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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 𝑣 𝑥 = 5 𝑥 , 𝑣 𝑦 = − 5 𝑦 , 𝑣 𝑧 = 0 , determine the acceleration vector. Also, determine whether this velocity profile has a local and/or convective acceleration. 𝑎 ⃗ = 𝜕 𝑣 ⃗ 𝜕𝑡 + 𝑣 𝑥 𝜕𝑣 𝜕𝑥 + 𝑣 𝑦 𝜕𝑣 𝜕𝑦 + 𝑣 𝑧 𝜕𝑣 𝜕𝑧 𝑣 ⃗ = 5 𝑥 𝑖 ̂ − 5 𝑦 𝑗 ̂ 𝜕𝑣 𝜕𝑡 = 0 → 𝑛𝑜 𝑙𝑜𝑐𝑎𝑙 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝜕𝑣 𝜕𝑥 = 5 𝜕𝑣 𝜕𝑦 = − 5 𝜕𝑣 𝜕𝑧 = 0 𝑎 ⃗ = 0 + 5 𝑥 ( 5 ) 𝑖 ̂ − 5 𝑦 ( − 5 ) 𝑗 ̂ + 0 = 25 𝑥 𝑖 ̂ + 25 𝑦 𝑗 ̂ → 𝑜𝑛𝑙𝑦 𝑐𝑜𝑛𝑣𝑒𝑐𝑡𝑖𝑣𝑒 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 2.2: Consider a velocity vector 𝑣 = ( 𝑥𝑡 2 − 𝑦 ) 𝑖 ⃗ + ( 𝑥𝑡 − 𝑦 2 ) 𝑗 ⃗ . (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 = ( 𝑥𝑡 2 − 𝑦 ) 𝑖 ⃗ + ( 𝑥𝑡 − 𝑦 2 ) 𝑗 ⃗

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    𝑑𝑣 𝑑𝑡 = 2 𝑥𝑡 𝑖 ̂ + 𝑥 𝑗 ̂ → 𝑛𝑜𝑡 𝑠𝑡𝑒𝑎𝑑𝑦 ( 𝑏 ) 𝜕𝑣 𝜕𝑥 + 𝜕𝑣 𝜕𝑦 = 0 ( 𝑡 2 − 1 ) 𝑖 + ( 𝑡 − 2 𝑦 ) 𝑗 ̂ → 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑏𝑙𝑒

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    2 .3: Given the velocity 𝑣 = ( 2 𝑥 − 𝑦 ) 𝑖 ⃗ + ( 𝑥 − 2 𝑦 ) 𝑗 ⃗ , determine if it is irrotational. 𝜉 = ( 𝜕 𝑣 𝑧 𝜕𝑦 − 𝜕 𝑣 𝑦 𝜕𝑧 ) 𝑖 ̂ + ( 𝜕 𝑣 𝑥 𝜕𝑧 − 𝜕 𝑣 𝑧 𝜕𝑥 ) 𝑗 ̂ + ( 𝜕 𝑣 𝑦 𝜕𝑥 − 𝜕 𝑣 𝑥 𝜕𝑦 ) 𝑘 ̂ = ( 0 − 0 ) 𝑖 ̂ + ( 0 − 0 ) 𝑗 ̂ + ( 1 − ( − 1 ) ) 𝑘 ̂ = 2 𝑘 ̂ → 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, 4 x y = . 𝑑𝑢 𝑑𝑡 = 4 𝑥 𝒊 ⃗ + 4 𝑦 𝑥 2 𝒋 ⃗ 𝑢 = 4 𝑡 𝑥 𝒊 ⃗ + 4 𝑡𝑦 𝑥 2 𝒋 ⃗ 4 𝑡 𝑥 4 𝑡𝑦 𝑥 2 = 1 4 𝑥 𝑦 = 1 4 𝑜𝑟 4 𝑥 = 𝑦 4 𝑡 1 = 1 𝑠𝑜 𝑡 = 0 . 25 𝑠 Check 4 𝑡 ( 4 ∗ 1 ) 1 2 = 4 𝑠𝑜 𝑡 = 0 . 25 𝑠 Time at 1cm is equal to 0.25s; for time at 4cm

