Ok, so convective acceleration is the change in velocity of a particle due to change in position, correct? If I have a 1-D flow where the velocity is a linear function of position, the convective acceleration is constant? Say v= k(1-x), the convective acceleration is -k?
I have an exam in a few days and this is the only thing I’m a little fuzzy on.
I’ve never run across the term convective acceleration specifically before, but it sounds to me like it’s referring to the, ah, convective part of the total derivative, written in 1D as Dv/Dt = dv/dt + vdv/dx. (I’m using the lower-case d to represent the partial derivative, del.) The “convective part” of that is vdv/dx.
If I’m right about that, then I don’t think the convective acceleration in your case is constant. dv/dx is constant, yes, but that’s a spatial velocity gradient. The acceleration felt by a particle of fluid (rate of velocity change with respect to time) in this case is v*dv/dx, or k[sup]2/sup.
If dv/dt = 0, then that term - v*dv/dx, represents the entire acceleration felt by the particle - which is what the quantity Dv/Dt represents.
If this isn’t clear, try this: Imagine a car accelerating from a standing start down a race track. Work out how specifying that dv/dt = const differs from saying that dv/dx = const. Remember the chain rule - dv/dt = dv/dx * dx/dt, where, you’ll note, dx/dt = v.
Ah, yes that makes sense. Thanks.
Yes, right. Constant acceleration happens when the speed is a linear function of time, not of distance. Just to confirm brad_d’s good answer.