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

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Chapter 22.1: For the velocity distribution𝑣𝑥=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𝜕𝑧𝑣⃗=5𝑥𝑖̂−5𝑦𝑗̂𝜕𝑣𝜕𝑡=0→𝑛𝑜𝑙𝑜𝑐𝑎𝑙𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛𝜕𝑣𝜕𝑥=5𝜕𝑣𝜕𝑦=−5𝜕𝑣𝜕𝑧=0𝑎⃗=0+5𝑥(5)𝑖̂−5𝑦(−5)𝑗̂+0=25𝑥𝑖̂+25𝑦𝑗̂→𝑜𝑛𝑙𝑦𝑐𝑜𝑛𝑣𝑒𝑐𝑡𝑖𝑣𝑒𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛2.2: Consider a velocity vector𝑣=(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

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𝑑𝑣𝑑𝑡=2𝑥𝑡𝑖̂+𝑥𝑗̂→𝑛𝑜𝑡𝑠𝑡𝑒𝑎𝑑𝑦(𝑏)𝜕𝑣𝜕𝑥+𝜕𝑣𝜕𝑦=0(𝑡2−1)𝑖+(𝑡−2𝑦)𝑗̂→𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑏𝑙𝑒

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2.3: Given the velocity𝑣=(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→𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙2.4The velocity vector for a steady incompressible flow in the xy plane is given by, wherethe coordinates are measured in centimeters. Determine the time it takes for a particle to movefrom x = 1cm to x = 4cm for a particle that passes through the point()(),1, 4x y=.𝑑𝑢𝑑𝑡=4𝑥𝒊⃗+4𝑦𝑥2𝒋⃗𝑢=4𝑡𝑥𝒊⃗+4𝑡𝑦𝑥2𝒋⃗4𝑡𝑥4𝑡𝑦𝑥2=14𝑥𝑦=14𝑜𝑟4𝑥=𝑦4𝑡1=1𝑠𝑜𝑡=0.25𝑠Check4𝑡(4∗1)12=4𝑠𝑜𝑡=0.25𝑠Time at 1cm is equal to 0.25s; for time at 4cm

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4𝑡4=4𝑠𝑜𝑡=4𝑠Therefore, it takes 3.75 seconds for this particular particle to movefrom 1cm to 4cm2.5:Flow in a 2-dimensional channel of width W has a velocity profile defined by:𝑣𝑥(𝑦)=𝑘[(𝑊22)−𝑦2]where y=0 is located at the center of the channel. Sketch the velocity distribution and find theshear stress/unit width of the channel at the wall.𝜏𝑥𝑦=𝜇(𝜕𝜕𝑦(𝑘(𝑊22−𝑦2)))=𝜇𝑘(𝜕𝜕𝑦(𝑊22−𝑦2))=−2𝑦𝜇𝑘VelocityDistribution (k=1, W=10):-6-4-202460102030405060Channel LocationVelocity

