Instructor’s Solutions Manual for Transport Phenomena in by George A. Truskey, Fan Yuan, David F. Katz

By George A. Truskey, Fan Yuan, David F. Katz

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Dc(µ)a(Vf)b(dt)c (l)(m l-1t-1)a(lt-1)b(l)c For Dc For mass a=0 For length 1+b+c = 0 For time b=0 c = -1 Thus the dimensionless group with ρ is: Dc/dt which is the diameter ratio. 11. (a) For two-dimensional flow of an incompressible fluid, the conservation of mass yields: ∂ vx ∂ vy + =0 ∂x ∂y When the fluid contacts the sold surface the no slip condition results in fluid deceleration in the x-direction. Since the partial of vx with respect to x is negative for 0 ≤ y ≤δ, then ∂ vy >0 ∂y Since the derivative is positive within the boundary layer, vy is positive throughout this region.

The second dimensionless group can be found, using the same approach, to be RcRP-1. 10. The force F acting on the leukocyte is a function of the following variables: F =f(dc,Dt, vc, vf, Hct, ρ, µ) While there are eight variables, the hematocrit Hct is dimensionless. So there are 7 dimensional variables and three characteristic dimensions, mass, length and time. Thus, in addition, there are four other dimensionless groups. If we choose the basis group as (µ)a(Vf)b(dt)c, the four other variables to make dimensionless are F, ρ, Dc and vc.

We do this assuming that the turbulent profile is uniform. If the flow is laminar, we need to recalculate assuming that the centerline is two times the average. 0286 cm2/s) = 8,741 turbulent A more accurate estimate of the Reynolds number can be obtained by computing the average velocity for turbulent flow. R ⎛ ra ⎞ v max 2π ∫ ⎜ 1 − a ⎟ rdr R ⎠ 2 ⎛ R2 R2 ⎞ ⎛1 1⎞ 0⎝ vz = v = − ⎟ = 2v max ⎜ − ⎟ max 2 2 ⎜ πR R ⎝ 2 a+2⎠ ⎝2 7⎠ 55 ⎛ 7 2 ⎞ 5 v z = 2v max ⎜ − ⎟ = v max ⎝ 14 14 ⎠ 7 So the Reynolds number at position 1 is 5,000 and the Reynolds number at position 2 is 6,244.

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