Fluid flow acceleration in a channel

AI Thread Summary
In a fluid mechanics problem involving a two-dimensional converging channel, the discussion centers on determining flow acceleration along the x-axis with known cross-sectional areas and discharge. Participants clarify that discharge (Q) can be expressed as Q = A.v, where A is the area and v is the velocity. The acceleration can be derived using the relationship vdv/dx, assuming a flat velocity profile due to the gradual convergence of the channel. There is consensus that the problem likely expects the use of these equations, despite initial ambiguities regarding the terms. The conversation concludes with acknowledgment of the approach to solve for the velocity field and acceleration.
Proletario
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MODERATOR NOTE: THIS HOMEWORK WAS SUBMITTED TO A NON-HOMEWORK FORUM SO THERE IS NO TEMPLATE

Hi everyone,

I'm stuck on a fluid mechanics problem and maybe you could help me.

As shown in figure, there is a two-dimensional converging channel, in which cross-section area varies linearly with x. If Q is the discharge, determine the flow acceleration along the x-axis (A1, A2 and L are known).
6119a243-1265-4e02-b2cd-6078d0bd7c5a.jpe


I got for area: A(x)=(A2-A1/L).x+A1

I'm not sure what to do about Q, should Q=A.v be used? Any hint about how to get the velocity field?

Thanks in advance!
 
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You don't have any other indication about what they want you to use for discharge? That's an ambiguous term.
 
At steady state, the acceleration is vdv/dx. Are you familiar with this term in the momentum balance equations?

Chet
 
Chestermiller said:
At steady state, the acceleration is vdv/dx. Are you familiar with this term in the momentum balance equations?

Chet

Do you mean Navier-Stokes eq.? I still don't know how to get the velocity field... :frown:
 
boneh3ad said:
You don't have any other indication about what they want you to use for discharge? That's an ambiguous term.

No, that's all the information. I think discharge here means volume per time. That's why I tried to use Q=A.v
 
Proletario said:
Do you mean Navier-Stokes eq.? I still don't know how to get the velocity field... :frown:
What about getting v(x) from the two equations in your original post:
A(x)=(A2-A1/L).x+A1
and
Q=A.v

Chet
 
Chestermiller said:
What about getting v(x) from the two equations in your original post:
A(x)=(A2-A1/L).x+A1
and
Q=A.v

Chet

I thought the same, but isn't v the mean velocity? If I can use v as the v-field, then the acceleration is quite simple via material derivative.
 
Proletario said:
I thought the same, but isn't v the mean velocity? If I can use v as the v-field, then the acceleration is quite simple via material derivative.
It's pretty safe to say that this is what they expected you to do. They just forgot to mention that the channel is converging very gradually and that locally you can assume a flat velocity profile.

Chet
 
Chestermiller said:
It's pretty safe to say that this is what they expected you to do. They just forgot to mention that the channel is converging very gradually and that locally you can assume a flat velocity profile.

Chet

Wow, so obvious. Thanks Chet. :bow:
 
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