Worksheet for Exploration 15.1: Blood Flow and the Continuity Equation

Transcription

1 Worksheet for Exploration 15.1: Blood Flow and the Continuity Equation Blood flows from left to right in an artery with a partial blockage. A blood platelet is shown moving through the artery. How does the size of the constriction (variable from 1 mm to 8 mm from each wall) affect the speed of the blood flow? Restart. Assume an ideal fluid (position is given in millimeters and pressure is given in torr = mm of Hg). We can use the continuity equation and Bernoulli's equation to understand the motion: Continuity: Av = constant Bernoulli: P + (1/2) ρv 2 + ρgy = constant. With a 2.0-mm constriction: a. What is the platelet's speed before and after it passes through the constriction? v before = v after = b. What is the platelet's speed while it passes through the constriction? v constriction =

2 Set the constriction to 8.0 mm. c. Does the speed of the platelet before it reaches the constriction increase, decrease, or not change? d. With the 8-mm constriction, is the speed of the platelet in the constriction faster, slower, or the same as with the 2-mm constriction? e. Assume that the blood vessel and the blockage are cylindrical (circular cross-sectional area for both). Measure the radius of the artery and the radius of the flow area where the blockage is. Verify the equation of continuity to compare the 2-mm and 8-mm cases. A out = A in = Now compare the 2-mm and 8-mm cases. f. What is the pressure inside of and outside the constriction (use the white box to measure pressure)? P in = P out = g. Does the pressure decrease or increase in the region where the blockage is?

3 h. This result, (g), is surprising to many students so let's figure out why: At the instant the platelet travels from the wide region to the narrower constricted region, what is the direction of acceleration? i. In addition to considering net acceleration of the platelet, also consider which side of the platelet accelerates more (or first) as it enters the constriction region. i. What, then, is the direction of the force that the platelet feels? i. For parts (i) and (h) you should consider what causes the force on the platelet. Sketch the platelet and see if the force on the left side of it or right side of the platelet is greater. ii. Also, as the platelet enters the constriction which side has a greater force acting on it? What does this do to the shape of the platelet (not shown in animation). j. What region should have a larger pressure? i. To determine this you should carefully consider the answers above and note that the pressure on the side of the platelet that has not yet entered the constriction remains constant. So what must happen on the constriction side to cause the acceleration you note? k. Do the same analysis for the platelet as it leaves the constricted region and goes back to the unblocked artery (sketch a diagram to show the direction of acceleration and force). l. Verify that Bernoulli's equation holds inside and outside the constricted region for the 2-mm and 8- mm cases (760 Torr = 760 mm of Hg = 1.01 x 10 5 Pa). The density of blood is 1050 kg/m 3.

4 Worksheet for Exploration 15.2: Bernoulli s Equation Bernoulli s equation describes the conservation of energy in an ideal fluid system. Assume an ideal fluid (position is given in meters and pressure is given in pascals). The dark blue in the animation is a section of water as it flows into the region marked by the horizontal line and the corresponding water that must move out of the region in the top right. We will explore the connection between Bernoulli's equation and conservation of energy. Restart. Note: The format of the pressure is written in short hand. For example, atmospheric pressure, 1.01 x 10 5 Pa, is written as 1.01e+005. The relationship between the speed and dimensions of the water going in compared with the water leaving is governed by the continuity equation (what flows in must flow out unless there is a leak in the pipes!): Av = constant, where A is the cross-sectional area and v is the speed of the liquid. Assume the pipes are cylindrical. a. What is the volume of both (darker) blue regions (should be the same)? V left = V right = b. What is the speed of the water in the left pipe? Speed left= c. What is the cross-sectional area of the left pipe? A left =

5 d. What is the speed of the water leaving the region (in the right pipe)? Speed right= e. What is the cross-sectional area of the right pipe? A right = f. Does the continuity equation hold? As the water travels through this pipe system, work must be done on the fluid to raise it up and to increase its speed. The work done must be equal to the change in kinetic plus potential energy. The upper blue region is effectively the same as the lower blue region but at a later time (after traveling up the pipe). So when you make measurements on that region, you are determining how the properties of that mass of water change. g. Given the pressure (you can move the red pressure indicators), find the force (from the water behind it) on the lower left dark blue region. P left = F left = h. What is the work done by that force for the duration of the animation? W left = i. Similarly, find the force on the upper right dark blue region that opposes the motion. P right = F right =

6 j. What is the work done by that force for the duration of the animation (note that the displacement and force are in opposite directions, so this is negative work)? W right = k. What is the net work done, then, during the duration of the animation on the water in the middle region? i. This really refers to the net work done to move the same volume of water that is in either dark blue region from the bottom to the top. W net = l. Calculate the difference in kinetic energy of the dark blue regions. Note that since the volume is the same, the mass is the same. (The density of water is 1000 kg/m 3.) i. This is the difference in kinetic energy between a dark blue volume of water at the low left region and the kinetic energy of the same mass when it is up on the right. KE right-left = m. Calculate the difference in potential energy of the center of mass of the dark blue regions. Does the net work equal the change in kinetic energy plus the change in potential energy? i. Again this change is really the difference between the right and left dark blue masses of water. PE right-left =

7 This is all described by Bernoulli's equation. n. Show that the net work is equal to (P left - P right ) Avt. o. Show that the net change in kinetic energy is (1/2) ρavt (v right 2 - v left 2 ). p. Show that the net change in potential energy is ρgavt (y right - y left ). P is the pressure, ρ is the density of the fluid, v is the speed of the fluid flow, A is the cross-sectional area, t is the time, and y is the height of the fluid. Combining these three terms, we get (P left - P right ) = (1/2)ρ(v right 2 - v left 2 ) + ρg(y right - y left ), or Bernoulli's equation, P + 1/2ρv 2 + ρgy = constant, so that Bernoulli's equation is simply another way to restate conservation of energy.

8 Worksheet for Exploration 15.3: Application of Bernoulli's Equation Adjust the height of the water in the reservoir and notice what happens to the water flow out of the opening. Assume an ideal fluid (position is given in meters). We can use Bernoulli's equation (i.e., conservation of energy for fluids) to understand what happens, P + 1/2ρv 2 + ρgy = constant, where P is the pressure, ρ is the density of the fluid, v is the speed of the fluid flow, and y is the height of the fluid (you can, of course, pick any point to be y = 0 m). Restart. The amount of water leaking out is small during the animation. So the height effectively stays constant during the time this animation is running (to a good approximation). a. Use Bernoulli's equation to find the pressure at the bottom of the reservoir. Pick a height of water in the reservoir. The pressure above the water is atmospheric pressure (1.0 x 10 5 Pa). What is the pressure of the water at the bottom of the reservoir? (Note that for both of these cases, v = 0 m/s). height= P bottom = b. Use Bernoulli's equation at the bottom of the reservoir to find the speed of water flow out of the reservoir. Equate Bernoulli's equation somewhere in the middle of the bottom of the reservoir (where v = 0 m/s) to the water flowing out of the opening (where P is atmospheric pressure) and note that the value of y is the same for both cases. v bottom =

9 c. Using this value for the initial x velocity of the water, calculate where the water will land and verify that it does land at the proper spot. Repeat this procedure for another value of the reservoir height. First height= Landing Coordinates= Second height= Landing Coordinates=