 John drives a 2013 Honda Accord with a lot of character. It features Honda's K24W2 engine, which produces 185 horsepower at peak operating conditions, which is about 6,400 rpm. Full of stuff and with John at the wheel, the car weighs 3,600 pounds. If it were able to output peak horsepower all the time and everything else were ideal, that is, there are no mechanical losses, no air resistance, perfect combustion all the time, whatever. How long would it take for John's car to accelerate from rest to 60 miles per hour? Well, our system is going to be John and his car, and we are considering an energy conversion process. That process involves an increase in kinetic energy. I'm going to define the process that we are analyzing as being from state 1 to state 2. Those are different points in time for our analysis here. Again, state 1 and state 2, different points in time. I'm saying that the mass of the car is going to be the same, mass 2 is equal to mass 1, and I'm going to include the given information, v1, the velocity at state 1 is 0 miles per hour, and velocity 2 is 60 miles per hour. Furthermore, I was told that the car weighs 3,600 pounds, that would be pounds of force, and I was told that the engine produces 185 horsepower. So I'm going to call that an E.in term, E.in, the rate of energy entering the car. For the purposes of this analysis, we're saying that the energy is entering the car as power, even though in reality it's really entering as chemical energy, but again, depending on how you define your scope, you could consider lots of different starting points. Eventually, it all goes back to the sun, and I guess before that it goes back to the big bang, but we can't start every analysis to the big bang, or we wouldn't get anything done. So E.in is 185 horsepower. Now I'm going to write out my energy balance, and that energy balance has to be performed on something. I am defining my something, my system, as the car, and as a result of having a constant mass, that system is going to be a closed system. The other thing that we have to consider here is whether or not to treat the analysis as a steady process or a transient process. Is time important is really what that comes down to. In a steady process, all effects of time are neglected. The situation is the same regardless of when you look at it. In a transient process, time is important. Is this a steady process or transient process? It is transient, that's right. We know time is important because the problem is asking us for a duration, trying to figure out how long it would take to accelerate from rest to 60 miles per hour. A steady state analysis would be something like if the engine were sitting on a dyno and had been running long enough that it was as hot as it was going to get, that would be a steady state analysis of the engine. Does that distinction make sense? It has completely warmed up. Or it might be steady state two if we were considering a situation like how hot does the engine get when it's running? If we were considering a situation where it was just constantly running at the same rate, regardless of when you looked at it, and we were considering the temperature increasing over time. As the temperature is increasing, the amount of heat rejected by the car is increasing as well. And eventually, it reaches a point where the temperature levels out. So before that point, we would have transient analysis. After that point, it would be steady analysis because time doesn't matter anymore. Again, I think that's one of those things that makes more sense the more practice you get. So this is a closed system undergoing a transient process. Good habit to get into. Just think through that every time. Now that we have a system defined, we can perform our energy balance. By the way, I abbreviate energy balances EB in my analyses. I will write this out for your sake. The first couple of times energy balance. We are performing an energy balance on the car. And as all energy balances is going to begin with delta E is equal to E in minus E out. The energy change of the vehicle is equal to the energy entering the vehicle minus the energy exiting the vehicle. For our purposes right now, I'm saying that the energy could enter as heat transfer or work. But it isn't important that we distinguish that. Because we know the energy is entering at a rate of 185 horsepower. We don't really care if it's heat or work. I mean, we know it's work because it's mechanical power. But we don't have to break this apart. All we have to do is recognize that we have a rate of energy in. We can consider any opportunities for energy exiting, like maybe the car is hotter than its surroundings and is emitting some energy to its surroundings, things like that. But for our purposes here, we are considering ideal operation. So we're saying all of the power of the engine is going into the kinetic energy of the car. So the proper thing to do here would be to assume that no energy was exiting our system. In fact, I will go as far as to start a list of formal assumptions. Generally speaking, you don't have to establish assumptions that are already listed. So I don't have to say there are no mechanical losses as an assumption because it was already given to me in the problem. I don't have to assume that there's no air drag because that's already given to me in the problem, etc. So my first assumption, I guess, is that I have a closed system. The mass of the car doesn't change throughout this process. Two, I'm saying energy out is zero. Then this term disappears and I'm left with just energy in. Now, is the energy in 185 horsepower? No, it's not. Because 185 horsepower was E dot in. And E dot in is the rate of change of the energy entering with respect to time. That means it is energy entering over duration divided by that