Page 13 - Bio Engineering Approaches to Cancer Diagnosis and Treatment

10 CHAPTER 1 Introduction

exposed to a shear force. However, different fluids reveal different rates of deforma-
tion when a specified shear force acts on them. According to Newton’s law of viscos-
ity, for common fluids such as water, oil, and gasoline the ratio of shear stress existed
in fluid layers to the velocity gradient is equal to a property of fluid named dynamic
viscosity (for more details, readers are referred to Ref. [47]):
τ = µ
/
τdu/dy=µ du dy (1.1)
τ τ is shear stress, µ is dynamic viscosity, and du/dy is velocity gradient.
Such fluids, that shear stress linearly varies with velocity gradient, are catego-
rized as Newtonian fluids. On the other hand, for non-Newtonian fluids, shear stress
does not vary linearly with velocity gradient.
In order to have a comprehensive understanding of fluid mechanics problems,
differential analysis of fluid flow is necessary. Therefore, fluid mechanics governing
equations, conservation of mass, and momentum must be analyzed.
Conservation of mass: (also called continuity equation) in vector notation is [47]:
∂ ρ +∇ ρ ( ⋅ V) = 0
∂ρ∂t+∇⋅ρV=0 t ∂ (1.2)
ρ is the fluid density, V is velocity vector, and ∇ is gradient vector. First and second
terms in Eq. (1.2) denote the rate of change in fluid density and net rate of mass
outflow per unit volume, respectively. Density for incompressible fluids is constant.
Therefore Eq. (1.2) becomes:

∇⋅V=0 ∇⋅ V = 0 (1.3)
Conservation of momentum: based on Newton second law the conservation of
momentum is [47]:
∂V
ρ + V ⋅∇ V + =∇⋅−PI + τ ] (1.4)
g
)
(
[
ρ∂V∂t+ρ(V⋅∇)V+g=∇⋅[−PI+τ] ∂t
In which P is pressure and τ is stress tensor and I is identity matrix. The first
and the second terms on the left-hand side of Eq. (1.4) denotes local and convective
acceleration and the inertia force per unit volume, respectively. Based on Stokes
theorem, shear stress for a Newtonian fluid is expressed as [48]:
2
τ = 2 µ − µ V ( ⋅∇ I ) (1.5)
S
τ=2µS−23µV⋅∇I 3
S is the strain-rate tensor and is [1]:
1
S = (∇ +∇V ( V) T ) (1.6)
S=12∇V+∇VT 2
By plugging Eqs. (1.5), (1.6) into the momentum equation (Eq. 1.4), it becomes [47]:
 2  ∂V
)
)
V
g
∇⋅−PI + µ(∇ +∇V) T )− µ ( ⋅∇ I += ρ + ρ ( ⋅∇ V (1.7)
V
(
V
∇⋅−PI+µ∇V+∇VT−23µV⋅∇I+g=ρ∂V∂t+   3   ∂t
ρV⋅∇V