### Video Transcript

Fluid flows through a tube at a rate of 1.00 times 10 to the two cubic centimeters per second. The pressure difference across the tube increases by a factor of 1.50. What is the flow rate through the tube if no other changes are made? A fluid with viscosity 3.00 times greater than that of the original fluid flows through the tube. What is the flow rate through the tube if no other changes are made? The length of the tube is increased by a factor of 4.00. What is the flow rate through the tube if no other changes are made? The radius of the tube is reduced by a factor of 10.0. What is the flow rate through the tube if no other changes are made? The radius of the tube is reduced by a factor of 10.0; the length of the tube is reduced by a factor of 2.00; and the pressure difference across the tube is increased by a factor of 1.50. What is the flow rate through the tube if no other changes are made?

In this exercise in each of the five scenarios, we want to solve for the flow rate capital πΉ. We imagine a cylindrical tube with fluid flowing through it. In scenario one, weβre told that the final pressure difference from one end of the tube to the other Ξπ sub π is equal to 1.50 times the initial pressure difference from one end of the tube to the other, Ξπ sub π. In scenario two, weβre told that the final viscosity of the fluid flowing to the tube, π sub π, is equal to 3.00 times the initial fluid viscosity. In the third scenario, the final length of the tube, πΏ sub π, equals 4.00 times the initial tube length, πΏ sub π. In our fourth case the final tube radius, π sub π, is equal to the initial tube radius, π sub π, divided by 10.0. And finally, in the fifth case, where π sub π the final radius equals the initial radius divided by 10.0, and the final length equals the initial length divided by 2.00, and the final change in pressure across the tube is equal to 1.50 times the change in initial pressure across the tube.

For each of these five different scenarios, we assume an initial flow rate, πΉ sub π, of 1.00 times 10 to the two cubic centimeters per second. As we approach each of these scenarios, letβs recall a mathematical relationship for the flow rate of fluid through a cylindrical tube. The flow rate πΉ is equal to π times Ξπ, the change in pressure from one end of the tube to the other, multiplied by the radius of the tube to the fourth power all divided by eight times the viscosity of the fluid flowing through the tube, π, multiplied by πΏ, the length of the tube. If we write the equation for our initial flow rate, πΉ sub π, that equals π times the change in pressure initially across the tube times the initial radius of the tube, π sub π, to the fourth divided by eight times the initial viscosity of the fluid flowing through to the tube multiplied by the initial tube length, πΏ sub π.

And this equals 1.00 times 10 to the two cubic centimeters per second. If we look at our initial scenario, where we see a pressure difference change from Ξπ sub π to 1.50 times Ξπ sub π, if we replace Ξπ sub π in our flow equation for πΉ sub π with 1.50 Ξπ sub π, now weβre solving not for πΉ sub π but for πΉ. And that will equal 1.50 times our initial flow rate. Calculate it out; that equals 150 cubic centimeters per second. Thatβs the new flow rate when the pressure difference across the tube goes up by 50 percent. And now letβs move on to look at our second scenario, when the viscosity of the fluid flowing through the tube increases by a factor 3.00.

Once again we write our equation for the initial flow rate, πΉ sub π. Now we replace π sub π in the denominator of this equation with 3.00 π sub π. When we do that to solve for the new flow rate, πΉ, we see that new flow rate is equal to the original flow rate divided by 3.00. This is equal to 33.0 cubic centimeters per second. Thatβs the new flow rate if the viscosity of the fluid flowing goes up by a factor of 3.00. Now letβs move on to the third scenario, where the length of the tube that the fluid flows through increases by a factor of 4.00. In our equation for initial flow rate, weβll replace πΏ sub π, the length in the denominator, with 4.00 πΏ sub π in order to solve for the new flow rate πΉ.

Thatβs equal to the original flow rate divided by 4.00, which equals 25.0 cubic centimeters per second. We see that when the length of the tube goes up by a factor of 4.00, the flow rate drops by that same factor. Now onto the fourth scenario, where the radius of our tube shrinks by a factor of 10.0. Now as we look at our equation for πΉ sub π, weβll replace π sub π, the initial radius, with π sub π divided by 10.0. With this change to our flow rate, the new flow rate πΉ is equal to the original flow rate divided by 10.0 to the fourth power. This is equal to 0.0100 cubic centimeters per second.

The flow rate decreases by a factor of 10000 when the radius decreases by a factor of 10.0. Now letβs move on to our last scenario, where instead of just changing one factor in the flow rate equation, we change three. In our equation for πΉ sub π, the original flow rate, weβll replace π sub π with π sub π divided by 10.0. Next, weβll replace πΏ sub π with πΏ sub π divided by 2.00. And finally, weβll replace Ξπ sub π with 1.50 times Ξπ sub π. With these three changes, our new flow rate πΉ is equal to our original flow rate multiplied by 1.50 divided by 10.0 to the fourth over 2.00. When we calculate this out, we find a new flow rate πΉ of 0.0300 cubic centimeters per second. Thatβs the new flow rate through the tube with these three changes made.