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2.6:A velocity filed is given by𝑣(𝑥,𝑦,𝑧)=20𝑥𝑦𝑖⃗−10𝑦2𝑗⃗, calculate the acceleration, the angularvelocity, and the vorticity vector at thepoint (-1,1,1), where the units of the velocity equation aremm/s.𝑎⃗=[200𝑥𝑦(20𝑦)𝑖̂−10𝑦2(20𝑥𝑖̂−20𝑦𝑗̂)]𝑚𝑚𝑠2=[200𝑥𝑦2𝑖̂+200𝑦3𝑗̂]𝑚𝑚𝑠2𝑎⃗(−1,1,1)=[200(−1)(1)2𝑖̂+200(1)3𝑗̂]𝑚𝑚𝑠2=[−200𝑖̂+200𝑗̂]𝑚𝑚𝑠2𝜔⃗⃗⃗=12[(0−0)𝑖̂+(0−0)𝑗̂+(0−20𝑥)𝑘̂]𝑟𝑎𝑑𝑠=12[−20𝑥]𝑘̂𝑟𝑎𝑑𝑠=−10(−1)𝑘̂𝑟𝑎𝑑𝑠=10𝑟𝑎𝑑𝑠𝑘̂𝜉=2𝜔⃗⃗⃗=20𝑟𝑎𝑑𝑠𝑘̂2.7:Considering one-dimensional fluid (density,𝜌; viscosity,𝜇)flow in a tube with an inletpressure of𝑝𝑖, and outlet pressure of𝑝𝑜,and tube radius of𝑟and length𝑙.The density of thefluid can be represented as𝜌.Express the wall shear stress as a function of these variables.Hint: balance Newton’s second law of motion.∑𝐹=𝑚𝑎→𝑎𝑠𝑠𝑢𝑚𝑒𝑚𝑖𝑠𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑖𝑛𝑡𝑢𝑏𝑒𝑜𝑓𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡∑𝐹=𝑝𝑖𝑖̂−𝑝𝑜𝑖̂−𝜏𝑖̂=(𝑝𝑖−𝑝𝑜)𝑖̂−𝜏𝑖̂𝑚=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 anddetermine if the fluid is pseudoplastic,Newtonian or dilatants. Approximate the viscosity (or theapparent 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)0102030405060708090100-200020040060080010001200Shear StressShear RateShear stress (dyne/cm2)59234895Shear rate (s-1)0.11101001000

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2.9: Consider a red blood cell that originates from the origin of our𝑥𝑦-coordinate system. Thevelocity of the fluid is unsteady and is described by:𝑣={𝑣𝑥=1𝑐𝑚𝑠𝑣𝑦=0.5𝑐𝑚𝑠0𝑠≤𝑡≤2𝑠𝑣𝑥=0.25𝑐𝑚𝑠𝑣𝑦=1𝑐𝑚𝑠2𝑠<𝑡≤5𝑠Plot the location of this red blood cell at time zero, 1s, 2s, 3s, 4s, and 5s (enters from the originat time zero). Also plot the location of the red blood cells that enter (from the origin) at time 1s,2s, 3s, 4s.Red BloodCell Enters atTimet=0t=1t=2t=3t=4Time inFluid1s(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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00.511.522.533.544.500.511.522.53Enters at t=0Enters at t=1Enters at t=2Enters at t=3Enters at t=4

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*2.10: A flow isdescribed by the velocity field𝑣=(𝑦𝑠)𝑖⃗+0.75𝑚𝑠2𝑡𝑗⃗At𝑡=2𝑠, what are the coordinates of the particles that passed through the point (1,5) at𝑡=0𝑠.What are the coordinates of the same particle at𝑡=3𝑠.𝑣=𝑑𝑥𝑑𝑡→𝑥=∫𝑣𝑑𝑡=∫(𝑦𝑠)𝑖̂𝑑𝑡+∫0.75𝑚𝑠2𝑡𝑗̂𝑑𝑡𝑥=(𝑦𝑡𝑠+𝑥0)𝑖̂+((0.75𝑚𝑠2)𝑡22+𝑦0)𝑗̂𝑥0=1,𝑦0=5𝑥(𝑡=2)=(𝑦𝑡𝑠+1)𝑖̂+(0.75𝑚𝑠2(𝑡22)+5)𝑗̂𝑦=0.75𝑚𝑠2(4𝑠22)+5=6.5𝑚𝑥=6.5𝑚(2𝑠)𝑠+1=14𝑚𝑥(𝑡=2)=(14𝑖̂+6.5𝑗̂)𝑚𝑥(𝑡=3)→𝑦=0.75(322)+5=8.375𝑚𝑥=8.375(3𝑠)𝑠+1=26.125𝑚𝑥(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 representedas𝑢𝑢𝑚𝑎𝑥=1−(2𝑦ℎ)2whereℎis the separation distance between the two flat plates and the origin is located half-waybetween the plates. Consider the flow of blood at 37C (𝜇=3.5𝑐𝑃)with maximum velocity of 25cm/s and a separation distance of 10mm. Calculate the force on a 0.25 m2section of the lowerplate.𝑢(𝑦)=𝑢𝑚𝑎𝑥−𝑢𝑚𝑎𝑥(2𝑦ℎ)2𝜏𝑦𝑥=𝜇𝜕𝑢𝜕𝑦=𝜇𝜕𝜕𝑦(𝑢𝑚𝑎𝑥−𝑢𝑚𝑎𝑥(2𝑦ℎ)2=𝜇[−2𝑢𝑚𝑎𝑥(2𝑦ℎ)(2ℎ)]=−8𝜇𝑢𝑚𝑎𝑥𝑦ℎ2=𝐹𝐴𝐹=−8𝜇𝑢𝑚𝑎𝑥𝑦𝐴ℎ2𝐹=−8(3.5𝑐𝑃)(25𝑐𝑚𝑠)(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 bedescribed by𝑢=𝜌𝑔𝜇(ℎ𝑦−𝑦22)𝑠𝑖𝑛𝜃if the coordinate axis is aligned with the inclined plant. Determine a function for the shear stressalong this fluid. Plot the velocityprofile and shear stress profile, if the fluid’s density is 900kg/m3and its viscosity is 2.8cP. The fluid thickness, h, is equal to 10mm and the plane isinclined at an angle of 40.𝜏=𝜇𝜕𝑢𝜕𝑦=𝜇𝜕𝜕𝑦(𝜌𝑔𝜇(ℎ𝑦−𝑦22)𝑠𝑖𝑛𝜃)=𝜌𝑔𝑠𝑖𝑛𝜃[𝜕𝜕𝑦(ℎ𝑦)−𝜕𝜕𝑦(𝑦22)]=𝜌𝑔𝑠𝑖𝑛𝜃[ℎ−𝑦]00.0020.0040.0060.0080.010.012024681012Velocity (m/s)Velocity Profile

