Biofluid Mechanics: An Introduction to Fluid Mechanics, Macrocirculation, and Microcirculation, 3rd Edition (2021)

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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

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, 4x y = .
𝑑𝑢
𝑑𝑡 = 4
𝑥 𝒊⃗ + 4𝑦
𝑥2 𝒋⃗
𝑢 = 4𝑡
𝑥 𝒊⃗ + 4𝑡𝑦
𝑥2 𝒋⃗
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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4𝑡
4 = 4 𝑠𝑜 𝑡 = 4𝑠
Therefore, it takes 3.75 seconds for this particular particle to move from 1cm to 4cm
2.5: Flow in a 2-dimensional channel of width W has a velocity profile defined by:
𝑣𝑥(𝑦) = 𝑘 [(𝑊
2
2
) − 𝑦2]
where y=0 is located at the center of the channel. Sketch the velocity distribution and find the
shear stress/unit width of the channel at the wall.
𝜏𝑥𝑦 = 𝜇 ( 𝜕
𝜕𝑦 (𝑘 (𝑊2
2 − 𝑦2))) = 𝜇𝑘 ( 𝜕
𝜕𝑦 (𝑊2
2 − 𝑦2)) = −2𝑦𝜇𝑘
Velocity Distribution (k=1, W=10):-6
-4
-2
0
2
4
6
0 10 20 30 40 50 60
Channel Location
Velocity

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2.6: A velocity filed is given by 𝑣(𝑥, 𝑦, 𝑧) = 20𝑥𝑦𝑖⃗ − 10𝑦2𝑗⃗, calculate the acceleration, the angular
velocity, and the vorticity vector at the point (-1,1,1), where the units of the velocity equation are
mm/s.
𝑎⃗ = [200𝑥𝑦(20𝑦)𝑖̂ − 10𝑦2(20𝑥𝑖̂ − 20𝑦𝑗̂ )] 𝑚𝑚
𝑠2 = [200𝑥𝑦2𝑖̂ + 200𝑦3𝑗̂ ] 𝑚𝑚
𝑠2
𝑎⃗ (−1,1,1) = [200(−1)(1)2𝑖̂ + 200(1)3𝑗̂ ] 𝑚𝑚
𝑠2 = [−200𝑖̂ + 200𝑗̂ ] 𝑚𝑚
𝑠2
𝜔⃗⃗⃗ = 1
2 [(0 − 0)𝑖̂ + (0 − 0)𝑗̂ + (0 − 20𝑥)𝑘̂ ] 𝑟𝑎𝑑
𝑠 = 1
2 [−20𝑥]𝑘̂ 𝑟𝑎𝑑
𝑠 = −10(−1) 𝑘̂ 𝑟𝑎𝑑
𝑠 = 10 𝑟𝑎𝑑
𝑠 𝑘̂
𝜉 = 2𝜔⃗⃗⃗ = 20 𝑟𝑎𝑑
𝑠 𝑘̂
2.7: Considering one-dimensional fluid (density, 𝜌; viscosity, 𝜇) flow in a tube with an inlet
pressure of 𝑝𝑖, and outlet pressure of 𝑝𝑜, and tube radius of 𝑟 and length 𝑙. The density of the
fluid can be represented as 𝜌. Express the wall shear stress as a function of these variables.
Hint: balance Newton’s second law of motion.
∑𝐹 = 𝑚𝑎 → 𝑎𝑠𝑠𝑢𝑚𝑒 𝑚 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑖𝑛 𝑡𝑢𝑏𝑒 𝑜𝑓 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡
∑𝐹 = 𝑝𝑖𝑖̂ − 𝑝𝑜𝑖̂ − 𝜏𝑖̂ = (𝑝𝑖 − 𝑝𝑜)𝑖̂ − 𝜏𝑖̂
𝑚 = 𝜌𝑉 = 𝜌𝜋𝑟2𝑙
(𝑝𝑖 − 𝑝𝑜)𝑖̂ − 𝜏𝑖̂ = 𝜌𝜋𝑟2𝑙 (𝑎𝑖̂)

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𝜏 = (𝑝𝑖 − 𝑝𝑜 − 𝜌𝜋𝑟2𝑙𝑎)
2.8: Data obtained from one experiment are presented in the table below, plot the data and
determine if the fluid is pseudoplastic, Newtonian or dilatants. Approximate the viscosity (or the
apparent viscosity for this fluid).
The fluid is pseudoplastic. Fitting this with a power fit, we get: 𝜏 = 10.147 𝜕𝑣
𝜕𝑦
−0.6715
(𝜕𝑣
𝜕𝑦)
The apparent viscosity is 47.6 (0.1 s-1), 10.1 (1 s-1), 2.2 (10 s-1), 0.46 (100 s-1), 0.098 (1000 s-1)0
10
20
30
40
50
60
70
80
90
100
-200 0 200 400 600 800 1000 1200
Shear Stress
Shear Rate
Shear stress (dyne/cm2) 5 9 23 48 95
Shear rate (s-1) 0.1 1 10 100 1000

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2.9: Consider a red blood cell that originates from the origin of our 𝑥𝑦-coordinate system. The
velocity of the fluid is unsteady and is described by:
𝑣 = {
𝑣𝑥 = 1 𝑐𝑚
𝑠 𝑣𝑦 = 0.5 𝑐𝑚
𝑠 0𝑠 ≤ 𝑡 ≤ 2𝑠
𝑣𝑥 = 0.25 𝑐𝑚
𝑠 𝑣𝑦 = 1 𝑐𝑚
𝑠 2𝑠 < 𝑡 ≤ 5𝑠
Plot the location of this red blood cell at time zero, 1s, 2s, 3s, 4s, and 5s (enters from the origin
at time zero). Also plot the location of the red blood cells that enter (from the origin) at time 1s,
2s, 3s, 4s.
Red Blood
Cell Enters at
Time
t=0 t=1 t=2 t=3 t=4
Time in Fluid
1s
(1,0.5) (0,0) --- --- ---
2s (2,1) (1,0.5) (0,0) --- ---
3s (2.25, 2) (1.25, 1.5) (0.25, 1) (0,0) ---
4s (2.5, 3) (1.5, 2.5) (0.5, 2) (0.25, 1) (0,0)
5s (2.75, 4) (1.75, 3.5) (0.75, 3) (0.5, 2) (0.25, 1)

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*2.10: A flow is described by the velocity field
𝑣 = (𝑦
𝑠) 𝑖⃗ + 0.75 𝑚
𝑠2 𝑡𝑗⃗
At 𝑡 = 2𝑠, what are the coordinates of the particles that passed through the point (1,5) at 𝑡 = 0𝑠.
What are the coordinates of the same particle at 𝑡 = 3𝑠.
𝑣 = 𝑑𝑥
𝑑𝑡 → 𝑥 = ∫ 𝑣𝑑𝑡 = ∫ (𝑦
𝑠) 𝑖̂ 𝑑𝑡 + ∫ 0.75 𝑚
𝑠2 𝑡𝑗̂ 𝑑𝑡
𝑥 = (𝑦𝑡
𝑠 + 𝑥0) 𝑖̂ + ((0.75 𝑚
𝑠2) 𝑡2
2 + 𝑦0) 𝑗̂
𝑥0 = 1, 𝑦0 = 5
𝑥(𝑡 = 2) = (𝑦𝑡
𝑠 + 1) 𝑖̂ + (0.75 𝑚
𝑠2 (𝑡2
2 ) + 5) 𝑗̂
𝑦 = 0.75 𝑚
𝑠2 (4𝑠2
2 ) + 5 = 6.5𝑚
𝑥 = 6.5𝑚(2𝑠)
𝑠 + 1 = 14𝑚
𝑥(𝑡 = 2) = (14𝑖̂ + 6.5𝑗̂ )𝑚
𝑥(𝑡 = 3) → 𝑦 = 0.75 (32
2 ) + 5 = 8.375𝑚
𝑥 = 8.375(3𝑠)
𝑠 + 1 = 26.125𝑚
𝑥(𝑡 = 3) = (26.125𝑖̂ + 8.375𝑗̂ )𝑚

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2.11: The velocity distribution for laminar flow between two parallel plates can be represented
as
𝑢
𝑢𝑚𝑎𝑥
= 1 − (2𝑦
ℎ )
2
where ℎ is the separation distance between the two flat plates and the origin is located half-way
between the plates. Consider the flow of blood at 37C (𝜇 = 3.5𝑐𝑃) with maximum velocity of 25
cm/s and a separation distance of 10mm. Calculate the force on a 0.25 m2 section of the lower
plate.
𝑢(𝑦) = 𝑢𝑚𝑎𝑥 − 𝑢𝑚𝑎𝑥 (2𝑦
ℎ )
2
𝜏𝑦𝑥 = 𝜇 𝜕𝑢
𝜕𝑦 = 𝜇 𝜕
𝜕𝑦 (𝑢𝑚𝑎𝑥 − 𝑢𝑚𝑎𝑥 (2𝑦
ℎ )
2
= 𝜇 [−2𝑢𝑚𝑎𝑥 (2𝑦
ℎ ) (2
ℎ)] = − 8𝜇𝑢𝑚𝑎𝑥𝑦
ℎ2 = 𝐹
𝐴
𝐹 = − 8𝜇𝑢𝑚𝑎𝑥𝑦𝐴
ℎ2
𝐹 = − 8(3.5𝑐𝑃) (25 𝑐𝑚
𝑠 ) (5𝑚𝑚)(0.25𝑚2)
(10𝑚𝑚)2 = −0.0875𝑁

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2.12: A biofluid is flowing down an inclined plane. The velocity profile of this fluid can be
described by
𝑢 = 𝜌𝑔
𝜇 (ℎ𝑦 − 𝑦2
2 ) 𝑠𝑖𝑛𝜃
if the coordinate axis is aligned with the inclined plant. Determine a function for the shear stress
along this fluid. Plot the velocity profile and shear stress profile, if the fluid’s density is 900
kg/m3 and its viscosity is 2.8cP. The fluid thickness, h, is equal to 10mm and the plane is
inclined at an angle of 40.
𝜏 = 𝜇 𝜕𝑢
𝜕𝑦 = 𝜇 𝜕
𝜕𝑦 (𝜌𝑔
𝜇 (ℎ𝑦 − 𝑦2
2 ) 𝑠𝑖𝑛𝜃) = 𝜌𝑔𝑠𝑖𝑛𝜃 [ 𝜕
𝜕𝑦 (ℎ𝑦) − 𝜕
𝜕𝑦 (𝑦2
2 )] = 𝜌𝑔𝑠𝑖𝑛𝜃[ℎ − 𝑦]0
0.002
0.004
0.006
0.008
0.01
0.012
0 2 4 6 8 10 12
Velocity (m/s)
Velocity Profile

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For film thickness, velocity at the wall is 0 and acceleration at contact is a
𝜏 = 𝜇 𝜕𝑣
𝜕𝑦|
𝑦=ℎ
Assume that the velocity is linear because it is a small film layer.
𝑉𝑦 = (𝑦
ℎ 𝑉) → 𝜏𝑦=ℎ = 𝜇𝑉
ℎ
∑ 𝐹𝑦
⬚
𝑚2
= 𝑇 − 𝑚2𝑔 = −𝑚2𝑎 → 𝑇 = 𝑚2(𝑔 − 𝑎)
∑ 𝐹𝑥
⬚
𝑚1
= 𝑇 − 𝐹𝑓 = 𝑚1𝑎 → 𝑇 = 𝑚1𝑎 + 𝐹𝑓
𝑚1𝑎 + 𝐹𝑓 = 𝑚2(𝑔 − 𝑎)
𝑎(𝑚1 + 𝑚2) = 𝑚2𝑔 − 𝐹𝑓 = 𝑚2𝑔 − 𝜇𝑉𝐴
ℎ
𝑑𝑉
𝑑𝑡 (𝑚1 + 𝑚2) = 𝑚2𝑔 − 𝜇𝑉𝐴
ℎT
W=m2g
a
T,a
Ff=A

