 Steam enters a nozzle at 400 degrees Celsius and 800 kilopascals with a velocity of 10 meters per second and leaves at 300 degrees Celsius and 200 kilopascals while losing heat in a rate of 25 kilowatts. For an inlet area of 800 square centimeters, determine the velocity and volumetric flow rate of the steam at the nozzle exit. So let's parse that into pieces. First of all, we recognize that we were given a working fluid. It's steam. I will remind you that when we talk about water, we typically describe it as water, but it's very common in the thermodynamic industry to refer to both liquid and vaporous water as steam, especially during the transition period. In applications like steam power plants, steam is the description for all of the water regardless of phase. So don't read too much into the fact that the working fluid is described as steam. Don't let that imply to you that it is a superheated vapor, even though it probably is considering it's 400 degrees Celsius and 800 kilopascals. I was told in inlet temperature and pressure, those two independent intensive properties will fully define that state point. And at the inlet, I also known a velocity and a cross-sectional area. I was told at the outlet, the temperature and pressure, which again fully define state two. And I was told that throughout the process, we are rejecting heat at a rate of 25 kilowatts. All this is occurring inside of a nozzle. And a nozzle is one of our steady flow devices. If we jump over to our list of steady flow devices, we can see that a nozzle's goal is to convert enthalpy to kinetic energy. I mean, if you think about the nozzle on the end of a garden hose, for example, the goal of that nozzle is to increase the velocity of that stream of water. That increase in kinetic energy has to draw its energy from somewhere. The place that it gets its energy to accomplish that increase in velocity is the enthalpy of the fluid. Remember that we define enthalpy as the combination of internal energy and what we call flow energy, which is pressure and specific volume. So anything within that could decrease to increase the kinetic energy of the fluid. However, if we think about the fact that over the course of a nozzle, we aren't drastically changing the temperature, we're probably unlikely to change the density as much as the pressure. Therefore, the primary source of the energy is going to be the pressure of the fluid. So we're decreasing the enthalpy of the fluid to increase the kinetic energy. In this problem statement, we were told enough information to figure out how much the enthalpy decreased across the process. And with that information, we can figure out how much the kinetic energy had to increase. So that's going to be the beginning of the problem. The second half of the problem asks us for the volumetric flow rate, which we can use the velocity at the outset, the outlet of the nozzle, as well as the specific volume of the water, which we can look up using the temperature and pressure to figure out what that volumetric flow rate has to be. So let's get into it. I'll start by drawing a rough diagram of the nozzle. We traditionally draw our nozzles and diffusers as simply increasing or decreasing their cross-sectional area. And we typically draw them from left to right. So I can indicate an inlet on the left and an outlet on the right. And I will describe those two state points as one and two respectively. So those state points are referring to different positions in space, not necessarily different positions in time. Since I'm commonly analyzing nozzles and diffusers as being steady state, because there's unlikely to be much interesting change over time, especially after the beginning of the process, the transient period of the process, these two state points could be described at the same point in time. We could be talking about different water at the inlet, then at the outlet at the same point in time. So different positions in space. Those two state points have a temperature and pressure. State ones temperature and pressure are 400 degrees Celsius and 800 kilopascals, which we can use with our steam tables to look up any other property that we need. Like for example, the enthalpy and the specific volume. We also know the velocity and the cross-sectional area at state one. Those are 10 meters per second. I'm pretty sure. Yep. And 800 square centimeters. Then at state two, I know the temperature and pressure are 300 degrees Celsius and 200 kilopascals. Then again, those two independent intensive properties fully define my state point from which I can determine any other piece of information I want to know. Probably enthalpy, probably specific volume. Generally speaking, I would encourage you to get as far in the problem as you can symbolically before you start going into the property table lookups, because it can be easy to waste time and effort looking at properties that you don't necessarily need. So even though I haven't drawn out a mass balance or an energy balance yet, I've drawn out H1 and V1, H2 and V2 as properties that I'm looking up. If I go through the process and change my mind about those properties, I can always go back and remove them from my lookup list. While we're here, it's a good idea to start listing our assumptions as well. I'll start by considering the fact that I'm treating this process as steady state. So I'm neglecting the effects of time in my