Introduction to biomedical engineering: biomechanics and bioelectricity - part I

Intended as an introduction to the field of biomedical engineering, this book covers the topics of biomechanics (Part I) and bioelectricity (Part II). Each chapter emphasizes a fundamental principle or law, such as Darcy's Law, Poiseuille's Law, Hooke's Law, Starling's law, levers and work in the area of fluid, solid, and cardiovascular biomechanics. In addition, electrical laws and analysis tools are introduced, including Ohm's Law, Kirchhoff 's Laws, Coulomb's Law, capacitors and the fluid/electrical analogy. Culminating the electrical portion are chapters covering Nernst and membrane potentials and Fourier transforms. Examples are solved throughout the book and problems with answers are given at the end of each chapter. A semester-long Major Project that models the human systemic cardiovascular system, utilizing both a Matlab numerical simulation and an electrical analog circuit, ties many of the book's concepts together.

Introduction to Biomechanics PartiEditor John D. Enderle, University of Connecticut Introduction to Biomedical Engineering: Biomechanics and Bioelectricity - Part I Douglas A. Christensen 2009 Basic Feedback Controls in Biomedicine Charles S. Lessard 2008 Understanding Atrial Fibrillation: The Signal Processing Contribution, Volume II Luca Mainardi, Leif Sornmon, and Sergio Cerutti 2008 Understanding Atrial Fibrillation: The Signal Processing Contribution, Volume I Luca Mainardi, Leif Sornmon, and Sergio Cerutti 2008 Introductory Medical Imaging A. A. Bharath 2008 Lung Sounds: An Advanced Signal Processing Perspective LeontiosJ. Hadjileontiadis 2008 An Outline of Information Genetics Gerard Battail 2008 Neural Interfacing: Forging the Human-Machine Connection Thomas D. Coates, Jr. 2008Quantitative Neurophysiology Joseph V. Tranquillo 2008 Tremor: From Pathogenesis to Treatment Giuliana Grimaldi and Mario Manto 2008 Introduction to Continuum Biomechanics Kyriacos A. Athanasiou and Roman M. Natoli 2008 The Effects of Hypergravity and Microgravity on Biomedical Experiments Thais Russomano, Gustavo Dalmarco, and Felipe Prehn Falcao 2008 A Biosystems Approach to Industrial Patient Monitoring and Diagnostic Devices Gail Baura 2008 Multimodal Imaging in Neurology: Special Focus on MRI Applications and MEG Hans-Peter Muller and Jan Kassubek 2007 Estimation of Cortical Connectivity in Humans: Advanced Signal Processing Tecliniques Laura Astolfi and Fabio Babiloni 2007 Brain-Machine Interface Engineering Justin C. Sanchez and Jose C. Principe 2007 Introduction to Statistics for Biomedical Engineers Kristina M. Ropella 2007 Capstone Design Courses: Producing Industry-Ready Biomedical Engineers Jay R. Goldberg 2007 BioNanotechnology Elisabeth S. Papazoglou and Aravind Parthasarathy 2007 Bioinstrumentation John D. Enderle 2006V Fundamentals of Respiratory Sounds and Analysis Zahra Moussavi 2006 Advanced Probability Theory for Biomedical Engineers John D. Endcrlc, David C. Farden, and Daniel J. Krause 2006 Intermediate ProbabilityTheory for Biomedical Engineers John D. Enderle, David C. Farden, and Daniel J. Krause 2006 Basic ProbabilityTheory for Biomedical Engineers John D. Enderle, David C. Farden, and Daniel J. Krause 2006 Sensory Organ Replacement and Repair Gerald E. Miller 2006 Artificial Organs Gerald E. Miller 2006 Signal Processing of Random Physiological Signals Charles S. Lessard 2006 Image and Signal Processing for Networked E-Health Applications Ilias G. Maglogiannis, Kostas Karpouzis, and Manolis Wallace 2006Copyright © 2009 by Morgan & Claypool All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means-electronic, mechanical, photocopy, recording, or any other except for brief quotations in printed reviews, without the prior permission of the publisher. Introduction to Biomedical Engineering: Biomechanics and Bioelectricity - Part I Douglas A. Christensen www.morganclaypool.com ISBN: 9781598298444 paperback ISBN: 9781598298451 ebook DOI 10.22 00/S000182 EDI V01Y200903BME028 A Publication in the Morgan & Claypool Publishers series SYNTHESIS LECTURES ON BIOMEDICAL ENGINEERING Lecture #28 Series Editor: John D. Enderle, University of Connecticut Series ISSN Synthesis Lectures on Biomedical Engineering Print 1932-0328 Electronic 1932-0336Introduction to Biomechanics Parti Douglas A. Christensen University of Utah SYNTHESIS LECTURES ON BIOMEDICAL ENGINEERING #28 MORGAN CLAYPOOL PUBLISHERSABSTRACT Intended as an introduction to the field of biomedical engineering, this book covers the topics of biomechanics (Part I) and bioelectricity (Part II). Each chapter emphasizes a fundamental principle or law, such as Darcy's Law, Poiseuille's Law, Hooke's Law, Starling's law, levers and work in the area of fluid, solid, and cardiovascular biomechanics. In addition, electrical laws and analysis tools are in­troduced, including Ohm's Law, Kirchhoff's Laws, Coulomb's Law, capacitors and the fluid/electrical analogy. Culminating the electrical portion are chapters covering Nernst and membrane potentials and Fourier transforms. Examples are solved throughout the book and problems with answers are given at the end of each chapter. A semester-long Major Project that models the human systemic cardiovascular system, utilizing both a Matlab numerical simulation and an electrical analog circuit, ties many of the book's concepts together. KEYWORDS biomedical engineering, biomechanics, cardiovascular, bioelectricity, modeling, MatlabTo Larainexi Contents Synthesis Lectures on Biomedical Engineering.........................................................................iii Contents............................................................................................................................................xi Preface..............................................................................................................................................xv \ Basic Concepts: Numbers, Units and Consistency Checks.......................................................1 1.1 Introduction........................................................................................................................1 1.2 Numbers and significant figures........................................................................................1 1.2.1 Scientific Notation 2 1.2.2 Accuracy and Precision 3 1.2.3 Significant Figures in Calculations 3 1.3 Dimensions and units..........................................................................................................5 1.3.1 SI Units 6 1.3.2 Keeping Track of Units in Equations 8 1.3.3 English and Other Units 8 1.4 Conversion factors................................................................................................................8 1.4.1 The Use ofWeight to Describe Mass 10 1.5 Consistency checks............................................................................................................10 1.5.1 Reality Check 11 1.5.2 Units Check 11 1.5.3 Ranging Check 12 1.6 Organization of the remaining chapters.........................................................................12 1.7 Problems............................................................................................................................13 2 Darcy's Law: Pressure-Driven Transport Through Membranes...........................................15 2.1 Introduction - Biological and Man-Made Membranes.............................................15 2.1.1 Man-Made Membranes 17xii CONTENTS 2.2 Darcy's Law........................................................................................................................18 2.2.1 Ideal and Nonideal Materials 21 2.3 Mechanical Filtration (Sieving)......................................................................................22 2.4 Problems............................................................................................................................25 3 Poiseuille's Law: Pressure-Driven Flow Through Tubes.........................................................27 3.1 Introduction - Biological Transport..............................................................................27 3.2 Poiseuille's Law..................................................................................................................30 3.2.1 Simplified Version of Poiseuille's Law 33 3.2.2 Assumptions for Poiseuille's Law 34 3.3 Power Expended in the Flow..........................................................................................36 3.4 Series and Parallel Combinations of Resistive Elements...........................................36 3.4.1 Series 37 3.4.2 Parallel 37 3.5 Problems............................................................................................................................42 4 Hooke's Law: Elasticity ofTissues and Compliant Vessels...................................................45 4.1 Introduction.......................................................................................................................45 4.2 The Action of Forces to Deform Tissue.........................................................................45 4.3 HOOKE'S LAW AND ELASTIC TISSUES...........................................................46 4.4 Compliant Vessels..............................................................................................................50 4.5 Incompressible Flow of Compliant Vessels...................................................................53 4.6 Problems............................................................................................................................55 5 Starling's Law of the Heart, Windkessel Elements and Conservation of Volume............61 5.1 Introduction - Compliance of the Ventricles............................................................................................................................61 5.2 Pressure-Volume Plots: the pv Loop..............................................................................62 5.3 STARLING'S LAW OF THE HEART.....................................................................64 5.4 Windkessel Elements........................................................................................................67 5.5 Conservation of Volume in Incompressible Fluids.......................................................68 5.6 Problems............................................................................................................................70CONTENTS xiii 5 Euler's Method and First-Order Time Constants...................................................................73 6.1 Introduction - Differential Equations..........................................................................73 6.2 Euler's Method....................................................................................................................74 6.3 Waveforms of Pressure and Volume..............................................................................75 6.4 First-Order Time Constants............................................................................................76 6.5 Problems.............................................................................................................................84 7 Muscle, Leverage, Work, Energy and Power.............................................................................87 7.1 Introduction - Muscle......................................................................................................87 7.2 Levers and Moments........................................................................................................87 7.3 Work....................................................................................................................................92 7.4 Energy................................................................................................................................92 7.5 Power..................................................................................................................................94 7.5.1 Power in Fluid Flow 94 7.6 Problems.............................................................................................................................95 A Conversion Factors........................................................................................................................97 Material Constants.......................................................................................................................99 R.l Viscosity................................................................................................................................99 B.2 Density and Specific Gravity............................................................................................99 R.3 Permeability......................................................................................................................100 R.4 Young's Modulus and Ultimate Stress..........................................................................100 Ribliography................................................................................................................................101Preface NOTE ON ORGANIZATION OFTHIS BOOK The material in this book naturally divides into two parts: 1. Chapters 1-7 cover fundamental biomechanics laws, including fluid, cardiovascular, and solid topics (1/2 semester). 2. Chapters 8-15 cover bioelectricity concepts, including circuit analysis, cell potentials, and Fourier topics (1/2 semester). A Major Project accompanies the book to provide laboratory experience. It also can be divided into two parts, each corresponding to the respective two parts of the book. For a full-semester course, both parts of the book are covered and both parts of the Major Project are combined. The chapters in this book are support material for an introductory class in biomedical en­gineering1 They are intended to cover basic biomechanical and bioelectrical concepts in the field of bioengineering. Coverage of other areas in bioengineering, such as biochemistry, biomaterials and genetics, is left to a companion course. The chapters in this book are organized around several fundamental laws and principles underlying the biomechanical and bioelectrical foundations of bio­engineering. Each chapter generally begins with a motivational introduction, and then the relevant principle or law is described followed by some examples of its use. Each chapter takes about one week to cover in a semester-long course; homework is normally given in weekly assignments coordinated with the lectures. The level of this material is aimed at first-semester university students with good high-school preparation in math, physics and chemistry, but with little coursework experience beyond high school. Therefore, the depth of explanation and sophistication of the mathematics in these chapters is, of necessity, limited to that appropriate for entering freshman. Calculus is not required (though it is a class often taken concurrently); where needed, finite-difference forms of the time- and space-varying functions are used. Deeper and broader coverage is expected to be given in later classes dealing with many of the same topics. Matlab is used as a computational aid in some of the examples in this book. Where used, it is assumed that the student has had some introduction to Matlab either from another source or from a couple of lectures in this class. In the first half of the cardiovascular Major Project discussed below, ^ At the University of Utah, this course is entitled Bioen 1101, Fundamentals of Bioengineering I.xvi PREFACE Matlab is used extensively; therefore, the specific Matlab commands needed for this Major Project must be covered in class or in the lab if this