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𝑑𝑉
𝑚2𝑔 − 𝜇𝑉𝐴
ℎ
= 𝑑𝑡
𝑚1 + 𝑚2
− 1
𝜇𝐴
ℎ
ln (𝑚2𝑔 − 𝜇𝑉𝐴
ℎ ) = 𝑡
𝑚1 + 𝑚2
ln (𝑚2𝑔 − 𝜇𝑉𝐴
ℎ ) = − 𝜇𝐴𝑡
ℎ(𝑚1 + 𝑚2)
𝑚2𝑔 − 𝜇𝑉𝐴
ℎ = 𝑒− 𝜇𝐴𝑡
ℎ(𝑚1+𝑚2)
𝜇𝑉𝐴
ℎ = 𝑚2𝑔 − 𝑒− 𝜇𝐴𝑡
ℎ(𝑚1+𝑚2)
𝑉 = 𝑚2𝑔ℎ
𝜇𝐴 − ℎ
𝜇𝐴 𝑒− 𝜇𝐴𝑡
ℎ(𝑚1+𝑚2)

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2.14: For the following velocity field
𝑣 = (𝑥𝑡2 + 2𝑦 − 𝑧2)𝑖⃗ + (𝑥𝑡 − 2𝑦𝑡)𝑗⃗ + (3𝑥𝑡 + 𝑧)𝑘⃗⃗
calculate the shear rate tensor, the vorticity, the hydrostatic pressure (use a viscosity of 𝜇) and
the acceleration of the fluid.
𝜕𝑣𝑥
𝜕𝑥 = 𝑡2 𝜕𝑣𝑦
𝜕𝑥 = 𝑡 𝜕𝑣𝑧
𝜕𝑥 = 3𝑡
𝜕𝑣𝑥
𝜕𝑦 = 2 𝜕𝑣𝑦
𝜕𝑦 = −2𝑡 𝜕𝑣𝑧
𝜕𝑦 = 0
𝜕𝑣𝑥
𝜕𝑧 = −2𝑧 𝜕𝑣𝑦
𝜕𝑧 = 0 𝜕𝑣𝑧
𝜕𝑧 = 1
𝜕𝑣
𝜕𝑡 = (2𝑥𝑡)𝑖̂ + (𝑥 − 2𝑦)𝑗̂ + 3𝑥𝑘̂
𝑑 = 1
2 [
2𝑡2 2 + 𝑡 3𝑡 − 2𝑧
2 + 𝑡 −4𝑡 0
3𝑡 − 2𝑧 0 2
]
𝜉 = (0 − 0)𝑖̂ + (−2𝑧 − 3𝑡)𝑗̂ + (𝑡 − 2)𝑘̂
𝑃ℎ𝑦𝑑𝑟𝑜 = − 1
3 (𝑡2 − 2𝑡 − 1)
𝑎 = 2𝑥𝑡𝑖̂ + (𝑥 − 2𝑦)𝑗̂ + 3𝑥𝑘̂ + (𝑥𝑡2 + 2𝑦 − 𝑧2)(𝑡2𝑖̂ + 𝑡𝑗̂ + 3𝑡𝑘̂) + (𝑥𝑡 − 2𝑦𝑡)(2𝑖̂ − 2𝑡𝑗̂ )
+ (3𝑥𝑡 + 𝑧)(−2𝑧𝑖̂ + 1𝑘̂)

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2.15: A velocity field is given by
𝑣 = 0.2𝑠−1𝑥𝑖⃗ − 0.35(𝑚𝑚𝑠)−1𝑦2𝑗⃗
where x and y are given in millimeters. Determine the velocity of a particle at point (2,4) and the
position of the particle at t=6s of the particle that is located at (2,4) at t=0.
𝑣 = (0.2)(2)𝑖̂ − 0.35(4)2𝑗̂ = 0.4𝑖̂ − 5.6𝑗̂ 𝑚𝑚
𝑠
𝑣𝑥 = 𝑑𝑥
𝑑𝑡 = 0.2𝑥
∫ 𝑑𝑥
𝑥
𝑥
𝑥0
= ∫ 0.2𝑑𝑡
𝑡
0
ln ( 𝑥
𝑥0
) = 0.2𝑡 → 𝑥 = 𝑥0𝑒0.2𝑡
𝑣𝑦 = 𝑑𝑦
𝑑𝑡 = −0.35𝑦2
∫ 𝑑𝑦
𝑦2
𝑦
𝑦0
= ∫ −0.35𝑑𝑡
𝑡
0
− 1
𝑦 + 1
𝑦0
= −0.35𝑡
− 1
𝑦 = −0.35𝑡 − 1
𝑦0
= −0.35𝑡𝑦0 − 1
𝑦0
𝑦 = 𝑦0
0.35𝑡𝑦0 + 1
𝑥 = 2𝑚𝑚(𝑒0.2∗6) = 6.64𝑚𝑚