analysis. Now the context clues for that are the fact that steady is in the description of a steady flow device. We treat them as being steady state. Second of all, I recognize that there's not really anything interesting for the nozzle to do over time. We have an inlet, we have an outlet, and we're described properties of the inlet and outlet regardless of when we look at it. Furthermore, we know the rate of heat rejection, which is independent of time. It's 25 kilojoules of heat rejected every second, 25 kilowatts, regardless of if that's the first second or the 100th second. So steady state is the first assumption that we make. Next, I could establish that I'm treating this as an open system. That's not really an assumption that I'm making. Remember, the default state of the universe is an open system. When we simplify an open system to a closed system or an isolated system, that's really the assumption that we're making. We are neglecting some aspects of reality in our model. When we treat this as an open system, what we're really saying is we aren't neglecting mass crossing the boundary nor energy crossing the boundary. But just in the interests of being extra explicit about my assumptions, I'll list it just in case. To this list, I could add the fact that we were told a heat rejection and there's unlikely to be heat transfer going in multiple directions. And if there was a situation that was complicated enough to warrant heat transfer in multiple directions, I would probably have enough information to describe that in the problem. That special circumstance would warrant description. So I'll add to my list of assumptions that there is likely to be only heat transfer in the outward direction, therefore no QN. While I'm here, I can also probably pretty safely assume that because we weren't told any work occurring in the process, there is unlikely to be any work occurring in the process, therefore no work of any sort. Next up, I recognize that in my process, I am unlikely to have any substantial change in potential energy. Now I drew my nozzle as being horizontal, which pretty explicitly means no change in elevation. But that wasn't actually given in the problem statement. We weren't told a change in elevation. And the fact that we weren't told a change in elevation implies to us that there isn't a substantial change in elevation. So we shouldn't have to try to account for that, because we have no ability to do that anyway. So I'll add to my list, no change in potential energy. And let's see how far that gets us. So we have stay one properties, stay two properties. We've started some assumptions. Now let's set up an energy balance. The energy balance will allow us to relate the change in enthalpy to the change in kinetic energy. Let's start with our energy balance. My energy balance has to be performed on something. I'm defining my system as a control volume here, instead of a control mass, because I'm accounting for mass crossing my boundary. So I will indicate on my nozzle, my control volume, and recognize that there is mass crossing the boundary in the inward direction at state one. There is mass crossing the boundary in the hour direction at state two. And then there is also a heat rejection term. That's actually a rate of heat rejection. So I will indicate that as a capital Q dot out to indicate total rate of change of energy. If I wanted to, I could add to this, the actual amount is 25 kilowatts, but I don't necessarily have to. Lastly, while I'm here, it's pretty common to indicate the working fluid on your system diagram. So we've indicated a nozzle as evidenced by the fact that our cross-sectional area is decreasing. We have an inlet and outlet state point. We've indicated that it's water and that there's heat rejection. Now, energy balance, delta E of our control volume is equal to E in minus E out. I recognize that because I'm describing this process as being steady state, I can probably use the rate form of my energy balance. So I'm going to divide everything by dt, at which point I have the rate of change of the energy of the control volume with respect to time is equal to E dot in minus E dot out. My first simplification will come by recognizing that if I'm assuming steady state analysis, nothing can change with respect to time. The temperature at any given point can't change with respect to time. The density of the water at any given point can't change with respect to time. And in fact, the energy of the control volume cannot change with respect to time. So I will indicate that I'm neglecting the rate of change of energy of the control volume with respect to time because of assumption number one. That means the rate of energy entering has to equal the rate of energy exiting. Next, I recognize that because I have an open system, energy can cross the boundary in either direction in the form of heat transfer, work, or the energy associated with a moving mass. So E dot in could be heat transfer in it could be work in the rate of work in and it could also be the sum in of m dot theta heat transfer work and the energy associated with a moving mass. I will move that a little bit more to the left so that my equal signs remain lined up. E dot out, likewise, could be u dot out, it could be work dot out, and it could be the energy associated with the exiting mass. Next, I'll neglect terms that aren't relevant to the problem at hand. I recognize