particular part of the project is implemented. A Major Project accompanies these chapters at the end of the booklet. The purpose of the Major Project, a semester-long comprehensive lab project, is to tie the various laws and principles together and to illustrate their application to a real-world bioengineering/physiology situation. The Major Project models the human systemic cardiovascular system. The first part of the problem takes approximately one-half of a semester to complete; it uses Matlab for computer modeling the flow and pressure waveforms around the systemic circulation. Finite-difference forms of the flow/pressure relationships for a lumped-element model are combined with conservation of flow equations, which are then iterated over successive cardiac cycles. The second half of the problem engages a physical electrical circuit to analyze the same lumped-element model and exploits the duality of fluid/electrical quantities to obtain similar waveforms to the first part. This Major Project covers about 80% of the topics from the chapter lectures; the lectures are given "just-in-time" before the usage of the concepts in the Major Project. Although the Major Project included with this book deals with the cardiovascular system, other Major Project topics maybe conceived and substituted instead. Examples include modeling human respiratory mechanics, the auditory system, human gait or balance, or action potentials in nerve cells. These projects could be either full- or half-semester assignments. ACKNOWLEDGEMENTS The overarching organizational framework of these chapters around fundamental laws and princi­ples was conceived and encouraged by Richard Rabbitt of the Bioengineering Department at the University of Utah. Dr. Rabbitt also provided much of the background material and organization of Chapters 2 and 4. Angela Yamauchi provided the organization and concepts for Chapter 3. David Warren contributed to the initial organization of Chapter 8. Their input and help was vital to the completion of this booklet. Douglas A. Christensen University of Utah March 2009CHAPTER 1 Basic s: Numbers, Units 1.1 INTRODUCTION Welcome to biomedical engineering, a very rewarding field of study! Biomedical engineering is the application of engineering principles and tools for solving problems in health care and medicine. Of all the engineering specialties, it is arguably the most interdisciplinary, requiring the study of biology, physiology and organic chemistry in addition to mathematics, physics and engineering topics. This makes the field particularly challenging as well as engrossing. Many professionals in biomedical engineering have chosen this field because they strive to improve the lives of their fellow humans and society. The goal of this book is to introduce you to some basic concepts of biomedical engineering by covering several fundamental physical laws and principles that underlie biomedical engineering. This book is focused on the biomechanical and bioinstrumentation (electrical) aspects of the field. An optional Major Project dealing with modeling the human cardiovascular system ties many of the book's topics together. It is left to other texts to cover other important areas in the field, such as biochemical, molecular and biomaterial topics. We start our studies with some basic concepts involving numbers and calculations. 1.2 NUMBERS AND SIGNIFICANT FIGURES Numbers and values are the stock-in-trade of the engineering profession. Engineers are often in­volved in various measurements (for example, a bioengineer may design a device to measure the level of blood glucose in a new and novel way), and one of the distinguishing features of a success­ful engineer is that s/he strives to be quantitatively correct. This requires careful attention to the manipulation and display of numerical values. An important characteristic of any numerical value is the number of digits, or significant figures, it contains. A significant figure is defined as any digit in the number ignoring leading zeros and the decimal point. For example: 821 has 3 significant figures 160.6 has 4 significant figures 160.60 has 5 significant figures 0.0310 has 3 significant figures 1.5 x 10J has 2 significant figures2 CHAPTER 1. BASIC CONCEPTS How many significant figures should you use when writing a number? A commonly accepted rule in specifying a number is to use as many, but not more, significant figures as can be reasonably trusted to be accurate in value. For example, let's say you measure a tall classmate's height with a meter-long ruler. Which one of the following ways do you think has the appropriate number of significant figures to report the result of the height measurement? a. 2.0131 m (with 5 significant figures) b. 2.01 m (with 3 significant figures) c. 2 m (with 1 significant figure) Answer a is unrealistic, because itgives the impression that your classmate's heightis measured with a precision or accuracy1 of one-tenth of a millimeter, or 0.0001 m-about the thickness of a human hair, which is obviously not justified2 since it depends upon the posture of the classmate at the time of the measurement, how much his hair was pushed down, how well the ruler was aligned with his spine, etc. Besides, it's impossible to read a meter ruler to this fine an increment anyway. Answer c is also unreasonable for the opposite reason: it doesn't convey enough information about the accuracy of the measurement you took. It could describe anyone with a height between 1.50 m (even shorter than I am) and 2.49 m (taller than any NBA player). Surely you measured within a tighter range than that. Answer b is the most reasonable choice. It tells the reader that in your best judgement, your measurement can be trusted to within about a centimeter, which is consistent with the uncertainties of the measurement. Therefore you report a value to the nearest 0.01 m (which is the 2.01 m answer), with three significant figures. But what if your classmate measured exactly 2 m (within a fraction of a centimeter)? Do you report answer c? No. Your number should make it clear that you have confidence to the centimeter level, so in this case you should report a measurement of 2.00 m, with three significant figures. 1.2.1 SCIENTIFIC NOTATION In light of the preceding discussion, numbers you come across which are multiples of 10 need to be interpreted with care. Consider the number 200. It is not clear whether the trailing zeros in this number show the confidence level of a measurement, or are simply decimal place holders. For instance, the length of the 200-m dash, a track-and-field event, is actually specified by the sport's rules to a high accuracy-to within 1 cm-so it really is the 200.00-m dash. On the other hand, a "200-m high cloud" only means a cloud whose distance off the ground is about 200 m, give-or-take approximately 10 m. So, to avoid confusion and to be more specific about the intended precision, scientific notation is preferred by engineers in these situations. Using scientific notation, the examples in the previous 1The concepts of accuracy and precision are discussed later in this chapter. 'Whether any accuracy is justified or not is object dependent. The diameter of a small metal stent whose purpose is to keep open a coronary artery can (and should) be specified to one-tenth of a millimeter.paragraph would be written as 2.0000 x 102 m and 2.0 x 102 m, respectively, indicating the intended precision. 1.2.2 ACCURACY AND PRECISION These terms are often used interchangeably, but they really describe different things. Accuracy is a measure of how close a value is to the "true" value (as determined by some means). Precision is an indication of the repeatability of the measurement, if done again and again under the same circumstances. Thus a golfer would rather be accurate than precise if she repeatedly hooks her drives into the trees. Accuracy and precision are sometimes specified in terms of a ± range indicating the uncer­tainties of a measurement, such as 56 ±3 mmlig or 18.5 ±0.8% of full scale. If unspecified, the uncertainty is assumed to be approximately one unit of the last significant digit of the number, as explained above. 1.2.3 SIGNIFICANT FIGURES IN CALCULATIONS Modern computers allow the manipulation and display of numbers with many significant figures (up to 10 or more). Rut you must be careful not to mislead your reader when reporting your final answer. Use only the number of significant figures justified by the reasoning above. Here are some generally accepted rules that are consistent with that reasoning: Addition and Subtraction - After the operations of addition and subtraction, the final answer should contain digits only as far to the right as the rightmost decimal column found in the least precise number used in the calculation. For example, consider the following: 31.5 + 2.8925 34.3925 (as displayed on your calculator) Following the rule, you should round off 3 this answer to 34.4 before you report it. This is because the first number, 31.5, is the least precise and has significant digits on the right out to the 0.1 decimal column, so your final answer should be rounded to this same column. Here's a thought problem explaining why: Suppose you were asked the value of your personal assets for an application for a scholarship. You own a computer (estimated value of $730, accurate to within about $10), clothes (estimated value of $23-you're a sharp dresser), and a bank account (balance of $63.21). Therefore you sum your total assets: Round off as follows: Round up the last digit kept in the answer by 1 if the next digit (the first one to be dropped) is 5 or greater; otherwise don't change the last kept digit. Usually the displays of calculators do this rounding automatically when set to the appropriate number of significant figures. 1.2. NUMBERS AND SIGNIFICANT FIGURES 3 4 CHAPTER 1. BASIC CONCEPTS $730 + 23 + 63.21 $816.21 (as displayed on your calculator) You might be tempted to put down $816.21 on the application form, but with a little thought that would be misleading. It would imply that you know the worth of your computer and clothes to the penny. More realistically, the implied accuracy of the answer should be no better than the least accurate of the parts (i.e., the estimated value of your computer). Using the rule above, $820 is the best answer. (Actually, $8.2 x 102 is an even better answer, because it clearly shows that you're confident only to the $10 level). More Examples: 0.1035 + 0.0076 0.1111 The answer should be reported as 0.1111. 16.732 - 0.11 16.622 The answer should be reported as 16.62. 5.5 + 17.83 + 2.11 25.44 The answer should be reported as 25.4. 2.0 - 0.0006 1.9994 The answer 4 should be reported as 2.0. Multiplication and Division - After the operations of multiplication and division, the answer should contain only as many significant figures as found in the number with the fewest significant figures. 4This rounded-off answer may appear strange since it appears as though no subtraction took place, but remember that since the original 2.0 number is assumed precise only to one decimal place, the answer can be no more precise than to one decimal place. (If, however, the original 2.0 number is really known to four decimal places, it would have been written as 2.0000.Then the final answer would be 1.9994.)1.3. DIMENSIONS AND UNITS 5 For example, when you multiply (83.6) x (10,858) the calculator display may show 907,728.8 but the final answer should be rounded to three significant figures: 908,000 (or even better, 9.08 x 105). As a division example, 563 / 6.2 = 90.81 should be reported as 91. Here's a thought problem to help explain why: Suppose you are given the task of finding the weight of one full Pepsi can. You decide to weigh a six-pack of Pepsi cans on a bathroom scale, and divide by 6 to get an "average" answer-not a bad idea. On the scale, you measure the weight of the six-pack (without any packaging material) at 4.7 Ibf. (Note that you have recorded a reasonable number of significant figures-two-for this measurement since the bathroom scale you used can be read only to about the nearest 0.1 Ibf.) But when you use your calculator to divide 4.7 Ibf by 6, you get 0.78333 Ibf as the weight of a single can. Obviously you can't report the weight of a Pepsi can to a precision of 0.00001 Ibf (the weight of a grain of sand) when you originally measured the total six-pack to a precision of only 0.1 Ibf. It is more logical to use the degree of precision of the least precise number in the division, which in this problem is two significant figures. So the final answer should be reported as 0.78 Ibf. (But what about the divisor, the number 6? Doesn't it have only one significant figure? No; in this case, it is an exact number and therefore can be considered to have an infinite number of zeroes after the decimal point-i.e., an infinite number of significant figures. Thus the number with the fewest significant figures is the number 4.7.) More Examples: (0.431)(0.002) = 0.000862 should be reported as 0.001. 163.4 / 16.555 = 9.87013 should be reported as 9.870. (4.3 x 10s) / (6.241 x IO""'3) = 6.890 x 101(1 should be reported as 6.9 x IO10. More Complex Operations - When an operation of multiplying or dividing is combined with adding or subtracting to achieve the final answer, or when nonlinear operations such as logarithmic or trigonometric operations are performed, it's harder to determine the correct number of significant figures in the answer. Intermediate answers must be rounded at each step, which is often difficult and awkward to implement. Therefore, there is no simple rule to apply here, and you should apply commonsense using the spirit of the previous rules as much as possible. 1.3 DIMENSIONS AND UNITS Specifying dimensions is a scheme for grouping and labeling similar physical quantities. Thus the letter L represents the dimension of any length, M the dimension of the mass of any body, T the dimension of time, etc. Dimensions are not the same as units, since there are several possible alternative units for measuring any dimension. For example, the length (of dimension L) of an object can be measured in units of meters, inches, yards, chains, miles, or even light-years. There are two major systems of units in use in the world today:6 CHAPTER 1. BASIC CONCEPTS * A metric system known as the Systeme International d'Unites, or SI, which is the international standard used in most scientific work, and * A non-metric system called the English system, which is in widespread use in the United States and Britain, especially for non-scientific applications. Engineers need to be familiar with both major systems of units and be able to readily convert any value back and forth between the two systems. Biomedical engineers in particular must be agile with unit conversions because they must deal with units that are traditional in clinical settings but which are outside even the two major systems, such as blood pressure measured in mmlig and cardiac output measured in L/min. Any number reported without units is meaningless. For example, if your systolic blood pres­sure is 2.3 (with no units), are you in good shape or bad shape? Without specifying the units of measurement, it's impossible to tell. If in units of psi, the pressure is fine; if in units of mmlig, it's not good at all. The importance of specifying units (and also of close communication within an engineering team) was painfully demonstrated by the following incident: On September 23,1999, contact was abruptly lost with the NASA Mars Climate Orbiter. It was later determined that due to an information error on earth, the spacecraft failed to enter a proper orbit around Mars and was lost. An investigation uncovered the cause: Two teams, one in California and one in Colorado, were responsible for coordinating the correct maneuvering of the spacecraft as it neared Mars. As unbelievable as it sounds, one team used English units (e.g., inches, feet and pounds) while the other used metric units (centimeters, newtons), and each group was unaware that the other was using different units! NASA has since taken steps to make sure that this type of error doesn't happen again. 