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*2.13: A block with mass𝑀1is in contact with a thin fluid layer of thicknessℎ. The block isconnected to a second mass,𝑀2, via a cable that goes over an ideal pulley. Determine arelationship for the velocity of theblock, when the second mass is released from rest.00.0020.0040.0060.0080.010.0120102030405060Shear Stress (N/m^2)Shear ProfileM1Viscosity,Thickness, hM2

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For film thickness, velocity at the wall is 0 and acceleration at contact is a𝜏=𝜇𝜕𝑣𝜕𝑦|𝑦=ℎAssume that the velocity is linear because it is a small film layer.𝑉𝑦=(𝑦ℎ𝑉)→𝜏𝑦=ℎ=𝜇𝑉ℎ∑𝐹𝑦⬚𝑚2=𝑇−𝑚2𝑔=−𝑚2𝑎→𝑇=𝑚2(𝑔−𝑎)∑𝐹𝑥⬚𝑚1=𝑇−𝐹𝑓=𝑚1𝑎→𝑇=𝑚1𝑎+𝐹𝑓𝑚1𝑎+𝐹𝑓=𝑚2(𝑔−𝑎)𝑎(𝑚1+𝑚2)=𝑚2𝑔−𝐹𝑓=𝑚2𝑔−𝜇𝑉𝐴ℎ𝑑𝑉𝑑𝑡(𝑚1+𝑚2)=𝑚2𝑔−𝜇𝑉𝐴ℎTW=m2gaT,aFf=A

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𝑑𝑉𝑚2𝑔−𝜇𝑉𝐴ℎ=𝑑𝑡𝑚1+𝑚2−1𝜇𝐴ℎln(𝑚2𝑔−𝜇𝑉𝐴ℎ)=𝑡𝑚1+𝑚2ln(𝑚2𝑔−𝜇𝑉𝐴ℎ)=−𝜇𝐴𝑡ℎ(𝑚1+𝑚2)𝑚2𝑔−𝜇𝑉𝐴ℎ=𝑒−𝜇𝐴𝑡ℎ(𝑚1+𝑚2)𝜇𝑉𝐴ℎ=𝑚2𝑔−𝑒−𝜇𝐴𝑡ℎ(𝑚1+𝑚2)𝑉=𝑚2𝑔ℎ𝜇𝐴−ℎ𝜇𝐴𝑒−𝜇𝐴𝑡ℎ(𝑚1+𝑚2)

