In: Chemistry
Use the general integral balance equation to derive an integral balance equation for the concentration, c, of some substance that diffuses through boundaries of the control volume. The substance may also be produced or consumed within the control volume by chemical reactions
(a)Define the surface fluxes and volumetric source/sink and explain your choice
(b)Write out the final form of the integral balance
(c)Explain what additional information or models would be required in order to solve this equation for the volume-averaged concentration, {c}.
Can someone help me with this, not sure how to
attempt.
Integral Balance Equation - Basically Integral Balance Equation is the equation for conservation of mass,energy and momentum.
As we know, Mass can neither be created nor be destroyed , it can only be conserved. So, this paricular principle is applied to a fixed volume in the space of an arbitrary shape that contains fluid. This fixed volume is called as Control volume and the fluid is permitted to enter or leave the control volume.
In short , Integral Balance Equation is defined for fluid filled body.
So, here in this question , we have been asked to derive an intergral balance equation for the substance ( of concentration C) , that diffuses through the boundaries of control volume or we can say which is permitted to leave the control volume.
The equation for such a substance is as followed - gghjfysfsdgd
Let's say we have a fixed control volume V1of arbitrary size and shape. And as we know that fluid flows across the control volume. Then a general balance law for a quantity G ( G can be any quantity like Mass, momentum , Energy) can be written as:
Accumulation of = Transfer of G in V1 through the surface of V1 + Other effects that transfer G into V1 G into V1 by virtue of fluid flow independent of the fluid flow
Now, If G is total mass, then:
Accumulation of mass in V1 = transfer of mass into V1 by fluid flow + 0 (no other effects)
across the surface of V1 ......eq no. 1
Note: The external effects for mass transfer are zero because mass can only be transferred by virtue of the fluid flow.
The physical meaning of
equation this above mentioned equation no.2 is:
the rate of accumulation of mass inside a control volume V1 (left hand side) is equal to the net influx of mass through the surface bounding V1 (right hand side). The parts of the surface of V1 on which - v • n is positive correspond to fluid inflow , while the parts over which - v • n is negative correspond to fluid outflow.
Since, the velocity is normal to the area (steady state) A1 of the entry port, - n • v simply evaluates to - v*cos180o = v*,
where v* is the magnitude of the fluid velocity across the entry port.
NOTE -The 180o comes about because v and n point in opposite directions, so that the angle between these two vectors is 180o .
Since v* and the density were assumed constant across the inlet,
the resultant integral
simply evaluates to
1v*A1 ............... eq 5
Evaluating the integral over the exit port in an identical manner results in
1v1A1
-
2v2A2 = 0 ...........................eq 6
Integration over the exit port gave the negative term -
2v2A2 since - n • v evaluated to -v2. Note -the
outlet port, n and v are pointing in the same direction.
1v1A1
is the mass flow rate into the control volume across area A1, while
2v2A2 is the mass flow rate out of the control volume across area
A2.
Equation (6) states the obvious result that, at steady state, the mass flow rate of fluid into the control volume is exactly counterbalanced by the mass flow rate out of the control volume, so that no accumulation occurs.
This is the Integral Balance Equation for Mass for a substance ( concentration C) , which flows across the control volume.