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Fluid mechanics - convective acceleration

  • Thread starter stuey777
  • Start date
  • #1
3
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Hey guys,

I've been at this question for a while so I thought someone might be able to help me.

A 3m converging duct converges linearly, the inlet and outlet diameter's are 0.46m and 0.15m respectively. Also the volume flow rate (Q) is constant at all points in the duct and its value is 0.3m^3/s. Steady flow.
The question is to find the convective acceleration halfway along the duct.

It would be great if someone could get me on the right track. This should be an easy question.

Thanks, Stuart
 

Answers and Replies

  • #2
3
0
I know convective acceleration is v(dv/ds):

Q=u1*A1=u2*A2

Therefore u1=Q/A1 and u2=Q/A2, however I'm not sure if thats the correct way to find out velcities considering they both use Q.
In other texts for example u2=u1(A1/A2), but as for as i can see this is useless because not enough knowns are known.

Also I assume A1 should be in regards to 0.46m diamater and A2 should be halfway along the converging duct.

Therefore convective acceleration: v(dv/ds) = u2((u2-u1)/s2-s1)). But I compute this values and dont seem to get the correct answer at all.
 
  • #3
70
0
hi Stuey, is this duct having circular cross section or a rectangular cross section? the breath of the duct seems important. Then you should be able to find (v*dv/dx)@mid by finding dv/dx@mid, since v@mid is known.

[tex]\frac{dQ}{dx}=A\frac{dv}{dx}+v\frac{dA}{dx}=0[/tex]

where A stand for the area. try to solve for the value of dv/dx at mid point.
hope this help. :smile:
 
  • #4
3
0
Yes it's a circular cross section
 

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