that I had assumed that heat transfer was only in the outward direction so we can get rid of Q in. That was on account of assumption number three. I'm neglecting work altogether because of assumption number four and I can simplify my summation of mass flow rate entering and exiting by accounting for all of the places where mass is entering that is just state one. So m dot one theta one and then all of the locations where mass is exiting my system which is just state two. Then I can expand theta. Theta is the combination of the enthalpy of the fluid plus the kinetic energy of the fluid on a specific basis plus the potential energy of the fluid on a specific basis. So that would be h one plus v one squared over two plus g z one. For theta two that would be h two plus v two squared over two plus gravity times z two. So plugging those substitutions in I have m dot one times h one plus v one squared over two plus g z one is equal to q dot out plus m dot two times h two plus v two squared over two an excellent looking square plus g z two. And again I will move that over to keep my equal signs aligned. My next simplification will come from recognizing that we had neglected any changes in potential energy. That was I believe assumption number five I'm just gonna write down a five all confident like and scroll over to find that I was correct. Next I recognize that for there to be an equal rate of mass entering and exiting which comes from the mass balance by the way can write that out in the same way I did my energy balance I can say the change in mass of my control volume is equal to the mass entering my control volume minus the mass exiting my control volume because I'm assuming steady state analysis it's going to be more convenient for me to use the rate form of the mass balance so I will divide all three terms by dt at which point I have the rate of change of mass of the control volume with respect to time the rate of mass flow rate entering and exiting and then I will neglect dm dt on account of the steady state assumption therefore all of the mass flow rate entering has to equal all of the mass flow rate exiting I only have one inlet state point that's state one I only have one exiting state point that is state two so m dot one has to equal m dot two which I will just start writing as m dot the only mass flow rate so back in my energy balance if m dot one is equal to m dot two it might be convenient for me to combine these terms and these terms together I could do that in a couple of ways I could subtract m dot two times h two plus v two squared over two from both sides of the equation and factor out the mass flow rate I could also divide all three terms by m dot like this at which point I would just have h one plus v one squared over two is equal to q dot out divided by mass flow rate plus h two plus v two squared over two in this equation I know v one I have the ability to look up h one I have the ability to look up h two and I know q dot out so I could calculate v two from this equation if I knew m dot luckily for us we have enough information to describe m dot we could describe m dot one by relating mass flow rate to volumetric flow rate that would be density times volumetric flow rate and then I could recognize that volumetric flow rate itself could be described as velocity times cross-sectional area then mass flow rate could be written as density times velocity times cross-sectional area or more conveniently for us velocity times cross-sectional area divided by a specific volume so we could calculate the mass flow rate at state one which is the same as the mass flow rate at state two if we knew velocity at state one which we do we were given it area at state one which we do we were given it and specific volume at state one so if we look up specific volume at state one we can calculate the mass flow rate at state one which we can plug into our energy balance along with the h one and h two which we look up at states one and two and calculate the velocity at state two so rearranging this equation I could write v two squared over two is equal to h one minus h two minus q dot out divided by mass flow rate plus v one squared over two so the kinetic energy term at state two is going to be equal to the change in enthalpy plus the specific kinetic energy term at state one and then we also account for the fact that we are losing energy over the course of the process as a result of heat transfer out so the presence of heat transfer out is decreasing how much kinetic energy there is left at the end which makes sense I could go one step further and write this as v two is equal to the square root of two times the quantity h one minus h two minus q dot out over m dot plus v one squared over two now all I need to do is look up h one h two and specifically one calculate m dot and plug everything into my equation I will also point out while I'm here in some circumstances it's easier to calculate v two squared as a quantity in meter squared per second squared and then square root that term sometimes it can be difficult to keep track of the unit conversion underneath the radical so pros and cons I prefer typically to perform as few calculations as possible so we'll leave that the way it is for now next up I can recognize that at the end of the process I'm going to be looking for the volumetric flow rate at the exit and to figure out the volumetric flow rate at the exit