 A young engineer is designing a 1500 watt hair dryer and hopes to sell it on the open market. He or she calculates, using science and a lot of testing, that the outlet conditions for optimum hair drying occur with a temperature and velocity of 50 degrees Celsius in 20 meters per second respectively. Any hotter or faster than that, despite doing a better job of drying, might cause the customer to burn themselves or their hair, which the legal department suggests is a bad thing. The engineer assumes that under normal conditions, the ambient air will be about 22 degrees Celsius and 100 kPa. Because of the stagnant condition of the ambient air and the large size of the intake, the inlet velocity will be negligibly small. Additionally, the body of the hair dryer will be well insulated so as to avoid the whole burning the customer thing. Determine first the mass flow rate of the air through the device, that's how much mass flow rate this can process while consuming 1500 watts, and accomplishing the change in kinetic energy and enthalpy required. Secondly, determine the volumetric flow rate at the outlet of the device, and thirdly, the diameter of the outlet which would yield optimum hair drying conditions. So the first thing I'm going to do is draw an approximate diagram and indicate what I know. So I know conditions at the inlet of the device, and I know conditions at the outlet of the device. I'm going to treat the internal mechanisms as two separate steady flow devices. First a fan, which I'm assuming handles all the change in kinetic energy, and then an electric resistance heater. I'm assuming the fan handles all the change in kinetic energy, and the electric resistance heater handles all the change in enthalpy. Now, of course, in reality it's a little bit more complicated than that. They aren't going to be that exclusive of one another, but this is sort of like when you're analyzing kinematic motion and you figure out the path of a projectile by considering the x-axis and the y-axis individually. It's not as though it actually traverses only the x-axis and then the y-axis, but breaking it apart into pieces allows you to analyze one piece at a time and bring them together. In the same way, if we assume all the change in kinetic energy happens in the fan and all the change in enthalpy happens in the heater, then we end up at the same result. By treating the work input to the fan as going directly into kinetic energy, we are also accounting for any pressure increases that are actually converted back into kinetic energy inside of a nozzle, say, at the end of the shape of the hairdryer. If we draw a control volume around the entire thing, it eventually becomes kinetic energy regardless of which individual conversions happen within the internal mechanisms. Since we're assuming 100% operating efficiency, essentially, it doesn't matter how many conversions occur. Does that make sense? Excellent. So since I have two study flow devices here, I am going to indicate a state point between them, and that gives me the ability to analyze the fan and the heater individually if I want, but I can also draw a control volume around the entire body. So I have electrical power input to the fan mechanism, and I have electrical power input to the heater. Now, I know what you're thinking. You're thinking, but John, surely the heater should be a heat transfer in, right? Well, it depends on how you draw your control volume. Since I'm drawing my control volume around the body, that means that the energy is crossing my boundary in the form of electrical work. If I drew an energy balance around the heating mechanism itself, then it's entering my control volume in the form of heat transfer. Since both of them are energy rate inputs in my energy balance, it doesn't really matter whether I call it a W or a Q, because it appears as the same term anyway. Then I know the power input of the fan and the power input of the heater have to sum together to 1500 watts. So I'm assuming the velocity at state one is pretty close to zero, from the words, negligibly small. Now, of course, the velocity at the inlet can't actually be zero, or else there would be no mass flow rate. What we're saying here is that the velocity is so close to zero that when we calculate the kinetic energy at the inlet, we're saying a small number squared relative to the outlet velocity squared is basically zero. Another way to think about it is that the kinetic energy at the inlet actually still came from the fan. The fan is pulling the air, so if you expand your control volume out far enough, you're actually grabbing stagnant air. So the change in kinetic energy happens as a result of the fan anyway. Next, we are assuming the temperature at the inlet is about ambient conditions, which we were told were 22 degrees Celsius, and that the pressure is about 100 kilopascals. At the outlet, we have a velocity and a temperature, and then we can assume the pressure at the outlet is still about ambient conditions. Then at state two, the temperature at state two is going to be assumed to be essentially T1, because we're assuming only kinetic energy is increased across the fan. We're assuming all the enthalpy increase happens across the heating element. Similarly, we're saying V2 is approximately equal to V3. Remember, state two doesn't really exist. It's the intersection of the Y and X components of my projectile motion. I have the