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    4 𝑡 4 = 4 𝑠𝑜 𝑡 = 4 𝑠 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: 𝑣 𝑥 ( 𝑦 ) = 𝑘 [ ( 𝑊 2 2 ) − 𝑦 2 ] 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. 𝜏 𝑥𝑦 = 𝜇 ( 𝜕 𝜕𝑦 ( 𝑘 ( 𝑊 2 2 − 𝑦 2 ) ) ) = 𝜇𝑘 ( 𝜕 𝜕𝑦 ( 𝑊 2 2 − 𝑦 2 ) ) = − 2 𝑦𝜇𝑘 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 ) = 20 𝑥𝑦 𝑖 ⃗ − 10 𝑦 2 𝑗 ⃗ , 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. 𝑎 ⃗ = [ 200 𝑥𝑦 ( 20 𝑦 ) 𝑖 ̂ − 10 𝑦 2 ( 20 𝑥 𝑖 ̂ − 20 𝑦 𝑗 ̂ ) ] 𝑚𝑚 𝑠 2 = [ 200 𝑥 𝑦 2 𝑖 ̂ + 200 𝑦 3 𝑗 ̂ ] 𝑚𝑚 𝑠 2 𝑎 ⃗ ( − 1 , 1 , 1 ) = [ 200 ( − 1 ) ( 1 ) 2 𝑖 ̂ + 200 ( 1 ) 3 𝑗 ̂ ] 𝑚𝑚 𝑠 2 = [ − 200 𝑖 ̂ + 200 𝑗 ̂ ] 𝑚𝑚 𝑠 2 𝜔 ⃗ ⃗ ⃗ = 1 2 [ ( 0 − 0 ) 𝑖 ̂ + ( 0 − 0 ) 𝑗 ̂ + ( 0 − 20 𝑥 ) 𝑘 ̂ ] 𝑟𝑎𝑑 𝑠 = 1 2 [ − 20 𝑥 ] 𝑘 ̂ 𝑟𝑎𝑑 𝑠 = − 10 ( − 1 ) 𝑘 ̂ 𝑟𝑎𝑑 𝑠 = 10 𝑟𝑎𝑑 𝑠 𝑘 ̂ 𝜉 = 2 𝜔 ⃗ ⃗ ⃗ = 20 𝑟𝑎𝑑 𝑠 𝑘 ̂ 2. 7 : Considering one - dimensional fluid (density, 𝜌 ; viscosity, 𝜇 ) flow in a tube with an inlet pressure of 𝑝 𝑖 , and outlet pressure of 𝑝 𝑜 , 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 = 𝑝 𝑖 𝑖 ̂ − 𝑝 𝑜 𝑖 ̂ − 𝜏 𝑖 ̂ = ( 𝑝 𝑖 − 𝑝 𝑜 ) 𝑖 ̂ − 𝜏 𝑖 ̂ 𝑚 = 𝜌𝑉 = 𝜌𝜋 𝑟 2 𝑙 ( 𝑝 𝑖 − 𝑝 𝑜 ) 𝑖 ̂ − 𝜏 𝑖 ̂ = 𝜌𝜋 𝑟 2 𝑙 ( 𝑎 𝑖 ̂ )

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    𝜏 = ( 𝑝 𝑖 − 𝑝 𝑜 − 𝜌𝜋 𝑟 2 𝑙𝑎 ) 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/cm 2 ) 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 = { 𝑣 𝑥 = 1 𝑐𝑚 𝑠 𝑣 𝑦 = 0 . 5 𝑐𝑚 𝑠 0 𝑠 ≤ 𝑡 ≤ 2 𝑠 𝑣 𝑥 = 0 . 25 𝑐𝑚 𝑠 𝑣 𝑦 = 1 𝑐𝑚 𝑠 2 𝑠 < 𝑡 ≤ 5 𝑠 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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    0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 0 0.5 1 1.5 2 2.5 3 Enters at t=0 Enters at t=1 Enters at t=2 Enters at t=3 Enters at t=4

