Power Output of a Gasoline Engine

by Patrick Ford
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Hey guys, so I've got a great sort of real life example problem that's gonna pull together a lot of equations that we've seen from heat and temperature. And even earlier stuff in physics and also with heat engines. So just kind of bear with me here, there's lots of parts to this problem, but I think it's a really, really great one. So let's get started. We have a gasoline engine that takes in this amount of heat and it does this by combusting gasoline and then it does 3700 jewels of work per cycle. So whenever I get these sort of numbers here, I'd like to draw the energy flow diagram. So I've got my energy flow diagram, this is my hot reservoir that pulls in heat to the cold reservoir and then it does some work and then it expels the rest out to the cold reservoir. Now, what we're told is that it takes in 1.6 times 10 to the four jewels of heat from combustion gasoline. Remember that's going to be Q. H. That's the heat that you draw in from the hot reservoir. And really this comes from combustion of gas, then it does some amount of work, right? And this work here, we're told is equal to 3700 jewels. The heat that it takes in is 1.6 times 10 to the fourth. And then what happens is it's going to expel some heat out to the cold reservoir. So this is just basically the waste heat, that's how much heat that doesn't get used by the engine. So let's get started with our first part here, we're asked to calculate the heat that is expelled each cycle. So which variable is that? We'll remember. It's gonna be heat. So it's gonna be one of the queues. And we're really looking for the waste heat. How much heat doesn't get used by the engine. So if we're looking for Q. C. Here, we're gonna start off with our equation that relates works and heats for heat engines. This equation over here. So we have the absolute value of the work is equal to Q. H. Minus Q. C. And everything sort of absolute values. Everything is positive numbers. Just so you don't get confused. All right. So what happens here is we're looking for the Q. C. We're gonna rearrange this equation and we're gonna move the QC to the other side and then the work over to the other side so that it becomes positive. Right? So what happens is that the QC is just gonna be Q. H minus W. And that should make some sense. Remember what happens here? Is that a heat engine takes in some heat from the heat engine from the hot reservoir? It does some work right? It sort of produces some mechanical work and then whatever it doesn't use gets sort of thrown out to the waste reservoir the cold reservoir as waste heat. So really what happens is you're gonna take in 1.6 times 10 to the fourth. And then you're gonna do 3700 jewels of work? So you're gonna subtract that and what you end up with is waste heat is 1.23 times to the fourth jules. Alright, so that is the first answer here. That's how much he gets expelled out. So now let's take a look at the second part of our problem. So, in the second part of our problem, we are asked to calculate what mass of fuel is burned each cycle. So what variable is that? Well, for mass, we've always use em. So that's basically what we're looking for here. So where does em come into play with our heat engine? Well, if you look at what happens, we're told that gasoline has a latent heat of combustion. Remember we looked at latent heat when we looked at kalorama tree problems, we're told that this latent heat is this number over here, Remember what this means? Is that for every one kg of mass that you've burned for every one kg of of gasoline, you basically can extract this amount of energy from it. So, whenever we used to sort of phase change here, combustion is sort of one of those phase changes, we used Q equals M. L. So, really, what happens here is that this qh remember the heat that comes from combusting gasoline is going to equal mass times however much sometimes latent heat of combustion. Right? So we're looking for here is this little m So what happens here is I'm gonna take my cue H which is 1.6 times 10 to the fourth, and I'm gonna divided by the latent heat of combustion 4.6 times 10 to the seventh. So, if you look at the exponents here, we have 10 of the 4, 10 of the seven, we should expect to get a number that's pretty small here, Right? Because We've only generated 1.6 times 10-4 when you work this out, What you're gonna get is that the mass that gets combusted is 3.5 times 10 to the minus four kg. In other words, uh, if you sort of convert this two g, that's about 0.35 g of fuel and that kinda makes sense. Right? Every time you have an explosion in the gasoline engine, it actually doesn't require a whole lot of gasoline because it mixes with air and the cylinders are small and things like that. So, it doesn't require a whole lot of gasoline and you can generate a large amount of work from it. Alright, so now let's move on to our last part over here. Our last part asks us to calculate, well, if this engine here completes 60 cycles per second, in other words, 3600 rpm. For those of you who are familiar with cars, then what is going to be the power output in kilowatts. So essentially we're looking for here is what is p remember that is the equation for power? Well, we're gonna have to use an old equation that relates power with the amount of energy and time. Right? So we're told this is the this is 60 cycles per second. What's its power output in kW? So we're gonna have to use this equation over here. That power is equal to work done, divided by Delta T. In other words, it's also equal to the amount of energy that's produced, divided by delta T. So what is the energy that gets produced here? Well, really what happens here, the work that's done is it's actually just gonna be the 3700 jewels, right? Every single cycle you're producing 3700 jewels of work. So then what's the delta T. While the delta T. Is going to be a little bit trickier to solve because we have to relate it to the frequencies. I'm going to sort of go over here. So we're looking for delta T. Remember that frequency and time are related by in verses of each other. So frequency is equal to one over T. We're told that the frequency is the 60 cycles per second. So remember this cycles per second here is going to be frequency. Alright, So this is gonna be 60/1. So what that means here, is that your delta t. The amount of time that it takes for every single second of this engine is actually 1/60. So this is actually where we can now plug back into this equation here. Okay, so the power is going to be 3700 divided by 1/60. And when you work this out, what you're gonna get is you're gonna get 2.22 times 10 to the fifth, and that's in watts. So this is also equal to about 222 kilowatts. And again, for those of you who are very familiar with cars, this is actually equal to about horsepower. Okay, so this is the correct answer. That's just sort of like a bonus here. So it's about 300 horsepower for this gasoline engine. So that's sort of like a real life example about how all of these numbers sort of work out. Alright, so that's it for this one. Thanks for bearing with me, and I'll see you in the next one.