Derivation and Discussion of Loss Equation

Okay, now we're going to talk about the special cases for the energy equation. Okay, special cases for conservation of energy. Okay, what we did first when we were covering conservation of mass and conservation of momentum was the steady. So I’ll go and do the same here as well. Okay, so simply what it means is there's no change with respect to time. Okay, if I go up in here, okay, you will see that the first term on the right-hand side, this term, is a function of time. It looks at how this integral changes over time. The answer is it doesn't. Okay, so then I will simply go ahead and write my equation as Q dot minus W dot will be equal to control surface, rho, V square over 2 plus gz plus h, careful, not u, V dotted with n dA. And the reason why we get this equation is this term turns out to be the first term on the right-hand side turns out to be zero. V square over 2 plus gz plus u, not h, d volume, is equal to zero. Right. Okay, fine. So this was my first special case. Let's go ahead and discuss a kind of unique special case. Okay, so I want to discuss this one, and you will understand why in a minute. Okay, I want to look at steady, I want to look at a uniform flow, and I want to look at one inlet and one outlet. What I mean by one inlet, one outlet is, think of this as a single pipe or a bend. Okay, but it's not like a tee fluid component where two inlets or two exits, just a single inlet and outlet. Okay, so if I look at this equation, what will happen if this. Q dot minus W dot, so I'm going to start with this equation, and right-hand side, by having uniform flow, I will be able to get rid of the double integral and I will be going to the summation sign. Right, however, let’s note that, there is only one inlet and one outlet, so I don't really need the summation sign. Okay, so for that reason, I'm simply going to have the density times V square over 2 plus gz plus h at the exit, V n exit, A exit, minus rho, V square over 2 plus gz plus h at the inlet, V ni, A i. I have only one inlet, I only have one outlet. Okay, so we can call this exit and we can call this inlet as well. It doesn't have to be constant density, I didn't make that assumption. Right. Okay, so you may have remember, and I discussed this before as well, so what is the combination of these three multiplication terms? There is something called m dot, right, mass flow rate, and the unit of it in SI is kilogram per second. Right, m dot exit times V square over 2 plus gz plus h at the exit minus m dot inlet, V square over 2 plus gz plus h at the inlet. Let me ask you a question. I'm looking at these two terms, m dot exit and m dot inlet. Can I obtain a relationship between m dot exit and m dot inlet? And actually, I'm going to tell you the answer. It is, actually, they are the same. But let's think for a minute why they're the same. Okay, well, it's actually coming from conservation of mass. So if I write my conservation of mass for this particular case, what it will do for one inlet, one outlet, it's going to be rho exit, V n exit, A exit, will be equal to rho inlet, V n inlet, A inlet. And if you see over here, these are the mass dot exit, this is the m dot inlet. Right, so that's why they are the same. So, simply I'm going to go ahead and say that m dot exit is equal to m dot inlet is equal to m dot. I will not be treating them separately. Okay? So then if I go ahead and write Q dot minus W dot will be equal to m dot times, V square over 2 plus gz plus h at the exit, and actually, bring a large bracket so I don't have to rewrite this again, V square over 2 plus gz plus h at the inlet, and the large bracket as well. Nice. So now the question is, can I divide both sides by m dot? Yeah, why not, right? So I can. So let's do it. And I'm going to call this lowercase q, there's no dot, I'll explain why, minus w, lowercase w, will be equal to V square over 2 plus gz plus h at the exit, minus V square over 2 plus gz plus h at the inlet. Now let's talk about these two terms. What is the units? Okay, so I usually when I explain a concept, I want to talk about units as well, so that kind of helps us. So in order to do that, first we have to understand what unit of these are. Okay, if you remember from the last video, you would remember that these are in watts. Okay, this is power. Remember, it was a total power, I separated the flow power component, so this was everything else. Right, so that is in watts. Okay, and if you think about it, what is joule per second? Right, in SI. What is m dot? It is kilogram per second. Right, and I'm dividing both sides of the equation by kilogram per second. What is joule per second divided by kilogram per second? Joule per second divided by kilogram per second. Well, seconds cancel, I get myself joule per kilogram. Right. Okay, remember, work per unit mass. Okay, so you can understand from that, then my q needs to be heat transferred per unit mass. Okay, and this lowercase w will be the work, not a power, per unit mass. Okay. Now if I go ahead and express h in terms of u, I will get something useful for me. Okay, so let's look into that. So it's gonna be q minus w, okay, will be equal to V square over 2 plus gz plus P over rho plus u, right? That was the