1.3.1 SI UNITS As mentioned, metric SI units are used throughout the world. The SI is an outgrowth of the older MKS system, where MKS stands for meter, kilogram and second (the units for length, mass and time respectively). These units are retained as three of the base units of the SI. The other base units are the ampere, mole, kelvin, and candela. In addition, various combinations of these base units can define many other derived units defined for often-used quantities. Table 1.1 lists the base units and some important derived units in the SI. Prefixes - Sometimes the size of a unit system does not match well the scale of the quantity being measured. For example, a meter is okay as it stands to use for measuring the length of a building, but the diameter of a virus is much smaller than a meter, and the diameter of the earth is much larger. So standard multiplying factors represented by prefixes are used to scale the meter (or any other unit) to better match the measurement. Some common prefixes to scale the root unit by multiples of ten are given in Table 1.2.1.3. DIMENSIONS AND UNITS 7 For example, the diameter of a flu virus is approximately d = 0.00000010 m = 1.0 x 10-7 m = 0.10 jim. Notice how much more convenient the prefix notation is when specifying this value. Also note that the proper use of prefixes, like scientific notation, can avoid the ambiguity of how many significant figures are intended in a number that is a multiple of 10. Table 1.1: Some Base and Derived SI Units Quantity Unit Symbol Equivalent to length meter m (base unit) mass kilogram kg (base unit) time second s (base unit) electrical current ampere A (base unit) material amount mole mol (base unit) temperature kelvin K (base unit) light intensity candela cd (base unit) force newton N kg-m/s2 pressure pascal Pa N/m2 = kg/(m-s2) energy joule J N-m = kg-m2/s2 power watt W J/s = kg-m2/s3 frequency hertz IIz 1/s Table 1.2 : Common Prefixes Prefix Symbol Multiplier femto- f io-15 pico- P io-12 nano- n ]Cry micro- M 1CT6 milli- m 1CT3 centi- c icr2 kilo- k io3 mega- M 106 g'ga- G 109 tera- T 1012 Rules for the Use of SI Units - When writing SI units, some generally accepted rules apply. These rules are intended to improve the consistency of SI unit usage and reduce the chance of misinterpretation.8 CHAPTER 1. BASIC CONCEPTS A good summary of these rules can be found at the National Institute of Standards and Technology (NIST) website (http: //physics .nist.gov/cuu/Units/checklist .html). 1.3.2 KEEPING TRACK OF UNITS IN EQUATIONS It is very helpful, whenever possible, to carry units along in the numerical solution of equations. For example, consider the problem of finding the force needed to accelerate a 5.0 kg mass to an acceleration of 1.6 cm/s2. The physical law that applies here is Newton's second law: F= ma (1-1) where F is the value of the force, m is the value of the mass,5 and a is the value of the acceleration. To solve for F, of course, you put values for m and a into (1.1). If possible, you should also put in their respective units. Re careful not to mix units. That is, make sure you consistently use the same unit system throughout the entire equation. In SI units, (1.1) gives F= (5.0 kg)(1.6 cm/s2) = 8.0 kg-cm/s2. (1-2) You may need to convert prefixed units (such as cm here) to the root unit (m) in order to get understandable equivalent units in the final answer. In this case (1.2) becomes F= 8.0 kg-cm/s2 = 8.0 x 10-2 kg-m/s2 = 8.0 x 10-2 N = 0.080 N, (1-3) where the equivalence between the units kg-m/s2 and N has been used (see Table 1.1) to put the final answer in the usual SI unit for force, the newton N. [In fact, fundamental equations such as (1.1) are the means for determining the equivalency between derived units and their base-unit form.] 1.3.3 ENGLISH AND OTHER UNITS Any student living in the United States is familiar with English units. They are commonly used in dealing with everyday objects and measurements (a "ten-pound sack," a "30-inch waist"). In addition, health-care practitioners are accustomed to dealing with hybrid units that have been long accepted and traditionally used in the medical field ("systolic blood pressure of 120 mmHg," and "glucose concentration of 100 mg/dL"). Table 1.3 lists some commonly used English units and medical units. 1.4 CONVERSION FACTORS Because of the large variety of units encountered, an engineer must be able to quickly and accurately convert any value between different units. Many of the conversion factors needed for this class are tabulated in Appendix A. The only tricky part of doing a units conversion is making sure you don't use the inverse of the correct factor by mistake. For example, let's say you want to convert the measurement of the length ~SNote that the symbol m is used here for the value of the mass. But m is also the symbol for the SI unit meter. Be careful not to confuse these two different uses.1.4. CONVERSION FACTORS 9 Table 1.3: Some Common English and Medical Units Quantity Unit Symbol length foot ft mass slug slug time second s force pound-force lbf pressure pound-force per square inch psi power horsepower hp energy big calorie or food calorie Cal volume liter L cardiac output liters per minute L/min mass concentration milligrams per liter mg/L blood pressure millimeters of mercury mmHg bladder pressure centimeters of water cmEbO / of the tibia from centimeters to inches. A typical measurement gives / = 42.1 cm. (1.4) You know that the required conversion factor, from Appendix A, relating the length of one inch to its equivalent length in centimeters is: 1 in = 2.54 cm. (1-5) Dividing both sides of (1.5) by 2.54 cm gives 6 1 in \ . ' 1. (1.6) „ 2.54 cm, The left-hand side of (1.6)-in the parenthesis-is now equal to unity, so it is a conversion factor you can use to multiply the right-hand side of (1.4) without changing the equality of (1.4): / 1 in \ . . / = 42.1 cm (---------- I = 16.6 in. (1-7) V 2.54 cm J The key to knowing that you have multiplied by the correct factor in this case is to notice hozu the original centimeter units cancel in the numerator and denominator of(1.7), leaving the answer in the desired inch units. b In H .5), the 1 on the left-hand side is an exact number so it can be considered to contain an infinite number of significant figures. As a consequence the right-hand side of (1.5) sets the significant figure limit of the conversion factor. Now to avoid unduly letting the conversion factor determine the significant figures in the final answer rather than the value being converted, make sure the conversion factor is specified to at least as many significant figures as the number being converted. [Note that in (1.10), the value being converted-not the conversion factor-sets the precision of the final answer, as is proper.]If the opposite direction of conversion is desired, the procedure is similar but now the manipu­lation of the conversion factor (1.5) must be inverted. Say you wanted to convert s = 6.1 in (1.8) into centimeter units. Divide both sides of the conversion factor (1.5) by 1 in to get ( 2.54 cm \ 1 = -- • (1.9) V > 111 / The right-hand side of (1.9) is a new unity conversion factor you can use to multiply (1.8): / 2.54 cm \ s = 6.1 in I -----;---- 1 = 15 cm. (1.10) V 1 'n / Again note that the original inch units cancel, leaving the answer in the desired centimeter units. By keeping careful track of the units in all equations, important verification is provided that the con­version factors-(1.6) and (1.9)-are in the correct order, respectively, to accomplish their intended direction of conversion. 10 CHAPTER 1. BASIC CONCEPTS 1.4.1 THE USE OF WEIGHT TO DESCRIBE MASS In the English system, it is common to describe the mass of some object, say a person, bywords such as "a one-hundred-and-seventy-pound man." But this description is actually based on weight (the same as force-175 lbf in this case) not mass, so it is only an indirect measure of the man's mass. In fact, weight is a somewhat untrustworthy way to describe mass since the same mass will have different weights depending upon the local constant of gravitational acceleration. For example, on the moon the man will weigh only 29 lbf. To convert from weight to mass, one needs to specify the assumed acceleration of gravity. At the earth's sea level, the acceleration of gravity is g = 32.174 ft/s" = 9.8067 m/s". (1-11) (These values are normally the constants assumed for gravitational acceleration if no explicit value otg is given.) Newton's second law, (1.1) withg replacing a, can be used to find the mass of the man in question in English units: m = F/g = (175 lbf)/(32.174 ft/s2) = 5.44 1bfs2/ft = 5.44 slug. (1.12) 1.5 CONSISTENCY CHECKS Mistakes in calculations happen. Of course, careful students and engineers try to keep errors to a minimum, but when they do happen it is very useful to have a set of quick checking tools that alert that something is amiss in the answer. Consistency checks are one class of such error-catching1.5. CONSISTENCY CHECKS 11 tools. These procedures will not necessarily tell you the correct answer, but once you become familiar with their use they will flag mistakes in calculations and equations. Three consistency checks are described below. 1.5.1 REALITY CHECK Some students call this the "duh" test. It is very quick and easy to perform. You simply think about the value of the answer you just derived, and based on commonsense you make a judgement about whether the answer could possibly be right or not. For example, let's say you calculate the blood pressure in your capillary bed based upon realistic physiological values of capillary compliance and blood volume. After converting to English units, you get the answer P = 475 psi. (1.13) If you think about this value for a moment, it is way out-of-line physiologically. The air pressure in your car tires is about 30 psi, so a blood pressure of 475 psi would undoubtedly blow out all your capillaries. You probably made a conversion factor error (incorrectly inverting the conversion factor?). Of course, the more familiar you are with the units being used, the more errors this method will pick up. SI units, especially newtons and pascals, are still not a part of everyday life. Also, the utility of this method will improve as you gain more experience with values normally encountered for the case being studied. 1.5.2 UNITS CHECK When you carry units throughout a calculation, the units of the final answer must be consistent with those expected for the dimension of the solved quantity. This is a powerful check. For example, let's say you've solved for the length increase A / of a 10-cm long bone when a pressure of 18 Pa is applied longitudinally to one end. You come up with an answer of Al = 0.016 N. (1.14) Obviously something is wrong, because newton is not an appropriate unit for a length (or a length change); it must be given in units of meter, or millimeter, or inch, or similar. You need to go back and recheck the steps leading to this answer. Care must be taken when applying this check, though, to make sure you aren't fooled by equivalent units. For example, in another calculation of a length, you may get an answer of /= 2.32N-s2/kg. (1.15) The units here are actually okay for length, because when the base-unit equivalent to a newton, namely kg-m/s2 (see Table 1.1), is substituted into the units above, the result is N-s2/k g = kg-m-s2/(s2-kg) = m, (1.16) 12 CHAPTER 1. BASIC CONCEPTS which is appropriate for a unit of length. Units checking is especially helpful in keeping track of which conversion factor to use. You must make sure that the original units cancel, leaving the desired units as discussed in Section 1.4 above. [A units check might have caught the mistake leading to the wrong answer of (1.13).] 1.5.3 RANGING CHECK This check is most useful for finding errors in derived equations. It involves letting one of the variables on the right-hand side of the equations (an independent variable) take on increasing or decreasing values, even to the limit of infinity or zero, and noting whether the effect on the variable on the left-hand side (the dependent variable) is in the right direction and, in the case of a limit, converges to the correct value. Usually there are several independent variables; any one can be selected for "ranging" in this manner and checked for the expected effect on the dependent variable. For example, let's say you have solved for the pressure drop AP across a blood vessel whose diameter is D and whose length is / by using Poiseuille's Law (to be discussed in Chapter 3) and you get Z) til , . , . AP = ---------, (error) (1-17) where Q is the volumetric blood flow and fx is the blood viscosity. To quickly check for a major error, do a ranging check on (1.17) by imagining that the independent variable for diameter D takes on larger and larger values. From (1.17), pressure drop AP will increase for this case. Rut it doesn't make sense that a larger vessel diameter will produce a larger pressure drop (all other parameters staying constant); instead, the pressure drop should get smaller. This is a strong indication that a mistake involving D was made in the derivation of the equation. Doing more with this example, let the value of the vessel length / go to the limit of zero, and by examining (1.17) find the resulting limit of AP: lim AP = 0. (1.18) 1-rO This result does make sense: If the length of a blood vessel gets vanishly short, the pressure needed to force blood flow through it should also vanish. You can conclude, at least, that the location of the variable / in the numerator of (1.17) is correct. [Going further, ranging checks done on the other independent variables shows that the behavior of AP upon the ranging of the viscosity variable fx seems reasonable, but the position of the flow variable Q in the equation is questionable. A units check on (1.17) would also have shown that there is an error in the equation.] 1.6 ORGANIZATION OFTHE REMAINING CHAPTERS The remainder of this book is organized in chapters. Each chapter covers an important law or prin­ciple relating to biomechanics (fluids and solids), cardiovascular mechanics, electricity, or biosignals. Each chapter has a homework set that will be assigned. Many, but not all, of the chapters relate1.7. PROBLEMS 13 to the Major Project, and the skills learned in doing the homework will be essential in solving the Major Project. 