Introduction to Biological Transport Phenomena Transport Phenomena

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Introduction to Biological Transport Phenomena

Adapted From:

Transport Phenomena Byron Bird, Warren Stewart, and Edwin Lightfoot Chapter 3

Bioengineering Fundamentals Ann Saterbak, Ka-Yiu San, Larry McIntire Chapter 4

John P. Fisher

Transport Phenomena

Introduction

• Mass Transport • Solute diffusion through membrane bound channels • Renal separation of ionic components

• Momentum Transport • Blood flow through the arterial system • Interstitial fluid flow

• Energy Transport • Cardiac energy • Skeletal muscle energy

1 Transport Phenomena

Basic Laws and Connectivity Between Phenomena

• Mass: Fick’s Law of Diffusion J = −D∇C

• Momentum: Newton’s Law of Viscosity τ = −µ∇v

• Energy: Fourier’s Law of Heat Conduction q = −k∇T

Transport Phenomena

∇ Operator

• The vector differential operator, ∇, known as del, is a vector operator and it cannot stand on its own, but must operate on a scalar, vector, or tensor

∂ ∂ ∂ rectangular coordinates ∇ = δ +δ +δ x ∂x y ∂y z ∂z

∂ 1 ∂ ∂ cylindrical coordinates ∇ = δ +δ +δ r ∂r θ r ∂θ z ∂z ∂ 1 ∂ 1 ∂ spherical coordinates ∇ = δ +δ +δ r ∂r θ r ∂θ φ r sinθ ∂z

2 Transport Phenomena

∇ Operator

• The vector differential operator, ∇, known as del, is a vector operator and it cannot stand on its own, but must operate on a scalar, vector, or tensor

∂v ∂vy ∂v rectangular coordinates ∇ ⋅v = x + + z ∂x ∂y ∂z

1 1 v v cylindrical coordinates ∂ ∂ θ ∂ z ∇ ⋅v = (rvr )+ + r ∂r r ∂θ ∂z 1 1 1 ∂v spherical coordinates ∂ 2 ∂ φ ∇ ⋅v = 2 (r vr )+ (vθ sinθ )+ r ∂r r sinθ ∂θ r sinθ ∂φ

Transport Phenomena

∇2 Operator

• The vector differential operator, ∇2, is known as the Laplacian operator

2 2 2 2 ∂ s ∂ s ∂ s rectangular coordinates ∇ s = + + ∂x2 ∂y2 ∂z 2 1 ∂ ∂s 1 ∂2s ∂2s cylindrical coordinates 2 ⎛ ⎞ ∇ s = ⎜r ⎟ + 2 2 + 2 r ∂r ⎝ ∂r ⎠ r ∂θ ∂z 1 s 1 s 1 2s spherical coordinates 2 ∂ ⎛ 2 ∂ ⎞ ∂ ⎛ ∂ ⎞ ∂ ∇ s = 2 ⎜r ⎟ + 2 ⎜sinθ ⎟ + 2 2 2 r ∂r ⎝ ∂r ⎠ r sinθ ∂θ ⎝ ∂θ ⎠ r sin θ ∂φ

3 Mass Transport

Law of Conservation

• Consider a stationary volume element

• Rate of mass Acc = Rate of mass In - Rate of mass Out

• Mass unit: ρ x v x a • ρ = density (m/l3) • v = velocity (l/t) • a = area (l2) • ρ x v x a [=] m/t

Mass Transport

Law of Conservation

• Consider a stationary volume element

• Rate of mass Acc = Rate of mass In - Rate of mass Out

• Mass unit: ρ x v x a • ρ = density (m/l3) • v = velocity (l/t) • a = area (l2) • ρ x v x a [=] m/t Δy

Δz

y z x Δx

4 Mass Transport

Law of Conservation

• Consider a stationary volume element

• Rate of mass Acc = Rate of mass In - Rate of mass Out ρvy ρv • Mass unit: ρ x v x a y+Δy z • ρ = density (m/l3) z+Δz • v = velocity (l/t) • a = area (l2) • ρ x v x a [=] m/t

