 When we simplify the Reynolds transport theorem for energy, we are going to be plugging in a beta value of the specific energy of the system and a b value of the total energy of our system. That looks like this. And this is too much to talk about all at the same time, so let's start with the term on the left. This dE dt of the system is a relationship that we already know. That is the same dE dt that we encountered in thermal 1. So we can write the dE dt of our system as being equal to the rate of energy entering minus the rate of energy exiting. And I know what you're thinking. You're thinking, John, hey, why did you only write heat transfer and work here? Why didn't you account for the mass entering and exiting and the energy that that brings with it? Well, because I'm already accounting for that on the right hand side of our equation here. So I'm just looking at the dynamic energy terms that don't include mass. That's heat transfer and work for our purposes. Therefore, I'm writing the rate of energy entering as the sum of heat transfer rate entering plus power entering, and exiting energy rate is written as Q dot out plus work dot out. And then because our focus in fluid mechanics is often on what's happening within the fluid and the movement of the fluid and less on heat transfer and work, we can kind of shorthand those a little bit. We're only going to be keeping track of one type of heat transfer at a time, if any, and likely only one or two types of work. So instead of keeping track of Q dot in and Q dot out and work dot in and work dot out separately, we will instead just keep track of net heat transfer in and net work out. That's net heat transfer in is Q dot in minus Q dot out and net work out is work dot out minus work dot in. So plugging those in to our simplification of the Reynolds transport theorem for energy yields net heat transfer in as a rate minus net work out as a rate of our system is equal to the EDT of the time rate of change of our integral of specific energy of our system times density with respect to volume plus the integral across the control surface of specific energy of our system times density times the velocity vector with respect to area. Remember that specific energy of our system, just like it was in thermo, is internal energy plus specific kinetic energy plus specific potential energy. The only difference here is that in order to try to keep track of all of the different U's and V's and stuff, we add hats to the internal energy and a hat to the enthalpy. So U hat is internal energy within fluid mechanics. Therefore, lowercase e is going to be U hat plus one half times velocity squared plus gravity times z, that's height measured relative to something, and enthalpy will be H hat and that's U hat plus pressure times specific volume, which again in order to try to avoid the confusion that comes along with so many different types of V terms, we often write as internal energy plus pressure divided by density. Furthermore, we are only going to be considering three types of work, so we'll make that substitution now. For our purposes, the only work that exists within a fluid mechanics problem is shaft work, pump work, and what we call viscous work. Shaft work is separate, so we'll leave it, but we can plug in the pump work for now. That pump work is the integral across the control surface of pressure times velocity with respect to area, and when I plug those all in and then break out the pump work as it's actually represented, bringing it over to the right-hand side of the equation allows me to group together my control surface integrals, and then I will plug in what e actually represents, that's U plus kinetic energy plus potential energy, and remember that that term on the right, e plus pressure over density, is our enthalpy term, we get this. I can then start to simplify it for some common situations that we're going to encounter. If I have uniform flow, that is also steady state, I can write the net rate of heat transfer in minus the shaft power out minus the viscous power out is equal to the sum in the outward direction of mass flow rate times the quantity specific enthalpy plus specific kinetic energy plus specific potential energy minus the summation in the inward direction of mass flow rate times specific enthalpy plus specific kinetic energy plus the specific potential energy. This looks a lot like something you would see in a thermal problem, doesn't it? For uniform flow that is steady state and only has one inlet and one outlet, then the mass flow rates are all the same and I can get rid of the summations and substitute in those quantities with respect to just two state points. For now, I'm arbitrarily defining my state points as state one and state two, whereas state one is the inlet state point and two is the outlet state point. Then I'm left with the specific enthalpy plus the specific kinetic energy plus the specific potential energy of state one is equal to the same term for state two minus specific heat transfer. Remember that's a net heat transfer in plus the specific shaft work plus the specific viscous work. Then for convenience, you'll see later I'm going to divide everything by gravity and remember that enthalpy represents internal energy plus pressure times our specific volume term. Therefore I can write p1 over density times gravity plus internal energy over gravity plus specific kinetic energy divided by gravity plus z1 is equal to p2 over density times gravity plus internal energy as state two divided by gravity plus v2 squared over two times g plus z2 and then I subtract the specific heat transfer divided by gravity plus a quantity that I call head. That head term is the specific work divided by gravity so that would be plus the shaft head and viscous head. The quantity of head that's the specific work divided by gravity is a way of kind of normalizing the specific work that allows you to keep track of it in one dimension and that one dimension is useful when you're also talking about other quantities in one dimension like how you can represent a pressure in terms of just a height of a column of fluid. So you might hear things like the pump head is 10 meters or something like that. Again that'll make more sense once we start getting into examples with context. Then we define another term for convenience that's another head term that's friction head which for now is just going to be something that we either are given or something we neglect. Later on we will actually learn how to quantify it separately but that friction head is internal energy at state 2 that's the end state point remember minus the internal energy at state 1 the endless state point minus the specific heat transfer that's a net term in the inward direction all divided by gravity. Okay then the next simplification that we make is to ignore viscous work entirely for now and we will split the shaft work into two terms. We have the inlet direction and the outlet direction which we are calling pump work or pump head and turbine work or turbine head which allows us to write this. Then the last simplification we make is to account for what happens when our flow is not a perfectly distributed profile. When it is laminar it has a parabolic profile that affects our kinetic energy term therefore we add in a quantity called alpha to account for that. So our conservation of energy equation is pressure over density times gravity plus alpha times velocity squared over 2 times gravity plus z1 is equal to pressure over density times gravity plus alpha times velocity 2 squared over 2 times gravity plus z2 plus the friction head plus the turbine head minus the pump head. And when we have laminar flow we are plugging in an alpha value of 2 and for turbulent flow we are plugging in an alpha value of 1. That is also a simplification. I mean technically it's anything that isn't laminar isn't 2 but for now it's just 1. If it isn't laminar it's turbulent and we call it an alpha value of 1. Furthermore we haven't actually talked about how to determine whether the flow is laminar or not so for now while we're in chapter 3 anything that doesn't specify just assume it's turbulent flow. Furthermore neglect friction head unless you have it in the problem or are asked to calculate it. So if it helps here's our three generalizations of the conservation of mass, conservation of momentum and conservation of energy term so far. You can think of these as our equations that we're going to be applying to our control volumes.