
#1
Jan1510, 06:08 PM

P: 5

1. The problem statement, all variables and given/known data
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? 2. Relevant equations Poiseuille's law: 8nLI/(pi*r^4) while n is viscosity L is the length R is radius 3. 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?? 



#2
Jan1510, 06:22 PM

HW Helper
P: 2,324

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:
ΔP_{1}=8nLI_{1}/(pi*r_{1}^4) ΔP_{2}=8nLI_{2}/(pi*r_{2}^4) If you assume the two ΔP's are the same and divide one equation by the other, you'll see your mistake. 



#3
Jan1510, 06:32 PM

P: 5

Thank you alot :)



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