The Atmosphere, the Ocean, and Environmental Change

GG 140 - Lecture 5 - Earth Systems Analysis (Tank Experiment)

Chapter 1: System Analysis (Tank Experiment) [00:00:00]

Professor Ron Smith: We're going to be talking about a simple example of systems analysis. I put the term up here. Almost everything we deal with in science, engineering, even everyday life, has some elements of system analysis to it. The simplest kind of a system, like the one I have here, consists of a reservoir, where you store things, and inputs and outputs.

With that definition, I think you can imagine a wide variety of things that might resemble this. For example, the heat budget of the Earth, which we'll be speaking about next week, you can store heat in the Earth's solid parts and the atmosphere. You can have heat coming from the Sun, which is the input. You can have heat radiating to space. And it's the nature of that systems analysis that will help to understand the climate of the Earth. If you look at the hydrologic cycle--for example, let's look at the atmospheric part--how much water vapor do you have stored in the atmosphere? What are the inputs? What are the outputs? Only by understanding that way can we really understand how water vapor is handled in the Earth's atmosphere. The same thing for lakes. A lake has a storage capability. It has inputs. It has outputs. The carbon cycle, there's carbon in the atmosphere stored there, inputs and outputs.

In other words, almost everything we deal with in life has some elements of a simple system like the one we're dealing with today. So this is meant to be--nothing special about it. It's just an analogue for many of the things that we'll be dealing with in this course and you may run into in other circumstances as well.

So we have to first understand those three components, the storage--the reservoir itself--the input, and the output. So this is a thin tank made of Plexiglas. And the dimensions you will need, the dimensions are here. The inside dimensions of that are 59 centimeters in length, two centimeters in width, and 25 centimeters in depth. So if you need to compute the volume of this, you can do it easily from those dimensions.

It has a little pipe—a little hole in the end at this end--you can't see it from where you're sitting--a short little pipe, a valve, and then it flows out into the sink and down into the drain. We'll need to know that that valve is about six centimeters below the bottom of the tank. I'll explain why that's important in just a moment. The input is this little rubber hose right here. It's controlled by a faucet, and I can change the rate at which water is flowing in just by twisting that valve. And then the output comes out the valve. And that's a controllable valve, but I'm going to leave it set where it is for right now, and we're going to just leave it and deal with it in the form that it's in.

Any questions yet about this simple system I have here?

Now, the inputs and the outputs, we're going to use capital Q for this. It's the volume flow rate. And it'll have units of milliliters per second. Or if you like, a milliliter is a cubic centimeter. It's a centimeter by a centimeter by a centimeter. And the way we're going to measure that--and Patrick has agreed to help me with this. If you can come on up, try that stopwatch, try starting it, stopping it, and then resetting it over here. OK, is that going to work?
So I'm going to pull out this pipe. This is a 400-milliliter container, 400-milliliter container. So we're going to time how long it takes to fill that up, and then we will just get the flow rate from that. So I will give you a countdown. Three, two, one, start. And as soon as that reaches the top there, you can stop.
So the flow rate coming out of the valve right now is--Q is 400 milliliters divided by--

Student: 7.41

Professor Ron Smith: 7.41. Can you do that calculation for me? 400 by 7.41.

Student: 53.98.

Professor Ron Smith: 53.98. And the units on that are going to be milliliters per second. So we're going to be doing this over and over again with different flow configurations. Now, you can sit for a few minutes. It'll take a while now to get this set.

I'm going to try to determine the nature of the valve. How easily is the valve letting water out of the tank? That's going to be a critical part of our understanding of this system. And I'm going to tell you that Qout will take the form of some constant times the square root of Zeffective, where Z is the depth of the water. I think you would expect that the deeper the water, the more pressure there is trying to push water through the valve, and so the faster Qout will be, the larger Qout will be.

So we have to determine the--well, this constant. This constant will depend on how I have that exit valve set. And if I were to change it, I'd have to redo the calculation of K again. But I'm going to leave it like it is. We're going to let this come to a steady state for a minute, and then we'll do a calculation.

Now, what is Zeffective? If the water depth is here, we measure Z as the height from the bottom of the tank up to the water top, and then we add six centimeters to it. So Zeffective is Z plus six centimeters. Because that gives us the depth all the way to the valve itself, and that's what's going to be pushing the water out. The point is, even when the tank is empty, if there's a little bit of water in that pipe, that vertical pipe, you're still going to be getting some water running out until all the water in the pipe is gone. And then of course, Zeffective would be zero, and then of course Qout would be zero.

