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Skip to 0 minutes and 4 secondsIn this worked example, we're going to look at a vehicle powered by a four cylinder 1,600 cc engine, with a cylinder stroke and bore of 85 millimetres and 88 millimetres, respectively. And it's going along on a level road at a constant speed of 60 kilometres per hour, with an engine speed of 2,000 revs per minute. And the vehicle has a fuel consumption of 12 litres per 100 kilometres, using a fuel that has a density of 711 kilogrammes per metre cubed, that releases 33 megajoules per kilogramme of energy during combustion.

Skip to 0 minutes and 44 secondsAnd we're told that 35% of the energy consumed is lost as heat transfer through the cylinder block, and assuming the inlet and outlet valves are closed during the combustion stroke, we're asked to find the power output of each cylinder. So I'm going to start out by just trying to decompose a little bit all that information about the amount of fuel it's using. So what we're told is that it's consuming 12 litres per 100 kilometres, at the speed of 60 kilometres per hour. And so we can do a little bit of calculation here to figure out how much it's using per minute. So that'll be 12 times 60 divided by 100. And that will give us 0.12 litres per minute.

Skip to 1 minute and 53 secondsSo that's its fuel consumption per minute. And then we're told that it does 2,000 revs per minute, so we can say at 2,000 revs per minute, that implies that if we take this number here-- 0.12-- and divide by the 2,000 we get that it uses 0.00006 litres per rev. And we have four cylinders, so with four cylinders we can work out the consumption per cylinder. So fuel used is - so it's going to be 0.000015 litres per rev per cylinder. And we're told the density of the fuel, so we can say now fuel density equals 0.711 kilogrammes per litre. And so we can now write the mass of fuel used is going to be 0.711 times this number here. 0.000015.

Skip to 3 minutes and 36 secondsAnd if we calculate that, it comes out at 10.665 milligrammes per rev per cylinder. OK? So, we're going to use this number in the next part of the question. And I'm just going to clear my board so I can do that. So, moving on to the core of this problem, we need a picture of the situation. So I'm going to draw a schematic over here of our cylinder with a piston in it.

Skip to 4 minutes and 15 secondsOK? And we're told that this has a diameter across here of 85 millimetres. And that it's got a total stroke length here of 88 millimetres. And we've got some internal energy that we're putting into this, so we're going to have a delta U equal to 33 megajoules per kilogramme. And a heat transfer out of delta Q equal to 0.35 times delta U, we're told. And so I'm going to define a system boundary just inside, like this. So my system boundary will move with my piston, up and down. So now I can apply the first law of thermodynamics to this system. So we can write over here apply first law of thermodynamics to the system.

Skip to 5 minutes and 42 secondsAnd this is the system boundary here. And so we can write the normal equation. Q in minus Q out plus the work in minus the work out, plus the energy transfer that occurs as a consequence of mass flow in, minus the energy transfer that occurs as a consequence of mass flow out. And that's all equal to the change in the internal energy plus the change in the kinetic energy of the system, plus the change in the potential energy of the system. And so we can take a look at these terms and decide which of them will have a value in this scenario here.

Skip to 6 minutes and 32 secondsAnd so if we assume that our car is going along a level road, that would mean that there's no change in the potential energy. And we're told it's doing a constant speed, and so that means there'll be no change in the kinetic energy either. So this term will go to zero. We're also told that the inlet and outlet valves are closed, so the gases are trapped inside here. And so that means that these two terms here also go to zero, like that. So we're just left with our heat transfer terms, our work term, and our internal energy term.

Skip to 7 minutes and 13 secondsSo we can rewrite the equation down here as the change in the work will be equal to the change in the internal energy minus the change in the heat. And we've been told that the heat loss - the heat transfer - is equal to 35% of the internal energy generated by combustion that we've written here. So we can substitute this term down into here. And so we can rewrite this as delta W is going to be equal to delta U into 1 minus 0.35. And so now we need to find out what this going to be. And we know that we've got heat or combustion of 33 megajoules per kilogramme.

Skip to 8 minutes and 9 secondsAnd in the previous part we calculated that we were using 10.6 milligrammes per rev per cylinder. So inside here we've got a mass of fuel, that we found from the previous part was 10.6 milligrammes per rev per cylinder. So we can substitute those values over here. And we can say that delta W is going to be equal to 33 times this value here. 10.6. And then we can multiply that by 1 minus 0.35. This is 33 megajoules per kilogramme, and this is 10.6 milligrammes per rev per cylinder. So I've got a 10 to the minus 6 here, and a 10 to the 6 here. So they cancel each other out. So I've got consistent units here.

Skip to 9 minutes and 11 secondsI'm going to end up with an answer in terms of joules. And if you get your calculator out and you do that calculation, then it comes out at 229 joules per rev per cylinder. So if we want to convert that into more sensible value in terms of the power output, then that would be equivalent to 229 times the number of revs we're doing, which is 2,000, divided by 60, for the number of seconds in a minute. And that then gives you 7,625 watts per cylinder. OK. And so that's the final answer. And if you convert this into brake horsepower, and multiply it up by 4 for the number of cylinders, it comes in at about 41 brake horsepower.

Skip to 10 minutes and 14 secondsAnd that compares to a Ford Focus, which has a brake horsepower of typically about 80 brake horsepower. So it's a reasonably sensible answer, given some of the simplifying assumptions that we've made.

Combustion in cylinder (worked example)

A vehicle is powered by a 4-cylinder 1600cc engine with a cylinder stroke and bore of 85mm and 88mm respectively. On a level road at a constant speed of 60km/hr the vehicle has an engine speed of 2000 rev/min and a fuel consumption of 12L/100km. The fuel has a density of 711 kg/m and releases 33MJ/kg of energy during combustion.

If 35% of the energy of combustion is lost as heat transfer through the cylinder block and assuming the inlet and outlet valves are closed during the combustion stroke, calculate the power output of each cylinder.

Try solving this problem yourself and then watch the video or check your answers in the solution PDF at the bottom of the page.

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This video is from the free online course:

Energy: Thermodynamics in Everyday Life

University of Liverpool

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