AP Physics 2 : Poiseuille Flow
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Example Questions
Example Question #1 : Poiseuille Flow
The mass flow rate through a cylindrical pipe of cross sectional area 
Assume
We will use Poiseuille's law to solve this problem:
We can determine the volumetric flow rate from the mass flow rate:
Now to determine an expression for radius:
Plugging this into the first expression, we get:
Now plugging in all of our values:
Example Question #2 : Poiseuille Flow
Water enters a circular tube at a velocity of 
Let's begin with the generic form of Poiseulle's law:
Where:
Substituting this in:
And:
Substituting this in:
Rearranging for diameter:
Plugging in values:
Example Question #3 : Poiseuille Flow
Dynamic viscosity of water:
Find the pressure change due to the Hagen-Poiseuille law in a 
None of these
Using
Where 
Plugging in values:
Example Question #4 : Poiseuille Flow
Dynamic viscosity of water:
Find the pressure change due to the Hagen-Poiseuille law in a 
None of these
Using
Where
Plugging in values:
Example Question #5 : Poiseuille Flow
Suppose two pipes made out of an identical material have the same length and the same liquid flowing through them. If pipe A has a cross-sectional diameter that is twice as great as pipe B, how does the flow rate in pipe A differ from the flow rate in pipe B?
The flow rate in pipe A is greater than pipe B by a factor of
The flow rate in pipe B is greater than pipe A by a factor of
The flow rate in pipe B is greater than pipe A by a factor of
The flow rate is the same in both pipes
The flow rate in pipe A is greater than pipe B by a factor of
The flow rate in pipe A is greater than pipe B by a factor of
For this question, we're asked to consider two pipes. Each pipe has the same length, is made out of the same material, and has the same fluid moving through it. The only difference is the cross-sectional diameter of these pipes. We're asked to find how the flow rate will differ between the two pipes.
In order to solve this question, we'll need to use Poiseuille's equation.
This equation tells us that the volume flow rate is directly proportional to two things: the pressure gradient between the ends of the pipe and the radius of the pipe raised to the fourth power. Moreover, the volume flow rate is inversely proportional to the viscosity of the fluid and also to the length of the pipe.
Now we need to ask the question - which of the variables in this equation is different in pipes A and B? Both pipes have the same length. Since each pipe has the same fluid moving through it, the viscosity will also be the same. Furthermore, we can assume that the pressure gradient at the end of each pipe is the same. The only thing that's left is the radius.
We're told that the diameter of pipe A is twice as great as pipe B. Since the radius is just half of the diameter, this also means that the radius of pipe A is twice as great as pipe B. Because the volume flow rate is dependent on the radius of the pipe raised to the fourth power, we can see that doubling the radius will result in a 
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