I'm going to recognize that I have enough information to calculate the mass flow rate at the exit so just like we did with m dot one we can look up the specific volume at state two and relate the mass flow rate at state two to the volumetric flow rate at state two with the specific volume so m dot two is going to equal density times volumetric flow rate at state two which is the same as volumetric flow rate at state two divided by specific volume at state two then the volumetric flow rate at state two would be the mass flow rate at state two times the specific volume at state two we will know the mass flow rate at state two because we will have calculated the mass flow rate at state one at this point and we recognize that they are the same quantity because the mass has nowhere else to go so once we know m dot one we know m dot two we multiply that by the specific volume at state two and we have the answer for the volumetric flow rate at the exit of the nozzle so all that's left to do really is look up h one specific volume one h two and specific volume two for that we will jump into our property lookup tables we're going to be using the properties of water so we're going to pay attention to tables a two through about a five on this list the first step is going to be fixing the phase at state one so in state one i had a temperature of 400 degrees celsius and a pressure of 800 kilopascals the easiest way to fix state one is probably to look up the saturation temperature corresponding to the pressure or the saturation pressure corresponding to my temperature and compare one to the other so i will use my pressure to look up a saturation temperature at that pressure and i will compare my temperature to that saturation temperature so in the tables if i go to my saturation tables listed in order of pressure remember that table a two and a three contain the same information or rather are referring to the same information one is given in even increments of pressure one is given an even increments of temperature so what i want is table a three on table a three i find eight bar because 800 kilopascals would be eight bar and i see that the saturation temperature corresponding to a bar is 170.4 degrees celsius my temperature is 400 degrees celsius state one 400 degrees celsius is higher than the saturation temperature therefore this must be a superheated vapor so i will go now into my superheated vapor tables in my superheated vapor tables i want to find eight bar so that i can look up my enthalpy and specific volume at that pressure unfortunately i see that i don't have happened to have a pressure sub table for eight bar instead i have a pressure sub table at seven bar and a pressure sub table at ten bar so in order to look up the property at 400 degrees celsius and eight bar i'm going to have to interpolate between the two pressure sub tables so if i go back into my property look-ups i will draw that out maybe down here property look-ups it's day one i'm using a temperature of 400 degrees celsius and a pressure of eight bar and then from my superheated vapor tables which are tables a four i'm interpolating between seven bar and ten bar so my interpolation will be eight minus seven divided by ten minus seven is equal to h one minus the specific enthalpy at 400 degrees celsius and seven bar which is 3268.7 3268.7 divided by the specific enthalpy at 10 bar and 400 degrees celsius which would be 3263.9 minus the same enthalpy at seven bar and 400 degrees celsius which is 3268.7 that proportion of the way we are between the two pressure sub tables is assumed to be the same as the proportion we are between the enthalpy at the two pressure sub tables and it is also the same as the proportion we are between these specific volumes at the two pressure sub tables so adding to this i could write v one minus the specific volume at seven bar and 400 degrees celsius which would be 0.4397 0.4397 divided by the specific volume at 400 degrees celsius and 10 bar which would be 0.3066 0.3066 minus the same specific volume from the seven bar sub table and that gives me my two proportions so i'm going to use this one equation with one unknown to calculate h one and i will use this other one equation with one unknown calculate v one so if i pull up my calculator here i am going to cheat a little bit by instead of doing the algebra i will just type out the equation symbolically and let the calculator solve for it 3268.7 divided by 3263.9 minus 3268.7 solving that for x getting parentheses okay so eight minus seven divided by 10 minus seven is equal to x minus 3268.7 divided by 3263.9 minus 3268.7 and this is incorrect because i typed the negative sign instead of minus i'm using a keyboard here to type in on a ti 89 emulator and some of the keys aren't exactly perfect so look forward to more mistakes like that 3267.1 is my enthalpy 3267.1 kilojoules per kilogram and now the same thing with specific volume so i will get rid of 3268.7 and replace that with 0.4397 i will get rid of 3263.9 replace that with 0.3066 like the same substitution for 0.4397 and i get a specific volume at state one of 0.3953 and then just for good measure let's throw a couple extra threes on we are paying no attention to significant figures okay now that i have h1 and v1 about halfway through my property lookups i can repeat the same general process at state two at state two i had a temperature of 300 degrees