So we're going to determine this. How are we going to do it? We're going to simply solve this for K. So it's going to be Qout divided by the square root of Zeffective. So, let’s do,Patrick, let's do this calculation again now. Let's do the Qout first. Are you ready? Are you set? One, two, three, start.

Student: 8.37.

Professor Ron Smith: So what is Qout? You do the--you do that.

Student: 47.79.

Professor Ron Smith: 47.79. Could you now measure the depth of this, please, in centimeters. Come around and measure it from the top of the plastic there--not from the table, but from the top of the plastic--and get the depth.

Student: It's about 10 centimeters.

Professor Ron Smith: 10 centimeters. So that means Zeffective is going to be 16. So if you could do one more calculation for me, then. I need to get 47.79 divided by the square root of 16. Well, that's easy, isn't it? The square root of 16 is four, so it's--

Student: 11.95.

Professor Ron Smith: 11.95. So the valve constant K--thanks, we're done. So the valve constant K for that particular valve setting is 11.95. The units on that are weird. You'd have to work out the units from the flow rate here and the fact that the Z is inside a square root sign, but we'll just leave it as a number without the units on it. So we've characterized the inflow controlled by my faucet. We've characterized the reservoir, characterized by its dimensions. And we've characterized the outflow by finding this valve constant K. So now we know for any fluid depth, we can compute what the outflow will be. So we're off to a good start.

Now, we have to do a little bit of calculation to protect against disaster here. So let's try to compute from these formulas what flow rate--what's the maximum flow rate I could use before the tank would overflow? So let's see. We're going to use—we’re going to assume a steady state, in which case Qin will equal Qout. And so I'm going to specify some particular Qin. And then the Qout is going to be given by this, the valve constant times the square root of Zeffective.
Now, what's the maximum value Zeffective could be? It's the depth of the tank, 25 centimeters, plus the six centimeters to get down to the valve. So this is going to be K times the square root of 31, if I've done that calculation correctly. And could you do that for me, Patrick? What's that? Use the valve constant here and then times the square root of 31.

Student: 66.53.

Professor Ron Smith: 66.53. So we're going to have to watch out for that number. We're not going to want to put more water in there than that value. That's in milliliters per second. We predict a faster flow rate than that would make the whole system overflow.

So now let's see if we can predict the level that the water will come to when I change the input flow rate. So let’s--I'll take this out of here. Let's see. I'll decrease it a little bit. We'll do a measurement, Patrick, on the new flow rate here. You ready? One, two, three, start. What's the time on it?

Student: 10.75.

Professor Ron Smith: Say again?

Student: 10.75.

Professor Ron Smith: 10.75. What's that flow rate?

Student: 37.21.

Chapter 2: Steady State Calculations [00:13:35]

Professor Ron Smith: 37.21. So here's a proposal. If I stick that flow back in the tank, what water depth will we get when it reaches a steady state? Let's do the calculation first. Well, no, I guess we can plug it in there and get it running. It'll take a while for this to work out. So let's do the calculation. We're going to now solve for Zeffective.

So again using the steady state assumption, I'm going to say Zeffective will be Qin over K squared. Is that right? Yeah, I've just taken this and solved for Zeffective. So that's going to be 37.21 over 11.95. And we will square that and see what we get for our prediction for the water depth. Can you do that for me?
Student: 9.70.

Professor Ron Smith: 9.70 centimeters. Well, now, that doesn't look like it's going to work out too well, does it? Let's do a quick measurement here. I'll do it. Has that reached a steady state? Yeah. Right now, we've got three centimeters in there, so our prediction looks pretty poor at this stage. Oh, wait. That was Zeffective. So Z we predict would be six centimeters less than that, so that would be 3.7. My Z Measured was three. That's a pretty good prediction, actually. I forgot about that offset.

Well, let's do it one more time just to be sure this system works. I'm going to crank up the flow rate a little bit, and we're going to repeat the experiment. One, two, three.

Student: 6.06.

Professor Ron Smith: And what's that for a flow rate?

Student: 66.0.

Professor Ron Smith: 66.0. I remind you of our warning. Well, let's give it a try, shall we? It looks like that's right on the margin of what's going to happen. So we will take a look and see what happens with this 66. The units again on this are milliliters per second.

