3
$\begingroup$

I am working through David Andrews' "An introductuion to atmospheric physics". (2nd Ed)

Question 4.8 concerns finding the relationship between the angle a weather front makes with the ground and the temperatures and along front winds either side of it.

Atmospheric fronts are narrow regions of large horizontal temperature gradient. A simple model takes the front to be a sloping surface across which the temperature and along-front wind are discontinuous (with warm air overlying cold), but the pressure and cross-front wind are continuous. Take the y-axis along the front and the x-axis pointing towards the cold air and apply the hydrostatic and geostrophic wind relationships to the region AB in Figure 4.17. Hence show that the slope α at any level is related to the temperatures T1 and T2 and along-front winds v1 and v2 at that level by $(T_2-T_1)g\tan\alpha = f(v_1T_2 - v_2T_1).$

Figure 4.17:

Diagram of an idealised front

My progress:

Hydrostatic balance: $\frac{1}{\rho}\frac{\partial p}{\partial z} = -g$ Geostrophic balance (x direction): $\frac{1}{\rho}\frac{\partial p}{\partial x} = fv$

Integrating hydrostatic balance with respect to z:

$\int^{p(z+dz)}_{p(z)} \frac{1}{\rho} dp = -g dz$

Since $p = R_a \rho T$:

$-gdz = \int^{p(z+ dz)}_{p(z)} \frac{R_a T}{p} dp = R_a \ln p (T_2 - T_1)$ (Since pressure is assumed continuous and dz is assumed small.)

Integrating geostrophic balance similarly gives me:

$f dx = R_a \ln p (\frac{T_1}{v_1} - \frac{T_2}{v_2})$

Hence $\tan \alpha = \frac{dz}{dx} = \frac{f (T_2 - T_1) v_1 v_2}{g (T_2 v_1 - T_1 v_2)}$

This is not the same as the form we are told to aim for.

I would really appreciate knowing what I am doing wrong here!

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy