4.10

## University of Liverpool

Skip to 0 minutes and 4 secondsSo we're going to look at the performance of a spark ignition engine that operates on an ideal auto cycle. We're told in each cycle that it uses seven milligrams of gasoline, that combusts at its flash point of 232 degrees C, and has a net calorific value of 44,400 kilojoules per kilogram. And if the engine discharge is its exhaust at an ambient temperature of 20 degrees C, then what will be its maximum power output at 3,000 RPM, assuming Carnot efficiency. So I'm going to start this problem in the same way I have all the other ones we've done, by drawing a diagram.

Skip to 0 minutes and 49 secondsExcept that this time, I'm going to draw a plot of the volume on the x-axis versus pressure on the y-axis. And we're going to start the cycle down in the bottom corner here at 0. And we're going to suck in gases into the engine at a constant pressure till we get to location 1 here. And then we're going to adiabatically compress those gases up to 2 here. So we're going to go up here in adiabatic conditions. So that means there's no heat transfer. And then we're going to combust the fuel and move to here. And in the process, there'll be a heat input into the engine.

Skip to 1 minute and 47 secondsAnd having done that, we're going to do some work and allow the gases to expand. And so we'll progress back down here, adiabatically again. But now we're going to get some work out of this engine as we expand. So it takes us to 4. Then we'll open the exhaust valves those gases will begin to escape and then the final step will sweep them back out as we go from 0 back from one back down to zero. So that's the schematic representing an ideal cycle and a way of being asked to look at its thermal efficiency.

Skip to 2 minutes and 33 secondsSo we can say that by definition, thermal efficiency, eta TH, is going to be equal to what we want out, which in this case is work. So this is work output divided by what we have to provide in order to get that, and that's the heat in here. So this is heat is input. So we can express that in terms of the symbols I've used here as w out divided by Q in. So we can rearrange that and say that w out will be equal to the thermal efficiency times the heat in. So we're told that this is auto engine's operating at Carnot efficiency. So we can find this efficiency in terms of Carnot's expression.

Skip to 3 minutes and 38 secondsSo we can say, assuming Carnot efficiency-- and Carnot efficiency is essentially the best that we can ever hope to achieve. It's right on the edge of what the second law of thermodynamics will allow us to get. So we can effectively say this is the maximum thermal efficiency that's possible. And it's defined in terms of the temperature range over which the engine is operating. So in this case, that's going to be between the cold temperature it's operating, which is ambient temperature in this case, and the temperature combustion of the other gases. I'm going to use TC and TH for those. And so if we substitute in some numbers into that, it's exhausting its gases at 20 degrees C.

Skip to 4 minutes and 29 secondsSo that's going to be at 293 Kelvin. And the combustion of the fuel happens at 232 degrees C. So that's 505 at Kelvin. And if you get your calculator out and do that little sum, it comes out at 0.419. So that's given us a value that we can use here. So now we need a value for the heat inwards. And we've been given the calorific value of the fuel that we're burning here, and so we can say that Q in is going to be equal to the amount of fuel we use as a mass times NCV. And NCV is the Net Calorific Value of the fuel. And we're told that's equal to 44,400 kilojoules per kilogram.

Skip to 5 minutes and 35 secondsSo that means we can substitute the numbers into here. We're going to get Q in is equal to the mass, which is 7 milligrams. That's 7 times 10 to the minus 6 in order to get into kilograms, times this value here. That's 44.4 times 10 to the 6, because we've got three powers of 10 here and another three buried in this kilojoules. And so now, if you do that sum, the answer comes out at 311.3 joules. And so finally, we can work out the w out. And so we can say, hence w out will be equal to this efficiency value we found here, 0.419, times-- we need the Q in that we've just calculated. So that's 311.

Skip to 6 minutes and 37 secondsAnd if you do that little calculation, it comes out at 130.5 joules. So that's the amount of work we can get out per cycle as we go around this cycle here. Now, we're asked to find the power. So that's the amount of work per second. So we can say, so at 3,000 RPM, the power output-- so w dot out, in other words, it's the work per unit time, per second-- will be equal to this 130 times number of revolutions per second. So that's 3,000 divided by 60, because it's defined in revs per minute. And this is a four-stroke cycle, so that means actually we get two revolutions each time the engine goes around this cycle.

Skip to 7 minutes and 36 secondsSo we need to put a factor of 2 in here to account for that. And when you do that calculation, it comes out at 3,263 watts. And that's the power output of this engine.

# Spark ignition engine (worked example)

A spark ignition engine operates in an ideal Otto cycle. In each cycle it uses 7mg of gasoline that combusts at its flash point of 232°C and has a Net Calorific Value of 44,400 kJ/kg. If the engine discharges its exhaust at an ambient temperature of 20°C then what will be its maximum power output at 3000 rpm, assuming Carnot efficiency.