Question 4.1: Given the eulerian velocity vector field V = 3ti + xzj + ty^...

Given the eulerian velocity vector field
V = 3ti + xzj + ty^2k
find the total acceleration of a particle.

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• Assumptions: Given three known unsteady velocity components, u = 3t, v=xz, and w=ty^2.

• Approach: Carry out all the required derivatives with respect to (x, y, z, t), substitute into the total acceleration vector, Eq. (4.2), and collect terms.

a=\frac{dV}{dt}=\frac{\partial V}{\partial t}+\left(u\frac{\partial V}{\partial x}+v \frac{\partial V}{\partial y}+w\frac{\partial V}{\partial z}\right)=\frac{\partial V}{\partial t}+(V \cdot \nabla)V               (4.2)

                            Local                            Convective

• Solution step 1: First work out the local acceleration \partial V/\partial t:

\frac{\partial V}{\partial t}=i\frac{\partial u}{\partial t}+j\frac{\partial v}{\partial t}+k\frac{\partial w}{\partial t}=i\frac{\partial}{\partial t}(3t)+j\frac{\partial}{\partial t}(xz)+k\frac{\partial}{\partial t}(ty^2)=3i+0j+y^2k

• Solution step 2: In a similar manner, the convective acceleration terms, from Eq. (4.2), are

u\frac{\partial V}{\partial x} = (3t) \frac{\partial}{\partial x}(3ti + xzj + ty^2k) = (3t)(0i + zj + 0k) = 3tz  j

 

v \frac{\partial V}{\partial y} = (xz)\frac{\partial}{\partial y}(3ti + xzj + ty^2k) = (xz)(0i + 0j + 2tyk) = 2txyz  k

 

w \frac{\partial V}{\partial z} = (ty^2) \frac{\partial}{\partial z}(3ti + xzj + ty^2k) = (ty^2)(0i + xj + 0k) = txy^2  j

• Solution step 3: Combine all four terms above into the single “total” or “substantial” derivative:

\frac{dV}{dt}=\frac{\partial V}{\partial t}+u\frac{\partial V}{\partial x}+v \frac{\partial V}{\partial y}+w\frac{\partial V}{\partial z}=(3i + y^2k) + 3tzj + 2txyzk + txy^2j=3i + (3tz + txy^2)j + (y^2 + 2txyz)k

• Comments: Assuming that V is valid everywhere as given, this total acceleration vector dV/dt applies to all positions and times within the flow field.

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