Transfer function of flow measurement system HELP

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
4 replies · 3K views
cabellos2
Messages
13
Reaction score
0
Transfer function of flow measurement system HELP!

1.

I have attached the question, which is to derive the transfer function of a volumetric flow measurement system...

2.

I know of the following relevant equations:

F = Ma = MD^2.x

Flow f = area x velocity

and f through a restriction f = C(p2-p1)

Also Force F = pressure x Area

3.

When considering the Mass alone I have

Sum of forces F = F - kx - BDx = MD^2.x

therefore x/F = 1 / K + BD + MD^2

and finally, x/F = (1/k) / 1 + (B/k)D + (M/k)D^2




However I am struggling to break the problem down further. Which parts of the system do I also need to consider and how do I go about manipulation of the equations to derive the transfer function...??

Your help is very much appreciated.
 

Attachments

Physics news on Phys.org


You have an equation relating pressure difference to volumetric flow. The pressure on each side of the transducer would have a contribution to force on the transducer plate.
 


I am still struggling with this one I am afraid. I can't figure out where the 2Mf/k . A^2/a^2 term originates from...

Some guidance would be very much appreciated...
 


Ok this is where I have solved to thus far:

I have the equation x/F = 1 / (k + BD + MD^2)

but F = Pressure x Area

so Pressure difference on transducer mass P2 - P3 x Area = Force (F)

therefore,

x / A (P2-P3) = 1 / (k + BD + MD^2)

and then,

x / (P2 - P3) = A / (k + BD + MD^2)

then finally divide rhs of equation by k to give,

x / (P2 - P3) = A/k / ((1 + B/k(D) + M/k(D^2))

My next equation of (P2 - P3)C = f

so substituting this in gives,

x/flow = A/CK / ((1 + B/k(D) + M/k(D^2))

BUT where does the A^2 / a^2 term come in...??

Any pointers?

Thanks
 


I would guess this.
You may consider the smaller diameter pipes to be obstructions. Whenever the plate would be in motion [itex]\left( \frac{dx}{dt} \neq 0 \right)[/itex] then fluid would be flowing through these pipes. Thus p1 would not be equal to p2 and p3 would not be equal to p4.

Even though fluid would not pass through the plate, you may consider the path through both smaller pipes as one flow.