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2.16: A fluid is placed in the area between two parallel plates. The upper plate is movable and
connected to a weight by a cable. Calculate the velocity of the plate for a) the fluid between the
plates is water, b) the fluid between the plates is blood (assumed to have no yield stress) and c)
the fluid between the plates is blood with a yield stress of 50mPa. For all cases take the mass
to be 10g, the height between the plates as 10mm and the contact area to be 0.5m2.
Force on the plate is
𝑚𝑔 = 10𝑔 (9.81 𝑚
𝑠2) = 0.098𝑁
𝜏 = 0.098𝑁
0.5𝑚2 = 0.196𝑃𝑎
𝜏 = 𝜇 (Δ𝑉
Δ𝑦)
(𝑎)𝑉 = 𝜏
𝜇 ℎ = 0.196𝑃𝑎
1𝑐𝑃 (10𝑚𝑚) = 1.96 𝑚
𝑠
(𝑏)𝑉 = 0.196𝑃𝑎
3.5𝑐𝑃 (10𝑚𝑚) = 0.56 𝑚
𝑠
(𝑐)𝜏𝑎𝑐𝑡𝑢𝑎𝑙 = 𝜏 − 𝜏𝑦 = 0.146𝑃𝑎M
h

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2.18: A plate is moving upward through a channel filled with water as shown. The plate velocity
is 10cm/s, and the spacing is 50 mm on each side of the plate. The force required to move the
plate is equal to 5N. Determine the surface area of the plate.
Velocity at the wall is zero, the force must balance the shear stress acting on both sides of the
plate. Assume that the velocity profile is linear.
𝜏 = 𝜇𝑉
ℎ
𝐹 = 2𝐴𝜇𝑉
ℎ
𝐴 = 5𝑁(50𝑚𝑚)
2(1𝑐𝑃) (10 𝑐𝑚
𝑠 ) = 1250 𝑚2F
50mm 50mm

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2.19: Use the same figure for problem 2.18, except that the distance on the right hand side is
increased to 100mm. The fluid on the right side is blood. Determine the viscosity of the fluid on
the left side. The plate is moving with a constant velocity.
Using the same analysis as for 2.18
𝐹 = 𝜏𝑙𝑒𝑓𝑡 + 𝜏𝑟𝑖𝑔ℎ𝑡
𝐹 = 𝜇𝑙𝑒𝑓𝑡𝑉𝐴
ℎ𝑙𝑒𝑓𝑡
+ 𝜇𝑟𝑖𝑔ℎ𝑡𝑉𝐴
ℎ𝑟𝑖𝑔ℎ𝑡
ℎ𝑟𝑖𝑔ℎ𝑡 = 2ℎ𝑙𝑒𝑓𝑡
𝐹 = 𝑉𝐴
ℎ𝑙𝑒𝑓𝑡
(𝜇𝑙𝑒𝑓𝑡 + 𝜇𝑟𝑖𝑔ℎ𝑡
2 )
For equilibrium
𝜇𝑙𝑒𝑓𝑡 = 𝜇𝑟𝑖𝑔ℎ𝑡
2 → 𝜇𝑙𝑒𝑓𝑡 = 1.75𝑐𝑃