celsius and a pressure at state two of two bar right two bar yep two bar and from that i can perform a property lookup so just like at state one the first step is to fix the phase it's likely to still be a superheated vapor but we should check just to be sure so i will jump back into my saturation tables and again i could look up the saturation temperature corresponding to my pressure and then compare that to my temperature or i could look up the saturation pressure corresponding to my temperature and compare that to my pressure so at two bar i see that the saturation temperature is 120.2 degrees celsius my temperature at state two is higher than that because 300 is higher than 120.2 therefore it must be a superheated vapor i will jump into my superheated vapor tables again and i will scan optimistically for a superheated vapor sub table at two bar and unfortunately i don't find one instead i see that i have a pressure sub table at 1.5 bar and a pressure sub table at three bar i have a direct row for 400 degrees celsius which means that my interpolation is only going to be one-dimensional so that interpolation will look the same as it did before i will start off with 2 minus 1.5 divided by 3 minus 1.5 and that's going to be equal to the enthalpy at 1.5 and 400 that enthalpy is going to be the third column on my sub table so the enthalpy at 1.5 bar and 400 degrees celsius is 32 77.4 so h2 minus 32 77.4 divided by the enthalpy at three bar and 400 degrees celsius which would be 32 75.0 32 75.0 minus the enthalpy at 1.5 bar and 400 degrees celsius which was 32 77.4 and the proportion of the way we are between the 1.5 pressure sub table and the three bar pressure sub table is assumed to be the same for both enthalpy and specific volume so you can interpolate for specific volume using the same proportion that would be v2 minus the specific volume at 1.5 bar and 400 degrees celsius which is 2.067 generally speaking we should pay attention to the specific volume column to see if we are multiplying it by 10 to the negative third or not and we are not here so that's just going to be 2.067 divided by the specific volume at three bar and 400 degrees celsius which is 1.032 minus the same specific volume from earlier 2.067 and now can interpolate for h2 and v2 so I will start with the same solve function as earlier and I will say two correct minus sign 1.5 divided by three correct minus sign 1.5 is equal to the thing that we're solving for minus 32 77.4 divided by 32 75.0 correct minus sign 32 77.4 we're solving for x so h2 is going to be 32 76.6 kilojoules per kilogram and v2 2.067 1.032 and 2.067 specific volume to say 2 is 1.722 1.722 cubic meters per kilogram and that gives us what the property lookups that we need to perform our energy balance and calculate volumetric flow rate at state 2 let's just double check that we grab the correct numbers 400 degrees celsius and 8 bar 400 degrees celsius and 8 bar and 300 degrees celsius and 2 bar 300 degrees celsius and 2 bar then let's look at the changes in our numbers to see if that matches what we would expect we see that our specific enthalpy is going from 32 67.1 to 32 76.2 so we are increasing our specific enthalpy that is a bad sign that is an indication that we are increasing our kinetic energy and our enthalpy while rejecting heat throughout the process so we need to go back and double check our numbers to do that i will jump back into the property tables and let's first look at our superheated water vapor tables at 8 bar we were interpolating between 32 68.7 32 68.7 and 32 63.9 32 63.9 32 63.9 yeah so 32 67.1 seems reasonable i see that i should be about a third of the way between 68.7 and 63.9 the difference between those two is roughly five a third of the way of five would be about one and a half so we should be about one and a half less than 68.7 which we are which means each one is probably correct then at state two we were interpolating between 1.5 and three 1.5 and three our two values were 32 77.4 32 77.4 and 32 75.0 32 75.0 so our interpolation should be about a third of the way between 77.4 and 75.0 third of the way between the two should yield oh about one and a half are we decreasing by one and a half not quite let's just double check that interpolation here two minus 1.5 divided by three minus 1.5 is equal to x minus 32 77.4 divided by 32 75 32 75 minus 32 77.4 32 76.6 400 degrees oh right the property look up at state two is at 300 degrees celsius not 400 degrees celsius well that was a very humbling experience let's try that again this time we are looking at 280 and 320 so you know i do have the technology i could cut that out of the video but i think in the interests of demonstrating the the thought process behind verifying your numbers i'm going to leave that error in that totally intentional error for the purposes of learning yes 300 is not 400 okay so let's try that process again all of this is wrong instead i need to perform an interpolation between the 280 row and the 320 row to find our value at 300 degrees celsius and then i need to perform the same interpolation on the three bar table at 280 and 320 to find our value at 300 degrees celsius and then we're going to interpolate between those two resulting values so we are interpolating between 280 and 320 on one pressure subtable between 280 and 320 on the other pressure subtable and then we are interpolating between those two resulting