontal temperature gradient. The thermal wind therefore inherits the validity conditions of both parent relations, failing wherever either fails. Continuity, the thermodynamic equation and the vorticity and divergence equations play no part in the derivation, though they are of course members of the same governing set and are used elsewhere in this subject for other purposes.

Answer: B

The thermal wind is the vector difference between the geostrophic winds at the top and bottom of a layer, so it is a shear and not a wind at all. No parcel of air ever moves with it, and it cannot be measured by an anemometer. The name is unfortunate and misleads almost everyone on first acquaintance, which is why it repays being stated explicitly. Keeping the definition clear prevents the common error of adding the thermal wind to an observed wind as though the two were quantities of the same kind.

Answer: C

The thermal wind blows along the layer mean isotherms with cold air on its left in the northern hemisphere, exactly as the geostrophic wind blows along the isobars with low pressure on its left. The analogy is exact because thickness plays the role of geopotential height in the relation. Perpendicular flow would describe maximum thermal advection instead, and there is no latitude dependence in the geometry, only in the magnitude through the Coriolis parameter.

Answer: D

Because low thickness corresponds to cold air and the thermal wind blows with low thickness on its left, cold air lies to the left in the northern hemisphere. The rule reverses in the southern hemisphere along with the sign of the Coriolis parameter. It is a direct transcription of Buys Ballot's law with thickness substituted for pressure, so anyone who has the geostrophic rule firmly in hand gets the thermal one free. The shared properties explain why these belong to one family, and the differences explain why more than one exists.

Answer: A

With cold air to the north, a wind blowing from the west has the cold air on its left, so the thermal wind is westerly and the geostrophic westerlies strengthen with height. This is why the mid-latitude jet stream exists at all. The summer stratosphere reverses the temperature gradient, with warm air over the sunlit pole, and its winds are correspondingly easterly, which is a useful check that the rule has been applied in the right sense. Most errors here come from carrying a result across from a neighbouring case where it does not hold.

Answer: B

The equator to pole temperature contrast requires, through the thermal wind relation, a westerly shear throughout the troposphere, which is why the wind at 250 hectopascals is so much stronger than that at the surface. The trade winds arise from the Hadley circulation, the Ekman spiral from friction, and the diurnal wind variation from the daily cycle of boundary layer mixing. Only the first is a direct thermal wind consequence and it is the most important single application of the relation.

Answer: C

Westerly shear accumulates upward through the whole troposphere while the temperature gradient persists, so the wind reaches its maximum where that gradient reverses, which is at the tropopause. Above it the temperature gradient changes sign and the westerlies weaken again. The surface gradient may be sharp, but shear must be integrated over depth to produce a strong wind, which is precisely why the jet sits high rather than low. Checking the units of each quantity catches a wrong formula before the arithmetic begins.

Answer: D

In the lower stratosphere the polar region is warmer than the tropics, because the tropical tropopause is much higher and therefore much colder, so the thermal wind reverses to easterly. Integrating that reversed shear upward reduces the westerly wind above the jet core. The Coriolis parameter does not change sign within a hemisphere, and the geostrophic relation remains perfectly valid in the stratosphere.

Answer: A

The hypsometric equation gives the geopotential thickness between two pressure surfaces as proportional to the mean virtual temperature of the layer between them. It follows directly from integrating the hydrostatic relation with the ideal gas law substituted. Because thickness depends on temperature alone, a thickness chart is effectively a temperature chart, which is the fact that makes the whole practice of thickness analysis possible. Checking the units of each quantity catches a wrong formula before the arithmetic begins.

Answer: B

Virtual temperature is used because water vapour is less dense than dry air, so moist air behaves as slightly warmer dry air, and the mean is taken with respect to the logarithm of pressure. In practice the difference between virtual and actual temperature reaches a few kelvin in the tropics and is negligible in polar air. Using the base temperature or a mid-level value would be an approximation to the correct mass-weighted mean. Framing the question as which quantity the change acts upon makes the answer follow without calculation.