Along a streamline xy = 4, given u = 2x, v = -2y,(adsbygoogle = window.adsbygoogle || []).push({});

I was trying to calculate the acceleration at three different points A(1,4), B(2,2), C(4,1),

In my lecture notes, it is said that at A, the fluid is decelerating, at B, the acceleration is zero, and at C it is accelerating. However, i got some problem with my calculation at B.

V=ui+vj therefore the a = 4xi+4yj, when i plug in B, the acceleration is not zero. Why would it happen? Is there any problems with my calculation?

If i use another approach (i am not sure), V = sqrt(u^2+v^2), then

a = 2(u^2-v^2)/sqrt(u^2+v^2), when i plug in B, it is zero now. What is the difference between these two methods?

Another question is that in one of my practice problems, it is said that in a nozzle the cross-sectional area changes linearly from the base to the tip, does it mean that

A = A(in) + kx, where A is the cross-sectional area x m away from the inlet?

**Physics Forums | Science Articles, Homework Help, Discussion**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Fluid mechanics

**Physics Forums | Science Articles, Homework Help, Discussion**