duration. If we have three kilojoules entering the system per second over 30 seconds, then the total amount of energy entering is going to be three kilojoules per second multiplied by 30 seconds. We are considering the entire process when we run it out as E in over delta T as opposed to an instantaneous point. Does that make sense? Whatever the case, we can say E in is E dot in multiplied by duration. If I can make that substitution into my energy balance and write delta E, the static energy change of the system as being equal to the rate of energy entering the system multiplied by a duration. Okay, the right hand side of the equation has been considered. The left hand side we know is important because we have a transient process and I will go as far as to say this is the energy of our system at state two minus the energy of our system at state one. The energy of our system at state two would be the internal energy of our system at state two plus the total kinetic energy of our system at state two plus the potential energy of our system at state two minus the total energy of our system at state one which would be the total energy internal energy of our system at state one plus the total kinetic energy of our system at state one and I guess that minus is distributed so it really should be minus the total kinetic energy of our system at state one minus the total potential energy of our system at state one. This, I'm more commonly right in terms of u2 minus u1 plus k2 minus k1 plus p2 minus p1. I grouped them together, in which case I would be left with delta u of our system plus delta ke of our system plus delta pe of our system. That just means it's a little bit easier to write. But that doesn't change the meaning. Anyway, it's still equal to e dot in times the duration on the right. So we are simplifying our energy balance and we are making progress. We are left with five terms for now. Are there any other terms that you think it's reasonable to get rid of? Well, we can make a case for there being no internal energy change of our car. Presumably the car hasn't changed phase, presumably the charge, presumably the car hasn't undergone any sort of nuclear fission or nuclear fusion. I mean, we could make an argument that the temperature changed, but for the purposes of this analysis, we have ideal conversion from power into kinetic energy. So we can get away with assuming the change in internal energy of our system is zero. And while we're at it, I'm going to assume that the change in potential energy of our car is also zero. That is reasonable, because I am accelerating horizontally. How do I know that I'm accelerating horizontally because I'm in South Dakota and there's no hills? Whatever the case though, I'm assuming perfect ideal conversion of energy from power into kinetic energy anyway, so this is all part of that ideal umbrella. Anyway, change in kinetic energy of the car is equal to E dot N times delta T. I'm looking for duration, so at this point, I'm going to solve for duration and I'm going to write the change in kinetic energy of the car divided by the rate of energy entering the car. And delta KE is just KE2 minus KE1. Remember that one and two are referring to state points here. That's still divided by E dot N. And then total kinetic energy of the car can be written as mass times velocity squared times one half. So I could say one half times mass two times V2 squared minus one half times mass one times V1 squared divided by E dot N. And then I can recognize that because we are treating this as a closed system, m dot two, excuse me, m2 is equal to m1. And I could factor out one half in mass and write this as one half times mass divided by E dot N times the quantity V2 squared minus V1 squared. So so far, we have written out a full energy balance. We have neglected terms until we ended up with just the relevant energy changes in our system. And then we are rewriting that to solve for duration. While we're here, let me point out that a very common mistake among thermal one students is to simplify this change in velocity. That's V2 squared minus V1 squared as V2 minus V1 quantity squared. But that's not true. V2 squared minus V1 squared is not the same as V2 minus V1 quantity squared. I feel like there's a joke in there about you foiling your own analysis, but I'll leave that aside. We have E dot N, we have V2, we have V1. In fact, V1 was zero, was it not? It started from rest and is accelerated to 60 miles an hour. So I can get rid of V1 altogether. And we don't know the mass, but we know the weight of the car. And we could describe the mass in terms of the weight by recognizing that weight is going to equal the mass times gravity. Therefore, the mass of the car is going to be the weight of the car divided by gravity. And the cool thing about pounds of mass and pounds of force, by the way, is that the amount of mass in one pound of force under standard gravitational acceleration is one pound of mass. That's how we define it. So if we were to say 3200 pounds of force, 3,600 pounds of force, excuse me, and if we were to assume standard gravitational acceleration, 9.81, excuse me, be more logical for me to work this in a previous, 32.2 feet per second squared. And let me list that as a formal assumption. G is equal to 32.2 feet per second squared. And if we were then to go to our conversion factor sheet and recognize that one pound of force is 32.174 pounds of mass times feet per second squared, that actually should be what we're using for gravity, but rounded and infrapenny and propound that I'm saying one pound of force is equal to 32.174 pounds of mass feet per second squared. Then pounds of force cancels pound of force, feet cancels feet, second squared cancels second squared. And I'm left with theoretically 3,600 pounds of mass. Now