1.7 PROBLEMS 1.1. As team leader of an engineering group that has developed a new replacement aortic valve, one of your tasks is to calculate the manufacturing cost (parts and labor) of each valve. The cost of each valve has three components: a. Outside ring: cost = $134.31. b. Leaflets, made in Italy: 3 needed per valve. Each leaflet costs 88,221 Italian lira. (Exchange rate $1 = 2131.05 Italian lira) c. Labor: 0.40 hrs at a rate of $ 132/hr. Keeping the appropriate number of significant figures in each step, calculate the cost of each valve. fans: $312] 1.2. In a physics class, you have derived the following equation that gives the time t that a body takes, starting from rest, to fall a distance d under the influence of gravity: t = where g is the acceleration of gravity (9.8067 m/s"). a. Perform a units check on the above equation. b. Perform one ranging check on the above equation.CHAPTER 2 is 2.1 INTRODUCTION - BIOLOGICAL AND MAN-MADE MEMBRANES Cells are the building blocks of all living things, including plants and animals. Cells contain the components and perform the functions that allow life, such as energy conversion, chemical regulation, reproduction and repair. They occur in numerous varieties in different tissues of the human body depending upon the function performed by that tissue. During embryonic development, cells start undifferentiated, and then develop into specific differentiated types appropriate for the various roles they play in the tissues. Thus, the shapes and sizes of cells in the human body vary widely from very small (as thin as 0.2 jim across for the plate-like endothelial cells lining blood capillaries) to moderate (250 jim diameter for a spherical ovum) to very long (nearly 1 m for the threadlike axon projections of some central nervous system cells). The contents of the cells are packaged inside by the cell wall. A very simplified view of this compartmentalization is shown in Fig. 2.1. In human cells this wall is known as the plasma membrane and is composed of a sandwich of two layers of phospholipid molecules facing each other (a "bilayer"). The molecules in the phospholipid bilayer are arranged such that the "water- avoiding" (hydrophobic) ends of the molecules all face inward into the membrane center, while the "water-loving" (hydrophilic) ends all face toward the surface of the membrane. This means that both exposed sides of the plasma membrane are hydrophilic, consistent with the fact that both the fluid surrounding the outside of the cell and the material inside the cell (the cytoplasm) are composed mostly of water. Almost all cells in the human body (except for red blood cells) have a nucleus inside containing the DNA for cell replication; these cells are known as eukaryotic cells. More primitive cells such as bacteria do not have nuclei and are termed prokaryotic cells. The plasma membrane of the cell serves several purposes. It is foremost a mechanical barrier that compartmentalizes and protects the contents of the cell, separating it from its neighbors. But this barrier cannot be absolute and impenetrable. Otherwise nutrients (e.g., oxygen and glucose) needed to meet the energy requirements of the cell and keep it alive could not enter the cell, and the waste products (COj, urea) could not leave. Moreover, specific free ions (sodium, chloride, calcium and potassium to name a few, depending upon the cell function) must be able to travel through the membrane between the cytoplasm and the extracellular space in order to maintain electrical and chemical balance and to carry out the functioning of the cell. Water also passes through the16 CHAPTER 2. DARCY'S LAW extracellular fluid (mainly water) cytoplasm Figure 2.1: A simplified drawing of a generic human cell showing cell contents packaged inside a selectively permeable cell wall-the plasma membrane. membrane, either from inside to outside or visa versa, depending upon the direction of net force driving the flow. Therefore, the cell membrane does indeed selectively allow certain molecules (which are specific to the role of the cell) to pass through, thereby regulating cell contents. The cell membrane is said to be selectively permeable, or semipermeable. The ease with which each species can pass through varies depending upon the size, shape and electrical charge of the species, and upon the characteristics of the membrane. The presence of small channels (pores) through the plasma membrane-some of which are gated-accounts for some of the ability of substances to pass through. Proteins spanning across the membrane can also facilitate transport of selected molecules. There are three main ways that substances can be transported through a plasma membrane: 1. By fluid (or hydrostatic) pressure (the topic of this chapter)-This mechanism is relevant to liquid molecules such as water, and to gases. 2. By electro-chemical diffusion (covered in Chapter 14, Part II)-This mechanism applies to ions, small molecules and some macromolecules as well as to water. When water is driven across a semipermeable membrane due to a difference in water concentration on each side2.1. INTRODUCTION - BIOLOGICAL AND MAN-MADE MEMBRANES 17 (in turn due to a difference in solute concentrations), the water is said to be driven across by osmotic pressure. 3. By active transport by membrane-spanning proteins (also covered in Chapter 14, Part II)- This mechanism facilitates the transport of ions, small molecules and macromolecules. All three mechanisms are in play with cells, but some are more pronounced than others depending upon the cell type. For example, nerve cells rely upon the cooperation between electro­chemical diffusion and active transport of ions to maintain their nerve cell function and produce action potentials (Chapter 14, Part II). Water transport by hydrostatic pressure is minimal in these On the other hand, the cells that line the blood capillaries in the glomerulus of the kidney allow water to pass through their walls relatively easily by the action of hydrostatic pressure from the blood volume into urine-collecting spaces of the kidney. (Some water also leaks out through small gaps in the junctions between the cells.) Osmotic pressure opposes this water movement, but the hydrostatic pressure is greater here, so there is a net flow of water out of the blood in this part of the kidney (most is reabsorbed later in the kidney). The glomerulus thus plays a major role in the regulation of water in the body. Similarly, the endothelial cells that line the capillaries in the circulatory system also allow some water to pass outward from the blood into the interstitial space outside the vessels, again driven by hydrostatic pressure. The volume of water that leaks out depends upon how much greater the hydrostatic pressure is than the osmotic pressure opposing it. If water secretion is normal, only a small amount of water filters out; the lymphatic system collects it and returns it back into the veins. Rut if the hydrostatic pressure in the capillaries is abnormally high, caused for example by a weak left heart that doesn't empty the veins readily, the amount of water driven out of the capillaries can overwhelm the lymphatic system, leading to a pooling of water and swelling (edema) of the surrounding soft tissue. When this happens in the capillaries of the legs, water swells the leg's tissues, an early sign of a weak heart. In the lung capillaries, it leads to the collection of water in the lung alveoli (a symptom of "congestive heart failure"), with serious consequences on breathing ability. 2.1.1 MAN-MADE MEMBRANES In addition to naturally occurring membranes, there are several examples of man-made semiperme- able membranes. These can be as simple as the cellulose-fiber paper filters common in chemistry labs. Of more complexity, the first successful artificial kidney, or dialysis machine, was assembled by Dr. Willem Kolff1 in 1944 using long tubes made with thin cellophane walls. Blood drawn from the veins of a patient was passed through the inside of the tubes, which were immersed in a bath of fresh electrolyte solution. Water, ions and waste products were exchanged across the tubing, clearing 11 )r. Kolff (1911-2009) had a distinguished career in artificial organs, and joined the faculty of the University of Utah in 1967.18 CHAPTER 2. DARCY'S LAW the blood as it was continuously returned to the patient. Figure 2.2(a) is a picture of an early dialysis machine. a. b. Figure 2.2: Examples of the use of man-made membranes, (a) Dr. W. Kolff shown with an early artificial kidney machine, (b) Cross-sectional view of a small porous tube for encapsulating cells or drugs in host tissue. (Photos courtesy of the Dept, of Bioengineering, University of Utah.) More recently, tissue bioengineers have fabricated semipermeable membranes for encapsulat­ing collections of cultured cells or drugs. When implanted into body tissue, these membranes form a container keeping the cells together while selectively allowing exchange of nutrients and desirable products from the cells. When the membrane package contains drugs, it allows slow release of the drug in a measured fashion for predicable delivery to the patient. Figure 2.2(b) shows an example of a man-made semiporous membrane. 2.2 DARCY'S LAW The mathematical relationship between the flow of fluid through a porous obstruction and the pressure driving the flow was first derived by French hydraulic engineer Henry Darcy in 1856. Darcy was engaged in designing a water treatmen t system for the city of Dijon when he experimented with various flo w rates of water through different lengths of tubing filled with sand. The water was driven through the sand by gravity. He found a linear relationship between the flow rates and the driving2.2. DARCY'S LAW 19 pressure, and an inverse relationship with the length of the column of sand. Although Darcy's original work was done with water and sand, his findings can be applied to more general porous materials such as membranes. Darcy's relationships can be illustrated by the arrangement shown in Fig. 2.3. Here a membrane of thickness h (SI units: m) and area^ (m2) is inserted in a fluid-filled tube. A pressure P (Pa) is imposed on the fluid, causing it to flow with flow rate Q (m3/s) through the membrane. pressure P/ \ pressure P2 volume pi "► a cd volume flow rate -► X) flow rate Q G <D s face area A h ■*- thickness X x2 Figure 2.3: Arrangement in which the flow rate Q_through the porous membrane is measured as a function of the pressure drop AP = P\ - P2 across the membrane. Hydraulic pressure is defined as force per unit area: P = F/A. (2.1) where P (Pa) is the pressure, F (N) is the total force acting on the membrane face, and^ (m2) is the area of the membrane's face. It can be shown easily from (2.1) that the SI unit Pa is equivalent to N/m2. The fluid pressure on the left side of the membrane in Fig. 2.3 (the "entrance" side) has a value of Pi .The pressure drops linearly across the membrane (the distribution is shown by the upper solid line in the figure) to a lower pressure P2 on the right ("exit") side. The pressure drop across the membrane is defined as the difference between the entrance and exit pressures: A P=P\-P2. (2.2) If we now measure the relationship between flow rate Q and the variables in the setup, particularly the pressure drop, we get a plot similar to that shown in Fig. 2.4. The solid line, which20 CHAPTER 2. DARCY'S LAW volume flow rate Q (m3/s) 0 ideal membrane slope= I/R deformable material 0 pressure difference AP (Pa) Figure 2.4: Relationship between fluid flow rate and applied pressure difference for two different porous materials. The solid curve shows an ideal porous membrane following Darcy's Law; the dashed curve shows a deformable (nonideal) material. is representative of the behavior of an ideal membrane, shows a linear relationship between the flow rate and the pressure difference, and can be expressed by Darcy's Law: MAP Q = -r. H h Darcy's Law (2.3) The natural resistance of the fluid to flow is specified by the value of its fluid viscosity ji (SI units: kg/m-s). The ease or difficulty with which a particular membrane allows the fluid to pass is given by the permeability constant k (m2) of the membrane. The viscosities of various fluids and the permeability constants of certain materials, including biological materials and filters, are listed in tables in Appendix B. Note from (2.3) that the membrane properties k, A, and h along with the fluid viscosity \x determine the resistance of the membrane to allow fluid flow. It is reasonable, then, to think of the membrane as opposing the flow of the fluid with a hydraulic (fluid) resistance R, which we define as R= - kA' Inserting (2.4) in (2.3) gives an alternate and more compact form of Darcy's Law: AP 2= If Alternate Form of Darcy's Law (2.4) (2.5)2.2. DARCY'S LAW 21 Note from (2.5) that the higher the membrane's resistance is (for example, if the membrane thickness increases), the lower the flow rate is for a given pressure difference. 2.2.1 IDEAL AND NONIDEAL MATERIALS Darcy's linear relationship applies to many biological and man-made porous materials of interest, especially at low flow rates and pressures. In such ideal materials, the permeability /lisa constant and is independent of fluid flow rate, as shown by the solid curve of Fig. 2.4. This applies to many materials, but not all. For example, the dashed curve in Fig. 2.4 shows the relationship between the applied pressure difference and flow rate of a saline solution through articular cartilage in the knee. Although this material obeys a linear law for low flow rates, it is deformable such that as the pressure increases, the material compresses, narrowing or closing off some of the fluid microchannels. This causes its permeability constant k to decrease in the upper portion of the curve, consequently increasing its fluid resistance R to flow. It therefore exhibits a nonideal, nonlinear behavior as opposed to the linear behavior of an ideal material. Example 2.1. FlowThrough a Membrane A round membrane 2.00 mm thick has a permeability constant of k = 3.50 x 10_12/n2. Find the diameter that would allow water to flow at a volume flow rate of 17.5 cnrVmin with a pressure drop of 0.100 psi. Solution First convert the volume flow rate of 17.5 cnrVmin to SI units of nrVs: cm3 /1 x 10-6 m3\ / 1 min\ 7m3 Q = 17.5----- -----------r----- -------- = 2.92 x 10-7 -. min y 1 cm3 J \ 60 s / s Also convert the pressure drop to units of Pa: /6895 Pa \ AP = 0.100 psi ---;-J = 6.90 x 102 Pa. From (2.5) the resistance of the membrane needs to be „ AP 6.90 x 102Pa „ ^ ,^Pa-s R=----=----------------= 2.36 x 10}^-. Q 2.92 x 10 7m3/s m3 Putting this value in (2.4) and solving for^, using the value for /x of water from Appendix B, gives ,ih (0.0010 Pa-s)(2.00xl0-3m) _4 2 A - -- - - ~ r- - -p: - - 2.4x10 rn . Rk (2.36x 10 Pa • s/m3) (3.50x 10^12m2)22 CHAPTER 2. DARCY'S LAW Since A = jiD2/4, the diameter of the membrane is /(4)(2.4 xlCP4m2)\‘ „ D = I----------------------1 = 0.017 m = 1.7 cm. 2.3 MECHANICAL FILTRATION (SIEVING) In addition to partially resisting the flow of fluid through them, membranes also act as blocking filters, i.e., cutoff filters. This is where the filter completely excludes the passage of particles larger than a certain size regardless of the pressure applied.The maximum particle size allowed to pass corresponds roughly to the size of the channels or pores extending through the membrane, specified as the cutoff size or "effective" pore size of the filter. This mechanical filtration capability, also known as sieving, is perhaps the most common use of man-made membranes in the medical and bioengineering lab. Figure 2.5 is a photomicrograph of a typical man-made cellulose-fiber filter showing the intertwined fibers that provide the sieving action. The effective pore size is largely determined by the spacing between the fibers. Other filters can be made from glass fibers or polymer membranes such as nylon or Teflon. As an example, if the effective pore size of a certain filter is 0.1 /xm, no spherical particle with a diameter greater than 0.1 /im can pass through, regardless of the pressure applied-at least until the pressure is so great that it ruptures the membrane. However, for more realistically shaped biological particles that come in a variety' of three-dimensional shapes, the situation is more complicated. A long, cylindrical!