option of setting up a mass balance and an energy balance around a control volume on just the fan, around a control volume on just the heater, or around everything all at once. In the interest of really exploring the example problem to the best of our ability, why don't we try it both ways? So first we'll analyze the fan and the heater together, and then we'll analyze them individually. So I will call control volume control volume one, just the fan control volume two, just the heater. And maybe I'll color code that to make it a little bit easier to see. So blue control volume is control volume one, red control volume is control volume two, and then green maybe. Green control volume is the whole kit and caboodle. For those mass and energy balances, we're going to be setting up a simplification that looks pretty consistent. So let's just talk through our assumptions one time. First of all, we are assuming steady-state operation. The fan has been running long enough to reach steady-state. It's not as though we are analyzing the hairdryer being turned on initially. We're saying it's turned on. It has been running long enough that it has reached steady-state conditions. I could add to that that it's an open system, even though that's unnecessary, because that's not really a simplification we're making. The universe is open systems that we simplify in certain circumstances to closed or isolated systems. Next, I'll assume that the air is ideal. So I can use ideal gas properties for the air. Then I can explicitly say change in kinetic energy across control volume two is zero, and the change in enthalpy across control volume one is zero. That's what allows me to write these properties at state point two. That's not a very good-looking CV. Let me try that again. I'm going to assume the changes in potential energy are zero regardless of how you look at it. And then I will also explicitly assume that there is no power output and that the device is adiabatic. The assumption of adiabatic comes from the phrase well insulated. I was told that the hairdryer was well insulated, which means I'm neglecting any heat transfer through the body of the hairdryer. So with that set of assumptions in place, I can start my control volume analyses. So I will set up three parallel mass balances and energy balances, one on control volume one, one on control volume two, one on control volume three. Beginning with control volume three, for no reason in particular other than that's what I had said first. I have a mass balance and an energy balance. The mass balance starts with change in mass of control volume three is equal to the mass entering minus the mass exiting. The energy balance similarly begins with delta E on control volume three is equal to energy entering minus energy exiting. Because I have steady state operation, it's going to be most convenient for me to divide everything by dt or rather derivate everything with respect to time at which point I have dm dt is equal to m dot in minus m dot out. dm dt is zero because it's steady state and nothing can change with respect to time, including the mass of the control volume. Similarly on the energy balance, I will divide everything by dt, which again is really just derivating with respect to time at which point I have dedt, which is zero. Again, that's the energy of the control volume and that's zero because of assumption number one is equal to e dot in minus e dot out. Therefore the mass flow rate entering has to leave. I only have one inlet it's state one. I only have one outlet it's state three. So I can say m dot one has to be equal m dot three which I can just call m dot for simplicity's sake. Then on the energy balance, I can write e dot in is equal to e dot out. I can clear a little bit more space for myself by scooching over the mass balance. It is the technical term scooch and then I'm going to write e dot in as its full list of possibilities, which is e dot in plus work dot in plus the sum in of m dot theta because it's an open system. Any dot out could be q dot out could be work dot out and it could be the sum out of m dot theta because I have adiabatic operation. I can neglect the heat transfers because there's no opportunity for work output. I can neglect work out. That leaves me with power input power input plus the sum of all the mass full rates entering. There's just one eight state one times theta one. Remember that theta consists of enthalpy plus kinetic energy plus potential energy on the exit side. I have the sum of all the mass full rates exiting of which there is one eight state point three times theta three theta is enthalpy plus specific kinetic energy plus specific potential energy. Then I neglect changes in potential energy because of assumption number four and I'm going to factor out the mass full rates by bringing this term over to the left, excuse me, over to the right at which point I have the mass flow rate through the hairdryer multiplied by h3 minus h1 plus specific kinetic energy three minus specific kinetic energy one. Therefore the mass flow rate through the fan, excuse me, through the hairdryer can be written as the power input divided by h3 minus h1 plus specific kinetic energy three. Remember specific kinetic energy is one half times velocity squared. So I can write that as v3 squared over two minus v1 squared over two. I can figure out delta H because I know the properties of state one and the