Pressure gradient force

The gradient of the pressure field Vp(r, t) is very important in meteorology. Consider a parcel of air contained in the rectangular parallelepiped dx dy dz, and whose center is located at the point r. This volume element is embedded in a surrounding pressure field p(r) = p(x,y, z). Let us compute the x component of the net force on the volume element. As indicated in Figure 9.8, the left-most face experiences a force due to the external field to the right:

the right-most face experiences a force to the left fright face = —p(x + idx, y, z) dy dz. (9.28)

The net force on the volume element is dFxnet = — (p(x + 2 dx, y, z) — p(x — 1 dx, y, z)) dy dz d p d p = —- dxdydz = —- dV (9.29)

dx dx where dV is the volume of the infinitesimal material element. Newton's Second Law (force is mass times acceleration) tells us that

where ax is the x component of acceleration and dM is the mass contained in the parcel. We can divide each side by dV and obtain d p

d x where p is the density of the air in the parcel. Put in more conventional form we have:

If we evaluate the y and z components in a similar fashion we can summarize the result in vector form a = —Vp P

[pressure gradient force/mass].

The force per unit mass a as given here is called the pressure gradient force. We have encountered its vertical component earlier in establishing the hydrostatic equation. Above the atmospheric boundary layer its horizontal components are very nearly balanced by the Coriolis force in midlatitudes (called geostrophic balance).

Example 9.9: horizontal acceleration of a parcel in the tropics Suppose a parcel whose density is 0.7 kg m—3 is embedded in a field of pressure with a gradient 10hPa over 1000 km. What is the acceleration of the parcel (ignoring the Coriolis force) and what is its increase in speed from rest in passing over 1000 km? Answer: The acceleration is toward low pressure and is given by a = 1.43 x

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