 Next, what we will do, we'll take a look at a pump, and what we're going to do is assume that we're dealing with an incompressible fluid, which is quite often what you have with a pump. And if you recall, for steady flow devices, we came up with an equation for reversible work in, and it was specific volume times delta P. So if it is incompressible, I'm able to pull the specific volume out of the integral sign, and I end up with specific volume times P2 minus P1. That becomes an expression for reversible work in. Now using a very similar equation to what we just saw for the compressor for the component efficiency, we can say the component efficiency of a pump then becomes a ratio of the isentropic work P2 minus P1 divided by H2A minus H1. And so that would then be the component efficiency for a pump. And the last thing that we are going to look at is a nozzle. A nozzle is a device that accelerates a flow. It takes a high pressure and it converts that high pressure into kinetic energy. So looking at our first law, and so that's adiabatic, we do not have heat transfer there, the potential energy we can drop. However, I'm not going to be able to drop the kinetic energy out of this equation. Now one question, does a nozzle do work? It has fixed boundaries that aren't moving. The answer to that question is no, a nozzle does not do work, and consequently the work term drops out. And what we end up with is an expression. Now I have to make a couple of approximations here. One is that velocity on the inlet is approximately equal to zero. Velocity on the exit of a nozzle, because remember we're going from pressure and a stationary fluid and accelerating it, it does not equal zero. However, for a nozzle, for adiabatic, q dot is equal to zero and w dot is equal to zero. So with that, we cannot get rid of the kinetic energy. And what we find the first law translates into, the mass flow rate will cancel out because work has gone away on the left-hand side. But what we end up with is h1 minus h2 is equal to the velocity. And I have to be careful not to confuse it with specific volume. Velocity squared divided by 2. So that becomes the first law. Now looking at the component efficiency, you know, right over here is that we have room because I still want to put the Mollier diagram on. We have actual kinetic energy at the exit divided by isentropic kinetic energy at the exit. And it turns out the way to be able to express that is v2 actual squared divided by v2 isentropic squared. So that is our expression for the efficiency. Now what we have is we have the actual and the isentropic. We can write those out. So we have h1 minus h2a equals velocity 2a squared over 2. We have h1 minus h2 isentropic equals velocity 2 isentropic divided by 2 squared. And this only works if v1 is approximately zero. Now looking at this on the Mollier diagram, we're going from p1 to p2 in our nozzle. If it was an isentropic process, it would come down like this to state 2s from state 1. In reality, we increase an entropy and our process would look like this. Again, the entropy values, or sorry, the enthalpy would be h2s, h2a, and h1. And with that we can write out that the efficiency of our nozzle is v2a squared divided by v2s squared. And you can express that then in terms of the enthalpy change. And I'll box that off. So those are four different component efficiencies that you can use when you're doing cyclone analysis. And when you do the cyclone analysis, you evaluate the isentropic component for that particular step. And then you apply the component efficiency to modify the isentropic value to one that is the actual value.