properties of the state three. I can figure out the change in specific kinetic energy because I know the velocity of state three and the velocity of state one. Velocity of state one by the way is just zero and the power input was given 1500 watts. Therefore if I take 1500 watts divided by the specific change in enthalpy plus the velocity at the exit which was 20 meters per second squared divided by two and account for the units I will eventually get a mass flow rate. That's the result of the mass balance and energy balance on control volume number three. Since I'm going nowhere near control volume order let's consider control on two next. Control volume two was red and on control volume two I'm going to start with a mass balance and an energy balance. The first couple of steps are going to be identical from control volume three so I'm going to duplicate those over and in fact while I'm at it I will duplicate those over to control volume one as well. One was blue and I can go through and change all the subscripts. This is one, this is two, that was this is TOO not this is TWO. This is one also. This one, however, is two, TWO and this one is two. So as a result of the mass balance on control volume two I can say that the mass flow rate entering control volume two which is m dot two has to equal the mass flow rate at the exit which is m dot three. Similarly I will just abbreviate that as m dot. The energy balance on control volume two could consist of the same terms because it's still an open system. So I will duplicate those over as well. The only difference being that when I'm considering just the heating element I don't have to account for any changes in kinetic energy. Furthermore to clarify a little bit I will call this power input term the power into the heater. So the change in kinetic energy across the heating element is zero and therefore the power into the heating element itself is going to be m dot three times h three minus m dot two times h two because it's the same mass flow rate. I can factor out the mass flow rate and write that as m dot times h three minus h two. Two control volumes down. One more to go. The mass balance on control volume one has one inlet. It is state point one. It has one outlet. It is state point two. Therefore I can write m dot one is equal to m dot two. It's the same mass flow rate which I can abbreviate m dot. For the energy balance on control volume one I can write that in much the same way. Except this time I don't have any changes in enthalpy because I'm assuming all of that happens in the heater. Furthermore I will call this power input the power to the fan and then I have power to the fan. It is equal to m dot two k e two minus m dot one k e one which is just m dot times k e two minus k e one which is v two squared over two minus v one squared over two. And just like on control volume three the velocity of state one is zero. So I can simplify this down to just m dot times v two squared over two. So I have a power input to my heater. Of the mass flow rate through the heater times the delta h. And I have the power input to the fan as the mass flow rate through the fan times v two squared over two. And then because I know that both of those power terms have to come from the electrical socket they must sum together to 1500 watts. We are assuming perfect efficiency and that power factor doesn't exist. So I could combine these together and write this as 1500 watts is equal to the power input to the fan plus the power input to the heater. And then make the substitutions power input to the fan was m dot times v two squared over two. Power input to the heater was m dot times h three minus h two. I could simplify this by factoring out the mass flow rate at which point I would have v two squared over two plus h three minus h two. And then if I wanted to simplify it all the way down to parameters that I was given directly I could recognize that v two is the same as v three and write this as v three squared over two and that h two is equal to h one. Therefore this would be h three minus h one. And look we have the same equation if I solve this for the mass flow rate I would have mass flow rate times excuse me mass flow rate equal to 1500 watts divided by v three squared over two plus h three minus h one. Look it's the same. How neat is that? I think it's pretty neat. So regardless of which method we use to approach the problem we end up at the same result. And I think that we have effectively stalled as much as possible. Now we actually have to start doing math. For that I will jump on to a new page. And then just for simplicity sake I will copy this equation over. So on page two I can actually get to calculating things. The first thing I want to find is the mass flow rate. For that I'm going to have to actually evaluate a delta h. I know v three. I know the power input. I know v one is zero. So all I have to do is evaluate delta h. And I am home free to calculate the mass flow rate. For my evaluation of delta h remember that I have three options. The first option is to look up the h's in my property tables and subtract them directly. The second option would be to determine how the specific heat capacity of air changes as a function of temperature and evaluate it by using the definition for cp which is dh dt for ideal gases. Remember that when we're talking about changes in enthalpy of an ideal gas it is cp dt because that's