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

2. 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

3. 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

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

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

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

7. Momentum Transfer-fluid
A. Newtonian fluid
F, Force

8. 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

10. 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 = g.cm/s2/cm2
Velocity gradient = 20.0 s-1

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

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

13. 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

14. 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

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

16. 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）

17. 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

18. 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:

19. 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

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

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

22. Drag force FD
Reynold No.