Skip to main content
Ch 12: Fluid Mechanics

Chapter 12, Problem 12

BIO. Artery Blockage. A medical technician is trying to determine what percentage of a patient's artery is blocked by plaque. To do this, she measures the blood pressure just before the region of blockage and finds that it is 1.20×10^4 Pa, while in the region of blockage it is 1.15×10^4 Pa. Furthermore, she knows that blood flowing through the normal artery just before the point of blockage is traveling at 30.0 cm/s, and the specific gravity of this patient's blood is 1.06. What percentage of the cross-sectional area of the patient's artery is blocked by the plaque?

Verified Solution
Video duration:
7m
This video solution was recommended by our tutors as helpful for the problem above.
1199
views
Was this helpful?

Video transcript

Hey everyone. So this problem is working with fluid flow. Let's see what they're giving us and see what they're asking. So we know that there is a pipe in that pipe. Some amount of that pipe is blocked by scaling. The team measures the pressure at two different places. The first place, it measures a pressure of 2.6 times to the fifth pascal's and then the second pressure at the narrow region is 2.5 times 10 to the fifth pascal's. They use flow rates and determine that the flow at the kind of open area, the area before the scaling is 4m/s. And they have determined that the fluid has a specific gravity of .896. We are asked to calculate the percentage of the pipes area that is occupied by the scales. So the first thing that we need to do here is recognize that when we're working with the same fluid flowing along a pipe in two different areas, that's most likely the Bernoulli's equation type problem. So let's recall that equation first, See what we have and see what we need to find. So Bernoulli's equation is P one plus one half roe V one squared equals P two plus one half row V two squared. They don't tell us anything about height differentials or anything like that. So we can kind of ignore the height term of that. Bernoulli's equation. We also need to recall our continuity equation which is given as a one V one equals a two V two for flow at two different points in the same pipe and then they do talk about here, specific gravity. So let's just recall the specific gravity is the density of the fluid over the density of water, where the density of water is a given is a constant. Sorry, it's a 1000 kg per meters cubed. And so from there we can calculate the density of this fluid Is 897. kg Per meter. Cute! Alright, let's go back into the problem and see what else they tell us. So, P one is given to us as 2.6 times 10 to the 5th Pascal's. P two is 2. times 10 to the 5th Pascal's and V one. This is that region before the scaling Is given as four m/s. Okay, so um From there we can rearrange the Bernoulli equation to solve for V because in the end right there asking about area, so we're going to want to solve the continuity equation as our last step. So we know P 12. times 10 to the fifth pascal's plus one half. We now know our density of fluid. 897.6 kilograms per meters cubed. Times are V one which is four m per second. That term is squared Equals P two. That was also given two times 10 to the 5th Pascal's plus one half. Again, density. We already solved for times V two squared. So we're gonna plug all of that into our calculator and Soul for V two, B two is 11.8 m/s. So now we're gonna use that to solve in first to solve our area using our continuity equation. So A one B one equals a two B two. We actually don't have either area, but they're asking us for a percentage of area that's occupied by scales. So let's just solve a one in terms of A two and go from there. Okay, so A one equals a two times V two over V one. We know V two and V one now. So that comes out to a two equals 2.95, A one equals 2.95 times a two. So another way of looking at the percentage of the area that is blocked Is to say it's 1 - the percentage of the total area of the scales. So that's gonna be a two over 81. We already know that a one is 2.95 times a two. So we are going to use that and that comes out to one minus 10.339, That's .661. They're asking us for percent Multiply that by 100 and that's .661%. So if we look at our potential options, that aligns with B, The percentage of the pipes area that is occupied by the scales is 66.1%. That's all for this problem. We'll see you in the next video.
Related Practice
Textbook Question
A hollow plastic sphere is held below the surface of a freshwater lake by a cord anchored to the bottom of the lake. The sphere has a volume of 0.650 m^3 and the tension in the cord is 1120 N. (a) Calculate the buoyant force exerted by the water on the sphere. (b) (b) What is the mass of the sphere?
724
views
Textbook Question
Gold Brick.You win the lottery and decide to impress your friends by exhibiting a million-dollar cube of gold. At the time, gold is selling for $1282 per troy ounce, and 1.0000 troy ounce equals 31.1035 g. How tall would your million-dollar cube be?
434
views
Textbook Question
Water is flowing in a pipe with a varying cross-sectional area, and at all points the water completely fills the pipe. At point 1 the cross-sectional area of the pipe is 0.070 m^2, and the magnitude of the fluid velocity is 3.50 m/s. (c) Calculate the volume of water discharged from the open end of the pipe in 1.00 hour.
1587
views
Textbook Question
At one point in a pipeline the water's speed is 3.00 m/s and the gauge pressure is 5.00×10^4 Pa. Find the gauge pressure at a second point in the line, 11.0 m lower than the first, if the pipe diameter at the second point is twice that at the first.
822
views
Textbook Question
A 950-kg cylindrical can buoy floats vertically in sea-water. The diameter of the buoy is 0.900 m. Calculate the additional distance the buoy will sink when an 80.0-kg man stands on top of it.
1223
views
1
rank
Textbook Question
An ore sample weighs 17.50 N in air. When the sample is suspended by a light cord and totally immersed in water, the tension in the cord is 11.20 N. Find the total volume and the density of the sample.
1629
views