0:00 Welcome to Lesson 5a, Conservation of Mass. In this lesson we'll use the Reynolds transport theorem to generate the equation of conservation of mass for a control volume. We'll show some simplifications and we'll discuss mass flow rate, volume flow rate, and do some example problems. Conservation of mass for a system is dm system dt equals zero. The mass of a system has to stay constant since by definition mass does not enter or leave a system. Let's apply the RTT from the previous lesson. Let's consider a fixed control volume. for simplicity. Here's our Reynolds transport theorem for some arbitrary parameter B system where little b is capital B over M, or capital B per unit mass. For conservation of mass, let capital B be the mass itself and then little
0:45 b is 1. First we write the system equation, but since these two are equal when capital B is M, the Reynolds transport theorem also applies directly. Setting B equal 1, the control volume term becomes this and the control surface integral becomes this. All we've done is write the Reynolds transport theorem with capital B equal m and little b equal to 1. And we've inserted the system equation for conservation of mass. Right. The side of this equation is our conservation of mass equation for a fixed control volume. I'll call that equation 1. The left part of this equation is the conservation of mass equation for a system So the Reynolds transport theorem has helped us transform from a system equation to a control volume equation. And that's all there is to it. Now let's explain what these terms mean. The first term in blue is the
1:33 rate of change of mass within the control volume. You can also write it d dt of m control volume, where this integral is m control volume. It's the mass in the control volume, which can change with time in general. This term circled in green is the net rate of mass flow out of the system through the control surface. Let's think about the signs, if this term in blue is positive, the term in green must be negative. Why? Because these two terms have to add up to zero. If mass is increasing within the control volume, hence the plus term here, there must be a negative rate of mass flow out of the system. In other words, mass is flowing into the system. Similarly, if the blue term is negative, the green term must be positive. If the volume term here is zero, which means it's a steady flow, then the only way to keep these two terms adding up to zero
2:18 is for this control surface integral to also be zero. So for steady flow, there's no net mass flow out of the control volume. In other words, what goes in must come out. Let's talk about mass flow rate. If this is our control volume, and we're looking at some portion of that, I'll call it AC. We could study a differential element of that we'll call DAC, which has some outward unit normal n with some velocity passing through that little element DAC. We can write the mass flow rate through this whole area AC as the integral over that area, rho v dot n DAC. Note that m dot has dimensions of mass per time. It's a mass flow rate, in this case through this surface AC. And by our sign convention, m dot AC is the mass flow rate out of area AC. When you do these
3:05 vector dot products, the sign takes care of itself. If mass is flowing out, as in my diagram, the dot product of V dot N will be positive. Therefore, M dot AC will be positive also. So this is an outlet. If the mass flow rate is in, we have velocity flowing into the surface instead of out of it. But the outward normal is the same if we're talking about the same AC. In that case, V dot N is negative. And this is an inlet Now let's let AC be the entire control surface. And m dot across the control surface is integral over the control surface rho v dot n dA. Some authors put a circle around this integral to indicate that we're integrating over the entire control surface. We don't do that in the notation in our textbook. This is the second term in our conservation of mass equation for a control volume, which was equation one. It's this term that
3:55 we had circled in green. So following the logic of our element AC of area, this is the net rate of mass flow out of the control surface. Again, the signs will take care of themselves because of the nature of the dot product. It's a net mass flow rate out, so the portions where this integral is positive will be outlets, and the portion where it is negative will be inlets. In most of our problems, we usually have well-defined inlets and outlets. So we can split this integral of equation 2 into several parts. one for each inlet and outlet. For example, suppose we have a problem where we make a control volume cutting through these four pipes, where the top two are inlets and the bottom two are outlets. And this is some kind of a device that we're analyzing. In a case like this, the good news is that we don't have to integrate over the entire control surface. Instead, we
4:40 let m.cs equal the sum over all the outlets of m. minus the sum over all the inlets of m. Where we let m. be positive for both inlets and outlets, and the sign is taken care of by putting including in this negative sign for inlets. This is the net mass flow rate out of the control surface. Then our equation 1 simplifies to d dt of the volume integral equals sigma m dot in minus sigma m dot out, where we have put this control surface term over to the right side of the equation. Since this volume integral is the mass of the control volume, this equation simplifies to d dt of the mass of the control volume equal the sum of m dot in minus out. This form is the easiest one to understand. The term on the left is the rate of change mass within the control volume. And the terms on the right represent the net rate of mass
5:25 flow into the control volume. Be careful with the signs. Since we put our control surface integral on the right, this is now the net rate of mass flow into the control volume rather than out of the control volume. The negative sign is on the outlets now instead of the inlets. Let's look at the special simple case for steady flow for which the time derivative term goes to zero and then sigma m dot in must equal sigma m dot out. This is the steady flow conservation of mass equation for a fixed control volume. And this is the form we'll use most often in this course. Now let's talk about volume flow rate. Again for some little area AC and a fluid element DAC. We have our unit outward normal, some velocity vector passing through that little element, and the area of the element is DAC itself. We define V dot AC as the integral over area AC times V dot N
6:14 DAC. This is the volume flow rate out of area AC. Again the signs take care of themselves because of the nature of the dot product. In fact this equation is the same as what we had for m dot except the row is missing. Quick comment about our notation. Since V is volume, V dot is volume flow rate, analogous to M equal mass and M dot is mass flow rate. Some other authors use Q for volume flow rate. We will use this notation in these videos. Let's also define an average velocity through area AC, which is actually a speed, the magnitude of velocity. We define V average AC as 1 over AC times this integral up here. So this integral is V dot AC. So V average AC is V dot AC over AC. but you still have to do this integration to calculate v.ac. For
