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Chapter 14: Problem 5

An artery in a person has been reduced to half its original inside diameter by deposits on the inner artery wall. By what factor will the blood flow through the artery be reduced if the pressure differential across the artery has remained unchanged? The relationship governing flow rate, pressure differential, and opening radius is Poiseuille's Law, wherein \(J \propto r^{4}\). Therefore,

Short Answer

Expert verified
The blood flow is reduced by a factor of 16.

Step by step solution

01

Understanding Poiseuille's Law

Poiseuille's Law states that the flow rate of a fluid (J) through a cylindrical tube (like an artery) is proportional to the fourth power of the radius (r) of the tube: \( J \propto r^4 \). This means changes in the radius have a significant effect on the flow rate.
02

Determine the Original and New Radius

Since the artery's inside diameter has been reduced to half, the new radius is also half of the original radius. If the original radius is \( r \), then the new radius \( r' \) is \( r/2 \).
03

Calculate the Change in Flow

Using Poiseuille's Law, the original flow rate is proportional to \( r^4 \), while the new flow rate is proportional to \( (r/2)^4 \). This simplifies to \( J' \propto \frac{r^4}{16} \).
04

Finding the Reduction Factor

To find the factor by which the flow is reduced, divide the new flow by the original flow: \( \frac{J'}{J} = \frac{(r/2)^4}{r^4} = \frac{1}{16} \). Thus, the blood flow is reduced by a factor of 16.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Blood Flow
Blood flow refers to the movement of blood through a vessel in the circulatory system. When considering blood flow in terms of Poiseuille's Law, it's important to understand how various factors affect it. This law shows that the flow rate is sensitive to changes in the radius of the vessel.
In our given context, we're looking at an artery where the internal diameter has changed. Blood flow is not just a simple stream; it's influenced by viscosity and resistance within the vessel walls as well. Here's where Poiseuille's Law becomes incredibly helpful. - Blood flow rate is proportional to the fourth power of the artery radius. - This means even a slight change in the radius can cause significant changes in blood flow. - Understanding this sensitivity helps medical professionals predict how changes in artery size affect blood circulation. By comprehensively grasping the notion of blood flow as defined by Poiseuille鈥檚 Law, students can better appreciate cardiovascular dynamics.
Pressure Differential
Pressure differential is the difference in pressure between two points, which in this case, are two locations along the artery in question. It's a crucial driver for blood flow, powering the movement of blood throughout the body's arteries. - Poiseuille鈥檚 Law shows that blood flow is heavily reliant on both the pressure differential and the radius of the vessel. - Despite changes in radius, if the pressure differential remains constant, changes in blood flow primarily stem from the radius differences. With a constant pressure differential, knowing how changes in artery size impact flow helps in assessing cardiovascular health, aiding in detecting conditions like atherosclerosis.
Artery Radius Effect
The artery radius effect directly refers to how variations in the artery's diameter influence blood flow, primarily through the laws articulated by Poiseuille. - When the radius is halved, which is the case in our problem, the flow is reduced by a significant factor.- Specifically, by Poiseuille鈥檚 Law, the flow rate is proportional to the fourth power of the radius. Thus, halving the radius results in a \(16\) times reduction in blood flow, as calculated earlier.Understanding the artery radius effect is critical because:- Small changes in artery radius can drastically affect blood flow.- Medical conditions that change the radius could lead to significant cardiovascular issues.This insight helps students and professionals understand the importance of maintaining artery health and the severe implications of blockages or deposits in the arteries.

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Most popular questions from this chapter

What must be the gauge pressure in a large-diameter hose if the nozzle is to shoot water straight upward to a height of \(30.0 \mathrm{~m}\) ? To rise to a height \(h\), a projectile must have an initial speed \(\sqrt{2 g h}\). (We obtain this by equating to \(\frac{1}{2} m v_{0}^{2}\) to \(m g h\).) We can find this speed in terms of the difference between the pressures inside and outside the hose by writing Bernoulli's Equation for points just inside and outside the nozzle in terms of absolute pressure: $$ P_{\text {in }}+\frac{1}{2} \rho v_{\text {in }}^{2}+h_{\text {in }} \rho g=P_{\text {out }}+\frac{1}{2} \rho v_{\text {out }}^{2}+h_{\text {out }} \rho g $$ Here \(h_{\text {out }} \approx h_{i n}\), and because the hose is large, \(v_{\text {in }} \approx 0\); therefore, $$ P_{\text {in }}-P_{\text {out }}=\frac{1}{2} \rho v_{\text {out }}^{2} $$ Substitution of \(\frac{1}{2} m v_{0}^{2}\) to \(m g h\) for \(v_{\text {out }}\) yields \(P_{\text {in }}-P_{\text {out }}=\rho g h=\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(30.0 \mathrm{~m})=294 \mathrm{kPa}\) Since \(P_{\text {out }}=P_{A}\), this is the gauge pressure inside the hose. How could you obtain this latter equation directly from Torricelli's Theorem?

Exactly \(250 \mathrm{~mL}\) of fluid flows out of a tube whose inner diameter is \(7.0 \mathrm{~mm}\) in a time of \(41 \mathrm{~s}\). What is the average speed of the fluid in the tube? From \(J=A u\), since \(1 \mathrm{~mL}=10^{-6} \mathrm{~m}^{3}\) $$ v=\frac{J}{A}=\frac{\left(250 \times 10^{-6} \mathrm{~m}^{3}\right) /(41 \mathrm{~s})}{\pi(0.0035 \mathrm{~m})^{2}}=0.16 \mathrm{~m} / \mathrm{s} $$

Calculate the theoretical velocity of efflux of water, into the surrounding air, from an aperture that is \(8.0 \mathrm{~m}\) below the surface of water in a large tank, if an added pressure of \(140 \mathrm{kPa}\) is applied to the surface of the water.

Water is delivered at an average speed of \(4.0 \mathrm{~m} / \mathrm{s}\) via a pipe with an open cross-sectional area of \(25.0 \mathrm{~cm}^{2}\). How much water will arrive in \(0.50 \mathrm{~h} ?\)

Under the same pressure differential, compare the flow of water through a pipe to the flow of SAE No. 10 oil. \(\eta\) for water is \(0.801\) \(\mathrm{cP} ; \eta\) for the oil is \(200 \mathrm{cP}\). From Poiseuille's Law, \(J \propto 1 / \eta .\) Therefore, since everything else cancels, $$ \frac{J_{\text {water }}}{J_{\text {oil }}}=\frac{200 \mathrm{cP}}{0.801 \mathrm{cP}}=250 $$ The flow of water is 250 times as large as that of the oil under the same pressure differential.
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