Chapter 3 General Molecular transport Equation for Momentum, Heat and Mass Transfer

Transport process In molecular transport processes in general we are concerned with the transfer or movement of a given property or entire by molecular movement through a system or medium which can be a fluid ( gas or liquid) or a solid. Each molecular of a system has a given quantity of the property mass, thermal energy, or momentum associated with it. In dilute solution (gases) : the rate of transport very fast In dense solution: more slowly

General molecular transport equation For molecular transport or diffusion Where ΨZ = flux of the property being transferred (unit time per unit cross-sectional area perpendicular to the Z direction) δ= diffusivity (m2/s) Γ=concentration of the property

At steady state : ΨZ = constant Concentration of property, Γ Γ1 Γ2 Ψz flux Z1 Z2 Distance, Z

Unsteady state Unit area out = Ψz]z+△z In = Ψz]z z Z+△Z Rate of accumulation = Rate in + Rate out + Rate of generation

Dividing by △z and letting △z go to zero No generation General equation s for conservation of momentum, thermal energy, or mass

Momentum Transfer-fluid A. Newtonian fluid F, Force

Definition: Newtonian fluids For Newtonian fluids, the shear force per unit area (shear stress) is proportional to the negative of the local velocity gradient. That is, τ yx = flux (shear stress) of x-directed momentum in the y direction ν = μ/ρ = momentum diffusivity in m2/s μ = viscoisity in kg./m.s 1cp = 1×10-3kg/m.s = 1×10-3N.s/m2 = 1×10-3 Pa.s

Ex: referring to previous figure, y=0 Ex: referring to previous figure, y=0.5m, vx=10cm/s, and the fluid is ethyl alcohol at 273K having a viscosity of 1.77cp. Calculate the shear stress and the velocity gradient. τ yx = 0.354 g.cm/s2/cm2 Velocity gradient = 20.0 s-1

Types of fluid flow and Reynold number Re＜2300  Laminar flow Re＞2300  Turbulent flow Reynold number Re＝

Mass balance applied in fluid dynamics Normal to surface Consider a control volume shown below:

This is the net mass efflux from the control volume The rate of mass efflux through area dA: v(dA cos) Define mass velocity or mass flux: G＝v Take vector to be unit normal vector to area dA, the mass efflux may be expressed in the form of vector: v(dA cos)＝(dA) cos＝( )dA Integrating over the entire control surface This is the net mass efflux from the control volume

This is equation of continuity The rate of accumulation of mass with the control volume: Integral form of the law of conservation of mass: For a steady state flow, there is no accumulation in the control volume, i.e. This is equation of continuity

EXAMPE Steady one-dimensional flow At steady state : m=1A11= 2A22

Overall Momentum Balance Linear moment vector  = M (kg.m/s) F= force(N, kg.m.s2) Sum of forces acting on control volume = （rate of momentum out of control volume）－（rate of momentum into control volume）＋（rate of accumulation of momentum in control volume）

EXAMPLE :Finding the force exerted on a reducing pipe bend resulting from a steady state flow. As shown in the figure below: Specify control volume first

External forces imposed on the fluid include: Pressure forces at section  amd . Body force due to the weight of fluid in the control volume. The force due to pressure and shear stress, Pw and w, exerted on the fluid by the pipe wall, and is designated The external force: x-direction: ＝P1A1－P2A2cos＋Bx y-direction: ＝P2A2sin－W＋By Net momentum flux: x-direction: ＝(v2cos)(2v2A2)＋(v1)(1v1A1) ＝(v2sin)(2v2A2) y-direction:

Mass flow rate 1v1A1＝2v2A2＝ Perform momentum balance: x-direction: P1A1－P2A2cos＋Bx＝(v2cos)(2v2A2)＋(v1)(1v1A1) Bx＝ (v2cos－v1)－P1A1＋P2A2cos y-direction: P2A2sin－W＋By＝(v2sin)(2v2A2) By＝ (v2sin)－P2A2sin＋W

Force exerted on the pipe, say is the reaction to , ie Rx＝ (v1－v2cos)＋P1A1－P2A2cos Ry＝ (v2sin)＋P2A2sin－W

Momentum transfer application Flow past immersed objects and packed and fluidized beds

Drag force FD Reynold No.