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