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    *2. 10 : A flow is described by the velocity field 𝑣 = ( 𝑦 𝑠 ) 𝑖 ⃗ + 0 . 75 𝑚 𝑠 2 𝑡 𝑗 ⃗ At 𝑡 = 2 𝑠 , what are the coordinates of the particles that passed through the point (1,5) at 𝑡 = 0 𝑠 . What are the coordinates of the same particle at 𝑡 = 3 𝑠 . 𝑣 = 𝑑𝑥 𝑑𝑡 → 𝑥 = ∫ 𝑣𝑑𝑡 = ∫ ( 𝑦 𝑠 ) 𝑖 ̂ 𝑑𝑡 + ∫ 0 . 75 𝑚 𝑠 2 𝑡 𝑗 ̂ 𝑑𝑡 𝑥 = ( 𝑦𝑡 𝑠 + 𝑥 0 ) 𝑖 ̂ + ( ( 0 . 75 𝑚 𝑠 2 ) 𝑡 2 2 + 𝑦 0 ) 𝑗 ̂ 𝑥 0 = 1 , 𝑦 0 = 5 𝑥 ( 𝑡 = 2 ) = ( 𝑦𝑡 𝑠 + 1 ) 𝑖 ̂ + ( 0 . 75 𝑚 𝑠 2 ( 𝑡 2 2 ) + 5 ) 𝑗 ̂ 𝑦 = 0 . 75 𝑚 𝑠 2 ( 4 𝑠 2 2 ) + 5 = 6 . 5 𝑚 𝑥 = 6 . 5 𝑚 ( 2 𝑠 ) 𝑠 + 1 = 14 𝑚 𝑥 ( 𝑡 = 2 ) = ( 14 𝑖 ̂ + 6 . 5 𝑗 ̂ ) 𝑚 𝑥 ( 𝑡 = 3 ) → 𝑦 = 0 . 75 ( 3 2 2 ) + 5 = 8 . 375 𝑚 𝑥 = 8 . 375 ( 3 𝑠 ) 𝑠 + 1 = 26 . 125 𝑚 𝑥 ( 𝑡 = 3 ) = ( 26 . 125 𝑖 ̂ + 8 . 375 𝑗 ̂ ) 𝑚

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    2.1 1 : The velocity distribution for laminar flow between two parallel plates can be represented as 𝑢 𝑢 𝑚𝑎𝑥 = 1 − ( 2 𝑦 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 . 5 𝑐𝑃 ) with maximum velocity of 25 cm/s and a separation distance of 10mm. Calculate the force on a 0.25 m 2 section of the lower plate. 𝑢 ( 𝑦 ) = 𝑢 𝑚𝑎𝑥 − 𝑢 𝑚𝑎𝑥 ( 2 𝑦 ℎ ) 2 𝜏 𝑦𝑥 = 𝜇 𝜕𝑢 𝜕𝑦 = 𝜇 𝜕 𝜕𝑦 ( 𝑢 𝑚𝑎𝑥 − 𝑢 𝑚𝑎𝑥 ( 2 𝑦 ℎ ) 2 = 𝜇 [ − 2 𝑢 𝑚𝑎𝑥 ( 2 𝑦 ℎ ) ( 2 ℎ ) ] = − 8 𝜇 𝑢 𝑚𝑎𝑥 𝑦 ℎ 2 = 𝐹 𝐴 𝐹 = − 8 𝜇 𝑢 𝑚𝑎𝑥 𝑦𝐴 ℎ 2 𝐹 = − 8 ( 3 . 5 𝑐𝑃 ) ( 25 𝑐𝑚 𝑠 ) ( 5 𝑚𝑚 ) ( 0 . 25 𝑚 2 ) ( 10 𝑚𝑚 ) 2 = − 0 . 0875 𝑁

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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 ( ℎ 𝑦 − 𝑦 2 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/m 3 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 ( ℎ 𝑦 − 𝑦 2 2 ) 𝑠𝑖𝑛𝜃 ) = 𝜌𝑔𝑠𝑖𝑛𝜃 [ 𝜕 𝜕𝑦 ( ℎ 𝑦 ) − 𝜕 𝜕𝑦 ( 𝑦 2 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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    *2.1 3 : A block with mass 𝑀 1 is in contact with a thin fluid layer of thickness ℎ . The block is connected to a second mass, 𝑀 2 , via a cable that goes over an ideal pulley. Determine a relationship for the velocity of the block, when the second mass is released from rest. 0 0.002 0.004 0.006 0.008 0.01 0.012 0 10 20 30 40 50 60 Shear Stress (N/m^2) Shear Profile M 1 Viscosity,  Thickness, h M 2

