What Is the Flow Rate in the New Pipe?

[mary d]

A viscous fluid is moving through a pipe. The flow is 2 x 10^-3 m3/s
then you have a second tube with a fluid with twice the viscosity which is moving in a pipe whose length is 3 times the original with a radius 1.5 times the original. The pressure difference across this new pipe is 1/3 that of the original. What is the flow in the new pipe?
Again I have no idea where to start. What is the formula? I wonder if anyone could figure out how fast my professor could travel down that pipe of course using the same diameter? Sorry I just had to say it. Thanks for any help!

 

[arcnets]

mary d,
maybe I'm mistaken but it seems to me that this lecture goes thru hydrodnamics at a very quick pace, deriving some formulae for practical use by engineers or scientists. It looks to me like the lecture's intention is not a profound understanding of physical principles, but rather to give the students a 'toolbox' for further use. Please correct me if I'm wrong.

This problem looks like an exercise in Hagen-Poiseuille's law, which states

dp/dz = -8 eta R^-2 v

where
z is horizontal position along pipe,
p is pressure at position z,
eta is viscosity of liquid,
R is radius of round pipe,
v is mean velocity of liquid.

 

[M98Ranger]

To calculate the flow in the new pipe, we can use the equation for volumetric flow rate, Q = (πr^4ΔP)/(8μL), where Q is the flow rate, r is the radius, ΔP is the pressure difference, μ is the viscosity, and L is the length of the pipe.

Since we know the flow rate and viscosity of the original pipe, we can set up a ratio between the two pipes: (Q1/μ1) = (Q2/μ2). We can then rearrange the equation to solve for Q2, the flow rate in the new pipe.

Q2 = (Q1 x μ2)/μ1

We also know that the length of the new pipe is 3 times longer and the radius is 1.5 times larger than the original pipe. So, we can substitute these values into the equation and solve for Q2:

Q2 = (2 x 10^-3 m^3/s x 2μ1)/(μ1 x 1.5^4)

Q2 = 0.53 x 10^-3 m^3/s

Therefore, the flow rate in the new pipe is approximately 0.53 x 10^-3 m^3/s.

As for the professor traveling down the pipe, it would depend on the velocity of the fluid, which is not given in the problem. However, it is safe to assume that the professor would be able to travel faster in the new pipe due to the higher viscosity and smaller pressure difference.