While we're doing this, we might talk about a few other elements of the problem. Let's see. Would anything up here change if I made this tank wider? What if this tank, instead of being two centimeters wide, were 10 or 20 or 30 centimeters wide? Would, let’s see, would this calculation for the maximum possible flow rate change? It's a wider tank. The reservoir is much larger. It can hold a lot more water. Would that still be the maximum flow rate the system could handle, or would it change? Any thoughts on that? What do you think?

Student: [INAUDIBLE]?

Professor Ron Smith: Yeah, nothing would change. However, it would take a lot longer to reach the equilibrium state. If you get the water up to a new depth or to drain it down to a new depth, it would take--instead of just a few minutes, it could take maybe 20 or 30 minutes if that tank were wider. But in terms of the equilibrium states, for this particular system, it is independent of this depth.

Remember, the pressure that is pushing water out the bottom depends on the depth only. It doesn't depend on how wide the tank is. That's the key. This formula here does not need to take into account the horizontal dimensions of the box.

So the equilibrium states would be--now, that's a very interesting result and something that would be interesting to understand in all the other contexts that we'll approach in this course, like the water budget, the heat budget, the CO2 budget. What is the role of the size of the reservoir on the equilibrium states and/or the rate at which you approach the equilibrium states? In this case, the width does not influence the equilibrium states, but it does influence the rate at which you approach an equilibrium state.

Now, it looks like we've almost reached an equilibrium there. And I wonder--it doesn't look like we're in danger of overflowing. So let's do a quick test to see if maybe the flow rate changed. Now remember, there are errors in this calculation, everything we're doing. For example, there's a potential error in this valve constant, K. If that's off by a few percent--we just did one test. We didn't do a lot of tests and find the best value. We just did one quick test.

So when I see something that doesn't quite make sense like this, I'm wondering, well, maybe there's an error in the valve constant. But maybe also there's been some change in the rate at which we're putting water in. So, Patrick, let's just check that flow rate. In fact, let's do it off the exit. We'll do it off the exit. One, two, three.

Student: 6.34.

Professor Ron Smith: And what's that?

Student: 63.10.

Professor Ron Smith: So that's a little bit less than we thought. Maybe the system's not quite in steady state. Maybe it's still filling. Or maybe the line pressure in the building has changed slightly, and there's a little bit less water coming into the system. But it is pretty high. It may still be rising slowly. And so I think we're not too far off in our estimate that a 66 milliliter per second thing would bring it pretty much to the top of the tank. So I think I'd call that a success, even though we haven't quite proven that it's going to come right up to the very top.

I'm going to talk about a few other aspects of this. But before I do that, I want to see if you have questions on how this system works. We've looked at it from several points of view. There must be something, some part of it, you didn't fully grasp when I went through it. So let's have some questions about what's really going on here. Anything?

Student: I have a question.

Professor Ron Smith: Yeah, question.

Student: What, exactly, is K, again?

Professor Ron Smith: The question is what is K? I'm calling it the valve constant. I've got an exit valve here. Let me lift it up and show it to you, if I can do this. You see that valve there? Uh-oh. Sprung a leak. Sorry about that. I got you down there, didn't I? It's always a danger to sit in front like that. So there's a valve there that I could adjust by hand. And if I close that valve down, K would be smaller. If I open it up more, K would be larger. It has to do with the area of the pipe that's closed off by the valve.

I can derive this formula, by the way, from Bernoulli's Law. You may have heard of Bernoulli's Law. It relates pressure and velocity in moving fluids. And here I can compute the pressure by knowing the depth of the water. And then knowing something about the velocity and the area of the valve, I can compute the flow rate. So this formula can be derived. But without knowing the area, without going in there with a microscope and measuring how much area is open in that valve, I have to treat K as an empirical constant. I have to determine it experimentally, which is what we did. Other questions?