}' shaped bacterium oriented with its long axis parallel to the channel axis might slip through if its narrowest diameter is below the filter's cutoff size, even though its length may exceed this size. (Nevertheless, for simplicity' in calculation, biological particles are sometimes modeled as though they were spherical in shape, specified by a single diameter.) Biological particles are often electrically charged as well, further affecting their ability' to pass through the tortuous channels of a filter. Bioengineers usually characterize the "size" of biological particles or molecules by their mass in terms of molecular weight MW. The traditional (non-SI) units of molecular weight are Dal­tons (abbreviated Da) in honor of English chemist John Dalton. One proton or one neutron have essentially the same weight; that sets the size of one Dalton 2: lDa = 1.661 x 10 24 (2.6) Thus, a neutron or proton has a mass of ~ 1 Da while an electron is much lighter (9.1 x 10_2S g) and thus has a mass only about 1/2000 Da. Proteins are much larger. For example, the blood protein serum albumin has a molecular weight of 68 kDa, and the blood protein Immunoglobulin M has a molecular weight of 1000 kDa. 'The Dalton was initially defined as 1/16 the weight of an oxygen atom (nominally 8 protons, 8 neutrons and 8 electrons) in a mixture of isotopes of oxygen. The more recent atomic mass unit (amu or u) is only very slightly different in weight from a Dalton.2.3. MECHANICAL FILTRATION (SIEVING) 23 Figure 2.5: View under a microscope of the intertwined cellulose fibers in man-made filter paper. The pore size is roughly determined by the opening between fibers; for this filter, the effective pore size is 11 Proteins-biological particles commonly found in body fluids-come in a variety of sizes and shapes. Still, since the subunits that make up proteins take up about the same space independent of the particular protein they belong to, the mass density of all proteins is about the same, 1.37 g/cm3. In a similar fashion, the mass density of DNA is reasonably uniform among different DNA strands, about 1.6-1.7 g/cm3. Viruses, which are composed mostly of a DNA core enclosed in a thin protein coating, have about the same density as DNA. In summary, to a good approximation, the density of each of these particles is: Pprotein ^ 1-37 g/cm PDNA, virus ^ 1-6 - 1.7 g/cm . (2.7) Once the density and molecular weight of the particle are known (and assuming a spherical shape for simplicity), the radius r of the molecule can be estimated by the following steps. The total mass M of the particle is given by its molecular weight in Dalton units times the equivalent weight of one Dalton: The density p is total mass divided by volume V: M = MW • (1.661 x 1CF24 g / Da). (2.8) p = M / V , (2.9) 24 CHAPTER 2. DARCY'S LAW and the volume of a sphere is 1/ = ^jrr3. (2.10) Using (2.7), (2.8), (2.9), and (2.10) together, the radius of a particular particle can be estimated (see the example below, which estimates the diameter of serum albumin, a protein, to be 5.4 nm). Viruses can be as small as 10 nm in diameter, but are generally on the order of 100 nm in diameter; some forms of the human influenza virus are about 80-120 nm in diameter. Bacteria (which are prokaryotic cells) are usually much larger than viruses, with a diameter range of about 1-10 fim. Eukaryotic (plant and animal) cells are larger still, with typical diameters in the 10-100 fim range. A red blood cell is a biconcave disk with a disk diameter of about 8 fim and a thickness of about 2 fi m. As mentioned above, porous membranes are often used to filter out particles larger than a certain size. Thus, a backpacking water filter can be used to eliminate bacteria and spores from drinking water while still allowing the small water molecules (which are about 0.3 nm in diameter) to pass. A cutoff pore size of 0.4 fim (400 nm) will filter out bacteria such as E. coli, typhoid, and cholera. Similarly, the eggs and larvae of parasites are about 20-100 fim in size, giardia cysts are about 14 fim, and Cryptosporidium is about 4 fim in size, so these pathogens would be filtered out as well. Viruses and proteins, on the other hand, are so small it is difficult to make a practical water filter for them; the pore size would be so small that the filter's permeability to water molecules would be very low, making water passage difficult at reasonable pressures. For viruses, alternative treatment methods such as iodine or boiling the water are often used. Example 2.2. Diameter of a Protein Find the diameter of the protein serum albumin, given that its molecular weight is 68 kDa. Solution Since the density of all proteins is approximately 1.37 g/cm3, (2.8) and (2.9) together give the protein volume as /6.Sx^Da\ M , \l.37g/cm 3/\ 1 Da j Then using (2.10) the radius r is found to be -20 \ '/3 (3)(8.2xl0 cm ) ^ ^ in_7 ^ -------------------------- I = 2.7x10 cm = 2.7 nm. r: \ An Thus, the diameter 3 is about 5.4 nm. ^Remember, diameter is twice the radius. Also, this calculation assumes a spherical shape for the protein. Actually, serum albumin is more cylindrical in shape.2.4. PROBLEMS 25 2.4 PROBLEMS 2.1. a. Using Equation (2.4), find the SI units for fluid resistance. fans: kg/(s-m4) or Pa-s/nr'] b. Using Newton's second law, Equation (1.1), show that kg/(s-m4) is equivalent to Pa-s/nr'. c. Using the above results, do a units check on Equation (2.5) to show that the units on the left-hand side are consistent with the units on the right-hand side. d. A common non-SI unit of viscosity is the poise (P), which is equivalent to g/(cm-s). Using conversion factors and the results of part b, show that 1 P = 1 x IO-1 Pa-s. 2.2. A certain blood plasma filter has a permeability constant of 1.0 x 1CP1 J m". Its area is 1.0 cm" and its thickness is 1.0 mm. When a net pressure of 10.00 psi is applied across the filter, what is the volume flow rate of blood plasma through the filter? If spherical drops of blood plasma, each of diameter 4.0 mm, come out of the filter, how many drops per second will flow? fans: Q_= 5.7 x 10-7 m'Vs; 17 drops/s] 2.3. Estimate the molecular weight of a blood protein that is approximately spherical in shape with a diameter of 20.0 nm. fans: MW % 3460 kDa or 3.46 x 106 Dal 2.4. A certain backpacking water filter has an "effective" pore size (i.e., a pore diameter) of 2.0 fxm. a. Will this filter trap the hepatitis A virus, which has a total molecular weight (including protein envelope) of about 32,000 kDa? (Calculate the virus diameter assuming it is spherical and using a reasonable estimate of its density.) If it is not trapped, how could you kill the virus in the water when backpacking? fans: Diam % 40 nm, so it is not trapped.] b. Will this filter trap the giardia cyst? If not, how could you kill giardia in the water when backpacking? (Note: you may find it useful to search the internet or reference books for the answers to parts b and c of this problem.) c. Is the E. coli 0157:117 bacterium a "good" bacterium, or a "bad" bacterium? What effects does it have on the body? Will this filter trap the E. coli 0157:117 bacteria? If not, how could you kill these bacteria in the water when backpacking?26 CHAPTER 2. DARCY'S LAW 2.5. a. Estimate the total number of hepatitis A viruses that could fit (in one layer) inside the period at the end of this sentence. Use the virus diameter found in Prob. 2.4a. fans: Approx. 130,000,000. Your answer may vary depending upon your measurement of the size of the period.] b. Repeat this estimate for red blood cells. Assume they all lie flat in the plane of the paper. fans: Approx. 3100.]CHAPTER 3 27 Poiseuille's Law: Pressure-Driven Tubes 3.1 INTRODUCTION - BIOLOGICAL TRANSPORT Diffusion is a major mechanism for transporting vital molecules into and away from cells (as men­tioned in Chapter 2 and to be covered in more detail in Chapter 14, Part II). For single cell organisms, such as bacteria, or for multi-cell organisms with only a few cells, the distances from inside the cell to the outside environment are short enough that diffusion times for these organisms are suffi­ciently rapid to sustain life. For example, an oxygen molecule will travel across 100 jim (the size of a moderately large cell) in a water environment in only about 1 second. But for larger organisms, certainly for animals including humans, diffusion times are much too long to rely on diffusion alone to transport life-supporting molecules throughout the entire body. Einstein derived the following relationship describing the average time needed for a molecule to diffuse a distance d driven by a planar concentration gradient: t = d\/ad- (3-1) where t is the average time (s) for a molecule to diffuse a distance d (m), and D (mVs) is a diffusion constant that depends upon the molecular size, shape and charge, the viscosity of the surrounding medium and the temperature. Note that the diffusion time varies as the square of the distance traveled. Thus, for an oxygen molecule to travel from the human lung (where indeed it is rapidly absorbed into the blood across the thin alveolar walls by diffusion) to a far-reaching part of the body, say the foot, would take about 6 years to travel by diffusion alone! Therefore, some means of augmenting diffusion by a bulk fluid flow-i.e., blood flow-is necessary. This is a major role of the circulator}' system in larger organisms: to expeditiously carry oxygen and other nutrients from the outside world to the remote tissues of the organism, and then to transport waste products back out. (Blood flow also facilitates heat exchange and immunological defenses in the body.) But oxygen by itself is not readily dissolved in pure blood plasma (mostly water), so some molecular carrier is needed. This task falls to hemoglobin, which effectively binds oxygen in the lungs (becoming oxyhemoglobin), then releases it where needed in tissues at lower oxygen pressure. Hemoglobin molecules are packaged inside small biconcave disk-shaped red blood cells (erythro­cytes) about 8 jim in diameter; together the red blood cells and the blood plasma comprise whole28 CHAPTER 3. POISEUILLE'S LAW: PRESSURE-DRIVEN FLOWTHROUGHTUBES blood. The percentage of whole blood volume taken up by the red blood cells gives the hematocrit of the blood-normally between 40-45% for humans, a little higher in males than in females. Figure 3.1: Open circulation in insects. After it leaves the heart, the blood empties into the entire body volume. It is then collected through openings in the heart to be re-pumped. (After Withers, 1992.) To pump the blood throughout the body of the organism, some pressure source is needed (a heart), and some arrangement for directing outward flow and collecting the spent blood is required (a circulatory system). Different levels of insects and animals have developed various strategies for their heart and circulation. Figure 3.1 shows the circulatory arrangement common in insects.This is an open circulation, meaning that the blood pumped by the single tubular heart (which periodically expands and contracts its diameter by muscles attached to its circumference) is released into the open field of tissue filling the insect's body rather than being contained in arteries, capillaries or veins. It perfuses the open tissue and picks up new oxygen before reentering the heart through openings in its walls to be re-pumped. This system works well for small organisms, but in larger animals the open field of tissue would present an uneven and inefficient distribution for flow. In fish (see Fig. 3.2) the system is closed such that the blood is contained in a continuous network of vessels passing through the body. The fish heart is two-chambered (an atrium preceding the stronger ventricle) but the heart is one-sided. This means that blood expelled out of the fish heart passes first through the gills to pick up oxygen, then directly to the rest of the tissues without any intervening pressure boost, before returning back to the heart via the veins. In mammals, including man, the circulatory system is similarly closed but the circulation is divided into two segments connected in series. Thus, the heart is two-sided, each side being the pump for its respective segment of the system. As shown in Fig. 3.3, the right side of the heart (comprised of two chambers, the right atrium and right ventricle) sends blood through the lung capillaries at relatively low pressure where the red blood cells pick up oxygen; this part of the system is called the pulmonary circulation. The oxygenated blood then returns back to the left side of the heart (also comprised of two chambers) where it is pumped with much higher pressure (about five3.1. INTRODUCTION-BIOLOGICALTRANSPORT 29 Figure 3.2: Closed circulation of fish, in which the blood remains inside tubes. The fish heart has only one side. times higher than in the pulmonary circuit) through the remaining tissues of the body; this portion is the systemic circulation. Since the circulatory system contains both a heart and vessels, it is called the cardiovascular (CV) system. The network of tubing making up the human circulation is complex. After passing out of the outflow valves of the respective ventricles, the blood is initially directed through a large vessel (the aorta in the systemic circulation) before splitting progressively into smaller and more numerous arterial vessels.The arteries in turn split into arterioles (called the "gatekeepers" due to their smooth- muscle walls that can contract or expand in diameter), then into a vast number of thin-walled capillaries (as small as 6-7 /xm in diameter) where molecular exchange takes place by diffusion. Although it is not evident from the drawings, capillaries are ubiquitous throughout the body, passing to within about 100 /xm (the width of a human hair) of every living cell in the body. After the capillaries, the vascular tree begins to recollect the blood into progressively larger and fewer vessels, first the venules, then the veins, and finally into one or two large return ducts (the vena cava in the systemic circulation) that empty into the atrium on the other side of the heart. All of these vessels have some elasticity or compliance, especially in the venous portion of the systemic circulation where a good percentage (approximately two-thirds) of the body's total blood volume resides. This elasticity is especially beneficial in smoothing out the pulsing nature of the blood flow from the beating heart, as will be seen in the next chapter, as well as in helping propel the blood along. The volume of blood pumped around the system per minute is known as the cardiac output, or CO, with traditional clinical units of L/min. Thus, it can be seen that the vascular network is composed of tubes of various lengths, diameters and connections, and that blood is driven through the network by the pressure produced by the respective ventricles of the heart. To quantify the amount of flow for a given pressure, we need to study the flow of fluids through tubes, next.30 CHAPTER 3. POISEUILLE'S LAW: PRESSURE-DRIVEN FLOWTHROUGHTUBES Figure 3.3: A schematic diagram of the human circulation, a closed system. There are two sides of the heart: the right side delivers blood to the lungs for oxygenation (the pulmonary circulation); the left side receives this blood and pumps it at higher pressure through the remainder of the body (the systemic circulation). (After Guyton and Hall, 2000.) 