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2.14: For the following velocity field𝑣=(𝑥𝑡2+2𝑦−𝑧2)𝑖⃗+(𝑥𝑡−2𝑦𝑡)𝑗⃗+(3𝑥𝑡+𝑧)𝑘⃗⃗calculate the shear rate tensor, the vorticity, the hydrostatic pressure (use a viscosity of𝜇) andthe 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+3𝑥𝑘̂𝑑=12[2𝑡22+𝑡3𝑡−2𝑧2+𝑡−4𝑡03𝑡−2𝑧02]𝜉=(0−0)𝑖̂+(−2𝑧−3𝑡)𝑗̂+(𝑡−2)𝑘̂𝑃ℎ𝑦𝑑𝑟𝑜=−13(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 by𝑣=0.2𝑠−1𝑥𝑖⃗−0.35(𝑚𝑚𝑠)−1𝑦2𝑗⃗where x and y are given in millimeters. Determine the velocity of a particle at point (2,4) and theposition 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−5.6𝑗̂𝑚𝑚𝑠𝑣𝑥=𝑑𝑥𝑑𝑡=0.2𝑥∫𝑑𝑥𝑥𝑥𝑥0=∫0.2𝑑𝑡𝑡0ln(𝑥𝑥0)=0.2𝑡→𝑥=𝑥0𝑒0.2𝑡𝑣𝑦=𝑑𝑦𝑑𝑡=−0.35𝑦2∫𝑑𝑦𝑦2𝑦𝑦0=∫−0.35𝑑𝑡𝑡0−1𝑦+1𝑦0=−0.35𝑡−1𝑦=−0.35𝑡−1𝑦0=−0.35𝑡𝑦0−1𝑦0𝑦=𝑦00.35𝑡𝑦0+1𝑥=2𝑚𝑚(𝑒0.2∗6)=6.64𝑚𝑚

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𝑦=4𝑚𝑚0.35(6)(4)+1=0.426𝑚𝑚

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2.16: A fluid is placed in the area between two parallel plates. The upper plate is movable andconnected to a weight by a cable. Calculate the velocity of the plate for a) the fluid between theplates 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 massto be 10g, the height between the plates as 10mm and the contact area to be 0.5m2.Force on the plate is𝑚𝑔=10𝑔(9.81𝑚𝑠2)=0.098𝑁𝜏=0.098𝑁0.5𝑚2=0.196𝑃𝑎𝜏=𝜇(Δ𝑉Δ𝑦)(𝑎)𝑉=𝜏𝜇ℎ=0.196𝑃𝑎1𝑐𝑃(10𝑚𝑚)=1.96𝑚𝑠(𝑏)𝑉=0.196𝑃𝑎3.5𝑐𝑃(10𝑚𝑚)=0.56𝑚𝑠(𝑐)𝜏𝑎𝑐𝑡𝑢𝑎𝑙=𝜏−𝜏𝑦=0.146𝑃𝑎Mh

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𝑉=0.146𝑃𝑎3.5𝑐𝑃(10𝑚𝑚)=0.418𝑚𝑠

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2.17: What type of fluid can be classified by thefollowing shear stress-strain rate data? Plotthe data and classify. Determine the viscosity of the fluid.Newtonian:𝜇=4.2𝑐𝑃012345678910050100150200250Shear StressShear RateShear stress (N/m2)0.40.822.505.448.80Shear rate (s-1)01050120200

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2.18:A plate is moving upward through a channel filled with water as shown. The plate velocityis 10cm/s, and the spacing is 50 mm on each side of the plate. The force required to move theplate 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 theplate. Assume that the velocity profile is linear.𝜏=𝜇𝑉ℎ𝐹=2𝐴𝜇𝑉ℎ𝐴=5𝑁(50𝑚𝑚)2(1𝑐𝑃)(10𝑐𝑚𝑠)=1250𝑚2F50mm50mm

