TECHNISCHE UNIVERSITEIT EINDHOVEN Faculteit Biomedische Technologie, groep Cardiovasculaire Biomechanica Tentamen Cardiovasculaire Stromingsleer (8W090) blad /4 dinsdag 8 mei 2007, 9-2 uur Maximum score is 30 points. The partial scores are indicated in the margin of each question.. Answer the following questions with yes or no and give a short motivation. a. Is it correct that in the determination of the parameters of a 3-component windkessel model of the systemic circulation (resistance, inertance and compliance), the resistance can be obtained through division of the mean blood pressure by the cardiac output? b. Is it correct that the longitudinal impedance for flow in a straight ridid tube for large values of the Womersley parameter α becomes imaginary and linearly proportional to α? c. Is the inlet length to obtain fully developed tube flow for a stationary flow generally much smaller than the inlet length for a pulsating flow, in the case the latter has the same time-averaged flow rate? d. Is, in the analysis of entrance flow in a curved tube, it allowed to apply Bernoulli s law because the flow is friction dominated? e. Is it correct that in the case of a stationary fully developed flow in a curved tube the wall shear stress in the outer bend will increase with decreasing Dean-number? f. Does the non-linear stress-strain relation of collagen fibers in the arterial wall lead to a decrease of the compliance if pressure increases? g. Is it true that the imaginary part of the wave number k is a measure for the attenuation of the wave? h. In the figure on the right a part of the velocity profile of a layered flow of blood (high viscosity) and plasma (low viscosity) is given. r 2 Is layer the plasma-layer? z i. Two elastic tubes with admittance Y and Y 2 respectively are connected (end to end). Is it correct that the coefficient of transmission for pressure waves is invariant for exchange of the two tubes? j. The viscosity model of Carreau-Yasuda is given by: η η =[+(λ γ) a ] (n )/a. η 0 η Is it true that for lower values of the parameter a the viscosity decreases faster with increasing shear rate γ than for higher values of a?
Tentamen Cardiovasculaire Stromingsleer (8W090) blad 2/4 dinsdag 8 mei 2007, 9-2 uur 2. Endothelial cels that line the luminal side of the arterial wall appear to orient in the direction of the shear stress they experience from the blood flow. An experimental setup as depicted in the figure below in which a monolayer of endothelial cells on a rigid substrate is sheared with an oscillating shear stress τ w, is used to mimic the physiological situation. r A z a 0 z L 0 monolayer L Theset-upconsistsofarigidtubewithinternalradiusa 0 and length 2L. Atlocation z = L a piston is moving with an oscillating velocity given by v z = v z0 cos(ωt) and generates a flow q(t). At location z = 0, one can measure an harmonic pressure p m = p m0 cos(ωt + φ) independent from the radial position in the tube. For z>0 the flow is assumed to be fully developed. At z = L the tube is connected to a large reservoir with a constant pressure p z=l = 0. For fully developed flow in a tube, the Navier Stokes equations simplify to: v = ρ z + ν v (r r r r ) 0 = ρ r a. Give an expression for the pressure p(z, t) andtheflowq(t) intermsofv z0 and p m0 for z>0. b. Show that in complex notation (e ix =cosx + i sin x) we find: z = ˆp z eiωt with q =ˆqe iωt with ˆq = π v z0 ˆp z = p m0 L eiφ c. Scale the Navier-Stokes equation given above with v = v/v z0, r = r/a 0, z = z/l, t = ωt and p = p/p 0 and show that for p 0 = ρlνv z0 / the Womersley parameter α = a 0 ω/ν is the only dimensionless group. d. Show that for a frictionless (inviscid) flow the velocity profile can be assumed to be a flat profile given by: v(r, t) = p m0 ρωl cos(ωt) 2
Show also that in that case φ = π/2 andp m0 = ρωlv z0. What will be the wall shear stress for this situation. Is this rational for a frictionless flow? e. Show that for a friction dominated flow the velocity profile will be parabolic given by: v(r, t) = 4 ρν ( ) r2 z and show also that in that case we have φ =0andp m0 =8ρνLv z0 /. What, based on this profile, will be the wall shear stress? f. For arbitrary values of α (take α =5),forz>0aninstationary boundary layer with thickness δ a 0 α will exist. Show this and make a sketch of the corresponding velocity profiles at a few instants of time during the flow cycle. g. Show, making use of force equibrium on a slice with thickness dz that one can find: τ w = a ( ) 0 pm0 2 L cos(ωt + φ)+ρωv z0 sin ωt so that for given motion of the piston τ w can be determined by a pressure measurment at z =0. 3