3.2 POISEUILLE'S LAW Between 1838 and 1840, G. Hagen and J. L. Poiseuille independently obtained the relationship between fluid flow in a tube and the pressure required to produce this flow. This relationship is called the Hagen-Poiseuille law, or simply Poiseuille's Law. Figure 3.4 shows the arrangement analyzed.3.2. POISEUILLE'S LAW 31 radius volume flow rate Q pressure Pj length = / * pressure P Figure 3.4: Tube for illustrating Poiscuillc's Law. In Fig. 3.4, a volume flow rate Q of fluid passes through a tube of length / under a pressure difference AP = P\ - Pi- Hagen and Poiseuille found the following relationship: na4 AP Poiseuille's Law (3.2) Q 8/u- / where Q = volume flow rate (m3/s) AP = P] - Pz = pressure difference (PaorN/m" orkg/m-s") Pi = fluid pressure at entrance to tube (Pa) Pz = fluid pressure at exit from tube (Pa) a = tube radius (m) ix = fluid viscosity (kg/m-s or Pa-s), and / = length over which the pressure drop is measured (m). The existence of the various terms in (3.2) can be qualitatively explained by a ranging check using each of the variables in turn (except for the factor 8, which is only found by a mathematical derivation beyond the scope of this chapter). That AP appears in the numerator seems reasonable, since for a given length of tube, the higher the driving pressure, the higher the flow should be. The location of / in the denominator also makes sense, since the longer the tube, the less the flow for a given pressure drop (as anyone who has used a very, very long garden hose knows). The terms ntf4 and ix in (3.2) need a little more explanation. One portion of the n<74 term (namely na2) can be seen simply from the fact that the tube's cross-sectional area is given by na2 and volume flow is proportional to cross-sectional area for a given pressure drop. The remaining a2 dependence requires a look at the fluid flow velocity inside the tube. For Poiseuille's Law to hold, the flow profile is assumed to be laminar (or "layered"), as diagrammed in Fig. 3.5. A laminar profile is characterized by a "no-slip" condition at the walls; that is, the fluid velocity is zero where the fluid touches the walls, and then the velocity increases parabolically toward the peak velocity at the centerline of the tube. Note that the velocity must build in a parabolic manner from zero (at the walls) to a peak value (at the center). So the smaller the radius of the tube, the less distance the velocity has to build to a peak value, thus reducing this peak value and the volumetric flow rate. When the32 CHAPTER 3. POISEUILLE'S LAW: PRESSURE-DRIVEN FLOWTHROUGHTUBES two-dimensional nature of the cross-section is considered, the additional a2 term is found, leading to the overall na4 dependence. fluid velocity Figure 3.5: Velocity profile for laminar flow in a tube. This fourth power dependence on radius a is dramatic. It means that if the radius of a tube is reduced by 1/2, the flow rate will be reduced to l/16th of its original flow! This would severely limit the amount of blood flowing through the tiny capillaries were it not for the fact that there are many, many capillaries in parallel in the vascular network. The presence of the viscosity term jx in the denominator of (3.2) also needs some discussion. Viscosity is a measure of the resistance of the layers of the fluid to flowing past one another-"sliding friction" as it were. We first saw the term in Chapter 2 relating to flow through membranes. The laminar flow profile in Fig. 3.5 requires that neighboring layers of fluid must slide past each other going from the walls to the centerline, imposing a shearing nature to the flow, as shown for a local region of the fluid in Fig. 3.6. A u U---► velocity u- force F----------------► Figure 3.6: Layers of fluid sliding past each other, giving a shear rate Ah/A y, which requires a force F. The extent of the shear is expressed in terms of the shear rate: shear rate = Aw/Ay. (3.3) It takes a force F to overcome the friction of the layers sliding past one another. The force is described locally in terms of the shear stress r: where A is the area over which the force F acts. r = F / A, (3.4) 3.2. POISEUILLE'S LAW 33 a) shear stress X Newtonian b) Non-Newtonian blood plasma water shear rate Au/Ay Figure 3.7: (a) Plot of the shear stress vs. shear rate for three fluids that exhibit a linear relationship (Newtonian). The slope gives the fluid's viscosity fi, which is a constant, (b) Plot of whole blood. The red blood cells cause the fluid to be Non-Newtonian, where the viscosity varies with the flow rate. When the shear stress of a fluid is plotted as a function of the shear rate, a graph similar to Fig. 3.7(a) is obtained for many well-behaved fluids. It shows a linear relationship between shear stress and shear rate, such that the force required to drive the laminar flow is proportional to the velocity of the flow. When put in terms of shear stress r and shear rate, the fluid obeys the following equation: r = fi ■ (Au/Ay), (3.5) where the proportionality constant fi is the viscosity of the fluid. Now the presence of the viscosity term fi in the denominator of (3.2) seems reasonable: the more the fluid resists shear flow (i.e., the higher its viscosity), the more pressure AP it takes to cause a certain flow rate Q. Molasses takes more force to flow through a tube than water. When the viscosity value is a constant that is not dependent upon the flow rate, as in Fig. 3.7(a), the fluid is termed Newtonian. However, some fluids exhibit a viscosity that varies with flow rate. Such fluids are termed Non-Newtonian. An example is whole blood, where the addition of the red blood cells to blood plasma causes the viscosity to increase several fold from that of plasma (so "blood is thicker than water") and also to become variable, as shown in Fig. 3.7(b). 3.2.1 SIMPLIFIED VERSION OF POISEUILLE'S LAW As with Darcy's Law, we can put Poiseuille's Law (3.2) in a form that involves the resistance of the tube. For flow through tubes, the fluid resistance (or hydraulic resistance) is given by34 CHAPTER 3. POISEUILLE'S LAW: PRESSURE-DRIVEN FLOWTHROUGHTUBES R 8/xI n cr (3.6) With this definition of fluid resistance, Poiseuille's Law (3.2) has the following simplified form: A P Q = - R P1 - P2 R ' Alternate Form of Poiseuille's Law (3.7) It is often convenient in diagrams to show flow through pipes or tubes by replacing the physical shape of the tube with a symbol representing the resistance R of the tube. The symbol for a fluid element that has resistance to flow, such as a tube, is given in Fig. 3.8. P, - + R AAA Q -► P2 Figure 3.8: Symbol for a resistive fluid element, such as a tube. 3.2.2 ASSUMPTIONS FOR POISEUILLE'S LAW For Poiseuille's Law to hold, there are a number of assumptions that must apply to the flow through the tube: * The length of the tube must be much greater than the radius, * The flow must be steady in time and laminar in velocity' profile, * The fluid must be Newtonian, and * The tube must be rigid. Actually, in biological circulator}' systems, none of these assumptions are strictly met for flow through the entire organism. However, they hold to some degree under certain conditions. Let's examine the areas of validity' for each assumption for blood flow in the human circulation. * Length is much greater than radius - This approximation is valid in the aorta, in longer arteries and in some of the small-diameter, long capillaries, but it is not very valid in the rapidly branching, shorter networks found throughout the system. * Flow is steady and laminar - The blood flow in the aorta and arteries is pulsatile, so the approximation is not valid in those vessels. But when the flow reaches the capillaries and veins, it becomes increasingly steady due to the damping action of the compliant vessels (to be covered in Chapter 4).3.2. POISEUILLE'S LAW 35 Whether the flow profile is laminar or not can be predicted by the flow's Reynolds number % Re, which is a dimensionless (that is, unitless) number calculated by puD Re = -----, (3.8) IX where p is the fluid density (kg/m3), u is the fluid velocity' (m/s), D is the diameter of the tube (m), and fx is the dynamic fluid viscosity' (kg/m-s or Pa-s). Reynolds numbers of 2000­4000 are often considered to be the approximate dividing values between laminar and turbulent flow: when the Reynolds number is below 2000, the flow is likely laminar, while if the Reynolds number is above 4000, the flow will usually be turbulent. Between 2000 and 4000, the flow can be either depending upon other factors such as the time course of the flow and nearby boundary disruptions-e.g., branching-since that will alter the flow profile. Now almost all flows in the human body (with the exception of blood flow in the large aorta during peak ejection from the heart) are of such low flow velocities that the Reynolds number is much below 2000 and the flow is laminar, not turbulent. * Fluid is Newtonian-Without red blood cells, the blood plasma is reasonably Newtonian [see Fig. 3.7(a)], but when the red blood cells are added, the whole blood becomes Non-Newtonian, except over narrow ranges of flow rates [see Fig. 3.7(b)]. * Tube is rigid-All blood vessels are not rigid, but rather are distensible and flexible to some degree, with veins being more flexible than arteries. This proves beneficial to the circulation by smoothing out the flow. So it is clear that the assumptions inherent in Poiseuille's Law do not apply at all times and in all places in the blood circulation. Nevertheless, Poiseuille's Law in the form of (3.2) is still of value when applied to many of the individual vessels. Furthermore, when an entire ensemble of vessels is analyzed (as in calculating the peripheral resistance of the systemic circulation-see example below), detail about the vast number of individual vessels is not known so the simplified form (3.7) is used, in which case a single combined resistance characterizes the ensemble. On this scale, the validity' of the assumptions is less important and Poiseuille's Law in the form of (3.7) becomes even more applicable. Example 3.1. Peripheral Resistance of the Human Systemic System In physiology terms, the fluid resistance of the entire human systemic circulation, from the inlet at the aorta to the outlet of the vena cava, is called the "peripheral resistance." Given that a typical human cardiac output is 5.5 L/min, that the average pressure in the aorta is 100 mmHg, and that the average pressure in the right atrium is 7 mmHg, find the peripheral resistance in units of mmlig-s/L.Solution The peripheral resistance can be modeled with the simplified form of Poiseuille's Law: 36 CHAPTER 3. POISEUILLE'S LAW: PRESSURE-DRIVEN FLOWTHROUGHTUBES Q -► Pi*------W------• p2 Rpr Figure 3.9: Model of peripheral resistance used in Example 3.1 A P AP where Q = -----, so Rpr = -----. First find Q in units of L/s: Rpr Q L /I min\ L Q = 5.5 - ------- = 9.2 x 10 -. min \ 60 s / s Also AP = P\ - Pi = 100 mmlig - 7 mmlig = 93 mmlig.Then = «mmlig ip; mmlig-s = mm ; 9.2 x 10 L/s L S ; 3.3 POWER EXPENDED IN THE FLOW It takes energy to force the fluid through the tube against its resistance, and since power is defined as the energy expended per unit time, there is a power requirement for maintaining the flow. This power O is given by the product of the pressure drop and the volume flow rate: <t> = AP Q. (3.9) Relating pressure drop and flow to resistance by (3.7) gives alternate forms for the expended power: -> (AP)2 ® = Q-R = (3.10) These power concepts are addressed again in Chapter 7. 3.4 SERIES AND PARALLEL COMBINATIONS OF RESISTIVE ELEMENTS The circulatory network of vessels contains a great number of interconnections: some vessels can be considered connected in series (where the flow goes through each tube sequentially without splitting), others in parallel (where the flow splits into several branches before recombining). Using the concept of fluid resistances, a combination of several tubes either in series or in parallel can be3.4. SERIES AND PARALLEL COMBINATIONS 37 easily handled by defining an equivalent resistance Req that represents the effect of the resistances lumped together. These two situations are considered next. 3.4.1 SERIES This configuration can be illustrated using three hydraulic resistors in series: P1 P2 P3 P4 P1 P WWV--MV- equivalent to HWr-4 Rj R2 Rj R Q-- eq Figure 3.10: Resistive elements in series can be replaced by an equivalent resistor. Since the same flow Q must go equally through all elements, (3.7) can be applied to each resistance in turn: Pi-P2 = RiQ P2 - P} = R2 Q (3.11) and Pj, - P4 = Ri Q. Adding these three equations together and defining an equivalent resistance Req gives Ap = px _ p4 = (R{ + Rl+ = ReqQ, (3.12) where the equivalent series resistance can be seen to be Re q - R [ R2 "h R i Series (3.13) and the total pressure drop is related to the flow rate by (3.12). The general formula for combining N resistors in series is N Req = y ' R„. (3.14) n=l where E is the standard notation for summation. 3.4.2 PARALLEL The parallel case can be illustrated with two resistive elements in a branching configuration in Fig. 3.11. Here the flow is split-usually unequally-between the two branches. Flow Qi goes through resistor Ri and Q2 goes through R2. By the conservation ofvolume principle for incom­pressible fluids, the total flow Q entering (and exiting) the entire circuit must be equal to the sum of the flows in the two branches:38 CHAPTER 3. POISEUILLE'S LAW: PRESSURE-DRIVEN FLOWTHROUGHTUBES pj Q Q, R WW- vwv P2 equivalent to I) P -^wv2- R eq Figure 3.11: Resistive elements in parallel can be replaced by an equivalent resistor. 0= 01 + 02- The pressure drop is the same across each element, given by Py - P2 = AP. Using (3.7), the branch flows are related to the common pressure drop by (3.15) (3.16) and AP 0| = "kT' AP R2 Q2 Putting (3.17) and (3.18) into (3.15) and defining an equivalent resistance gives AP AP /I 1 0 =------1------= A P-------1----- * Pi p2 Ui r2 where the equivalent parallel resistance can be seen to be AP R, «i R, «i 1 1 Yy+~R^' Parallel The general formula for combining N resistors in parallel is N et) n=1 Rn (3.17) (3.18) (3.19) (3.20) (3.21) A more complex flow circuit containing both parallel and series combinations can be analyzed by combining each segment using the appropriate equivalent resistances, in order, until the entirecircuit is represented by an equivalent resistance. Application of this technique is given in the second example below. 3.4. SERIES AND PARALLEL COMBINATIONS 39 Example 3.2. Adding Tubing to a Membrane Filter Let the semipermeable membrane analyzed in the example of Chapter 2 (see p. 21) have inlet and outlet tubing added to each end of the membrane, as shown in Fig. 3.12. Assuming that the flow rate of water through the entire assembly is the same as in the example in Chapter 2 (Q = 2.92 x 1CP7rrrVs), and assuming that the inside diameter of the tubing is the same as that of the membrane (D = 1.7 cm), calculate how much additional pressure drop is incurred when the tubing is added. inlet tube membrane from Unit 2 outlet tube 11 = 30 cm l2 = 20 cm Figure 3.12: Drawing of configuration analyzed in Example 3.2. Solution The tubing can be considered to add additional fluid resistance to the overall assembly.The resistance of each tube can be found from (3.6): „ 8^/1 in 8^2 i = ------- and Ro = 4 izaZ in which = 112 = D/2 = 8.5 x 10 3m. Since the resistances of the two tubes are in series, their resistances add, as given by (3.14): TT 8/ti (/1 + h) 8(1.0x10 3Pa • s) (5.Ox 10 !m) ^ ( in5Pa- ' *(D/2)4 = n (8.5 x 10-3m)4 =2'4 X ^ where the viscosity of water /x = 1.0 x 10 3Pa-s has been used.The additional pressure drop that is added to the original pressure drop (0.100 psi) can be found from (3.7):40 CHAPTER 3. POISEUILLE'S LAW: PRESSURE-DRIVEN FLOWTHROUGHTUBES A^add = Q ■ ^tubing = (2.92 x 1(T7 m3/s) • (2.4 x 105 Pa • s/m3) = 7.0 x 10-2 Pa = 0.000010 psi. Note that the additional pressure drop due to the tubing is only 1/10,000th of the original pressure drop, so it is negligible compared to the drop across the membrane itself. This is because the tubing is relatively short in length and large in diameter. Example 3.3. Branching Vessels Whole blood flows through a single vessel before splitting into two identical parallel vessels in Fig. 3.13. The length of each segment is 2.0 cm. The combined cross-sectional area of the two parallel vessels is the same as that of the single vessel. Given a blood flow rate of 1.0 mL/min, what is the pressure drop across the entire configuration? (Assume that the viscosity of whole blood is 4.0 cP.) radius =1.0 111111 r - i-f Figure 3.13: Branching vessel arrangement analyzed in Example 3.3. Solution This arrangement can be analyzed as one resistance in series with a parallel combination of two other resistances in Fig. 3.14. We can use Poiseuille's Law to find each of the resistances, after we first find the radius ri of each of the identical parallel tubes. Given that the combined area of the two tubes is equal to that of the single tube, then3.4. SERIES AND PARALLEL COMBINATIONS 41 Q p, VW\r R. VWV VAV Figure 3.14: Schematic model of the branching tubes. 