Flow Rate Ratio of Two Tubes: Calculation

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SUMMARY

The discussion focuses on calculating the flow rate ratio of two tubes carrying an incompressible fluid with a viscosity of 1.5 Pl. Using Poiseuille's law, the initial calculations for flow rates F1 and F2 were incorrect due to a misunderstanding of the equation. The correct approach involves equating the pressure drops (ΔP) across both tubes and dividing the equations to find the flow rate ratio F1/F2, resulting in a final ratio of approximately 53.932.

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Homework Statement


Two tubes carry the same incompressible fluid with viscosity 1.5 Pl. They have lengths L1 = 6 and L2 = 22 m and diameters d1 = 1.2 and d2 = 4.5 cm. What is the ratio of their flow rates F1/F2?


Homework Equations


Poiseuille's law: 8nLI/(pi*r^4)
while n is viscosity
L is the length
R is radius


The Attempt at a Solution


F1= (8*1.5*6*I) / (pi*(1.2/2)^4)
F2= (8*1.5*22*I) / (pi*(4.5/2)^4)

The ratio F1/F2 is:
(8*1.5*6*I) / (pi*(1.2/2)^4)* (pi*(4.5/2)^4)/ (8*1.5*22*I)

F1/F2= 46.296*1.1649= 53.932.

And this is wrong, can u help??
 
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You need to write out your steps more clearly. 8nLI/(pi*r^4) is not Poiseuille's equation; Poiseuille equation is ΔP=8nLI/(pi*r^4), where I represents flow rate. You now have 2 equations:

ΔP1=8nLI1/(pi*r1^4)
ΔP2=8nLI2/(pi*r2^4)

If you assume the two ΔP's are the same and divide one equation by the other, you'll see your mistake.
 
Thank you a lot :)
 

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