Question 1.2: Derive the material (or Stokes) derivative, D/Dt operating o......
Derive the material (or Stokes) derivative, \frac{D}{Dt} operating on the velocity vector, describing the “total time-rate-of-change” of a fluid flow field.
Hint: For illustration purposes, use an arbitrary velocity field, \vec{v} =\vec{v} (x, y,z;t) , and form its total differential.
Recall: The total differential of any continuous and differentiable function, such as \vec{v} =\vec{v} (x, y,z;t), can be expressed in terms of its infinitesimal contributions in terms of changes of the independent variables.
d\vec{v} =\frac{\partial \vec{v}}{\partial x} dx+\frac{\partial \vec{v}}{\partial y} dy+\frac{\partial \vec{v}}{\partial z} dz+\frac{\partial \vec{v}}{\partial t} dt
• Dividing through by dt and recognizing that dx/dt = u, dy/dt = v and dz/dt = w are the local velocity components, we have:
\frac{\partial \vec{v}}{\partial t} =\frac{\partial \vec{v}}{\partial x} u+\frac{\partial \vec{v}}{\partial y} v+\frac{\partial \vec{v}}{\partial z} w+\frac{\partial \vec{v}}{\partial t}
• Substituting the “particle dynamics” differential with the “fluid element” differential yields:
\frac{d\vec{v} }{dt} \hat{=} \frac{D\vec{v} }{Dt} =\frac{\partial\vec{v} }{\partial t} +u\frac{\partial\vec{v} }{\partial x}+v\frac{\partial\vec{v} }{\partial y}+w\frac{\partial\vec{v} }{\partial z}\equiv \frac{\partial\vec{v} }{\partial t}+(\vec{v} \cdot \,∇ ) \vec{v} =\vec{a} _{local}+\vec{a} _{conv}.