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2.19: Use the same figure for problem 2.18, except that the distance on the right hand side isincreased to 100mm. The fluid on the right side is blood. Determine the viscosity of the fluid onthe left side. The plate is moving with a constant velocity.Using the same analysis as for 2.18𝐹=𝜏𝑙𝑒𝑓𝑡+𝜏𝑟𝑖𝑔ℎ𝑡𝐹=𝜇𝑙𝑒𝑓𝑡𝑉𝐴ℎ𝑙𝑒𝑓𝑡+𝜇𝑟𝑖𝑔ℎ𝑡𝑉𝐴ℎ𝑟𝑖𝑔ℎ𝑡ℎ𝑟𝑖𝑔ℎ𝑡=2ℎ𝑙𝑒𝑓𝑡𝐹=𝑉𝐴ℎ𝑙𝑒𝑓𝑡(𝜇𝑙𝑒𝑓𝑡+𝜇𝑟𝑖𝑔ℎ𝑡2)For equilibrium𝜇𝑙𝑒𝑓𝑡=𝜇𝑟𝑖𝑔ℎ𝑡2→𝜇𝑙𝑒𝑓𝑡=1.75𝑐𝑃

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2.20Velocity at the shaft wall can be defined as:𝑣=𝑟𝜔Where r can be defined from𝑟=ℎ𝑡𝑎𝑛(30)𝑣=ℎ𝜔tan(30)The velocity at the bearing wall is zero; therefore the slope of the velocity profile can be definedas:𝑣=𝜔ℎ𝑡𝑎𝑛(30)𝑎And the shear stress can be defined as𝜏=𝜔𝜇ℎ𝑡𝑎𝑛(30)𝑎Solving for the known values𝜏=25𝑟𝑒𝑣𝑠(0.05𝑁𝑠𝑚2)(35𝑚𝑚)tan(30)0.35𝑚𝑚=72𝑃𝑎2.21: Two data points on the rheological diagram of a biofluid are provided. Determine theconsistency index and the flow behavior index and the strain rate if the shear stress is increasedto 3 dynes/cm2. Assume that this is a two-dimensional flow.𝑑𝑉𝑑𝑦=15𝑟𝑎𝑑𝑠𝜏=0.868𝑑𝑦𝑛𝑒𝑠𝑐𝑚2

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𝑑𝑉𝑑𝑦=30𝑟𝑎𝑑𝑠𝜏=0.355𝑑𝑦𝑛𝑒𝑠𝑐𝑚2𝜏𝑥𝑦=𝑘(𝑑𝑉𝑑𝑦)𝑛0.868𝑑𝑦𝑛𝑒𝑠𝑐𝑚20.355𝑑𝑦𝑛𝑒𝑠𝑐𝑚2=(15𝑟𝑎𝑑𝑠)𝑛(30𝑟𝑎𝑑𝑠)𝑛→2.445=(12)𝑛𝑛=−1.289880.868=𝑘(15)−1.28988𝑘=28.545𝑑𝑦𝑛𝑒𝑠𝑐𝑚2𝑑𝑉𝑑𝑦=5.735𝑟𝑎𝑑𝑠,𝑤ℎ𝑒𝑛𝜏=3𝑑𝑦𝑛𝑒𝑠𝑐𝑚22.222.22Given that the volumetric flow rate for a fluid within a circular cross-section tubecan be represented by𝑄=𝜋𝑟4Δ𝑃8𝜇𝐿where r is the tube radius,is the fluid viscosity, P is the pressure drop across the tube and L isthe 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 sameconditions, what would the required pressure drop be for water (=1cP) and chocolate syrup (= 15,000 cP).BloodWaterChocolate Syrup

