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Solutions Manual for Biofluid Mechanics An Introduction to Fluid Mechanics, Macrocirculation, and Microcirculation 3rd Edition By David Rubenstein, Wei Yin, Mary Frame (All Chapters, 100% Original Verified, A+ Grade)$28.49
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Solutions Manual for Biofluid Mechanics An Introduction to Fluid Mechanics, Macrocirculation, and Microcirculation 3rd Edition By David Rubenstein, Wei Yin, Mary Frame (All Chapters, 100% Original Verified, A+ Grade)
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Course
Biofluid Mechanics An Introdu
Institution
Biofluid Mechanics An Introdu
This Is Original 3rd Edition of Solutions Manual From Original Author. All Other Files in the market are fake/old Edition. Other Sellers Have changed old Edition Number to new But solutions Manual is old Edition.
Solutions Manual for Biofluid Mechanics An Introduction to Fluid Mechanics, Macroci...
Test Bank For
Biofluid Mechanics An Introduction to Fluid Mechanics, Macrocirculation, and Microcirculation
3rd Edition
By
David Rubenstein
Wei Yin
Mary Frame Chapter 2 2.1: For the velocity distribution 𝑣𝑥=5𝑥,𝑣𝑦=−5𝑦,𝑣𝑧=0, determine the acceleration vector. Also, determine whether this velocity profile has a local and/or convective acceleration. 𝑎⃗=𝜕𝑣⃗
𝜕𝑡+𝑣𝑥𝜕𝑣
𝜕𝑥+𝑣𝑦𝜕𝑣
𝜕𝑦+𝑣𝑧𝜕𝑣
𝜕𝑧
𝑣⃗=5𝑥𝑖̂−5𝑦𝑗̂ 𝜕𝑣
𝜕𝑡=0→𝑛𝑜 𝑙𝑜𝑐𝑎𝑙 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝜕𝑣
𝜕𝑥=5 𝜕𝑣
𝜕𝑦=−5 𝜕𝑣
𝜕𝑧=0 𝑎⃗=0+5𝑥(5)𝑖̂−5𝑦(−5)𝑗̂+0=25𝑥𝑖̂+25𝑦𝑗̂→𝑜𝑛𝑙𝑦 𝑐𝑜𝑛𝑣𝑒𝑐𝑡𝑖𝑣𝑒 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 2.2: Consider a velocity vector 𝑣=(𝑥𝑡2−𝑦)𝑖⃗+(𝑥𝑡−𝑦2)𝑗⃗. (a) Determine if this flow is steady (hint: no changes with time). (b) Determine if this is an incompressible flow (hint: check if ∇∙𝑣=
0). (𝑎) 𝑣=(𝑥𝑡2−𝑦)𝑖⃗+(𝑥𝑡−𝑦2)𝑗⃗ Chapter 1: No Solutions for chapter 1. 𝑑𝑣
𝑑𝑡=2𝑥𝑡𝑖̂+𝑥𝑗̂→𝑛𝑜𝑡 𝑠𝑡𝑒𝑎𝑑𝑦 (𝑏) 𝜕𝑣
𝜕𝑥+𝜕𝑣
𝜕𝑦=0 (𝑡2−1)𝑖+(𝑡−2𝑦)𝑗̂→𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑏𝑙𝑒 2.3: Given the velocity 𝑣=(2𝑥−𝑦)𝑖⃗+(𝑥−2𝑦)𝑗⃗, determine if it is irrotational. 𝜉=(𝜕𝑣𝑧
𝜕𝑦−𝜕𝑣𝑦
𝜕𝑧)𝑖̂+(𝜕𝑣𝑥
𝜕𝑧−𝜕𝑣𝑧
𝜕𝑥)𝑗̂+(𝜕𝑣𝑦
𝜕𝑥−𝜕𝑣𝑥
𝜕𝑦)𝑘̂=(0−0)𝑖̂+(0−0)𝑗̂+(1−(−1))𝑘̂=2𝑘̂
→𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙
2.4 The velocity vector for a steady incompressible flow in the xy plane is given by , where the coordinates are measured in centimeters. Determine the time it takes for a particle to move from x = 1cm to x = 4cm for a particle that passes through the point ( ) ( ), 1,4xy= . 𝑑𝑢
𝑑𝑡=4
𝑥𝒊⃗+4𝑦
𝑥2𝒋⃗ 𝑢=4𝑡
𝑥𝒊⃗+4𝑡𝑦
𝑥2𝒋⃗ 4𝑡
𝑥
4𝑡𝑦
𝑥2=1
4
𝑥
𝑦=1
4 𝑜𝑟 4𝑥=𝑦 4𝑡
1=1 𝑠𝑜 𝑡=0.25𝑠 Check 4𝑡(4∗1)
12=4 𝑠𝑜 𝑡=0.25𝑠 Time at 1cm is equal to 0.25s; for time at 4cm
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