values so i would describe this as an h style triple interpolation h because we are essentially drawing an uppercase h between our pressure sub tables alternatively we could also look up a value at 280 degrees celsius at two bar a value at 320 degrees celsius at two bar and then interpolate between those two values which would create the shape of an uppercase i either way works they will produce the same result but h is what i'm going to do here so our first interpolation is going to be for a value at 300 degrees celsius and 1.5 bar so h at 300 degrees celsius and 1.5 bar then we're going to perform a second interpolation for h at 300 degrees celsius and 1.5 excuse me three bar and then we're going to interpolate between those two values and that third interpolation would be 2 minus 1.5 divided by 3 minus 1.5 is equal to the enthalpy at state 2 minus the enthalpy at 300 degrees celsius and 1.5 bar divided by the enthalpy at 300 degrees celsius and 3 bar minus the enthalpy at 300 degrees celsius and 1.5 bar so we perform two intermediate lookups to be able to perform our actual lookup for enthalpy so going back to our property tables we're going to start by interpolating for a value at 300 degrees celsius 1.5 bar we are building the left leg of our h so for that we'll take 300 minus 280 divided by 320 minus 280 is equal to x correct minus sign the value of h at 280 and 1.5 bar which would be 3032.8 divided by the value of h at 320 degrees celsius and 1.5 bar which would be 3113.5 minus 3032.8 we're solving for x so our first lookup is 3073.15 3073.15 and if you're doing this correctly you should hear a cat desperately wanting to go outside that is a good sign it means you're doing thermodynamics right then our interpolation for 3 bar will be the same proportion on the left but on the right that will be 3028.6 divided by 3110.1 minus 3028.6 we are solving for x that's 20.6 oh okay try that again 28.6 1100.1 2028.6 okay 3069.35 0.35 now we have everything we need to actually perform the lookup for h2 that would be h2 minus 3073.15 divided by 3069.35 minus 3073.15 a third interpolation here that's going to be 2 minus or act minus sign 1.5 divided by 3 minus 1.5 and that is equal to x minus 3073.15 divided by 3069.35 minus 3073.15 we're solving for x 3071.88 3071.88 kilojoules per kilogram so we are going from 32,067.1 kilojoules per kilogram at the inlet to 3071.88 kilojoules per kilogram at the outlet we are decreasing the enthalpy and that energy is going into heat transfer out as well as kinetic energy increase the fact that our enthalpy is decreasing means that we are more likely to have done that correctly okay now let's repeat that process for specific volume we are going to be determining a specific volume at 1.5 bar actually you know what just to demonstrate both aspects of interpolation process let's do this i style instead of h style so what i'm going to be doing this time is determining a specific volume at 2 bar and 300 degrees celsius excuse me 280 degrees celsius and a specific volume at 2 bar and 320 degrees celsius and then we are interpolating between those two results to come up with our specific volume at state 2 and that interpolation would be 300 minus 280 divided by 320 minus 280 is equal to specific volume at state 2 minus specific volume at 2 bar 280 degrees celsius divided by specific volume at 2 bar 320 degrees celsius minus specific volume at 2 bar 280 degrees celsius so two intermediate lookups in order to be able to actually evaluate the third lookup jumping back into our property tables the first lookup goes solve and then for 2 bar and 280 degrees celsius we're going to take 2 minus 1.5 correct minus sign is important divided by 3 minus 1.5 and that's equal to x minus correct minus sign the value for specific volume at 280 degrees celsius and 1.5 which would be 1.695 let's just double check that that's not taken times 10 to the negative third it isn't and then we are dividing that by the specific volume at 280 degrees celsius and 3 bar which would be 0.844 minus the same specific volume from earlier 1.695 we're solving that for x and our first value is 1.41133 1133 cubic meters per kilogram and then we will repeat that same process so that would be solve 2 minus 1.5 correct minus sign divided by 3 minus 1.5 is equal to x minus the value at 320 degrees celsius and 1.5 bar which would be 311 excuse me specific volume john 1.819 correct minus sign 1.819 divided by 0.907 minus 1.819 solving that for x we get 1.515 1.515 now 300 excuse me 300 minus 280 divided by 320 minus 280 is equal to v2 minus the specific volume at 2 bar and 280 degrees celsius which was 1.41133 divided by 1.515 minus 1.41133 and that will give us v2 so we have 300 minus 280 correct minus sign divided by 220 minus 280 is equal to x minus 1.41133 divided by 1.515 minus 1.41133 we get 1.46317 cubic meters per kilogram so we have an h1 of 3267.1 i'm going to highlight that to make it easier to grab from the other side a v1 of 0.395333 a specific volume of 1.46317 and and there'll be at state 2 of 3071.88 and our numbers seem reasonable we are increasing our specific enthalpy and we are increasing our specific volume excuse me decreasing our specific enthalpy increasing our specific volume now we can calculate a mass flow rate at state one so that we can plug that into our energy balance so we're going to take our specific volume at state one 0.395333 and put that