Download Chapter 3 General Molecular transport Equation for Momentum, Heat

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
driving force
rate of a transfer process 
resistance
d  For molecular transport or
 Z  
diffusion
dz
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
z2
1
1
2
 z z dz    d 
z 
 1  2 
z 2  z1
Concentration of property,
Γ
Γ1
Γ2
Ψz
flux
Z1
Distance, Z
Z2
Unsteady state
Unit area
out = Ψz]z+△z
In = Ψz]z
z
Z+△Z
Rate of accumulation = Rate in + Rate out
+ Rate of generation
  


z 1  z z 1  z z  z 1  Rz 1
t
Dividing by △z and letting △z go to zero
  z

R
t
z
2

 
 2  R
t
z

 2
 2
t
z
General equation s for
conservation of momentum,
thermal energy, or mass
d
 z  
dz
No generation
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,
 x  
d  x  
F
F
 
  
A
z
A
dz
dv x kg.m / s momentum
 yx   


2
dy
m s
m2  s
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=0.5m,
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
vD
Reynold number Re＝

Mass balance applied in fluid dynamics
 Rate of mass 


efflux
from


control volume


 Rate of mass 


flow into  

control volume


 Rate of accumulation


of mass with

0
 control volume 


Normal to surface
Consider a control volume shown below:
The rate of mass efflux through area dA: v(dA cos)
Define mass velocity or mass flux: G＝v
Take vector n̂ to be unit normal vector to area dA, the
mass efflux may be expressed in the form of vector:


v
v(dA cos)＝(dA) n̂ cos＝( v  n̂ )dA
Integrating over the entire control surface

 ( v  n̂ )dA
This is the net mass efflux from the control volume
C .S.
 Rate of mass 
 Rate of mass 




 efflux from    flow into  
control volume
control volume





 ( v  n̂ )dA
C .S.
The rate of accumulation of mass with the control volume:

t
 dV
C.V
Integral form of the law of conservation of mass:

t
 dV
C.V

   (v  nˆ )dA  0
C.S .
For a steady state flow, there is no accumulation in
the control volume, i.e.

 (v  nˆ)dA  0
C.S .
This is equation of continuity
EXAMPE Steady one-dimensional flow

  (v  nˆ)dA

  (v  nˆ)dA
A1
A1
    1 v1 dA1
    2 v 2 dA

At steady state    (v  nˆ )dA  0
cs
  dA   dA  0
A1
A2
At steady state : m=1A11= 2A22
2
Overall Momentum Balance
• Linear moment vector  = M (kg.m/s)
 
 d P d mv
 F  dt  dt
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）
 F    (  nˆ)dA 
A.

 dV

t
V
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 B
The external force:
x-direction: Fx ＝P1A1－P2A2cos＋Bx
y-direction: F ＝P2A2sin－W＋By
y
Net momentum flux:
x-direction:

 v x ( v  n̂ )dA ＝(v2cos)(2v2A2)＋(v1)(1v1A1)
C .S .
y-direction:

 v y ( v  n̂ )dA ＝(v2sin)(2v2A2)
C .S .

At steady state   v dV  0
C.V .
Mass flow rate 1v1A1＝2v2A2＝ m

Perform momentum balance:
x-direction:
P1A1－P2A2cos＋Bx＝(v2cos)(2v2A2)＋(v1)(1v1A1)
Bx＝ m
 (v2cos－v1)－P1A1＋P2A2cos
y-direction:
P2A2sin－W＋By＝(v2sin)(2v2A2)
By＝ m
 (v2sin)－P2A2sin＋W



Force exerted on the pipe, say R is the reaction to B, ie R   B
Rx＝ m
 (v1－v2cos)＋P1A1－P2A2cos
 (v2sin)＋P2A2sin－W
Ry＝ m
Momentum transfer application
• Flow past immersed objects and packed and
fluidized beds
• Drag force FD
• Reynold No.
 V02
FD  C D A p  
 2

N Re 
V0d p 


V0d