2 2 , 2 o 2 7tj*| = 7rr, + 7rr, = 27rr,, ri 1.0 mm _A so r? = -- =------- = 0.71 mm = 7.1 x 10 m. " sfl s/2 Then, using (3.6), 8 ill 8(4.0 x 10-3 Pa-s) (2.0xl0-2 m) xPa-s fa = fa = -----------------L = 8.Ox 10 - n(n) n (7.1 x 10-4 m) m_ 1 1 1 Now combine these two resistances in parallel using (3.20). See Fig. 3.15 where - =-----1-----. Rp R2 R3 VWV R2 equivalent to Figure 3.15: Two parallel tubes combined into one equivalent resistance. 1 2 fa 8.Ox 108 x Pa • s Since R2 = Ri. - = -,orRp = -^ =---------= 4.0x 108^. Rp Ri 2 2 mJ Hence, the entire configuration can be modeled as two resistors in series in Fig. 3.16 where Re^ = R\ + Rp. Again using (3.6), 8/il 8 (4.0 x 10-3 Pa • s) (2.0 x 10-2 m) Pa • s R1 =--------7 =---------------------------------4------------= 2.0 x 10 -r-. 7T(n) 7T (l X 10-3 m) m_ [Note that even though the combined cross-sectional area of the two parallel tubes is the same as the single tube, their equivalent combined resistance Rp is still twice that of the single tube R1. This42 CHAPTER 3. POISEUILLE'S LAW: PRESSURE-DRIVEN FLOWTHROUGHTUBES ^/VWAV- eqtalent -WW- R, R R Ip eq Figure 3.16: Two series resistances combined into one overall version. shows the large influence that the r4 term has in determining the resistance of an assembly, and why the arterioles can be so effective in controlling blood flow distribution by changing their diameter under smooth muscle control.] Thus, Req = R{ + Rp = 2.0 x 10s + 4.0 x 10s = 6.0 x 10s Pa • s/m3. The model is now simpli­fied to a single resistance. See Fig. 3.17 where mL 1 L (\ min\ (1.000 x 10-3 m3\ x mJ Q = 1.0- = 1.0 x 10-3- -- ------------------- = 1.7 x 10-8 -. mm mm \ 60 s / \ 1 L / s Q PI •--------W--------• p2 R 2 eq Figure 3.17: Branching tubes reduced to one overall resistance to find A P. Then using (3.7), AP = Q ■ Req = (1.7 x 10_s m3/s) • (6.0 x 10s Pa - s/m3) = 10 Pa. - tt n (1 mmlig\ Converting to units of mmHg, AP = 10 Pa ■ I I = 0.075 mmHg. \ Jr3. / 3.5 PROBLEMS 3.1. a. Given that an oxygen molecule in an aqueous environment will diffuse an average distance of 100 jim in 1.0 second, find its diffusion constant D from (3.1). [ans: D = 2.5 x 10^y m2/sl b. Using the value from part a, estimate how long it would take an oxygen molecule to diffuse from your lungs to your big toe. fans: Approx. 1.8 x 10s s, or 5.8 yrs for a typical height!] 3.5. PROBLEMS 43 3.2. Three tubes each have the same fluid resistance, R = 6060 Pa-s/m3 (to 4 significant figures). a. If all three are put in series, what is the equivalent overall resistance of the series combination? fans: Req =18,180 Pa-s/m3] b. If all three are put in parallel, what is the equivalent overall resistance of the parallel com­bination? Hint: Use Equation (3.21). fans: Req = 2020 Pa-s/m3] 3.3. a. In the human cardiovascular system, the blood pressure measured at the arterial side (input side) of the capillary bed is approximately 6650 Pa, while the blood pressure at the venous side (output side) is approximately 3325 Pa. These values are for a typical human volumetric blood flow of 92 mL/s. (Note the use of mL here, a non-SI unit.) Using the concept of fluid resistance, calculate the total equivalent resistance of the capillary bed in SI units. Remember that IL = 1.000 x 10-3 m3 = 1.000 x 103 cm3. fans: R = 3.6 x 107 Pa • s/m3] b. In the actual clinical environment, SI units are rarely used. In clinical practice, blood pressure is almost always measured in units of millimeters of mercury (mmHg), and blood flow is given in units of either liters per minute (L/min) or liters per second (L/s). In these units, the blood pressure across the capillary bed drops from 50 mmHg on one side to 25 mmHg on the other when the flow is 0.092 L/s. Using these clinical units, calculate the equivalent resistance of the capillary bed. fans: R = 270 mmHg-s/L] 3.4. a. A single capillary is about 8.0 fim in diameter and has a typical length of about 1.0 mm. Using Poiseuille's Law and values for whole blood from Appendix B, calculate the fluid resistance (in SI units) of a single capillary tube. fans: R 4.0 x 1016 Pa • s/m3 assuming a whole blood viscosity of 4.0 cP] b. Of course, the human capillary bed is composed of many, many capillaries arranged in parallel. Assuming that each one has the resistance value found in part a, calculate how many of them must be in parallel in the human CV system such that the overall equivalent resistance has the value found in Problem 3.3a. fans: about 1,100,000,000 !] c. In a sentence each, discuss how well this capillary configuration meets each of the require­ments found on pp. 34 to 35 for Poiseuille flow to be valid: 1) The length of the tube must be much greater than the radius, 2) The flow must be steady in time and laminar in velocity profile, 3) The fluid must be Newtonian, and 4) The tube must be rigid. 44 CHAPTER 3. POISEUILLE'S LAW: PRESSURE-DRIVEN FLOWTHROUGHTUBES CHAPTER 4 45 Hooke's Law: Tissues 4.1 INTRODUCTION Superman1 M may be the Man of Steel and his chest may stop bullets, but if his body were made entirely of steel-like material, it would not function nearly as well as yours or mine. For example, how could the skin around his elbow stretch and conform when it is bent, and how could the skin of his face show a smile? How could his bladder, if made of rigid material, expand to fill with urine, then contract as it is expelled? Without elasticity', his arteries and veins would not cushion and smooth out the pulses of blood flow from his heart, leading to punishing, pounding pressure waveforms throughout his body. And how could Superwoman1 M accommodate a pregnancy if her skin and abdominal organs could not expand or contract? In fact, all tissues in the body have elasticity', some more so than others depending upon their function. Even bones, which are relatively stiff and rigid in order to act as the skeleton framework for the body, have some elasticity'; otherwise they would not absorb shock and distribute stress appropriately. The soft tissues of the body are obviously more elastic than bone in performing their function. Cartilage is somewhere between bone and soft tissue in flexibility'. 4.2 THE ACTION OF FORCES TO DEFORM TISSUE There are three main classes of actions that any given external force can have on a tissue sample, depending upon the direction of the force and the direction of the distortion of the sample. These are summarized in Fig. 4.1, which considers a small stylized cube of the tissue. In the first action, the force components F push perpendicularly inward equally on all faces of the sample, causing it to compress its volume. The magnitude of the force is best described in terms of the pressure P = F/A, where A is the face area upon which the force is acting. We have already encountered the concept of pressure in the previous two chapters. In the second action, the force acts tangentially (sideways) to one or more faces of the sample, causing is to twist out of shape, but not changing its volume.The action of the force is best described here by the shear stress r = F/A , where A is again the area of the face over which the force is tangentially applied. We have seen how shear stress is related to fluid flow and viscosity' in Chapter 3. The third class of deformation is when the forces act only on two opposite ends of the sample, tending to pull it apart (or push it together). This puts the material in tension (or compression) and can best be related to the applied tensile stress a = F/A, where again A is the cross-sectional area46 CHAPTER 4. HOOKE'S LAW: ELASTI CITY OFTISSUES AND COMPLIANT VESSELS Pressure Shear Tension Figure 4.1: Three ways that a force can distort a tissue sample. of the face through with the force is applied perpendicularly. This is the deformation described by Hooke's Law below. Note that the dimensions of all three stresses in Fig. 4.1-pressure, shear stress and tensile stress-are the same (force per area) and therefore their units are the same (Pa in SI units). However, how they are applied is different in each case. Pressure and tensile stress are both applied normally (i.e., perpendicularly) to the faces, while shear stress is applied tangentially. Pressure acts equally on all sides of the sample, while tensile stress is applied only to opposite faces. 4.3 HOOKE'S LAW AND ELASTIC TISSUES Biomechanical engineers measure the extent to which tensile stress can distort various tissues by employing a testing machine similar to that shown in Fig. 4.2. A sample of the material to be tested (aligned such that the direction of stretch will be in a direction of interest in the sample material) is fabricated and clamped between a fixed base and a movable upper bar.The sample is usually fabricated in a "dog-bone" shape, the large ends for convenience in clamping and the narrower middle section to allow enough stretch for accurate measurement. Bones, ligaments, metals, man-made composites and some soft tissues can be tested in this manner. When the test is started, the upper bar moves slowly upward (or downward for compression) in a controlled fashion, exerting a force on the sample. The magnitude of the force F is measured by a gauge (a load cell) in series with the sample. Since the cross-sectional area A of the sample in its narrow region can be measured prior to the test (and is often assumed to remain approximately constant throughout the test1), the tensile stress a at any time during the test can be calculated from 1 When the cross-sectional area A is assumed to be constant, the stress is defined as "engineering" stress, as used here. "True" stress instead puts a changing area, which is more difficult to measure, in the denominator of (4.1).4.3. HOOKE'S LAW AND ELASTIC TISSUES 47 Figure 4.2: A tensile (or compressional) testing machine for determining the elastic constants of various materials. a = F/A. (stress) (4.1) The amount of stretch (or compression) of the sample during the test is measured by reference to two points separated by a distance / along the sample's axis. The measurement can be done with a number of different length gauges, or extensometers, including optical means and video cameras. With no applied force the points are a distance /q apart, the original spacing. As the force is gradually increased, their spacing changes to a new value /. The resulting strain s of the material can be determined from2 e = (/ - /())//()■ (strain) (4.2) Over the course of the test, values of both stress and corresponding strain are recorded. Note from (4.2) that strain is a dimensionless (therefore unitless) quantity, while from (4.1) stress has units appropriate for pressure. When plotted on a stress-strain curve, the results can appear as shown in Fig. 4.3 for two different materials. The solid line in Fig. 4.3 plots the results for a material that behaves in a linear fashion, at least up to the point where the material fractures. For a linear elastic material, the strain and stress are related by a constant coefficient and obey Hooke's Law: 'When strain is defined with the original distance /() in the denominator, as in (4.2), it is known as "engineering" strain, as used here. "True" strain instead puts the changing distance / in the denominator, but this is a little more difficult to employ in most analyses.48 CHAPTER 4. HOOKE'S LAW: ELASTI CITY OFTISSUES AND COMPLIANT VESSELS Figure 4.3: Results of a stress-strain test on two different materials. The material giving the solid line is linear with a slope equal to Young's modulus. The dashed line describes a non-linear material. E s. Hooke's Law (4.3) The constant of proportionality E in (4.3) is known as the elastic modulus, or Young's modulus or simply stiffness. It is found from the slope of the linear portion of the stress-strain plot. Since units of stress are the same as pressure (namely Pa) and strain is dimensionless, (4.3) shows that the units of E are also the same as pressure (Pa). Materials that follow the linear Hooke's Law are called Hookean. Bone and some biopolymers as well as biomaterials such as metal and ceramic orthopedic implants are approximately Hookean over a wide range of applied stress. Appendix B contains the Young's modulus for typical materials. The dashed line in Fig. 4.3 shows a material that is highly non-linear in behavior, being pliable at low force, stiffening up for moderate to high force, then again getting pliable just before if fractures. This response is typical of many soft tissues in the body, such as bladder walls, skin and the walls of blood vessels. However, every tissue is somewhat unique in its exact response to force, depending upon its function. Many materials have a limited range over which their response is approximately linear, and they can be specified by a local tangent elastic modulus that is valid when the material is within that range. As the force is applied larger and larger in Fig. 4.3, eventually all materials will fail by fracturing. The stress point at which this happens is called the ultimate stress, denoted by an asterisk (*) on the plot. It is the highest stress a material can tolerate without failure. Along with Young's modulus (if the material is linear), it characterizes the material. For example, with advancing age bone becomes more brittle and susceptible to fracture from falls.This is manifest in alowering of its ultimate stress. The bone structure loses density and becomes osteoporotic. The risk for this condition is especially prevalent in post-menopausal women.4.3. HOOKE'S LAW AND ELASTIC TISSUES 49 Example 4.1. Bone Splint When a splint (a bone support plate) is fixed along the side of a broken bone, it is important for the splint to be rigid enough to keep the bone at the fracture plane from moving under load, thus allowing the bone to stay in place and heal. Consider two different splint materials, polyethylene (a polymer) and stainless steel. Each splint has a cross section of 0.500 x 0.250 inches, and is 10.0 inches long. When a force of 200 pounds is applied to the top of the bone/splint combination, how much relative movement occurs between the two sides of the fracture for each material? rr splint bone 'fracture Figure4.4: Sketch ofbone splint configuration analyzed in Example 4.1. Solution Since the two faces of the bone at the fracture plane can slide and therefore will not resist any weight, all of the force will be absorbed by the splint. From (4.1) the tensile stress placed on the splint is <7 = F j A = 200 lbf/C0.125 in2) = 1600 psi = 1.10 x 107 Pa. (4.4) where a conversion factor from Appendix A was used to get the last value. The resulting strain is, from (4.3), s = a/E. (4.5) Since by the definition of (4.2) e = (I - lo)/lo = A//lo, solving for AI and using (4.5) gives AI = l Qe = I qo/E. (4.6) Now /() = 10.0 inches = 0.254 m, so using (4.4), /q<t = 2.79 x 106 Pa • m. We will next put this value into (4.6) to solve for Al for the two different materials with different Young's modulus. Polyethylene'. From Appendix B, Young's modulus for high-molecular-weight polyethylene is E = 1.00 x 109 Pa. Therefore, the compression of the splint (which is seen as relative movement between the two sides of the bone at the fracture plane) is, from (4.6), Al = (2.79 x 106 Pa - m)/(1.00 x 109 Pa) = 2.79 mm. This large amount of movement will certainly be disruptive to the healing process. Stainless steel'. From Appendix B, Young's modulus for stainless steel is E = 180 x 109 Pa. Thus, Al = (2.79 x 106 Pa - m)/(180 x 109 Pa) = 15.5 fim. This movement is much smaller, and can be tolerated by the bone healing process. 