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2.20
Velocity at the shaft wall can be defined as:
𝑣 = 𝑟𝜔
Where r can be defined from
𝑟 = ℎ𝑡𝑎𝑛(30)
𝑣 = ℎ𝜔 tan(30)
The velocity at the bearing wall is zero; therefore the slope of the velocity profile can be defined
as:
𝑣 = 𝜔ℎ𝑡𝑎𝑛(30)
𝑎
And the shear stress can be defined as
𝜏 = 𝜔𝜇ℎ𝑡𝑎𝑛(30)
𝑎
Solving for the known values
𝜏 =
25 𝑟𝑒𝑣
𝑠 (0.05 𝑁𝑠
𝑚2) (35𝑚𝑚)tan (30)
0.35𝑚𝑚 = 72𝑃𝑎
2.21: Two data points on the rheological diagram of a biofluid are provided. Determine the
consistency index and the flow behavior index and the strain rate if the shear stress is increased
to 3 dynes/cm2. Assume that this is a two-dimensional flow.
𝑑𝑉
𝑑𝑦 = 15 𝑟𝑎𝑑
𝑠 𝜏 = 0.868 𝑑𝑦𝑛𝑒𝑠
𝑐𝑚2

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𝑑𝑉
𝑑𝑦 = 30 𝑟𝑎𝑑
𝑠 𝜏 = 0.355 𝑑𝑦𝑛𝑒𝑠
𝑐𝑚2
𝜏𝑥𝑦 = 𝑘 (𝑑𝑉
𝑑𝑦)
𝑛
0.868 𝑑𝑦𝑛𝑒𝑠
𝑐𝑚2
0.355 𝑑𝑦𝑛𝑒𝑠
𝑐𝑚2
= (15 𝑟𝑎𝑑
𝑠 )
𝑛
(30 𝑟𝑎𝑑
𝑠 )
𝑛 → 2.445 = (1
2)
𝑛
𝑛 = −1.28988
0.868 = 𝑘(15)−1.28988
𝑘 = 28.545 𝑑𝑦𝑛𝑒𝑠
𝑐𝑚2
𝑑𝑉
𝑑𝑦 = 5.735 𝑟𝑎𝑑
𝑠 , 𝑤ℎ𝑒𝑛 𝜏 = 3 𝑑𝑦𝑛𝑒𝑠
𝑐𝑚2
2.22 2.22 Given that the volumetric flow rate for a fluid within a circular cross-section tube
can be represented by
𝑄 = 𝜋𝑟4Δ𝑃
8𝜇𝐿
where r is the tube radius,  is the fluid viscosity, P is the pressure drop across the tube and L is
the tube length, calculate the pressure drop across a tube of length 1m and diameter of 23mm.
The fluid is blood ( = 3.5cP) and has a volumetric flow rate of 4.5L/min. Assuming the same
conditions, what would the required pressure drop be for water (=1cP) and chocolate syrup (
= 15,000 cP).
Blood Water Chocolate Syrup

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Δ𝑃 = 8𝑄𝜇𝐿
𝜋𝑟4
=
8 (4.5 𝐿
min ⬚) (3.5𝑐𝑃)(1𝑚)
𝜋 (23𝑚𝑚
2 )
2
= 0.286𝑚𝑚𝐻𝑔
Δ𝑃 = 8𝑄𝜇𝐿
𝜋𝑟4
=
8 (4.5 𝐿
min ⬚) (1𝑐𝑃)(1𝑚)
𝜋 (23𝑚𝑚
2 )
2
= 0.0819𝑚𝑚𝐻𝑔
Δ𝑃 = 8𝑄𝜇𝐿
𝜋𝑟4
=
8 (4.5 𝐿
min ⬚) (15000𝑐𝑃)(1𝑚)
𝜋 (23𝑚𝑚
2 )
2
= 1229𝑚𝑚𝐻𝑔
2.23 At a stenosis, the energy in the flowing fluid is skewed from potential energy to kinetic
energy. Why? What effect may this have on locations downstream of the stenosis.
The energy is skewed to kinetic energy because the resistance to flow increases, thus, more
energy “taken” from the potential energy would be needed to maintain the flow conditions.
Downstream of the stenosis can be effected significantly, because there is an overall reduction
in the amount of energy available (since energy is not really conserved in the conversion of
potential energy to kinetic energy). Therefore, if there are future constrictions, the flow may not
be able to overcome these restrictions. Similarly, since there is a conversion to kinetic, that
would suggest 1) jet flow at a stenosis, 2) increased shearing forces at the stenosis and
immediately distal to the stenosis, a conversion of energy back from kinetic to potential.
2.24 The no-slip boundary wall condition for Newtonian fluids has been well established. No
known exceptions to this condition exist for Newtonian fluids. Blood is a non-Newtonian fluid.
What should the boundary condition for blood and a solid surface be.
Since blood is approximately 50% cells (semi-rigid elements) and approximately 50% fluid, the
boundary condition should be always changing based on what part of blood is interaction with
the wall. Plasma is very close to a Newtonian fluid and thus the no-slip boundary condition
should apply to this portion of blood. Cells can be considered rigid and thus the entire cell would
have a velocity that matches the surface of the cell that is opposing the solid surface. This