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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. 𝑉 𝑦 = ( 𝑦 ℎ 𝑉 ) → 𝜏 𝑦 = ℎ = 𝜇𝑉 ℎ ∑ 𝐹 𝑦 ⬚ 𝑚 2 = 𝑇 − 𝑚 2 𝑔 = − 𝑚 2 𝑎 → 𝑇 = 𝑚 2 ( 𝑔 − 𝑎 ) ∑ 𝐹 𝑥 ⬚ 𝑚 1 = 𝑇 − 𝐹 𝑓 = 𝑚 1 𝑎 → 𝑇 = 𝑚 1 𝑎 + 𝐹 𝑓 𝑚 1 𝑎 + 𝐹 𝑓 = 𝑚 2 ( 𝑔 − 𝑎 ) 𝑎 ( 𝑚 1 + 𝑚 2 ) = 𝑚 2 𝑔 − 𝐹 𝑓 = 𝑚 2 𝑔 − 𝜇𝑉𝐴 ℎ 𝑑𝑉 𝑑𝑡 ( 𝑚 1 + 𝑚 2 ) = 𝑚 2 𝑔 − 𝜇𝑉𝐴 ℎ T W=m 2 g a T,a F f =  A

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    𝑑𝑉 𝑚 2 𝑔 − 𝜇𝑉𝐴 ℎ = 𝑑𝑡 𝑚 1 + 𝑚 2 − 1 𝜇𝐴 ℎ ln ( 𝑚 2 𝑔 − 𝜇𝑉𝐴 ℎ ) = 𝑡 𝑚 1 + 𝑚 2 ln ( 𝑚 2 𝑔 − 𝜇𝑉𝐴 ℎ ) = − 𝜇𝐴𝑡 ℎ ( 𝑚 1 + 𝑚 2 ) 𝑚 2 𝑔 − 𝜇𝑉𝐴 ℎ = 𝑒 − 𝜇𝐴𝑡 ℎ ( 𝑚 1 + 𝑚 2 ) 𝜇𝑉𝐴 ℎ = 𝑚 2 𝑔 − 𝑒 − 𝜇𝐴𝑡 ℎ ( 𝑚 1 + 𝑚 2 ) 𝑉 = 𝑚 2 𝑔 ℎ 𝜇𝐴 − ℎ 𝜇𝐴 𝑒 − 𝜇𝐴𝑡 ℎ ( 𝑚 1 + 𝑚 2 )

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    2.1 4 : For the following velocity field 𝑣 = ( 𝑥 𝑡 2 + 2 𝑦 − 𝑧 2 ) 𝑖 ⃗ + ( 𝑥𝑡 − 2 𝑦𝑡 ) 𝑗 ⃗ + ( 3 𝑥𝑡 + 𝑧 ) 𝑘 ⃗ ⃗ calculate the shear rate tensor, the vorticity, the hydrostatic pressure (use a viscosity of 𝜇 ) and the acceleration of the fluid. 𝜕 𝑣 𝑥 𝜕𝑥 = 𝑡 2 𝜕 𝑣 𝑦 𝜕𝑥 = 𝑡 𝜕 𝑣 𝑧 𝜕𝑥 = 3 𝑡 𝜕 𝑣 𝑥 𝜕𝑦 = 2 𝜕 𝑣 𝑦 𝜕𝑦 = − 2 𝑡 𝜕 𝑣 𝑧 𝜕𝑦 = 0 𝜕 𝑣 𝑥 𝜕𝑧 = − 2 𝑧 𝜕 𝑣 𝑦 𝜕𝑧 = 0 𝜕 𝑣 𝑧 𝜕𝑧 = 1 𝜕𝑣 𝜕𝑡 = ( 2 𝑥𝑡 ) 𝑖 ̂ + ( 𝑥 − 2 𝑦 ) 𝑗 ̂ + 3 𝑥 𝑘 ̂ 𝑑 = 1 2 [ 2 𝑡 2 2 + 𝑡 3 𝑡 − 2 𝑧 2 + 𝑡 − 4 𝑡 0 3 𝑡 − 2 𝑧 0 2 ] 𝜉 = ( 0 − 0 ) 𝑖 ̂ + ( − 2 𝑧 − 3 𝑡 ) 𝑗 ̂ + ( 𝑡 − 2 ) 𝑘 ̂ 𝑃 ℎ 𝑦𝑑𝑟𝑜 = − 1 3 ( 𝑡 2 − 2 𝑡 − 1 ) 𝑎 = 2 𝑥𝑡 𝑖 ̂ + ( 𝑥 − 2 𝑦 ) 𝑗 ̂ + 3 𝑥 𝑘 ̂ + ( 𝑥 𝑡 2 + 2 𝑦 − 𝑧 2 ) ( 𝑡 2 𝑖 ̂ + 𝑡 𝑗 ̂ + 3 𝑡 𝑘 ̂ ) + ( 𝑥𝑡 − 2 𝑦𝑡 ) ( 2 𝑖 ̂ − 2 𝑡 𝑗 ̂ ) + ( 3 𝑥𝑡 + 𝑧 ) ( − 2 𝑧 𝑖 ̂ + 1 𝑘 ̂ )

