I am currently making an exercise from Paterson,1983, IV.3.ii. It goes as follows

Water flows steadily and incompressibly along a pipe whose area cross section $A(x)$ varies slowly with the coordinate $x$ along the pipe. Use the conservation of mass and/or incompressibility to calculate the mean velocity along the pipe at $x$, and calculate the acceleration of a moving particle moving with this mean velocity. Take the mean velocity to be along the pipe and depending only on $x$.

I would solve the first part of the question as follows.

Let $A(0)$ and $v(0)$ be the cross section and the velocity at the begin of the pipe respectively. Since the medium is incompressible, it holds that $A(0) v(0) = A (x) v(x)$ for every $x \in [0, L]$ where $L$ is the length of the pipe. The local velocity in the x-direction at any point $x$ therefore is $$v(x)=\frac{A(0) v(0)}{A(x)}$$ So the average velocity of a fluid particle therefore is $$\langle v(x)\rangle=\frac{1}{L}\int_0^L \frac{A(0) v(0)}{A(x)} dx $$ So far so good.

It's the second part of the question I don't get: How can I know where the particle attains its average velocity and what the local acceleration is. I only know that it exists (intermediate value theorem for differentiable functions).

  • $\begingroup$ I would guess that in the question mean velocity doesn't mean the average velocity between the start of the pipe and $x$, it means the velocity at $x$ i.e. $v(x)$ as you've calculated it above. It's described as a mean velocity because in a liquid there will be random thermal motion (e.g. Brownian motion) so not every small element of the fluid will be moving at the same speed in the same direction. Your $v(x)$ is the mean velocity you get by averaging out all this thermal motion. The acceleration is just $\dot{v}(x)$ as usual. $\endgroup$ – John Rennie Dec 20 '13 at 17:02
  • $\begingroup$ The mean velocity is the quasi-one dimensional velocity at any given section, not for the whole flow. In reality, the velocity of the real flow would be a function of location within the cross section. In other words, $v_{mean}=\iint_{S}v(y,z)dydz=f(x)$ $\endgroup$ – Bryson S. Sep 16 '14 at 0:17

Spatially speaking, to figure out where it reaches it's mean velocity you will need to know $A(x)$. Your first equation will give you the location in terms of the area $$A(x_{\langle v(x) \rangle}) = \frac{A_0v_0}{\langle v(x) \rangle}$$ If, for example, you took the area to be linearly varying $$A(x)=A_0 + \alpha x$$ you could solve for the position and get $$x =\frac{1}{\alpha}\left(\frac{A_0v_0}{\langle v(x) \rangle}-A_0\right)$$ The acceleration of the fluid anywhere in the pipe can be achieved from the chain rule $$a = \frac{dv}{dt} = \frac{dx}{dt} \frac{dv}{dx}= v\frac{dv}{dx} = v \frac{d}{dx}\left(\frac{A_0v_0}{A(x)}\right)$$


After reading the book question again, I don't think it ever asks for $\langle v(x) \rangle$. Instead, all it asks for is $$v(x)=\frac{A_0v_0}{A(x)}$$ and $$a(x) = v(x)\frac{d}{dx}\left(\frac{A_0v_0}{A(x)}\right)$$

  • $\begingroup$ The final line of SimpleLikeAnEgg's answer is the key to your problem. Since the flow is steady, what they are asking for is essentially the convective acceleration $v\frac{dv}{dx}$. Since you can find v(x) via conservation of volume (conservation of mass for an incompressible flow), just multiply the local velocity by its gradient to get the local convective acceleration $a(x)$. If $\frac{dA}{dx}>0$ then $\frac{dv}{dx}$ should be $<0$ and vice versa. $\endgroup$ – Bryson S. Sep 16 '14 at 0:33

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