Thermal Wind Balance Questions and Answers

Answer: A

The relation inherits the geostrophic breakdown near the equator, since dividing by a vanishing Coriolis parameter gives an unbounded shear for any temperature gradient. Mid-latitude free tropospheric flow is exactly where it works best. A barotropic layer simply gives zero thermal wind, which is a valid rather than an invalid result, and speed-only shear is an ordinary baroclinic case.

Answer: B

Because potential temperature surfaces slope where a horizontal temperature gradient exists on a pressure surface, the thermal wind may equivalently be written in terms of that isentropic slope. This form is convenient in isentropic coordinates and makes the connection with potential vorticity clearer. The vertical gradient of potential temperature is the stratification, which is a separate quantity entering the buoyancy frequency instead. The quantity that varies and the quantity that does not are worth naming before any substitution is attempted.

Answer: C

The Richardson number is the square of the buoyancy frequency divided by the square of the vertical shear, and values below one quarter indicate that shear instability is possible. Since the shear is given by the thermal wind, the Richardson number links the horizontal temperature gradient to the likelihood of turbulence. The Rossby, Froude and Burger numbers compare quite different pairs of effects.

Answer: D

Strong shear associated with the thermal wind can overcome the stratification, giving Kelvin-Helmholtz billows and the clear air turbulence encountered by aircraft near the tropopause. This is a direct practical consequence of thermal wind balance. Deep convection requires conditional instability rather than shear instability, and frontogenesis and wave breaking are horizontal processes on much larger scales. A limit that survives making every external component ideal is an intrinsic one, and worth knowing for that reason.

Answer: A

Although the surface temperature gradient may be sharp, the deep gradient through the middle and upper troposphere accumulates the greatest shear because the relation integrates over depth. This is why the strongest winds are found near the tropopause. Boundary layer gradients act over too shallow a depth, and the lower stratosphere has a reversed gradient that reduces the wind again.

Answer: B

The thermal wind is proportional to the logarithm of the pressure ratio across the layer, so a deeper layer with the same gradient accumulates more shear. That is why the wind grows steadily through the troposphere rather than jumping at any one level. The sign depends on the direction of the temperature gradient rather than on the position relative to the jet, and dilution is not how the relation works.

Answer: C

Integrating the hydrostatic relation gives a factor of the natural logarithm of the ratio of the bounding pressures, so the accumulated wind change depends on that logarithm. In geometric terms this is close to linear over a shallow layer but departs over a deep one because density falls with height. The relation does give a shear per unit pressure, but the question concerns the accumulated change across the layer. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

Answer: D

Because the westerly shear adds up through the whole troposphere, the wind reaches its maximum at the level where the temperature gradient reverses, which is the tropopause. The reason therefore explains the assertion completely. It also explains why the jet weakens again in the lower stratosphere, where the reversed gradient produces easterly shear that subtracts from the accumulated westerly wind below.

Answer: A

The reason gives the correct definition, and that definition is exactly why the assertion fails: a difference between two winds is a shear rather than a flow, and nothing moves with it. The misleading name is responsible for a great deal of confusion. Recognising that the quantity has the units of a wind without being one is the key step in using the relation confidently.

Answer: B

Veering places the thermal wind to the right of the lower level wind, and if the thermal wind has cold air on its left then the flow at that level must be coming from the warm side. The reason therefore explains the assertion exactly. The same argument run with backing gives cold advection, and both rules reverse in the southern hemisphere along with the sign of the Coriolis parameter.