definition, this was enthalpy, right? This is for the exit minus the inlet. V square over 2 plus gz plus P over rho plus u at the inlet. Okay, so I'm gonna rearrange this equation. So I'm gonna simply, what I'm gonna do is, I'm gonna pick up this u exit from here and I'm gonna move it to the other side of the equation. Okay, if I move the plus u exit to the other side of the equation, I will get myself a negative. Right. So it's gonna be q minus u exit. Okay, I'll do a similar treatment for u inlet as well. Look at the u inlet. It has a negative sign in front of it. Right. So then when I move it to the other side of the equation, it will be a plus u i. Okay? Minus w will be equal to V square over 2 plus gz plus p over rho at the exit minus the same quantity at the inlet. The reason why I moved these internal energy components to that side of the equation is this. Okay, I have a name for the combination of all these three. Okay. And I'm gonna call this loss. Okay, so it's simply known as loss, and typically this is the loss or energy per unit mass. We just discussed that. Unit is kilojoule per kilogram. So it is per unit mass. Okay. Or energy loss per unit mass due to the, most of the times, it's a friction. Okay, I will have a loss in my system. Okay, so now I'm gonna call this, let's call this l. Okay, and I'll get l minus w will be equal to V square over 2 plus gz plus P over rho at the exit minus V square over 2 plus gz plus P over rho at the inlet. Okay, so this is a very important equation. This relates for a one inlet, one exit only though. Okay, I want you to be clear on that. I derived this equation for only one inlet, one outlet. But let's say that there is no shaft work or turbine or a pump or propeller within my control volume. Right. So, and that's kind of assuming that this is zero. The first parenthesis that I have here is the energy at the exit, and this is energy at the inlet. And I'll ask a question. What do you think the relationship between energy at exit minus energy at the inlet? Okay? So if I go from state one to state two, am I gonna gain energy or I'm gonna lose energy? That's the question I'm asking you. Okay. And if you think about, I'll give you an example from driving. Right. When I start at some location, if I go to another location, obviously when I look at the fuel gauge, the fuel gauge will go lower. Right. Because the energy is being lost. So what I mean is that the exit energy will always be lower than inlet energy. And if my w is zero, what you will obtain, l is always equal to or less than zero the way we define it. There are other resources that define this equation another way around, such as inlet minus exit. You have to be careful which version you're using. Okay. But the way that I explain this concept, you will always get a negative value. Okay. Or equal to zero as well. We'll talk about that equal to zero case shortly. Note that I did this approach for one inlet, one outlet. Okay. If I had done the derivation for multiple inlet and multiple outlet cases, okay, I would get a modified version of this equation. And I'll write it down. I will not be deriving it. You can derive it yourself. It is very similar to the approach that we've taken. But what will happen is, if I have multiple inlets and multiple outlets, I will obtain an L dot. Okay. Minus W dot. Remember, this was Q dot minus W dot. That's why I put L dot minus W dot. I'll explain differences momentarily. Okay, will be summation over the exits, m dot exit times V square over 2 plus gz plus P over rho at the exit, minus over the inlets, I will get myself m dot inlet, V square over 2 plus gz plus P over rho at the inlet. So this is the other version of this formula. Okay, so careful that. Let's put it over here. Only for one inlet, one outlet. Okay. And the unit of L will be energy per unit mass. So let's look at the second equation and relate those two. Note that the only difference is I'm multiplying the inlets and exits by m dot so that we account for what kind of, what percentage of the flow is coming from what branch. Okay. So that's why I had to do this for multiple inlet, multiple outlet cases. Okay, let me be more specific. Let's say that I have this case. Right. Let’s for a change, let's do it this way. Okay, so there's two branches that is combined and forms one flow. Right. So what m dot is gonna enable me is, it's gonna account for what percentage is flow coming from here, what percentage of the flow is coming from here. Okay. It can be 50, 50, most likely not. Right. So I will have to account, let's say that ninety percent of the flow is coming from the upper branch and ten percent is coming from the lower branch. This m dot will account for that. The second point I want to make is, what is the unit of this? Okay, or the dimension of that. So what I'm gonna do is, up there, if you remember, l was, if I'm using SI units, right, that was joule per kilogram. Right. So now I'm multiplying both sides of the equation by m dot. And if you remember, m dot was kilogram per second. Again, I'm talking about SI only. So the kilograms cancel, I get myself joule per second. What is joule per second? It's watt. So what I obtain here, this is the power loss. Okay. And this is energy loss per mass, per unit mass.