it's slightly different as a result of my assumption that gravity is 32.2 as opposed to 32.174 which is what our conversion factor sheet is using. So since I'm doing that already, I might as well determine the mass of the car. Instead of 3,600 pounds of mass, it is 3,597.09 pounds of mass. That's the result of assuming gravity is 32.2. Anyway, even though I don't know mass, I have mass and that means I have everything I need to finish the question. So I will write this as horizontal line and then one half multiplied by the mass of the car, which we know is 3,597.09 pounds of mass. And then we are dividing by 185 horsepower. We are multiplying by V2 squared, which is 60 miles per hour. And then it's a squared term, so I square everything 60 squared miles an hour squared. And we are looking for a duration presumably in seconds. I mean, my car isn't exactly a sports car, but surely it doesn't take a number of minutes to accelerate from zero to 60. So let's shoot for seconds as our unit that we're targeting for an answer. In order to get to seconds, I'm going to take my secondary units here and break them apart into primary units, or rather my units of secondary dimension and break them apart into units of primary dimension. So I'm going to say that a mile per hour is a mile per hour. And I square everything. And I recognize that a mile is 5,280 feet. And again, I square everything. Then mile squared cancels an hour squared, excuse me, mile squared cancels a mile squared. And then I will break apart the horsepower. A horsepower as per the conversion factor sheet can be written as 2,545 BTUs per hour or 550 feet pounds of force per second or 0.7457 kilowatts. Most convenient one for me is 550 feet times pounds of force per second. So one horsepower is 550 feet times pounds of force per second. Then horsepower cancels horsepower, feet cancels one of the feet in the numerator. The other feet would come from our pound of force. So I'll break them apart as well. A pound of force, you'll remember, is defined as 32.174 pounds of mass per feet second squared. And again, that came from our conversion factor sheet specifically under force here. One pound of force is 32.174 pound of mass times feet per second squared. Okay, now feet and guys, it's pound of mass times feet per second squared. Now pound of mass divided by feet times second squared. So that should be in the denominator. What am I doing? Okay, let's try that again. One last time for all the means. Feet squared cancels feet and feet. Aha, we did it. Then pounds of mass is going to cancel pound of mass and pound of force cancels pound of force. I need to get rid of my hours and seconds. So I will recognize that one hour is 60 minutes. And 60 minutes, excuse me, one minute is 60 seconds. So far minute cancels minute and hour will cancel one of the hours to get rid of both of them. Then I need to square everything. One squared is boring. One squared is boring. Hour squared cancels hour squared. Minute squared cancels minute squared. Second squared cancels second square leaving me with an answer in seconds. And with all that established, now it's finally time to compute an answer. So calculator, if you would join us, please. We have one times 3597.09 times 60 squared 60 carat two times 5,280 squared two times one squared, which is one times one squared, which is one divided by two times 185 times 550 times 32.171, which gets rid of the error introduced by dividing by 32.2 and then multiplying by 32.174 in the previous step of our analysis. Up here. Up here. Isn't that neat? And then we are multiplying by, oh calculator, what are you doing? That's not at all what I said. 550 times 32.174. No. 32.174. Right? 174, not 171. 174, not 171. Told you, calculator. Okay, I'll go fix that. And then we multiply by 60 carat two times 60 carat two. Let me just double check that I typed all of those correctly. One times 3597.09 times 60 squared times 5,280 squared divided by two times 185 times 550 times 32.174 times 60 squared times 60 squared. Neat. We get 4.255 seconds. Not too shabby car. Now that we have an answer, let's have a conversation about it. Do you think that my car actually goes from zero to 60 in 4.255 seconds? No. Definitely not. This was an ideal analysis. We were ignoring mechanical losses. We were ignoring air resistance. We were ignoring every opportunity for energy to go anywhere else other than the place that we want to do. Does that mean this number is not useful? It doesn't. This number is still useful. It is useful because it represents a best case scenario. We can look at how long it actually takes my car to get from zero to 60 and compare that to 4.255, for example. Like if you were the design engineer working on this Honda Accord and you tested it and you determined that it took five seconds to go from zero to 60 and you know that the theoretical maximum is 4.255, that gives you an indication of the headroom left to improve performance. You could recognize how much of an effect those losses are having and see where it might make sense to try to optimize. You could reduce the mechanical losses associated with the gear train, the drive train. You could figure out how much losses were associated with air resistance, et cetera. But you know that all of those losses are making up the difference between 5 and 4.255. It gives you an idea of how much headroom there is for performance. You can think of it like if you took an advanced engineering math test and you got 45 points, that isn't necessarily a useful parameter of performance in and of itself. In order to evaluate how well you performed, you have to know how many points were available. If you got 45 out of 50 points, then that's a pretty good score. If you got 45 out of 100 points, that is not so good of a score, although maybe average.