7:00 a given inlet or outlet, we can make this approximation. This will make our algebra easier. m dot AC, the mass flow rate through AC, is approximately the average density times the average velocity times AC. Again, I use the word velocity when I really mean speed. And instead of writing that subscript AVG all the time, we often just write it this way without the subscripts. From the above equation, VAC is just V dot AC. So m dot AC is just rho times V dot AC. In other words, mass flow rate is just density times volume flow rate. Finally, we can simplify further if the flow is incompressible. In other words, is approximately constant and steady, our simplified steady conservation of mass equation for our control volume with well-defined inlets and outlets reduced to this equation. Well now we can write m dot as rho v dot,
7:48 but if the flow is incompressible, the densities cancel, and therefore sigma in v dot equals sigma out v dot. In other words, the net rate of volume flow in must equal the net rate of volume flow out. Fluid can't accumulate inside the control volume because it's incompressible and we have a fixed control volume with some inlets and some outlets. outlets. So again, whatever comes in must go out. This is in terms of both volume flow rate and mass flow rate. Now we're ready to do some examples. We'll do an unsteady example and then a steady example. Consider a tank of volume V. Air is pumped into this rigid tank. with a constant m.in, we'll assume isothermal conditions. In other words, the process is slow enough that the air in the tank remains at the same temperature. There are no outlets, just one inlet. So we know that density is going to have to go up. At time t equals zero,
8:35 density is rho-naught. We want to generate an equation for density in the tank as a function of time. First we draw an appropriate control volume. You'll get tired of me saying this. But this is really the most important step in any of these kind of control volume problems. In this case, I draw a control volume along the inside wall of the tank and slicing across the inlet. This is a fixed, non-moving control volume. Now we use an appropriate conservation of mass equation. In this case it's unsteady so we must keep the unsteady term. d dt of this integral over the control volume equals sigma m dot in minus sigma m dot out. Notice that I used the simplified version of this control surface integral since we have well-defined inlets and outlets. In fact, in this case we have no outlet, so that term goes away.
9:24 Let's assume rho is rho of t only. In other words, density changes with time, but density is the same anywhere inside this tank. It's not a function of space. Well, that allows us to take this density outside of the integral since density is not not a function of spatial variables. So the left hand side becomes d rho dt times the integral over the control volume dv. But since this is just a volume integral of dv, this integral is just v itself. And this volume is constant since the tank is rigid. Our equation reduces to d rho dt equal m dot n divided by v. In this particular problem, We also said that m.in was a constant. So everything on the right side of this equation is constant. That allows for easy integration. We separate variables and write the integral. Both of these are trivial integrals.
10:12 Rho minus Rho not equal m.in over v times t minus 0. Or finally, Rho equal Rho not plus m.in over v times t. Since these are constants, density increases linearly with time. This makes sense if you think about how this problem is set up. We have a constant mass flow coming in so the density has to increase linearly with time. If m dot were not constant, the integral would get more complicated. But the procedure is the same. Now let's do a steady flow problem. We have steady, incompressible two-dimensional flow of some liquid between two very long parallel plates. The plates extend a long distance into the page, making this two-dimensional. We have a nice rounded inlet, and the velocity profile at the inlet 1 is nearly uniform. We'll call it U1. And we give its value. So U1 is 3 meters per second.
11:01 V and W are both 0. At some downstream location 2, the flow is what we call fully developed, which we'll talk about in detail in another lesson. And I give this form of the equation for the velocity profile, where A is some constant and H is given, H is the height between the plates. We have to calculate this constant A and speed u max, which occurs at the middle. the plates. Note that U is only a function of Y. When it's fully developed it means that U is no longer a function of X. This profile shape stays the same if we continue further down field. And nothing is a function of Z into the page. Again the first step is to pick some kind of Y's appropriate control volume. I definitely want to slice through the inlet and I want to slice through 2 as my outlet. And we enclose the control volume between the walls. Now we pick an
11:49 appropriate conservation of mass equation. Here the flow is steady and incompressible, so we can use the volume flow rate form of conservation of mass. Sigma v dot in equals sigma v dot out. There's only one inlet, 1, and there's only one outlet, 2. At the inlet, the volume flow rate is the speed times the cross-sectional area, or u1 times b times h. where B is the width into the page. For the outlet we don't have uniform flow so we need to integrate. We integrate over the cross sectional area at 2 which we'll call A2. So the right hand side becomes integral over A2 V dot n times dA. Where dA is some little section of this outlet area. It has a unit outward normal and it has a local V vector which varies with y. This little distance is dy so dA is equal to
12:37 B dy. at any y location from the bottom wall. We have to evaluate v.n. The velocity vector is u2, 0, 0. And we're taking the dot product with n, which is 1, 0, 0. So this just gives us u2. dA, as I said, is b times dy. So this area integral becomes the integral from y equals 0 to h, u2, times b dy. Well, b is just a constant, so it cancels on both sides. And we're given an expression for u2. at this location too. So u1h equal the integral from y equals 0 to h, a y h minus y dy. You can integrate this and substitute in two limits. Do that on your own for practice. You should get u1h equal a h cubed over 6. One of the
13:24 h's cancel and we can solve for a. a is 6u1 over h squared. We plug in u1 and h, and we get our answer for a. u max occurs at y equal h over 2. We can see that from our diagram. u max is right at the center, halfway up where y equal h over 2. Using this equation again, u max is a times y, which is h over 2, times h minus h over 2, which becomes a h squared over 4. If you plug in this expression for a, and do a little algebra, you'll find that Umax is 3U1 over 2. Our final answers to three significant digits are A equal 1151 over meter second and Umax is 4.50 meters per second. We'll use conservation of mass over and over again in our control volume analyses.
