4.10

## University of Liverpool

Skip to 0 minutes and 4 secondsSo we're going to tackle a problem on a spark ignition engine that operates in an ideal Otto cycle. And the question tells us that in each cycle it uses seven milligrammes of gasoline that combusts at its flash point of 232 degrees centigrade. And has a net calorific value of 44,400 kilojoules per kilogramme. And we're asked, if the engine discharges its exhaust at an ambient temperature of 20 degrees C, then what will be its maximum power output, assuming Carnot efficiency. So I'm going to draw a pressure volume diagram first of all. And we'll put pressure on the vertical axis and volume on the x-axis.

Skip to 0 minutes and 57 secondsAnd we'll start down in this corner here at point 0. And we'll take air in at constant pressure and move us to point 1 here. And then we'll have adiabatic compression. Take us up to point 2 here. So this is adiabatic. That means that there's no heat transfer. And then we have the spark. And so we increase the pressure here with no change in volume. And we move to three here and so effectively, we've got a heat transfer in occurring during that part of the cycle. And then we have an adiabatic expansion back down to here, to point 4. So this is also adiabatic.

Skip to 1 minute and 53 secondsAnd in the process of doing that we get work out of the system. So we get W out occurring during that adiabatic expansion. And then the exhaust valve opens and the pressure drops back down to point 1. And then we sweep out the exhaust gases, and back to zero. And that's our complete Otto cycle. So now for seven milligrammes of gasoline with a net calorific value.

Skip to 2 minutes and 34 secondsSo net calorific value. We've got NCV for short of 44,400 kilojoules per kilogramme. We can say that the Qn value that we have here when we ignite it is going to be equal to the mass of fuel that we have times its net calorie value. So that's 7 times 10 to the minus 6 because we have milligrammes of it, multiplied by 44.4 times 10 to the 6 because we've got kilojoules of it. And that will give us an answer of 311 joules. And then assuming a Carnot efficiency.

Skip to 3 minutes and 36 secondsSo that's the very best we can have if we just satisfy the second law of thermodynamics. Then by definition that thermal efficiency is equal to 1 minus the cold temperature we're operating at over the hot temperature we're operating at. And so in this case, that will be equal to 1 minus 293 ambient temperature, and the temperature at which this ignites, which in Kelvin is 505. And if you get your calculator out and do that, it's going to come out at 0.42.

Skip to 4 minutes and 14 secondsAnd so we can also define thermal efficiency more simply.

Skip to 4 minutes and 26 secondsSimply what we want out, so that's the work output divided by what we have to put in, in order to get it, so that's the heat input. And so that implies that we can rearrange that and simply say that work out will be equal to the thermal efficiency times the Q in. And so we've already calculated that. That's this point 0.42 here times the work up here - or the heat in rather, I should say, is 311. And so if you get your calculator out and calculate that it comes out at 130.5 joules. So our engine is operating at about 3,000 r.p.m, and so we can say hence at 3,000 r.p.m.

Skip to 5 minutes and 27 secondsW dot, so work per second will be 3,000 revs divided by 60 times our energy output per cycle, and that comes out at 6525 watts - joules per second. And that's our final answer. That's the power output of our spark ignition engine operating on this ideal Otto cycle, assuming the very best efficiency you can hope for-- i.e. carnot efficiency, just satisfying the second law of thermodynamics.

# 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.

Work through the example, and then watch the video or check your answer with the solution at the bottom of this page.