For maximum efficiency of a carnot engine E = 1-(ratio of hot to cold operating temperatures) how does that relate to horsepower production?

Cause, we all know, lower temps means denser air, more combustion power. So what does more efficiency mean? Kinda confused here. If say we inject nitrous, would that change range of operating temperatures too? Cause Carnot is supposed to be the ideal thermodynamic engine, right?


Bleh, I'm confusing myself. Damn engineering fundamentals...

OK so yeah, how does that equation for efficiency relate to horsepower production?


And what IS a 4 cycle engine plotted on a pressure vs. volume graph? I'm under the impression that gases in automobile engines are assumed to go through adibatic compression/expansion because it happens so fast. So are the remaining two processes isothermal?

Kinda hard to explain, but if you've taken thermodynamics you know what I'm talking about. As in the graph of pressure vs. volume where work done is area of the graph.

Say start on top left corner and label that point A, and go clockwise so points are B, C, D. So which of the processes, AB, BC, CD, DA, correspond to the graph for a 4 stroke engine cycle? Hmmm......or maybe that's for a 2 stroke cycle cause like in the 4 stroke cycle you have to expand then compress twice right? ARGH. Ugh....hate school....

 

edit I meant, Adiabatic, not adibatic, sorry

 

The efficiency calculations used to depress the hell out of me. All that wasted power...

Essentially, take the number of energy units you wish to use (Joules or calories) and figure out how many are available through the complete combustion of your favorite hydrocarbon mixture of choice. Then figure out how much work that represents. Now look at how much work is available to us through the driveline of the car. We use 20% at best, on a good day! The rest is heat, either radiated or lost through friction. And don't forget we don't even HAVE complete combustion, thus the catalytic converter. Further, the typical intake charge will have compounds that actually ABSORB energy to combust (oxidize, more properly). I won't do the calculations, but I ballparked them once, got the 20% or so, and just went about my gas guzzling business...

Hmmm.. in your cyclical plot, AB is the intake stroke (piston goes down) of a piston engine, right? BC is the compression stroke (piston goes up), CD is the "power stroke" (piston goes down), and DA is the exhaust stroke (piston goes up), right? The crankshaft must complete two full revolutions per power pulse.

Rotary: With respect to a single face of the rotor (there are three total per rotor), the intake occurs during the first 120 degrees of rotation (roughly), the compression occurs during the next 90 degrees, the power stroke during the next 90 degrees, and the exhaust during the last 60 degrees (ROUGHLY, I said!). Thus, you have one power pulse per rotor face per revolution of the rotor. I'm not sure how many power pulses there are per eccentric shaft revolution (I think three?) given the two rotors. There was a hell of a discussion in a hijaaked thread about a month ago or so that was better at explaining some of this...

 

Good question. I think I'll take a break from studying for thermo and try and answer your question . First, a 4-stroke would be two complete cycles on a P-V diagram, you have 1. compression-power and 2. exhaust-intake. You can't really model a 4-stroke engine against a carnot engine using basic thermo for a couple of reasons. All the engines you study in thermo (including carnot) are quasiequiblibrium processes, so momentum doesn't play a role in the calculations. However, with a 4-stroke, the exhaust-intake cycle is driven completely by momentum carried over from the power stroke. Also, a lot of the work used up in each cycle is the result of pumping losses, which also are not modeled in quasiequ. processes. One thing you could do is calculate the work done during each stroke using W=IpDV, basically all the work is done on the power stroke, then calculate power from that based on rpms. Remember, you have one power stroke every 2 rpm. That is your theoretical maximum power. Then you can compare it to a published power figure at a given rpm and see how it matches up.

 

Ooops! Slap my *** and call me a donkey (huh?), I'd forgot the layout of the P-V diagram. Been awhile....