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    2.1 5 : A velocity field is given by 𝑣 = 0 . 2 𝑠 − 1 𝑥 𝑖 ⃗ − 0 . 35 ( 𝑚𝑚𝑠 ) − 1 𝑦 2 𝑗 ⃗ 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 ) 𝑖 ̂ − 0 . 35 ( 4 ) 2 𝑗 ̂ = 0 . 4 𝑖 ̂ − 5 . 6 𝑗 ̂ 𝑚𝑚 𝑠 𝑣 𝑥 = 𝑑𝑥 𝑑𝑡 = 0 . 2 𝑥 ∫ 𝑑𝑥 𝑥 𝑥 𝑥 0 = ∫ 0 . 2 𝑑𝑡 𝑡 0 ln ( 𝑥 𝑥 0 ) = 0 . 2 𝑡 → 𝑥 = 𝑥 0 𝑒 0 . 2 𝑡 𝑣 𝑦 = 𝑑𝑦 𝑑𝑡 = − 0 . 35 𝑦 2 ∫ 𝑑𝑦 𝑦 2 𝑦 𝑦 0 = ∫ − 0 . 35 𝑑𝑡 𝑡 0 − 1 𝑦 + 1 𝑦 0 = − 0 . 35 𝑡 − 1 𝑦 = − 0 . 35 𝑡 − 1 𝑦 0 = − 0 . 35 𝑡 𝑦 0 − 1 𝑦 0 𝑦 = 𝑦 0 0 . 35 𝑡 𝑦 0 + 1 𝑥 = 2 𝑚𝑚 ( 𝑒 0 . 2 ∗ 6 ) = 6 . 64 𝑚𝑚

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    𝑦 = 4 𝑚𝑚 0 . 35 ( 6 ) ( 4 ) + 1 = 0 . 426 𝑚𝑚

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    2.1 6 : 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 (assu med 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 0.5m 2 . Force on the plate is 𝑚𝑔 = 10 𝑔 ( 9 . 81 𝑚 𝑠 2 ) = 0 . 098 𝑁 𝜏 = 0 . 098 𝑁 0 . 5 𝑚 2 = 0 . 196 𝑃𝑎 𝜏 = 𝜇 ( Δ 𝑉 Δ 𝑦 ) ( 𝑎 ) 𝑉 = 𝜏 𝜇 ℎ = 0 . 196 𝑃𝑎 1 𝑐𝑃 ( 10 𝑚𝑚 ) = 1 . 96 𝑚 𝑠 ( 𝑏 ) 𝑉 = 0 . 196 𝑃𝑎 3 . 5 𝑐𝑃 ( 10 𝑚𝑚 ) = 0 . 56 𝑚 𝑠 ( 𝑐 ) 𝜏 𝑎𝑐𝑡𝑢𝑎𝑙 = 𝜏 − 𝜏 𝑦 = 0 . 146 𝑃𝑎 M h

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    𝑉 = 0 . 146 𝑃𝑎 3 . 5 𝑐𝑃 ( 10 𝑚𝑚 ) = 0 . 418 𝑚 𝑠

