August 30th, 2014, 01:10 AM  #1 
Member Joined: Aug 2014 From: India Posts: 88 Thanks: 0  what is the mean velocity of this?
The Maximum velocity of a one dimensional incompressible fully developed viscous flow, between two fixed parallel plates, is 6 m/s. The Mean Velocity ( in m/s) of the flow is (A) 2 (B) 3 (C) 4 (D) 5 
August 30th, 2014, 02:55 PM  #2  
Global Moderator Joined: May 2007 Posts: 6,416 Thanks: 558  Quote:
Mean could = maximum or less depending on what else is going on.  
September 1st, 2014, 04:23 AM  #3 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,084 Thanks: 699 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
There is enough information in the question; you just need to assume laminar flow through a 2D duct. Have a think about what the velocity profile looks like. Then you can use the information you've been given to solve the problem. Hint... The velocity of fluid at the edges of the pipe is zero. What about the velocity of the fluid as you go towards the centre of the pipe? 
September 5th, 2014, 06:24 AM  #4 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,084 Thanks: 699 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
Since there's not been a reply for a while... A velocity profile between parallel plates has a parabolic shape and can be described by $\displaystyle v(x) = a  bx^2$ where I have positioned the curve so that the maximum velocity is at $\displaystyle x = 0$ and the velocity drops to zero at the walls of the pipe/plates, situated at $\displaystyle x = \frac{L}{2}$ and $\displaystyle x = \frac{L}{2}$. $\displaystyle L$ is the distance between the plates. When $\displaystyle x = 0$, $\displaystyle v = 6$ m/s, so $\displaystyle a = 6$ When $\displaystyle x = \frac{L}{2}$, $\displaystyle v = 0$, so $\displaystyle 6  b\left(\frac{L}{2}\right)^2 = 0$ $\displaystyle 6 = b\left(\frac{L}{2}\right)^2$ $\displaystyle b = 6\left(\frac{2}{L}\right)^2$ $\displaystyle b = \frac{24}{L^2}$ So $\displaystyle v(x) = 6  24\frac{x^2}{L^2}$ is the velocity profile of the flow between the parallel plates. The average speed can be found by integrating under the profile and dividing it by the total width of the pipe: $\displaystyle \overline{v} = \frac{1}{L}\int^{\frac{L}{2}}_{\frac{L}{2}}v(x) dx$ $\displaystyle = \frac{1}{L}\int^{\frac{L}{2}}_{\frac{L}{2}} \left(6  24\frac{x^2}{L^2}\right) dx$ $\displaystyle = \frac{1}{L}\int^{\frac{L}{2}}_{\frac{L}{2}}\left(6  24\frac{x^2}{L^2}\right) dx$ $\displaystyle = \frac{1}{L}\left[ 6x  8\frac{x^3}{L^2}\right]^{\frac{L}{2}}_{\frac{L}{2}}$ $\displaystyle = \frac{1}{L}\left(3L  L  (3L + L)\right)$ $\displaystyle = \frac{1}{L}\left(4L\right)$ $\displaystyle = 4$ m/s so the answer is c) 

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