ρvx ρvx x x+Δx

ρvz z ρvy y

Mass Transport

Law of Conservation

• Solving gives us the Law of Conservation ∂ρ = −(∇⋅ ρv) ∂t ρvy ρv • Rearrangement and using the substantial derivative, D y+Δy z z+Δz Dρ = −ρ(∇⋅v) Dt

• Assuming constant density (∇⋅v)= 0

ρvx ρvx x x+Δx

ρvz z ρvy y

5 Mass Transport

Law of Conservation for a Binary Mixture

• Consider a stationary volume element

• Rate of mass Acc = Rate of mass In - Rate of mass Out

• Here, the fluid consists of a binary mixture of two components, A and B

• In addition, A may also may be produced through a 3 chemical reaction, rA [=] g/cm s

• Rate of mass A Acc = Rate of mass A In - Rate of mass A Out + Rate of mass A Production

Mass Transport

Law of Conservation for a Binary Mixture

• As before we can develop the Law of Conservation

∂ρ A + (∇⋅nA ) = rA ∂t

∂ρB + (∇⋅nB ) = rB ∂t

• Combining returns

∂ρ = −(∇⋅ ρv) ∂t

• Assuming constant density

(∇⋅v)= 0

6 Mass Transport

Diffusion Through the Lipid Bilayer

• Consider the case of a lipid soluble molecule diffusing through a cell’s lipid bilayer

• The Law of Conservation states that the rate of change of a species is equal to its accumulation due to transport, assuming no production and constant density (∇⋅v)= 0

• Using Fick’s Law (J=-D∇C) to describe our mass velocity then gives us (∇⋅(− D∇C))= 0

• Simplifying by assuming constant D provides Fick’s Second Law

D∇2C = 0

Mass Transport

Approximations for Solute Diffusion Coefficient

• The Stokes - Einstein equation is based upon Brownian motion of a particle with a Stokes frictional coefficient of f, which can be approximated for a spherical molecule kT kT D = = f 6πµa where k is Boltzmann’s constant (1.38x10-23 J/K), T is absolute temperature, π is the constant pi, µ is solution viscosity, and a is the radius of the solute

• The radius of a spherical molecular as well as its molecular density, ρ, may be used to approximate a molecule’s molecular weight 4 M = πa3ρ 3

thus providing D as a function of molecular weight

1/3 kT ⎛ ρ ⎞ D = ⎜ 2 ⎟ 3µ ⎝ 6π M ⎠

7 Mass Transport

Solute Diffusion Coefficients

Mw D water D plasma (g/mol) (cm2/s) (cm2/s) urea 184x10-7 146x10-7 creatinine 122x10-7 87x10-7 glucose 180 90x10-7 uric acid 121x10-7 75x10-7 sucrose 71x10-7 59x10-7 fluorescein 332 74x10-7 inulin 24x10-7 22x10-7 dextran 4,000 15x10-7 dextran 16,000 30x10-7 24x10-7 dextran 73,000 7x10-7 dextran 536,000 5x10-7 dextran 2,000,000 5x10-7 albumin 68,000 11x10-7 human IgG 150,000 6x10-7

Mass Transport

Diffusion Through a Membrane diffusion cell • With different assumptions, Fick’s Second Law can also be written as

∂C reservoir 1 = −D∇2C + p reservoir 2 ∂t

• Consider a steady state system where a membrane membrane separates two large reservoirs of a dilute chemical,

with a concentration of CL on the left (x = 0) and C = CL CR on the right (x = L), the above equation reduces to 2C

∂ concentration D = 0 C = C ∂t 2 R length • Assuming a “perfect source” and a “perfect sink”, C = CL we can solve the above to find

(CR −CL ) D C = CL + x and J = (CL −CR ) L L

C = CR

8 Momentum Transport

Conservation of Momentum

• In the same manner as we developed a mass conservation equation, one can also develop a momentum conservation equation

• Rate of momentum Acc = Rate of momentum In - Rate of momentum Out + Sum of Forces Acting on System