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    2.1 7 : What type of fluid can be classified by the following shear stress - strain rate data? Plot the data and classify. Determine the viscosity of the fluid. Newtonian: 𝜇 = 4 . 2 𝑐𝑃 0 1 2 3 4 5 6 7 8 9 10 0 50 100 150 200 250 Shear Stress Shear Rate Shear stress (N/m 2 ) 0.4 0.82 2.50 5.44 8.80 Shear rate (s - 1 ) 0 10 50 120 200

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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 = 2 𝐴𝜇𝑉 ℎ 𝐴 = 5 𝑁 ( 50 𝑚𝑚 ) 2 ( 1 𝑐𝑃 ) ( 10 𝑐𝑚 𝑠 ) = 1250 𝑚 2 F 50mm 50mm

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    2.1 9 : Use the same figure for problem 2.1 8 , 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 2.1 8 𝐹 = 𝜏 𝑙𝑒𝑓𝑡 + 𝜏 𝑟𝑖𝑔 ℎ 𝑡 𝐹 = 𝜇 𝑙𝑒𝑓𝑡 𝑉𝐴 ℎ 𝑙𝑒𝑓𝑡 + 𝜇 𝑟𝑖𝑔 ℎ 𝑡 𝑉𝐴 ℎ 𝑟𝑖𝑔 ℎ 𝑡 ℎ 𝑟𝑖𝑔 ℎ 𝑡 = 2 ℎ 𝑙𝑒𝑓𝑡 𝐹 = 𝑉𝐴 ℎ 𝑙𝑒𝑓𝑡 ( 𝜇 𝑙𝑒𝑓𝑡 + 𝜇 𝑟𝑖𝑔 ℎ 𝑡 2 ) For equilibrium 𝜇 𝑙𝑒𝑓𝑡 = 𝜇 𝑟𝑖𝑔 ℎ 𝑡 2 → 𝜇 𝑙𝑒𝑓𝑡 = 1 . 75 𝑐𝑃

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    2. 20 Velocity at the shaft wall can be defined as: 𝑣 = 𝑟𝜔 Where r can be defined from 𝑟 = ℎ 𝑡𝑎𝑛 ( 30 ) 𝑣 = ℎ 𝜔 tan ( 30 ) The velocity at the bearing wall is zero; therefore the slope of the velocity profile can be defined as: 𝑣 = 𝜔 ℎ 𝑡𝑎𝑛 ( 30 ) 𝑎 And the shear stress can be defined as 𝜏 = 𝜔𝜇 ℎ 𝑡𝑎𝑛 ( 30 ) 𝑎 Solving for the known values 𝜏 = 25 𝑟𝑒𝑣 𝑠 ( 0 . 05 𝑁𝑠 𝑚 2 ) ( 35 𝑚𝑚 ) tan ( 30 ) 0 . 35 𝑚𝑚 = 72 𝑃𝑎 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/cm 2 . Assume that this is a two - dimensional flow. 𝑑𝑉 𝑑𝑦 = 15 𝑟𝑎𝑑 𝑠 𝜏 = 0 . 868 𝑑𝑦𝑛𝑒𝑠 𝑐 𝑚 2