defined with respect to enthalpy. It has nothing to do with the pressure being constant or not constant. H goes to p, u goes with v. The udt is cv. Harry Potter ultraviolet. If I knew how cp varied as a function of temperature for air I could evaluate that integral and I could make it even easier by assuming it didn't change. In that case when I write dh is equal to cp dt and I integrate both sides and I can delta h on the left. If I assume cp is constant it comes out of the integral then I'm left with cp times delta t. That's even easier. So option one, look up h's and subtract them. Option two, I try to be a little bit organized about this. Option two, I guess that's actually h3 not h2. Option two would be to look up cp as a function of t and integrate option three is to assume the specific heat capacity doesn't change at which point it comes out of the integral. So those are the options that I could do. The real question is what I should do. Option one would yield the best answer under these circumstances. Option two would yield a pretty good answer. Option three would yield a worse answer. But I recognize that I have an ideal gas. I'm making a lot of simplifications for this problem. Like for example assuming that there's no heat loss and that all the energy conversion happens perfectly. Those simplifications end up with uncertainty in our answer that is much, much bigger than the difference between any of these options. As a result I might as well use the fastest one which is option three. But in the interest of character building and since we have effectively infinite time on the internet let's do all three shall we? Yeah I think that sounds like fun. So for option one I want to look up h1 and h3 using my property tables. For that I recognize that I actually have ideal gas properties of air on table a22. So I will jump over to table a22 and I can look up the enthalpy for a given temperature. Remember that for ideal gases u and h are only functions of temperature. So h1 will be the enthalpy of air at T1 and T1 was 22 degrees Celsius. 22 degrees Celsius plus 273.15 is going to be 295.15. You can verify that with a calculator just because I don't really trust my mental math at the moment. Yep 295.15. And in my ideal gas property tables I see that I don't have 295.15. I have 295. Now because I don't have an exact value the best thing to do would be to interpolate. Now 295.15 is like close enough to 295 to probably be okay with using 295. But you know I think in the interest of character building we should actually evaluate that interpolation. So what I'm going to do is take 295.15 minus 295 I'm going to divide that by 300 minus 295 and that's equal to the enthalpy at 295.15 which is what we're looking for minus the enthalpy at 295 which is 295.17 divided by the enthalpy at 300 which is 300.19 minus the enthalpy at 295 which is 295.17. And we're looking for x and we get 295.321. So h1 is about 295.321. We can repeat the same procedure for the enthalpy at t3. That was 50 degrees Celsius if I'm not mistaken. Yep 50. So we want 50 degrees Celsius. 50 plus 273.15 is like 323.15 I think. But let's double check that with the calculator 50 plus 273.15. Hey look at that. Surprising everyone the mental math was correct. So we find 323.15 surely we have a direct row for that. Oh man we don't. Oh no I guess we have to interpolate again. Well luckily for us this is a good opportunity for character building. So 323 that doesn't start with a two. 323.15 minus 320 divided by 325 minus 320 is equal to the thing that we're looking for. Nope that's not an equals command calculator. Keep up. The thing that we're looking for minus the enthalpy at 320 which was 320.29 divided by the enthalpy at 325 which was 325.31 minus the enthalpy at 320 which was 320.29. We're looking for x and we get 323.45 so 323.453. Let's use all those decimals just to be as certain as we possibly can be. So with h1 and h3 we can subtract the two if my calculator cooperates 323.453 minus 295.321 and we get 28.132. So our delta h using option one came out to be 28.132. For option two we are going to have to try to figure out a correlation between cp and temperature for ideal air. For that I can look at table A21. So table A21 gives me lines of best fit for the specific heat capacity divided by the molar gas constant for a variety of gases as a function of temperature. So I would have to solve for the molar cp by multiplying by the universal gas constant and then evaluating that integral. That seems like fun. So assuming that I remember to go in and speed that up you guys can see that I just copied over the constants these coefficients for air. Remember that these are not actually the constants the value given in the table is beta times 10 to the third so in order to write beta in our equation we actually have to write 1.337 times 10 to the negative third. Likewise for gamma instead of writing 3.294 we write 3.294 times 10 to the negative sixth. So with this fourth order polynomial I get an approximate function for cp of air as a function of temperature. We can integrate that in order to plug that into our delta H. But for that integral I would yield the molar specific heat capacity which means that I would get the molar specific enthalpy. So in order to actually write the mass specific heat capacity I have to take the molar specific heat capacity and divide it by the molar mass of air which means that