Download Biofluid Mechanics. An Introduction to Fluid Mechanics, by David Rubenstein, Wei Yin, Mary D. Frame PDF

By David Rubenstein, Wei Yin, Mary D. Frame

Content material:

, Pages i,iii

, Page iv

, Pages ix-x
Chapter 1 - Introduction

, Pages 3-9
Chapter 2 - basics of Fluid Mechanics

, Pages 11-48
Chapter three - Conservation Laws

, Pages 49-100
Chapter four - The Heart

, Pages 103-132
Chapter five - Blood move in Arteries and Veins

, Pages 133-178
Chapter 6 - Microvascular Beds

, Pages 181-215
Chapter 7 - Mass delivery and warmth move within the Microcirculation

, Pages 217-248
Chapter eight - The Lymphatic System

, Pages 249-261
Chapter nine - move within the Lungs

, Pages 265-288
Chapter 10 - Intraocular Fluid Flow

, Pages 289-303
Chapter eleven - Lubrication of Joints

, Pages 305-324
Chapter 12 - circulate throughout the Kidney

, Pages 325-345
Chapter thirteen - In Silico Biofluid Mechanics

, Pages 349-373
Chapter 14 - In vitro Biofluid Mechanics

, Pages 375-383
Chapter 15 - In vivo Biofluid Mechanics

, Pages 385-394
Further Readings Section

, Page 395

, Pages 397-400

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Additional resources for Biofluid Mechanics. An Introduction to Fluid Mechanics, Macrocirculation, and Microcirculation

Sample text

When there is a large volume of water with a small amount of sand, this fluid will have a lower viscosity. Under a low shear rate, this mixture behaves as water. As the shear rate increases, water will flow out of the mixture, leaving a similar quantity of sand and water. This mixture will have a larger viscosity (at the higher shear rate). At some time, most of the water will be removed from the mixture, and it will become very difficult to make this fluid flow. Although, under these large stresses, the fluid will marginally deform because the resistance against motion is extremely large.

The distance between the two plates is 10 mm and the velocity is 0 mm/s at the lower plate and 30 mm/s at the upper plate. 19 Figure for the Example Problem. Y Uupper = 30 mm/s d = 10 mm Ulower = 0 mm/s Solution τ upper 5 μ   @vy @vx @vx 1 5μ @y @x @y There is no velocity in the y-direction so the second term within the parenthesis drops out of the relationship. Since the velocity profile is linear in the y-direction, the shear stress equation simplifies to Uupper 2 Ulower Uupper Δvx 5μ 5μ Δy dupper 2 dlower dupper   1g=cm à s 30mm=s dyne τ upper 5 3:5 3 10 22 P à 5 0:0105Pa 5 0:105 1P 10mm cm2 τ upper 5 μ Similarly, the shear stress on the lower plate is τ lower 5 μ Ulower 2 Uupper Uupper Δvx 5μ 5μ Δy dlower 2 dupper dupper 1.

However, be cautioned that other body forces can be applied via a magnetic field (blood flow of a patient within an MRI) or by an electric field. The mass of the differential element, dm, is equal to the fluid density multiplied by the volume of the element (dm 5 ρdV 5 ρdxdydz, in Cartesian coordinates; for other coordinate systems the analysis is similar, note that V  volume and v  velocity ). Therefore, the force due to gravity becomes - - - d Fb 5 g dm 5 g ρdxdydz - where g is the gravitational constant.

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