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Δ𝑃=8𝑄𝜇𝐿𝜋𝑟4=8(4.5𝐿min⬚)(3.5𝑐𝑃)(1𝑚)𝜋(23𝑚𝑚2)2=0.286𝑚𝑚𝐻𝑔Δ𝑃=8𝑄𝜇𝐿𝜋𝑟4=8(4.5𝐿min⬚)(1𝑐𝑃)(1𝑚)𝜋(23𝑚𝑚2)2=0.0819𝑚𝑚𝐻𝑔Δ𝑃=8𝑄𝜇𝐿𝜋𝑟4=8(4.5𝐿min⬚)(15000𝑐𝑃)(1𝑚)𝜋(23𝑚𝑚2)2=1229𝑚𝑚𝐻𝑔2.23At a stenosis, the energy in the flowing fluid is skewed from potential energy to kineticenergy. 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, moreenergy “taken” from the potential energy would be needed to maintain the flow conditions.Downstream of the stenosis can be effected significantly, because there isan overall reductionin the amount of energy available (since energy is not really conserved in the conversion ofpotential energy to kinetic energy). Therefore, if there are future constrictions, the flow may notbe able to overcome these restrictions. Similarly, since there is a conversion to kinetic, thatwould suggest 1) jet flow at a stenosis, 2) increased shearing forces at the stenosis andimmediately distal to the stenosis, a conversion of energy back from kinetic to potential.2.24The no-slip boundary wall condition for Newtonian fluids has been well established. Noknown 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, theboundary condition should be always changing based on what part of blood is interaction withthe wall. Plasma is very close to a Newtonian fluid and thus the no-slip boundary conditionshould apply to this portion of blood. Cells can be considered rigid and thus the entire cell wouldhave 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 issmall (e.g. on the order of a capillary) these assumptions would need to change because bothsides of the cell would be in close contact with a solid surface, yet the cell would still flowthrough the container.

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Chapter 33.1 A two fluid manometer is used to measure the pressure difference for flowing blood in alaboratory experiment. Calculate the pressure difference between points A and B in thefluid.𝑃𝑤1=𝑃𝐴−(1050𝑘𝑔𝑚3)(9.81𝑚𝑠2)(0.5𝑚)=𝑃𝑎−38.62𝑚𝑚𝐻𝑔𝑃𝑤2=𝑃𝑤1+(1000𝑘𝑔𝑚3)(9.81𝑚𝑠2)(0.5−0.275)𝑚=𝑃𝑤1+16.55𝑚𝑚𝐻𝑔𝑃𝐵=𝑃𝑤2+(1050𝑘𝑔𝑚3)(9.81𝑚𝑠2)(0.275𝑚)=𝑃𝑤2+21.24𝑚𝑚𝐻𝑔𝑃𝐵=𝑃𝐴−38.62𝑚𝑚𝐻𝑔+16.55𝑚𝑚𝐻𝑔+21.24𝑚𝑚𝐻𝑔=𝑃𝐴−0.83𝑚𝑚𝐻𝑔ABBloodWaterd1=500mmd2=190mmd3=275mm

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3.2 NASA is planning a mission to a newly found planet and will monitor the density of the newplanet's atmosphere. Assume that NASA knows that atmosphere behaves as an ideal gasand that the planets gravitational force is a function of altitude (𝑔(𝑧)=18.7𝑚𝑠2(1−𝑧10,000𝑚),where z is inm). The temperature of the atmosphere is constant at 250K and the gasconstant 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.𝑑𝑝𝑑𝑧=−𝜌𝑔𝑧=−𝑝𝑅𝑇𝑔𝑧∫𝑑𝑝𝑝𝑝𝑝0=−∫𝑔𝑧𝑅𝑇𝑑𝑧𝑧0ln(𝑝)−ln(𝑝0)=−1𝑅𝑇∫18.7𝑚𝑠2(1−𝑧10000)𝑑𝑧𝑧0ln(𝑝𝑝0)=−18.7𝑚𝑠2𝑅𝑇∫(1−𝑧10000)𝑑𝑧=𝑧0−18.7𝑚𝑠2𝑅𝑇[𝑧−𝑧220000]0𝑧𝑝𝑝0=𝑒−18.7𝑚𝑠2𝑅𝑇[𝑧−𝑧220000]𝑝(𝑧)=𝑝0𝑒−18.7𝑚𝑠2𝑅𝑇[𝑧−𝑧220000]

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