Chapter 3: System Analogies [00:23:34]

So let's see. The formula's on the board there. Let's try to think about what some of the analogies might be with this system then. The most obvious one is with a lake. If you have rainfall coming into rivers that's providing a certain amount of input, where the lake exits into a river, there's probably some kind of a control. Maybe it's a low dam, maybe it's just rocks in the stream, but the outflow is going to be controlled by the depth of the water, typically, in a lake.
Therefore, it'll reach an equilibrium state very much the way this one does. The exit law will be a little bit different. It won't be this formula exactly. But there'll be some relationship between the depth of the water and the rate at which water leaves. Then equilibrium will be very much analogous to what we have here.
What about the transience? Well, if you get a drought, that would be equivalent to closing this valve down a little bit, and the water will continue to exit from the lake, but the lake level will drop. There's still water going in, but it's not enough to support the rate at which the water is flowing out. You might reach a new equilibrium, or depending how much I've closed that, this might actually come right down to the bottom. We'll watch and see what happens with that.
What's another analogy? How about water vapor in the atmosphere? Now, it's not a liquid anymore. It's a vapor. The input would be evaporation from lakes and from trees. The output would be rain. Now, the rain, which forms in clouds and falls to the Earth, probably the rain rate is going to be proportional roughly to how much water vapor you have.

So the loss of water vapor from the atmosphere is going to work a little bit like this too. You're going to have a constant input, which is going to be the evaporation, and then as the humidity in the atmosphere approaches 100% and you begin to form clouds with precipitation, that Qout is going to increase, and you're going to reach a steady state. And that's what we have most of the time in the Earth's atmosphere is a kind of a balance--yeah, that one almost bottomed out--kind of a balance between inputs and outputs.

Let's think about the heat budget of the Earth. So this is perhaps more difficult to imagine, but let the amount of water in that tank represent the amount of heat stored in the atmosphere and the ocean. It's getting heat from the Sun. The Earth is losing heat by radiating it to space. The hotter--of course, the amount of heat that the Earth is receiving is not particularly sensitive to its own temperature. That's just how hot is the Sun? How far is the Sun from the Earth? How much of the Sun's radiation reflects off the Earth? Let's say that's a given. Then the temperature of the Earth is going to determine, though, how much radiates to space. The hotter the Earth is, the faster will be the rate at which energy is lost to space.

So again, it's analogous to this. You're going to reach an equilibrium. The temperature of the Earth is going to rise until the rate of energy lost to space matches the rate of energy gained from the Sun. Then you'll have an equilibrium temperature for the Earth. And if the Sun were to increase its temperature--let's say that's the temperature of the Earth--if the Sun were to increase its temperature or for some other reason we were adding more heat to the Earth's system, suddenly, we'd go into a transient stage.

You wouldn't come to a new equilibrium right away. There would be a transient stage where the temperature builds up to a new value, where once again an equilibrium can be sustained. And we'll see that here. Every time I change this input valve setting, we go through a transient stage and then approach to a new steady state. So that's the way climate works on the Earth, as well.

Questions on that one? Yes.

Student: So you calculate the 66.53 for the maximum flow rate, is there a way to do that for the minimum—when it will bottom out?

Professor Ron Smith: Yes. Let's do that calculation. The question is whether we can compute the flow rate that would bring it to zero. Now, by zero, we mean empty the tank, but there'd still be some water in the little pipe down at the bottom. So Q--to minimize or to empty the tank, let's say, Qin to empty the tank would be K times the square root of the minimum value of Zeffective. When the tank is empty, Z is zero, so Zeffective is six. So I'll put six centimeters in there.
And, Patrick, could you do that calculation for me? It's the valve constant, 11.9, times the square root of six.

Student: 29.15.

Student: What is the K value?

Professor Ron Smith: Right there. So that's a good question. Now, whether that has any analogue in these other systems I'm talking about, I'm not so sure. For example, what would it mean--well, of course, a lake could empty out. We can imagine that. Could we imagine the Earth storing no heat? Not really. It'd have to be at absolute zero. So I don't think that minimum could be carried over to find an analogy in the Earth's heat budget. Probably not the maximum, either. I mean, there's no such thing as overflowing the Earth's heat budget. Where's the heat going to go? It's going to have to escape again.

So not everything here has an analogue with every system that we're talking about, but the basic elements of how an equilibrium is reached, I think that's the one that carries over fairly dramatically.

So let's imagine what would determine for the Earth's heat budget, what would determine the rate of approach to a new equilibrium? I think in this case it would be probably the heat capacity of the Earth. How much heat can you store? And so we'd have to do a calculation taking into account the thermal heat capacity of the oceans, the thermal heat capacity of the land surface, and that would determine the rate at which you'd approach the new equilibrium. Those things would not enter into what the new equilibrium state would be, but those would enter into the rate at which you reach that new equilibrium.

So these nuances--I call them nuances--they're actually of central importance. The difference between the nature of the steady state and the nature of the transient state is quite important in all of these systems.

Yes.

