OK, in this video we're going to continue with the result that we got last time for the
integral form of the conservation of mass.
So remember, here we're looking at an arbitrary volume like this potato.
In three dimensions. And we said that we can express
the conservation of mass in this arbitrary volume
in terms of two integrals.
And these two integrals represent the accumulation term, d(rho)/dt, integrated
over the volume, is the accumulation of mass
inside this volume.
And the integral over the surface of rho * (v dot n) * dA that represents that net flow of mass
through the surface area, into and out of the volume.
So this gives us an expression that we can apply to an arbitrary volume. So let let's apply it to a specific problem
that would be closer to the kinds of
things that we would be interested in
for chemical engineering fluid mechanics. And so in particular, a lot of the
problems that we deal with are related to flow through pipes, or different kinds of channels.
So imagine this situation. It's kind of a crazy pipe, but it's just a geometry.
And we're in two dimensions here, so we have the sidewalls.
And then these dashed lines are open surfaces. I'm going to call these two stations: #1, an entrance on the left hand side.
And the exit on the right hand side I'm going to call #2. And we're going to consider steady state incompressible flow
through this geometry.
So it's one dimensional flow in the sense that it doesn't vary.
The velocity is uniform over the cross section, as you see here.
So I draw
the flow going into this pipe.
The velocity is uniform over the cross sectional area, with some value V1.
And then the flow exits here at #2,
with some velocity V2.
And I also need to express the orientation of the surfaces. So surface #1 has a normal vector n1.
And surface #2 has a normal vector n2. By convention, we define the normal vectors such that they're pointing
from inside to outside
of the system, or the control volume, that we're examining. So these are outward pointing normal vectors.
OK, so again we are assuming that the velocity is constant over the cross sectional area.
So that's I've shown here. I tried to denote that the velocity profile is flat.
So let's apply the conservation of mass to this system.
So remember we have the accumulation term: integral of d(rho)/dt over the volume,
plus the integral over the area of rho * (v dot n) * dA, equals zero.
Steady state, nothing is changing with time. So this accumulation term goes to zero.
And so we can cancel that out.
And let me redraw the geometry here on the next slide so I can continue.
We have the flow coming in at some velocity V1, and uniform flow out at some velocity V2.
So we can look at the
remaining term, the flow term. So notice that we said this was incompressible flow.
So the density is constant, that factors out of the integral.
And since this term is equal to zero.
Then this cancels out. And so we're left with the integral over the area of (v dot n) * dA equals zero.
So let's evalute this now over the two surfaces
where we have inflow or outlflow. Or where we can have inflow or outflow.
And so that is surface #1 and surface #2. So we can take the integral od (v dot n) * dA over each of these surfaces.
And notice that for surface #1, the velocity V1
is in the opposite direction as the normal vector. So the flow is coming into the control volume.
And the normal vector is pointing outward. So they have opposite signs. The normal vector points in the opposite direction of the velocity.
For surface #2, notice that the velocity, the flow, is leaving the control volume at this surface.
And the normal vector also points outward.
So n2 and V2 are in the same direction. So the signs are the same in this term.
And since the velocity is constant over the cross section, then we can make
an important assumption that allows us to basically pull that out of the integral.
And so we just have minus V1 times the integral over surface #1 dA. Plus V2 times the integral over surface #2 dA.
So the integral of dA over the area is just the area, A1. And similarly for A2.
So minus V1 * A1 plus V2 * A2 equals zero.
So setting these two equal, I get an interesting result. I get that V1 * A1 equals V2 * A2 equals ** A
So remember that the product of the velocity
times the cross-sectional area is the volume flow rate. Cubic meters per second.
So conservation of mass
over this kind of control volume
that is like flow through a pipe gives me the result that the product
of the velocity times the cross-sectional area is constant.
For the case of incrompressible, steady state flow. Remember the assumption that we made.
So incompressible flow through a pipe
which is, remember we said that can
be assumed generally for most liquids,
the volume flow rate is constant.
OK, so what if the velocity is not uniform over the cross-sectional area. So instead of having a flat velocity profile, we could have a
parabolic profile, or some other kind of curve. Well, we can still use this relationship.
This relationship still holds. But instead of the uniform velocity, we can define an average velocity.
And take the average velocity over the cross section. So the mathematical definition of an average
is the integral over the cross-sectional area
of whatever quantity you are trying to average. So in this case we are averaging the normal component of the velocity.
So the average velocity is one over the cross sectional area times the integral of (v dot n) * dA.
Or the flow rate over the cross-sectional area.
So the relationship still holds if the velocity is not uniform. But instead of this constant velocity, we use the average velocity.
Calculated using the mathematical definition of an average.