I'm actually going to be taking the universal gas constant 8.3 and 4 divided by the molar mass for air which I can get from table A1. On table A1 I see that it's 28.97 and then I'm multiplying that by this function to get the mass specific heat capacity. So when I do that integral I can bring out the molar specific gas constant and the mass constant for air and I will get out a delta H term. Excited? I'm excited let's do some calculus and by we I of course mean let's make my calculator do some calculus for us. So if I pop up my calculator and I write out front I have come on calculator parentheses you can do it out front I have 8.314 kilojoules per kilogram kelvin excuse me kilojoules per kilo mole kelvin divided by 28.97 kilograms per kilo mole which will yield a result in kilojoules per kilogram kelvin and then I'm multiplying that by let's grab an integral we're going to integrate this lovely function 3.653 plus the quantity 1.337 times 10 to the negative third e negative 3 negative 3 much more better times temperature which is I'm going to call x for now x there we go much more better plus 3.294 3.294 e negative sign not minus times x squared plus negative 1.913 oh man did I remember the right negative let's jump out to that first term did I write negative or did I write I didn't write anything man okay negative sign not minus sign much more better look at all those negatives then over here I have negative 1.913 times 10 to the negative ninth negative sign not minus sign 9 times x carat 3 beautiful plus 0.2763 e to the negative sign 12 time x to the fourth that's my entire function close front see we are integrating with respect to x and we are integrating from 22 degrees celsius which remember that when we looked at these correlations what we are talking about is temperatures in kelvin so even though it doesn't really matter when we're talking about linear temperature differences it might matter in the exponentials here so I will write that is 22 plus 273.15 and then 50 plus 273.15 and we get nothing cool did I integrate with respect to 22 how did that happen come on calculator I don't want to integrate with respect to 22 I want to integrate with respect to x much more better okay calculator if you would be so kind we get 28.1295 and even the calculator thinks that we have questionable accuracy awesome so delta h of 28.1295 kilojoules per kilogram now option two is a lot of work it's actually more work than option one and it's less accurate because it's using a line of best fit I don't think that option two is really ever useful when we're talking about ideal gas properties of air if we actually want to be accurate we will use the actual enthalpies if we are approximating we will assume the specific heat capacity is constant option two is trying to be unnecessarily precise with an approximate method and I don't think it's worth our time to even try so on an exam with me at least I will never ask you to compute a delta h or a delta u using option two but if you're curious about how that would work now we've done it okay now for option three we're assuming cp is constant and we are going to look up the cp value for air on table a 22 no on table a 20 for air at a halfway point between 22 and 50 I see halfway between 22 and 50 is going to be like 37 here let's let the calculator do the math we get 36 that's more like it 36 plus 273.15 the halfway point between 20 and 50 is going to be 309.15 therefore we should be using a cp value of air at 309.15 no you could probably make an argument for the fact that we're approximating anyway we might as well just grab 300 instead of 309.15 but unnecessary precision is the name of the game at the moment I can evaluate an interpolation here by writing 309.15 minus 300 divided by 350 minus 300 is equal to the thing that we're looking for minus the value at 300 which is 1.005 divided by the value at 350 which is 1.008 minus the value of 300 which is 1.005 and we get we we get 1.00555 0.00555 and again this is more accurate than I would expect you to be on an exam if you were taking one of my exams it would be totally fine to assume the cp of air is constant at 300 because it's the closest row that you have since you're approximating anyway might as well just approximate there's no point in being that precise about an approximate method but you know we're really exploring all of our options here therefore delta h would be that cp value that we just calculated 1.00555 kilojoules per kilogram kelvin multiplied by our temperature change between states 3 and 1 which is going to be 50 minus 22 degrees celsius cancels kelvin here because the temperature change in degrees celsius is the same as the temperature change in kelvin I mean you could think of it like we can add 273.15 to both of these and we'll still get 38 or whatever I mean since we're here and we have infinite time we might as well just try that 50 minus 22 is 28 28 not 38 well done on the mental math john this is why we have a calculator and then 50 plus 273.15 minus 22 plus 273.15 see we still get 28 that's why degrees celsius and kelvin cancel when they are a temperature difference so 1.00555 times 28 value gives us 28.1554 1 no 28.1554 so being real extra accurate yielded a delta h of 28.132 being very approximate gives us 28.155 a difference of 200 ish that's well less than the amount of error that we introduced by other approximations that we made throughout