Tentamen Cardiovasculaire Stromingsleer (8W090) blad 3/4 dinsdag 8 mei 2007, 9-2 uur 3. Endotheleal cells also orient perpendicular to the direction in which they are stretched. To investigate this, a thin walled straight elastic tube with wall thickness h, radiusa 0 and length 2L is locally covered with a monolayer of endothelial cells. The Young s modulus of the tube is E and the Poisson ratio is μ. The elastic tube is connected to a rigid tube with a piston with cross-sectional area A 0 and an oscillatory motion given by the velocity v z = v z0 cos(ωt). A z a 0 r z h L L 0 monolayer At the location z = 0 one can measure a pressure according to p m = p m0 cos(ωt + φ). The end of the tube is terminated with a characteristic impedance so at that point we have q z=l =0. a. At the location of the pressure sensor z = 0 we find a wall motion in radial direction give by: u r = u r0 cos(ωt+ φ). Show, starting from force equilibrium, that in case we neglect inertia forces in the tube wall, we have: p = σ φφh and u r = ( μ2 ) p a 0 he Make use of the fact that for thin walled tubes the circumferential stress is given by σ φφ =(E/( μ 2 ))(u r /a 0 ). b. The compliance per unit of length of the tube is defined as C =(da/dp) witha the instantaneous cross-sectioal area. Show that: C = 2πa3 0 μ 2 h E c. Due to the motion of the piston pressure and flow waves will travel along the tube. Show, starting from conservation of mass for a slice with thicknes dz and cross-sectional area A that we have: C + q z =0 d. Show, starting from equilibrium of forces in axial direction that we have: ρ q + A 0 z =0 if friction can be neglected. e. Give for the experimental situation we consider here the wave equation for the pressure and an expression for the wave speed and the wave length. 4
TECHNISCHE UNIVERSITEIT EINDHOVEN Faculteit Biomedische Technologie, groep Cardiovasculaire Biomechanica Antwoorden by het tentamen Cardiovasculaire Stromingsleer (8W090) dinsdag 8 mei 2007, 9-2 uur. Answers: a. Yes. The mean pressure and flow (zeroth harmonic) are related according to p 0 = q 0 R.() b. No, the longitudinal impedance increases with α 2 since α 2 v ( pnt) = z for large α. c. No, when the time-averaged flow is not equal to zero the entrance length of pulsating flow will be determined by the steady component. Hence, it will be equal to the entance length in steady flow. ( pnt) d. No, this is because the boundary layer is not developed and the flow may be considered frictionless. ( pnt) e. No. The boundary layer thicikness of the scondary flow will decrease δ s /a = O(Dn /2 ). Hence the shear rate will increase. ( pnt) f. Yes. C = A decreases with increasing pressure. ( pnt) g. Yes. p =ˆpe i(ωt kz) =ˆpe iωt e ikrz e kiz. Attenuation e ikiz.() h. The stress η u is continuous. Layer shows the largest shear rate so the lowest r viscosity. Yes. ( pnt) i. No. Continuity of pressure and flow at the interfavce: p i + p r = p t q i + q r = q t Y p i + Y p r = Y p t 2Y p i =(Y + Y 2 )p t Y p i Y p r = Y 2 p t so T 2 = 2Y T 2.() Y + Y 2 j. No, the rate of decrease is related to (λ γ) n. The parameter a determines at what value of the shear rate the decrease of the viscosity is maximal. ( pnt) 2. Answers: a. Rigid tube so q = q 0 cos ωt anywhere. With q 0 = A z v z0. For z > 0wehave fully developd flow so =constant cos(ωt + φ) = (p z m0/l)cos(ωt + φ). So L z p = p m0 cos(ωt + φ). ( pnt) L b. = p m0 [cos(ωt + φ)+isin(ωt + φ)] = p m0 z L L ei(ωt+φ) = p m0 L eiφ e iωt = ˆp z eiωt q = q 0 [cos ωt + i sin ωt] =ˆqe iωt = π v z0. () v c. ωv z0 = p 0 ρl r r (r v + νv z z0 r r ). Dus p 0 = ρlνv z0 (r v ). Division by νv r r z0 / gives a2 0 ω v = p 0a 2 ν 0 ρlνv z0.() z + 5
d. α large so v =. substitution of v ρ z =ˆveiωt and p =ˆpe iωt gives ˆv = i ˆp = ρω z i ( p m0 ρω L eiφ )sov = p m0 sin(ωt + φ) =v ρωl z0 cos ωt. Soφ = π/2 andp m0 = ρωlv z0. ( pnt) z + ν r r e. α small so 0 = v (r ). Integration (2 times) + v(a) =0givesv(r, t) = ρ r ( 4 ρν r2 /a 2 ). Integration over cross-sectional area gives q = πa4 0 p m0 cos ωt. z 8ρν Also q = v z0 π cos ωt so p m0 = 8ρνLv z0.() f. The boundary layer thickness is determined by the transition where viscous forces are of the smae order of magnitude as the intertial forces. So O( νv z0 δ 2 )=O(ωv z0 ). Hence δ = O(a/α). (2 pnt) g. For the segment we have ΣF = ma so: [p(z) p(z + dz)]a 2πa 0 dzτ w = ρadz v or 2 z a 0 τ w = ρ v (2 pnt) 3. Answers: a. Equilibrium of forces yields 2pa 0 =2σ φφ h. Substitution of σ φφ gives the answer. (2 pnt) b. Gebruik da = π(a 0 + u r ) 2 π 2πa 0u r en antwoord b. (2 pnt) c. Mass conservation: [q(z + dz) q(z)]dt +[A(t + dt) A(t)]dz =0so A A + q = C + q z z (2 pnt) d. Equilibrium of forces: A 0 [p(z + dz) p(z)] = ρa 0 dz v e. Cross differentiation gives C 2 p + 2 q =0and 2 q + A 2 0 z z with c 0 = A 0 /(ρc). (2 pnt) 2 p ρ z 2 + q = z q so ρ + A 0 = 0 (2 pnt) z =0so: 2 p c 2 2 0 2 p =0 z 2 6