Ram pressure

An example of a ram air turbine (RAT). RATs generate power by rotation of the turbine via ram pressure.

The Eulerian form of the Cauchy momentum equation for a fluid is[1]

${\displaystyle \rho {\frac {\partial {\vec {u}}}{\partial t}}=-{\vec {\nabla }}p-\rho ({\vec {u}}\cdot {\vec {\nabla }}){\vec {u}}+\rho {\vec {g}}}$

for isotropic pressure ${\displaystyle p}$, where ${\displaystyle {\vec {u}}}$ is fluid velocity, ${\displaystyle \rho }$ the fluid density, and ${\displaystyle {\vec {g}}}$ the gravitational acceleration. The Eulerian rate of change of momentum in direction ${\displaystyle i}$ at a point is thus (using Einstein notation):

${\displaystyle {\begin{aligned}{\frac {\partial }{\partial t}}(\rho u_{i})&=u_{i}{\frac {\partial \rho }{\partial t}}+\rho {\frac {\partial u_{i}}{\partial t}}\\&=u_{i}{\frac {\partial \rho }{\partial t}}-{\frac {\partial p}{\partial x_{i}}}-\rho u_{j}{\frac {\partial u_{i}}{\partial x_{j}}}+\rho g_{i}.\end{aligned}}}$

Substituting the conservation of mass, expressed as

${\displaystyle {\frac {\partial \rho }{\partial t}}=-{\vec {\nabla }}\cdot (\rho {\vec {u}})=-{\frac {\partial (\rho u_{j})}{\partial x_{j}}}}$,

this is equivalent to

${\displaystyle {\begin{aligned}{\frac {\partial }{\partial t}}(\rho u_{i})&=-{\frac {\partial p}{\partial x_{i}}}+\rho g_{i}-{\frac {\partial }{\partial x_{j}}}(\rho u_{i}u_{j})\\&=-{\frac {\partial }{\partial x_{j}}}\left[\delta _{ij}p+\rho u_{i}u_{j}\right]+\rho g_{i}\end{aligned}}}$

using the product rule and the Kronecker delta ${\displaystyle \delta _{ij}}$. The first term in the brackets is the isotropic thermal pressure, and the second is the ram pressure.

In this context, ram pressure is momentum transfer by advection (flow of matter carrying momentum across a surface into a body). The mass per unit second flowing into a volume ${\displaystyle V}$ bounded by a surface ${\displaystyle S}$ is

${\displaystyle -\oint _{S}\rho {\vec {u}}\cdot \mathrm {d} {\vec {S}}}$

and the momentum per second it carries into the body is

${\displaystyle -\oint _{S}{\vec {u}}\rho {\vec {u}}\cdot \mathrm {d} {\vec {S}}=-\int _{V}{\frac {\partial }{\partial x_{j}}}\left[u_{i}\rho u_{j}\right]\mathrm {d} V,}$

equal to the ram pressure term. This discussion can be extended to 'drag' forces; if all matter incident upon a surface transfers all its momentum to the volume, this is equivalent (in terms of momentum transfer) to the matter entering the volume (the context above). On the other hand, if only velocity perpendicular to the surface is transferred, there are no shear forces, and the effective pressure on that surface increases by

${\displaystyle P_{\text{ram}}=\rho u_{n}^{2}}$,

where ${\displaystyle u_{n}}$ is the velocity component perpendicular to the surface.

What is the sea level ram air pressure at 100 mph?

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