Student: Would there a significance in an analogy to the length of the output valve?

Professor Ron Smith: Yeah. The question is is there analogue in the systems I mentioned to the out valve? And yeah, I think there would be one, and it's one that I should've thought about to mention. If we're talking about the heat budget analogue, the heat budget analogue is the Earth receiving heat from the Sun, storing some of it, and then radiating it to space. If I add-if I have an atmosphere with a greenhouse gas and I add more greenhouse gas, then for the same temperature, it's harder to get the radiation out. So adding a greenhouse gas like CO2 is essentially like closing down this valve a little bit.

So let me try that. We'll do a little greenhouse experiment here. We'll pretend that this is an analogue to the surface heat budget of our planet. And I'm going to add more absorbing gas to the atmosphere so that as the planet tries to radiate, it can't get that radiation out because of the greenhouse gas. So I'll close this valve down. And watch what happens then. I haven't changed the input. Notice, I'm not changing the Sun. I'm not changing what radiation is coming in, but I'm changing the output. I'm changing the output law, so I've just changed K by closing that valve down a little bit.

So we're out of equilibrium. We're in a transient stage. The climate is out of balance. And it looks like I may have made a tactical error here. It looks like I've closed that valve down too much, and we're going to--of course, we have a new value for the maximum now, because I've closed the valve. It looks like I'm going to exceed that.

But anyway, that's the transient. I'm going to have to open this up a bit. I put too much greenhouse gas in the atmosphere there. So it reached a new equilibrium. It took a while to do that. But we heated up the Earth. Now it can get rid of radiation at the same rate that it's coming in, and we've achieved a new balance.

So maybe that’s--thanks for that question, because that provides another way to think about the analogy between this and the Earth and the heat budget.
Was there a question back here? Yes.

Student: [INAUDIBLE]

Professor Ron Smith: No, that's right. That's the remarkable thing. So the gases we normally think of as greenhouse gases, like carbon dioxide, methane, and water vapor, they actually don't do very much to the Sun's radiation at all. Those molecules do not absorb those short wavelengths coming from the Sun, but they do absorb the longer wavelengths that the Earth is radiating at. We'll talk about this in class next week.

So this is a pretty good analogy of that, because adding greenhouse gas doesn't change the input, but it does change the output law. It changes the relationship between heat you've stored and the rate at which you're getting rid of heat by radiating it to space. So it's a pretty good analogy to the greenhouse effect.

Yeah.

Student: But so then once it leaves the equilibrium then, you'll still have the same radiation?

Professor Ron Smith: Yes. This radiation leaving will be equal to this once again, but now the Earth will be hotter. The Earth will have to be hotter to maintain that balance. That's exactly the point. Yes. You'll go back to a steady state again where Qin equals Qout, but in order to do--just like here, the new steady state, Qin equals Qout, but now the water depth is greater. There's more heat stored in the system for this analogue, and that means the Earth is a hotter place, a hotter place to be.

Other questions on this?

Chapter 4: Plotting Qin and Water Depth versus Time [00:35:59]

Well, let me just do one sketch then to make this transient thing a little bit more clear before we quit today. I want to get this in your notes. Thinking about this tank, if I plotted versus time Qin and I plotted beneath that the depth of the water in the tank versus time, let's say we established some flow rate, and it was in steady state, and so the water depth was constant as well.

Let's do a couple little experiments here. If I suddenly increase the inflow rate, what would happen to the water depth? Does it jump as well? Well, no, we did that experiment. Actually, it begins to rise immediately. There's a kink in the curve, but not a jump in the curve. So it rises like this and then ultimately approaches a new steady state.

And as we've talked about, the time it takes to come to a new steady state, the length of this transient period in this case would depend on the width of the tank. If the tank happened to be twice as wide, say four centimeters instead of two centimeters, this curve would look something like this. It would take about twice as long to get to the new steady state.

To make matters even more complicated, what if I slowly increased the Qin? Well then, this would be even slower. But that's because I've gone slow on changing the inputs. But eventually, sometime after the new steady state for Qin was reached, eventually the new steady state for the system would be reached. So we’re going to want to understand--in the case of global warming later in the course, we're going to want to understand the rate at which we're changing the properties of the system, Qin or Qout, and the nature of the transient response.

And I think that'll do it for the day. We'll finish a little bit early, and I'll see you on Monday. Be sure to leave your problem sets here on the table as you leave.

[end of transcript]