4.4 COMPLIANT VESSELS Tissue elasticity plays a major role in the proper functioning of the cardiovascular system. The vessels that carry blood from and to the heart are not rigid tubes, but rather have flexible walls that stretch in response to the blood pressure inside them, to a greater or lesser degree depending upon their elasticity and the pressure3. This vessel compliance has at least three important consequences on the nature of the circulation: • The pulses of blood discharged by the heart ventricles with each beat are propelled into compliant vessels (the aorta and the pulmonary artery) that temporarily store some of the energy of the ejected blood as potential energy in their stretched walls. After the ventricle output valves close-blocking the return of the blood to the ventricles-the vessel walls recoil and the stored potential energy is converted to additional blood velocity (kinetic energy), helping push the blood down the arteries. This augments the cardiac output of the heart during the diastolic phase while at the same time keeping the peak systolic pressure lower. The actions of wall recoil are diagramed in Fig. 4.5. • The pulsating blood pressure, very noticeable in the aorta and major arteries, is damped by the compliance of the arteries, arterioles, and capillaries in conjunction with their resistance, so that by the time the blood reaches the capillaries, the pressure waveform is almost flat. • The veins and venules are the most compliant of all the vessels, being thin-walled and very flexible, and they therefore store a good deal of the total blood volume (about 65%) of the J Compliant blood vessels are called "windkessel" vessels, and are discussed in more detail in Chapter 5. SO CHAPTER 4. HOOKE'S LAW: ELASTI CITY OFTISSUES AND COMPLIANT VESSELS 4.4. COMPLIANT VESSELS 51 a. systole b. diastole Figure 4.5: Diagram of the recoil action of the compliant walls of the aorta and arteries. The major effects of wall compliance are to lessen the peak pressure during systole (a), and to add flow down the circulation during diastole (b). (After Silverthorne, 1998.) entire circulation. The size of this pool of blood is regulated by the compliance (the "tone") of the venous vessels, which in turn is under partial control by the nervous system. This forms an important contributor in the control of the blood distribution and pressure in the systemic system. In the case of a major change in the tone of venous compliance, such as happens in anaphylactic shock, there is a large change in the distribution of blood, leading to severe alteration of the circulation's effectiveness, and possibly even death. A simple diagram of how the volume in a compliant vessel is related to the net pressure (i.e., inside pressure minus outside pressure) is shown in Fig. 4.6. Increasing the net pressure will linearly increase the vessel's volume-up to a point where the limit of elasticity is reached. In the linear region the proportionality constant between pressure P (Pa) and volume V (m^) is called the vessel compliance C (nrVPa).The larger the vessel's compliance, the larger the volume for a given pressure. Mathematically, the relationship shown in Fig. 4.6(b) can be stated as V = v4) +CP (4.7) where P = net pressure in the vessel V = volume of the vessel52 CHAPTER 4. HOOKE'S LAW: ELASTI CITY OFTISSUES AND COMPLIANT VESSELS Figure 4.6: When the pressure increases in a compliant vessel, its volume increases (a). The graph of the volume vs. pressure (b) usually has a linear region, whose slope is the vessel compliance C. C = compliance V$ = residual volume. Equation (4.7) is a generalized form ofHooke's law applied to compliant vessels.The residual volume V,p is the volume left in the vessel even with zero pressure; that is, it is the volume remaining when the vessel is completely relaxed. On the plot above, it is shown by the intercept of the straight line with the vertical axis. Arteries and arterioles have some residual volume (as evidenced by the fact that their diameters are still partially open when they are excised from the body), but venules and veins have much less residual volume and can collapse completely shut when unloaded. The pressure P in (4.7) is the transmural ("across the walls") pressure, which is the difference between the pressures inside and outside of the vessel. Usually the pressure outside -which is the background pressure-is constant, and is often considered zero as the reference pressure. Compliant vessels are found throughout the human body (blood vessels, bladders, lungs) as well as in other non-biological applications (hydraulic reservoirs, vibration dampers, automobile tires). Since they have the capacity for the storage of fluid volume, in diagrams they are given the symbol of a capacitor in Fig. 4.7. This symbol has certain significance.The two horizontal lines can be thought of as representing the outline of a storage container (with variable volume). The fact that there is no direct connection between the upper line and lower line indicates that there is no net leakage across the walls from inside the vessel to the outside. (Leakage would be represented by a resistive element in parallel with the capacitor to ground.)Flow rate 4.5. INCOMPRESSIBLE FLOW OF COMPLIANT VESSELS 53 0 / Pressure P Q . I net if Compliance C reference pressure = 0 Figure 4.7: The symbol for a compliant vessel. 4.5 INCOMPRESSIBLE FLOW INTO AND OUT OF COMPLIANT VESSELS To get fluid volume in and out of the vessels, there must be some fluid flow through one or more openings. There is a simple relationship between the volumetric flow rate Q (a variable introduced in previous chapters) and the volume V contained in the vessel. Since the liquids we are interested in (water and blood) are essentially incompressible, the volume of the fluid must be conserved when flowing from one region to another; this is the principle of conservation ofvolume. It means that during any increment of time At, a net volume flow rate Q into or out of the vessel will change the vessel's volume by the total amount of volume A V that has flowed, so av , s £?net = (4.8) Rearranging, AV = (2net A?. (4.9) Assuming that the volume of the vessel originally had a value of V0rig before flow was measured and time incremented, then (by the definition ofvolume change) the new volume Vnew after accounting for the flow Qnet is given by View - Vorjg + AV - Vorig + QnetAf. (4-10) If more fluid is entering the vessel than is leaving during At, the net flow is in the direction into the vessel and the sign of (?net is positive. Then (4.9) shows that the vessel volume change A V has a positive sign, and (4.10) shows in turn that the vessel volume increases. On the other hand, if more fluid leaves than enters, the net flow Qnet has a negative sign, A V has a negative sign, and the volume decreases. Now from (4.7), pressure P and volume V are related by C, so the new pressure and new volume are related by54 CHAPTER 4. HOOKE'S LAW: ELASTI CITY OFTISSUES AND COMPLIANT VESSELS Vnew - fy + C/W (4-11) Rearranging, Vn c (4.12) where is the residual volume of the vessel and shouldn't be confused with Vorjg.The value for Vnev/ is obtained from (4.10). These relationships are important in understanding the pressure/volume interaction in vessels of the human cardiovascular system. Extending (4.7) in another way, since pressure P and volume V are related by C, then changes in pressure and volume are similarly related. From (4.7), since the residual volume is a constant and n - 4 assuming Lis a constant , AP = -. (4.13) Substituting AV from (4.9) into (4.13) and rearranging gives AP 1 , - Onet- (4.14) For (4.14) to be accurate, Qnet and C must be steady during the interval At. If At is small enough, this is true. In fact, taking At to a limit approaching zero gives the differential form of (4.14): dP 1 - = ~Qnc f (4.15) dt C Equations (4.14) and (4.15) state that the more compliant a vessel is (i.e., the larger C is), the slower the pressure will change for a given flow rate in or out. Example 4.2. Capillary Compliance The human systemic capillary bed has a mean blood pressure of about 30 mmlig and holds about 6.0% of the total body blood volume. Its residual volume is about 1.0% of the total blood volume. Estimate its total compliance. Solution Since the total blood volume in a human is about 5.0 L, the systemic capillary bed has a residual volume Vtf, = 0.010 x 5.0 L = 0.050 L. Its total volume at a pressure of 30 mmlig is V = 0.060 x 5.0 L = 0.30 L. Thus, from (4.7) 4The compliance of most vessels can be considered constant over a short time period, but NOT that of the heart ventricles (see Chapter 5).C = (V - V^/P = (0.30 L - 0.050 L)/30 mmlig = 0.0083 L/mmHg. 4.6. PROBLEMS 55 Example 4.3. Pressure Increase Caused by Blood Flow If during a period of 5 seconds, the blood flow into the capillary bed of the previous example is 5.6 L/min, but the flow out is 5.2 L/min, how much would the capillary volume and blood pressure change during this period? Solution The net flow into the capillary bed is given by the inflow minus the outflow, or <2net = (5.6 - 5.2) L/min = +0.4 L/min = +0.007 L/s. According to (4.9), A V = QnetA? = (0.007 L/s)(5 s) = +0.04 L. From (4.10) and the values of the previous example, Vnew = 0.30 L + 0.04 L = 0.34 L and from (4.12), Pnew = (0.34 L - 0.05 L)/0.0083 L/mmHg = 35 mmHg. So the capillary blood pressure would rise from 30 mmlig to 35 mmlig. PROBLEMS m.5 Suppose that the following lines are typed in the command window of Matlab. After each line is typed, the "return" or "enter" key is hit. First, on a piece of paper without the aid of a computer, write down what you believe would be printed or plotted on the Matlab screen in response to each line. Then check your answers by actually entering these lines, one at a time, into the command window of a computer running Matlab. Correct your initial answers on the sheet of paper. (Notes: For this homework, just turn in your hand-written final answers; you don't need to print out the Matlab screen. Also, if you don't understand why you are getting certain responses, try using the help command or the help desk feature of Matlab.) A=ones(l,5) a=[5 6 4 2 1] a(3) Problems with the suffix "ni" have significant Matlab content. 4.6 4.1.56 CHAPTER 4. HOOKE'S LAW: ELASTI CITY OFTISSUES AND COMPLIANT VESSELS a(3)=0 b= [1:5] b(4:5) c=a + b; %careful, this one is tricky d=a + A pause(4) 4.2. m. Suppose that the following lines have been stored as an m-file under the name "work2.m." First, on a piece of paper without the aid of a computer, write down what you believe would be printed or plotted on the Matlab screen in response to typing work2 in the command window of Matlab. Then check your answers by actually storing these lines as an m-file with the name "work2.m," and then typing work2 in the command window. (Make sure yourworking Matlab path includes the directory that has this m-file inside so Matlab can find it. You can use a menu command to add the directory to Matlab's path if necessary.) Correct your initial answers on the sheet of paper. (Note: Just turn in your hand-written answers along with a rough sketch of the plot on your sheet; you don't need to print out the Matlab screen.) f=[6 3 2 8; 542 1; 9 0 0 3] f(3,1) f(2,: ) f ' clear f f = [3:2:12] f(4)=f(3) g=[l 2,3 4 5] plot(f); hold on; plot(g) xlabel('index number') ylabel('f and g') text(3,5,'great plot!') 4.3. m. Write a Matlab m-file (called a "script") that does the following tasks in order: a. Set up a row vector that has 21 elements ranging linearly in value from 0 to n. b. Using the vector from part a, produce a rowvector that has the shape of the positive half-cycle of a sine wave. It will have values ranging from 0 to 1. c. Using the vector from part b, produce a rowvector that has the same shape as part b, but has values ranging from 0 to 10.4.6. PROBLEMS 57 d. Using a for loop, search each element of the vector from part c to test whether the element has a value less than 7.0. If it does, set the element value to 0; if it doesn't, set the value to 12.0. (Note: use an if statement inside the for loop). e. Add up the values of all the elements in the vector as modified in part d. Write out this value to the screen. Write this Matlab program by hand first. Then type it in as an m-file, store it, and run it with Matlab. Print out a copy of your m-file only and turn it in along with the answer to part e. (Note: Do not print out or turn in your element values for the row vector. Also, if you don't have access to a printer, turn in a hand-written copy of the program.) fans: 132] 4.4. m. On a clean piece of paper, write down what would be displayed ("printed") on a computer screen in response to each of the lines below when they are typed one at a time (followed by an enter) in the command window of Matlab. (You don't need to repeat the commands on the paper.) Write down only what would be displayed; do all other calculations you might need on another piece of paper that you do not turn in. a = ones(1,6) b = [1:2:11] b(3) = 0 c = a+b d = 2*c for i = 1:6; if d(i) > 8; d(i) = 8; else d(i) = 0; end; end; d 4.5. Approximate your tibia as a long cylinder bone. Measure its approximate length and estimate its diameter. When you stand on one leg, how much total shortening occurs in the length of your tibia? (See Appendix B for the Young's modulus of bone.) fans for me: about 5.6 /u.m] 4.6. a. The pulmonary artery contains about 52 mL of blood (averaged over time) at an average pressure of 17 mmHg. Its residual volume is 21 mL. Calculate its compliance. fans:C 0.0018 L/mmHg] 58 CHAPTER 4. HOOKE'S LAW: ELASTI CITY OFTISSUES AND COMPLIANT VESSELS b. At the end of diastole (thus the beginning of systole), the pressure in the pulmonary artery has dropped to about 10 mmlig. How much blood is in the artery at that point in time? fans:V % 39 mL] c. During the first 0.10 s of systole, the flow rate of blood from the right ventricle into the pulmonary artery is 0.24 L/s, while the flow rate out of the artery toward the lungs is 0.15 L/s. At the end of this 0.10 s interval, what is the blood pressure in the artery? fans:P % 15 mmlig] 4.7. The heart ventricles are compliant vessels, but they have one very important difference from ordinary blood vessels: their compliance changes dramatically during each heart cycle, from a relatively large value during filling (diastole) to a small value during ejection (systole). a. The end-diastolic volume of blood, Vkd, at the very end of diastole in the left ventricle is about 135 mL at a filling pressure of 8.0 mmlig. What is its end-diastolic compliance? (Assume the residual volume of the left ventricle is negligible.) fans: C = 0.017 L/mmlig] b. At the peak of systole-which occurs about 175 ms after the end of diastole-the left ventricle compliance has a value of 0.000800 L/mmlig. The outflow of blood into the aorta averages 200 mL/s during systole. What is the systolic pressure in the ventricle at this moment? fans: P = 125 mmlig] 4.8. Tissues that are composed of randomly oriented cross-linked collagen fibers are in some ways analogous to rubber. The elasticity' of these types of tissues and materials can be measured using a standard tension test. Your homew