# Question Video: Applying Poiseuilles’s Law to Determine the Reduction in Radius of a Channel Corresponding to Particular Changes in Flow Rate and Pressure Difference

When physicians diagnose arterial blockages, they quote the reduction in flow rate. The flow rate in an artery has been reduced to 10.0% of its normal value by a blood clot and the average pressure difference between the ends of the artery has increased by 20.0%. By what factor has the clot reduced the radius of the artery?

05:18

### Video Transcript

When physicians diagnose arterial blockages, they quote the reduction in flow rate. The flow rate in an artery has been reduced to 10.0 percent of its normal value by a blood clot, and the average pressure difference between the ends of the artery has increased by 20.0 percent. By what factor has the clot reduced the radius of the artery?

Let’s begin our solution by drawing a diagram of the situation. We’re told about an artery that, due to a blood clot, experiences a 90 percent reduction in flow rate and a 20 percent increase in the pressure difference from one end of the artery to the other.

Let’s say that the artery drawn at the top labeled one is the initial artery in its original condition. Then the blood clot forms and the artery is effectively narrowed as shown by the number label two. Each artery has a flow rate — we’ll call them 𝐹 sub one and 𝐹 sub two — and a pressure difference across the artery from one end to the other — we’ll call them Δ𝑃 sub one and Δ𝑃 sub two, respectively.

The problem statement tells us that Δ𝑃 sub two is 20 percent more than Δ𝑃 sub one or that it equals 1.2 times Δ𝑃 sub one. And we’re told that 𝐹 two, the flow rate through the narrowed artery, is one-tenth as much as the original flow rate, 𝐹 sub one.

If we assume that each of these arteries is a perfect cylinder, then each can be described by a radius value 𝑟 sub one for the first artery and 𝑟 sub two for the second one. In our problem, we want to solve for the ratio 𝑟 sub two to 𝑟 sub one. That will tell us by what factor artery two is narrower than artery one.

Our topic in this segment is Bernoulli’s equation, and there is a helpful mathematical relationship for flow rate we can benefit from. this relationship is that 𝐹, the flow rate through an open cylinder, the flow rate 𝐹 is equal to 𝜋 times the pressure difference from one end of the cylinder to the other multiplied by the cylinder radius to the fourth power, all divided by eight times the Greek letter 𝜂, which represents the viscosity of the fluid in flow, multiplied by 𝐿, the length of the cylinder.

There are many factors involved in this relationship. Let’s see how it applies to our scenario by writing down the flow rate for artery two and the flow rate for artery one. Here we’ve written our fluid flow equations for artery two on top and artery one on bottom.

Since we’re moving toward finding the ratio 𝑟 sub two to 𝑟 sub one and we see those terms in our respective equations for 𝐹 sub two and 𝐹 sub one, let’s divide 𝐹 sub two by 𝐹 sub one on both sides. When we perform this operation, we see that there are a number of terms that cancel out. 𝜋 cancels out from the numerator of each fraction. Eight cancels out from the denominator, as does 𝜂, the viscosity, and 𝐿, the length of the artery.

This leaves us with a simplified ratio of 𝐹 two to 𝐹 one. Now let’s apply the relationships between the changes in pressure and 𝐹 two and 𝐹 one we’ve been given in the problem statement. Starting at the top, we can replace Δ𝑃 two with 1.2 Δ𝑃 one, and we can replace 𝐹 sub two with 0.1 𝐹 sub one.

We see with these substitutions that more cancelation is possible: 𝐹 one on the left side of our equation and Δ𝑃 one on the right side. Now our equation is simplified even more, and the only variables we’re left with are 𝑟 two and 𝑟 one.

Let’s first divide both sides by 1.2, canceling that term from the right side of our equation. And now as a final step, we can raise each side of our equation to the one-quarter power. Doing this cancels out the fourth power of each of our radii with the one-fourth, so we’re left with a simplified equation.

The ratio 𝑟 two to 𝑟 one is equal to 0.1 divided by 1.2 raised to the one-fourth power. When we evaluate this number, we find a result to three significant figures of 0.537. This is the ratio of the narrowed artery to the original wider artery. We see that the blood clot has caused roughly a 50 percent decrease in the artery’s radius.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.