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    𝑑𝑉 𝑑𝑦 = 30 𝑟𝑎𝑑 𝑠 𝜏 = 0 . 355 𝑑𝑦𝑛𝑒𝑠 𝑐 𝑚 2 𝜏 𝑥𝑦 = 𝑘 ( 𝑑𝑉 𝑑𝑦 ) 𝑛 0 . 868 𝑑𝑦𝑛𝑒𝑠 𝑐 𝑚 2 0 . 355 𝑑𝑦𝑛𝑒𝑠 𝑐 𝑚 2 = ( 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 𝑐 𝑚 2 𝑑𝑉 𝑑𝑦 = 5 . 735 𝑟𝑎𝑑 𝑠 , 𝑤 ℎ 𝑒𝑛 𝜏 = 3 𝑑𝑦𝑛𝑒𝑠 𝑐 𝑚 2 2.22 2.22 Given that the volumetric flow rate for a fluid within a circular cross - section tube can be represented by 𝑄 = 𝜋 𝑟 4 Δ 𝑃 8 𝜇𝐿 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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    Δ 𝑃 = 8 𝑄𝜇𝐿 𝜋 𝑟 4 = 8 ( 4 . 5 𝐿 min ⬚ ) ( 3 . 5 𝑐𝑃 ) ( 1 𝑚 ) 𝜋 ( 23 𝑚𝑚 2 ) 2 = 0 . 286 𝑚𝑚𝐻𝑔 Δ 𝑃 = 8 𝑄𝜇𝐿 𝜋 𝑟 4 = 8 ( 4 . 5 𝐿 min ⬚ ) ( 1 𝑐𝑃 ) ( 1 𝑚 ) 𝜋 ( 23 𝑚𝑚 2 ) 2 = 0 . 0819 𝑚𝑚𝐻𝑔 Δ 𝑃 = 8 𝑄𝜇𝐿 𝜋 𝑟 4 = 8 ( 4 . 5 𝐿 min ⬚ ) ( 15000 𝑐𝑃 ) ( 1 𝑚 ) 𝜋 ( 23 𝑚𝑚 2 ) 2 = 1229 𝑚𝑚𝐻𝑔 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. S imilarly, 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 - s lip 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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    assumes that the container is sufficiently large to contain the blood. If the container for blood is small (e.g. on the order of a capillary) these assumptions would need to change because both sides of the cell would be in close contact with a solid surfa ce, yet the cell would still flow through the container.

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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. 𝑃 𝑤 1 = 𝑃 𝐴 − ( 1050 𝑘𝑔 𝑚 3 ) ( 9 . 81 𝑚 𝑠 2 ) ( 0 . 5 𝑚 ) = 𝑃 𝑎 − 38 . 62 𝑚𝑚𝐻𝑔 𝑃 𝑤 2 = 𝑃 𝑤 1 + ( 1000 𝑘𝑔 𝑚 3 ) ( 9 . 81 𝑚 𝑠 2 ) ( 0 . 5 − 0 . 275 ) 𝑚 = 𝑃 𝑤 1 + 16 . 55 𝑚𝑚𝐻𝑔 𝑃 𝐵 = 𝑃 𝑤 2 + ( 1050 𝑘𝑔 𝑚 3 ) ( 9 . 81 𝑚 𝑠 2 ) ( 0 . 275 𝑚 ) = 𝑃 𝑤 2 + 21 . 24 𝑚𝑚𝐻𝑔 𝑃 𝐵 = 𝑃 𝐴 − 38 . 62 𝑚𝑚𝐻𝑔 + 16 . 55 𝑚𝑚𝐻𝑔 + 21 . 24 𝑚𝑚𝐻𝑔 = 𝑃 𝐴 − 0 . 83 𝑚𝑚𝐻𝑔 A B Blood Water d 1 =500mm d 2 =190mm d 3 =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 𝑠 2 ( 1 − 𝑧 10 , 000 𝑚 ) , 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 = − 𝜌 𝑔 𝑧 = − 𝑝 𝑅𝑇 𝑔 𝑧 ∫ 𝑑𝑝 𝑝 𝑝 𝑝 0 = − ∫ 𝑔 𝑧 𝑅𝑇 𝑑𝑧 𝑧 0 ln ( 𝑝 ) − ln ( 𝑝 0 ) = − 1 𝑅𝑇 ∫ 18 . 7 𝑚 𝑠 2 ( 1 − 𝑧 10000 ) 𝑑𝑧 𝑧 0 ln ( 𝑝 𝑝 0 ) = − 18 . 7 𝑚 𝑠 2 𝑅𝑇 ∫ ( 1 − 𝑧 10000 ) 𝑑𝑧 = 𝑧 0 − 18 . 7 𝑚 𝑠 2 𝑅𝑇 [ 𝑧 − 𝑧 2 20000 ] 0 𝑧 𝑝 𝑝 0 = 𝑒 − 18 . 7 𝑚 𝑠 2 𝑅𝑇 [ 𝑧 − 𝑧 2 20000 ] 𝑝 ( 𝑧 ) = 𝑝 0 𝑒 − 18 . 7 𝑚 𝑠 2 𝑅𝑇 [ 𝑧 − 𝑧 2 20000 ]

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