our solution again things like the assumption that we have perfect energy conversion that we have no losses whatsoever that there is no heat transfer out through the nacelle body of the hairdryer all of those are going to affect our answer way more than two hundredths of a kilojoule per kilogram which is why in these circumstances it would be downright encouraged to just use this value but you know sometimes worth looking stuff up in tables just to be extra certain as a result of having this number already we might as well just use it if you were doing this on an exam you could totally use this value or better yet you can actually just grab the cp value at 300 which is 1.005 and ignore that extra interpolation just for funsies here that produces a value of 28.14 so you could totally use 28.14 instead of 28.132 but you know what we have the number let's use the number now I'm plugging that into our mass flow rate calculation which I almost forgot we were doing it's been so long since we talked about it we can take 1500 watts and divide that by 28.132 plus the velocity at state three which was 20 meters per second yes 20 meters per second I'm going to square that and divide by two so I'm going to write 20 squared meter squared per second squared divided by two then I recognize that I need to add that two kilojoules per kilogram therefore I'm going to want to convert from meter squared per second squared to kilojoules per kilogram to do that I'm going to start with kilojoules and work backwards a kilojoule is a thousand joules a joule is a Newton times a meter and a Newton is a kilogram meter per second squared Newton cancels Newton joules cancels joules kilograms cancels nothing meters and meters cancel square meters second squared cancels second squared leaving me with kilojoules per kilogram so therefore if I take 28.132 plus 20 squared divided by 2000 I will get the denominator of this calculation generally speaking I am a proponent of doing arithmetic as few times as possible so I would probably continue to do the unit conversions after this resulting sum but just to make it a little bit easier to follow let's do it as two separate steps so I'm going to calculate just the denominator first 28.132 plus 20 squared 20 carat two divided by 2000 and we get 28.332 so we are taking 1500 watts divided by 28.332 kilojoules per kilogram and we are looking for an answer in kilograms per second because that's what part a is asking us for therefore we need to write a kilojoule is a thousand joules and that a watt is a joule per second joule cancels joules kilojoules cancels kilojoules watts cancels watts leaving me with kilograms per second so now I'm going to take in my calculator if it cooperates 1500 divided by 28.332 times 1000 and we get 0.053 look at that part a done done now part b asks for the volumetric flow rate at the outlet of the device for the volumetric flow rate we are going to involve the mass flow rate that we just calculated and a density term mass flow rate can be expressed as density times volumetric flow rate so because we're looking for volumetric flow rate we can take mass flow rate and divide it by density you could also think of that as taking mass flow rate times specific volume because specific volume and density are reciprocals of one another the density is going to come from our ideal gas law pv is equal to mrt remember this is the mass specific gas constant which is the universal gas constant divided by the molar mass so I can write density which is mass divided by volume is equal to pressure divided by gas constant times temperature and then in the interest of doing arithmetic as few times as possible I'm going to make all of these substitutions symbolically specific gas constant for air is equal to universal gas constant divided by the molar mass of air universal gas constant comes from the inside of the front cover of our textbook and is 8.314 kilojoules per kilo mole kelvin and the molar mass of air comes from table A1 which is 28.97 kilograms excuse me 28.97 kilograms per kilo mole so when I make this substitution for my volumetric flow rate I'm going to write this as that shiny new mass flow rate from part A divided by pressure times gas constant times temperature so I had what was that in 0.052944 you know just to be arbitrarily precise kilograms per second and then universal gas constant was 8.314 kilojoules per kilo mole kelvin and we're dividing that by molar mass which was 28.97 kilograms per kilo mole and then we are multiplying by temperature since we're talking about the volumetric flow rate at the outlet we're talking about the volumetric flow rate with the mass flow rate at the outlet and the density at the outlet which means that we're using the pressure at the outlet and the temperature at the outlet since the mass flow rate is the same everywhere it doesn't matter but it does matter for temperature and pressure the temperature at the outlet was 50 degrees celsius I'm pretty sure yeah and then the inlet temperature was 22 the outlet pressure and the inlet pressure are both 100 kilopascals so 50 plus 273.15 kelvin divided by 100 kilopascals now kilomoles cancels kilomoles kilograms cancels kilograms kelvin cancels kelvin or kilojoules and kilopascals to cancel I need to expand them a little bit a kilopascal can be expressed as a kilonewton per square meter and a kilodjoule can be expressed as a kilonewton times a meter