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Chapter 3
3.1 A two fluid manometer is used to measure the pressure difference for flowing blood in a
laboratory experiment. Calculate the pressure difference between points A and B in the
fluid.
𝑃𝑤1 = 𝑃𝐴 − (1050 𝑘𝑔
𝑚3) (9.81 𝑚
𝑠2) (0.5𝑚) = 𝑃𝑎 − 38.62𝑚𝑚𝐻𝑔
𝑃𝑤2 = 𝑃𝑤1 + (1000 𝑘𝑔
𝑚3) (9.81 𝑚
𝑠2) (0.5 − 0.275)𝑚 = 𝑃𝑤1 + 16.55𝑚𝑚𝐻𝑔
𝑃𝐵 = 𝑃𝑤2 + (1050 𝑘𝑔
𝑚3) (9.81 𝑚
𝑠2) (0.275𝑚) = 𝑃𝑤2 + 21.24 𝑚𝑚𝐻𝑔
𝑃𝐵 = 𝑃𝐴 − 38.62𝑚𝑚𝐻𝑔 + 16.55𝑚𝑚𝐻𝑔 + 21.24𝑚𝑚𝐻𝑔 = 𝑃𝐴 − 0.83𝑚𝑚𝐻𝑔A B
Blood
Water
d1=500mm
d2=190mm
d3=275mm

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3.2 NASA is planning a mission to a newly found planet and will monitor the density of the new
planet's atmosphere. Assume that NASA knows that atmosphere behaves as an ideal gas
and that the planets gravitational force is a function of altitude (𝑔(𝑧) = 18.7 𝑚
𝑠2 (1 − 𝑧
10,000𝑚),
where z is in m). The temperature of the atmosphere is constant at 250K and the gas
constant is 340Nm/kgK. Assume that the pressure at the planet's surface is 2atm.
Calculate the pressure and density at an altitude of 1km, 5km and 9km.
𝑑𝑝
𝑑𝑧 = −𝜌𝑔𝑧 = − 𝑝
𝑅𝑇 𝑔𝑧
∫ 𝑑𝑝
𝑝
𝑝
𝑝0
= − ∫ 𝑔𝑧
𝑅𝑇 𝑑𝑧
𝑧
0
ln(𝑝) − ln(𝑝0) = − 1
𝑅𝑇 ∫ 18.7 𝑚
𝑠2 (1 − 𝑧
10000) 𝑑𝑧
𝑧
0
ln ( 𝑝
𝑝0
) =
−18.7 𝑚
𝑠2
𝑅𝑇 ∫ (1 − 𝑧
10000) 𝑑𝑧 =
𝑧
0
−18.7 𝑚
𝑠2
𝑅𝑇 [𝑧 − 𝑧2
20000]
0
𝑧
𝑝
𝑝0
= 𝑒
−18.7𝑚
𝑠2
𝑅𝑇 [𝑧− 𝑧2
20000]
𝑝(𝑧) = 𝑝0𝑒
−18.7𝑚
𝑠2
𝑅𝑇 [𝑧− 𝑧2
20000]

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Mechanical Engineering

Biofluid Mechanics: An Introduction to Fluid Mechanics, Macrocirculation, and Microcirculation, 3rd Edition (2021)

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