Posted: 10/26/2004 4:45:55 PM EST
Question 1:
The velocity of a certain twodimensional flow field is given by the equation. V = 2xti  2ytj where the velocity is in feet per second when x, y, and t are in feet and seconds, respectively. Determine expressions for the local and convective componets of acceleration in the x and y directions. What is the magnitude and direction of the velocity and the acceleration at the point x = y = 1 ft and at time t = 0? Question 2: Two fixed, horizontal parallel plates are spaced 0.2 in. apart. A viscous liquid ( mu = 8 x 10^(3) lb*s/ft^(2), SG = 0.9) flows between the plates with a mean velocity of 0.9 ft/s. Determine the pressure drop per unit length in the direction of flow. What is the maximum velocity in the channel? 


PLEASE DO NOT MASTERBATE HERE



I'm pretty sure the V stands for vagina and not velocity.



oh HELL NO


"Guard thy airspeed, lest the ground arise to smite thee"

anyone?



bump



1. It's been a few years but don't you just take the partial derivatives and insert the appropriate values of x, y and t into the resulting equations? (the derivatives below are partial derivatives): Local acc = dv/dt = 2x i  2y j. Insert x=1 and y=1 and you get: 2i2j convective acc = (dv/dx)(dx/dt)+(dv/dy)(dy/dt) = 2t i (Vi)  2t j (Vj) = 2t(2xt)i2t(2yt)j=(4xt^2)i(4yt^2)j. At t=0, convective acceleration is zero. Total acceleration=local acc+convective acc=2i2j Velocity=0 I can't help with the second one  I'd have to look at my old textbooks, which are in my basement. 


To find the answer to the first problem, look to the acceleration of the flow, which is just the first derivative of the velocity.
Can't help you with the second, need a reference. Besides, the fluid leaks out the edges of a channel. If it's a rectangular duct, that's another story. 


bump



Because the velocity of the flow field is a function of position and time, you can't just take the first derivative of the velocity wrt time. You need to use partial differential equations. Using multivariable calculus, we can obtain the following: acc=DV/Dt (capital Ds used for total derivatives, small ds used for partial) if V=function(x,y,t), then DV/Dt=(dV/dx)(dx/dt)+(dV/dy)(dy/dt)+(dV/dt) The last term (dV/dt) is known as local acceleration, which is acceleration due to changes in time. The first two terms (there would be a third if V was also a function of z) are convective acceleration, which is acceleration due to changes in position. I believe the answer I gave above was correct. By the way, it's almost certain that you WILL get a problem on your final exam where the flow is acclerating solely due to gravity and you'll be asked to determine local and convective acceleration. Here's a hint  one of them is zero. Can you guess which one? P.S. If you'd give me a front pivot pin, detent and spring, I'd gladly pull out my books and solve the second problem for you. 


please check you email.



danno, what I said is precisely correct  I'm trying to get scottryan to understand the fundamental underlying principle, not work the problem for him.



Why don't you do your own damn homework?



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