kilonewton cancels kilonewtons kilodjoules cancels kilodjoules kilopascals cancels kilopascals I have cubic meters in the numerator I have seconds in the denominator which is a volumetric flow rate but remember the problem wants an answer in cfm which is cubic feet per minute so we have to convert from cubic meters into cubic feet and from seconds into minutes so first I can find the conversion for length one meter is 3.2 808 feet so I will write one meter 3.2 808 feet and then I will cube everything one cubed is one meters cubed cancels cubic meters and cubic meters and then I need the seconds to become minutes but there are 60 seconds in one minute seconds cancels seconds and then just to be consistent here I will write a cfm is a cubic foot per minute so that minute cancels minutes and cubic feet cancels cubic feet leaving me with cfm so zoom out a little bit here man that's going to be hard to see with a calculator okay well we can suffer through calculator if you would please turn on we have 0.052944 times 8.3 that's 8.314 multiplied by 50 plus 273.15 multiplied by 3.2 808 cubed multiplied by 60 and we are taking that entire numerator and dividing it by 2 8.97 multiplied by 100 multiplied by one cube which is one and I get 104.033 now where to write that I do a little bit of cleaning here scooch this up again technical term then I can write the volumetric flow rate at the outlet and write the volumetric flow rate at the outlet is 104.033 with part b done I can calculate part c for that I will go to a third sheet of paper part c wants to know what the diameter of the outlet should be so I have a velocity at the outlet I have a volumetric flow rate at the outlet I can use that information to calculate a cross-sectional area which if I have a circular outlet means I can calculate a diameter so my volumetric flow rate can be expressed as average velocity I can denote an average velocity as opposed to the velocity at any point if we were talking about a more representative velocity profile at the outlet it would be a little bit higher at the center and a little bit slower toward the edges but we're calling velocity here an average and then we are multiplying by cross-sectional area we can express the cross-sectional area of a circle as pi over four times diameter squared and then I can write that as velocity squared times pi times diameter squared divided by four I know the velocity at the exit it's 20 meters per second I know the volumetric flow rate at the exit it's 104.033 cubic meter excuse me cubic feet per second cubic feet per minute I know pi I know four so diameter is just going to be the square root of four times volumetric flow rate divided by velocity at the exit times pi so we could go back into our volumetric flow rate calculation and get rid of the conversion at the end to imperial I mean we could just knock off those two components at the end which were 3.2808 cubed n 60 or we could handle that unit conversion backwards or we can actually just substitute in the equation that we use to calculate volumetric flow rate in the first place inside of this calculation it doesn't really matter which method we use theoretically if I kept track of enough decimal places they should all be equivalent but just to be representative of the sort of work that someone might do if they had a limited amount of time on an exam and they had already calculated a volumetric flow rate in cfm let's do the unit conversions from cfm back to whatever it is we need to get an answer in inches so I have four I have 104.033 I was 104.033 John 3 3 and then with that I could put my calculator away for a moment and that was cubic feet per minute and I'm dividing by the velocity of the exit which was 20 meters per second I'm going to go double check that 20 meters per second and I'm going to divide by pi and then I'm going to perform whatever unit conversions I need so that when I take the square at the end I get inches out so in order to get inches out as the answer for diameter I need everything under the radical to be inches squared so I need to get cubic feet times seconds per minute times meters to become square inches so first of all I recognize that one minute is 60 seconds at which point minute cancels minute seconds cancel second next I represent one foot as being 12 inches and then I cube everything perhaps square everything is a little bit easier yeah because we have the conversion from meters to feet yeah let's do that square everything one square is one inches squared is what I want to get to square feet cancels two of the three feet in the cubic feet on the left and then I write one meter is 3.2808 feet which again comes from this conversion on the inside of the front cover and then meters cancels meters square feet and feet cancels cubic feet and I have square inches so when I take the square root of a quantity in square inches I will get out an answer in inches so calculator if you would be so kind I will write that as 4 times 104.033 times 12 squared all divided by 20 times 60 times one square inches is one times 3.2808 and then I take that entire quantity raised to the one half power and I get 3.9 let's double check that all of our numbers appear four appears 104.033 appears 12 squared appears 20 appears 60 appears 3.2808 appears awesome so for diameter at the outlet it should have 3.9 inches in diameter and that's everything we need to answer the question