• Note that momentum has two components • Convection • Molecular transport

Momentum Transport

Conservation of Momentum

• Using this approach, we can develop x, y, and z components of the momentum equation in rectangular coordinates

• For example, the x component is the following

⎛ ∂v ∂v ∂v ∂v ⎞ ∂p ⎛ ∂τ ∂τ yx ∂τ ⎞ x v x v x v x xx zx g ρ⎜ + x + y + z ⎟ = − −⎜ + + ⎟ + ρ x ⎝ ∂t ∂x ∂y ∂z ⎠ ∂x ⎝ ∂x ∂y ∂z ⎠

• Assuming a Newtonian fluid with constant ρ and µ ⎛ ∂v ∂v ∂v ∂v ⎞ ∂p ⎛ ∂2v ∂2v ∂2v ⎞ ⎜ x v x v x v x ⎟ ⎜ x x x ⎟ g ρ⎜ + x + y + z ⎟ = − + µ⎜ 2 + 2 + 2 ⎟ + ρ x ⎝ ∂t ∂x ∂y ∂z ⎠ ∂x ⎝ ∂x ∂y ∂z ⎠

9 Momentum Transport

Conservation of Momentum

• Alternatively, we can also develop the r, θ, and z components of the momentum equation in cylindrical coordinates

• For example, the z component is the following

⎛ ∂vz ∂vz vθ ∂vz ∂vz ⎞ ∂ρ ⎛ 1 ∂ 1 ∂τθz ∂τ zz ⎞ ρ⎜ + vr + + vz ⎟vx = − − ⎜ (rτ rz )+ + ⎟ + ρg z ⎝ ∂t ∂r r ∂θ ∂z ⎠ ∂z ⎝ r ∂r r ∂θ ∂z ⎠

• Assuming a Newtonian fluid with constant ρ and µ ⎛ ∂v ∂v v ∂v ∂v ⎞ ∂ρ ⎛ 1 ∂ ⎛ ∂v ⎞ 1 ∂2v ∂2v ⎞ z v z θ z v z ⎜ r z z z ⎟ g ρ⎜ + r + + z ⎟ = − + µ⎜ ⎜ ⎟ + 2 2 + 2 ⎟ + ρ z ⎝ ∂t ∂r r ∂θ ∂z ⎠ ∂z ⎝ r ∂r ⎝ ∂r ⎠ r ∂θ ∂z ⎠

θ r

Momentum Transport

Fluid Flow Between Parallel Plates

• Consider a volume element in a fluid between two parallel plates separated by a height, H. The bottom plate is stationary, while the top plate is pulled to the right at a velocity of V0. The fluid between the plates will also begin to move to the right due to the momentum imparted by the upper plate. Here, we wish to describe the velocity of the fluid as a function of position (height) between the plates.

V0 y

H x

• Our two boundary conditions state that the fluid next to the stationary plate is a rest (Vx = 0 at y = 0) and the fluid next to the plate in motion moves at the same velocity as the plate (Vx=V0 at y = H).

• Using this approach, we find that the velocity profile between the two plates is linear with position V V = 0 y x H

10 Energy Transport

Conservation of Energy

• To develop the conservation of energy equation, we will use a simpler approach and consider the modes of energy transport • Internal energy • Potential energy • Kinetic energy • Work flow • Heat added

• Grouping together the inlet and outlet flows of these modes of energy transport, allows the construction of a simple energy balance of a system

⎛ P P ⎞ m Eˆ − Eˆ + m Eˆ − Eˆ + m⎜ i − j ⎟ − m PdVˆ + W − f = 0 ( P,i P, j ) ( K,i K, j ) ⎜ ⎟ ∫ ∑ shaft ∑ ⎝ ρi ρ j ⎠

• From this, we may develop the Bernoulli equation

⎛ 1 P ⎞ ⎛ 1 P ⎞ W f ⎜ gh + υ 2 + i ⎟ − ⎜ gh + υ 2 + j ⎟ + shaft − = 0 ⎜ i 2 i ⎟ ⎜ j 2 j ⎟ ∑ m ∑ m ⎝ ρi ⎠ ⎝ ρ j ⎠  