in the denominator with the velocity of state one which was 10 meters per second and the area at state one which was 800 square centimeters so we have 10 meters per second multiplied by 800 centimeters squared divided by 0.395333 0.395333 cubic meters per kilogram we are likely looking for a mass flow rate in kilograms per second it doesn't really matter because that isn't explicitly asked for all we're doing is taking this number and plugging it into our energy balance so we can use whichever units we prefer but kilograms per second is probably going to be easiest to do that i will convert centimeters to meters for that i recognize that there are 100 centimeters in one meter then i square everything one squared is boring 100 squared centimeters squared cancel centimeters squared meters squared and meters cancel the cubic meters in the denominator which gives me kilograms per second so i am going to take 10 multiplied by 800 divided by the quantity 0.395333 times 100 squared 100 squared really i can't do the carrot okay i guess i will do that manually squared so 10 times 800 divided by 0.3395333 times 100 squared gives me a mass flow rate of 2.35618 and let's write that here 2.3535618 kilograms per second 2.35618 kilograms per second we'll go here i have 25 kilowatts as they queued out i have 10 meters per second as a v1 of h1 and h2 from our property lookups which means we have everything we need to calculate v2 so that will go where root and then two times the quantity h1 minus h2 h1 was 3267.1 3267.1 minus h2 which was 3071.88 3071.88 kilojoules per kilogram and now let's think about what units we want underneath our radical here so we're going to be taking a quantity in specific energy and we are going to take that to the power of one half as a result of that we want meters per second so the most convenient thing for us to use under the square root sign is going to be meter squared per second squared if we get this into meter squared per second squared then when we perform the square root we will end up with meters per second so i want to get from kilojoules per kilogram into meter squared per second squared for that i will recognize that one kilojoule is a thousand joules a joule is a newton times a meter and a newton is a kilogram meter per second squared newtons cancels newtons joules cancels joules kilojoules cancels kilojoules kilograms cancels kilograms leaving me with meters times meters or meters squared divided by second squared one quantity in two more to go then i'm subtracting i will carry that two over and write that as two times 25 kilowatts divided by mass flow rate which was 2.35618 and that was kilograms per second we're going to run out of room so i'll break that into the next line a kilowatt can be written as a kilojoule per second at which point we are back to the same unit conversion as earlier a kilojoule is a thousand joules a joule is a newton times a meter and a newton is a kilogram meter per second squared kilojoules cancels kilojoules joules cancels joules newtons cancels newtons seconds and seconds cancel kilograms and kilograms canceled kilowatts cancels kilowatts leaving me with meters squared per second squared and then i add to that two again multiplied by v one squared over two n squared meters squared per second squared divided by two i want meters squared per second squared and i have meters squared per second squared and cancel the two i will continue the radical just to hopefully not forget it and at this point you can perform that calculation so i will leave my square root as parentheses with a to the power of one half at the end just to hopefully make this as easy to read on the calculator as possible so we start with two times and actually we could write that factor or i guess we can leave it be interesting i wonder why it was b anyway 32 67.1 minus third 3071.88 multiplied by 1000 that'll give me my first quantity and then we are subtracting correct some subtraction sign john two times 25 times 1000 divided by 2.35618 and then we add to that a wonder why those letters appear that's interesting two times 10 squared can't use the carrot have to use to the to the power of two and then divided by two i will just get rid of the two and i close my parentheses and raise that to the power of one half at the end and i get 607.716 let's just double check two times 32 67.1 minus the 3071.88 times 1000 minus two times 25 times 1000 divided by 2.35618 plus 10 squared all inside parentheses raised to the one half power which gives me 607.716 so the velocity at the exit of the nozzle is 607.72 meters per second then our volumetric flow rate at state two will be the mass flow rate at state two multiplied by the specific volume at state two the mass flow rate at state two is the same as the mass flow rate at state one which is 2.35618 kilograms per second and then our specific volume at state two was 1.46317 1.46317 cubic meters per kilogram kilograms cancels kilograms leaving me with cubic meters per second 2.35618 times 1.46317 1.35618 times 1.46317 gives us 3.4475 3.4475 cubic meters per second the volumetric flow rate at state two just double check that we type those in correctly 2.35618 and 1.46317 we did that's a good sign and then I will box the velocity at state two as well and there we go the velocity at the exit is just over 600 meters per second the volumetric flow rate at the end is 3.45 cubic meters per second