Skip to 0 minutes and 4 seconds So we’re going to look at a problem involving a vehicle that’s powered by a four-cylinder 1,600 cc engine. And on a level road, we’re told, at a constant speed of 60 kilometers per hour, the vehicle has a fuel consumption of 12 liters per 100 kilometers. We’re told the fuel density is 711 kilograms per cubic meter, and it releases 33 megajoules per kilogram of energy during combustion. And we’re told that if 35% of the energy of combustion is lost as heat transfer through a cinder block, and to assume that the inlet and outlet valves are closed during the combustion stroke, then we’re asked to calculate the power output of the engine.
Skip to 0 minutes and 53 seconds So I’m going to start by dealing with those fuel consumption figures, which are expressed in a rather awkward form for doing thermodynamics calculations. So we’re told that we have a fuel consumption that is 12 liters per 100 kilometers at 60 kilometers per hour. And really, we want to express that in liters per second, and then ultimately in terms of kilograms of fuel, because that’s the way that the energy of combustion is expressed. So this is equivalent to– well, we can write this down as 12 over 100 and then multiply by the speed expressed in kilometers per second. So divide through by 3,600. And so that will give us a value of 0.002 liters per second.
Skip to 2 minutes and 8 seconds And we’re told the fuel density, which I’m going to use the symbol rho for, is 711 kilograms per meter cubed. And so we’ve got liters and meters cubed here, so we can divide through by 1,000 here and express this as kilograms per liter. And so finally, we can say that the fuel consumption in terms of mass flow, if you like, mass flow of fuel– put the dot on to indicate that it’s going to be per unit time– will be this 0.711 times the 0.002 up here. And so that comes out at 0.001422 kilograms per second. So that’s the amount of fuel that the vehicle’s using. Now, to do the next part, we need a diagram.
Skip to 3 minutes and 20 seconds So I’m going to have a very simplified picture here of the cylinder with a piston in it.
Skip to 3 minutes and 33 seconds OK. And we need to draw a system boundary, as before. So I’m going to have my system just inside the cylinder and coming down to the top of the piston, and contain it in that way. So this is my system boundary.
Skip to 3 minutes and 58 seconds And we’re going to put energy into this. And so we’re told that we have energy inflow here, delta u, which is 33 megajoules per kilogram that we put in. And that we’ve got a mass flow out– sorry, a heat transfer out, delta Q, that’s equal to 0.35 times the change in the internal energy. So I’ve used little u here because it’s per unit mass, and a big U here because it’s the total amount of energy. OK, so now we can apply the first law of thermodynamics to our system. So apply the first law to the system.
Skip to 4 minutes and 57 seconds And we’ve got our standard expression here. So Q in minus Q out, heat transfer in minus the heat transfer out, plus the work in minus the work out, plus the energy transfer as a consequence of mass flow in minus the energy transfer as a consequence of mass flow out. And that is all equal to the change in the internal energy plus the change in the kinetic energy plus the change in the potential energy.
Skip to 5 minutes and 38 seconds So we’re told that the car is moving at a constant speed, and so that means that there’s going to be no change in the kinetic energy. So this is equal to 0 due to constant speed.
Skip to 5 minutes and 56 seconds And we’re also told that it’s on a level road, so that means there’s no change in potential energy. So this is 0 due to level road. And then we’re told to assume that the inlet and outlet valves are closed. And so that means there can be no mass flow in or out. And so this term goes to 0. So this is 0 due to closed valves. OK. And so we’re left with heat transfer terms, the work, and the change in the internal energy.
Skip to 6 minutes and 37 seconds So we can equate these together, and we can say that the net work output– and I’m going to express this simply as delta W to save writing all that out again– is going to be equal to the change in the internal energy. And take this term over to the other side, minus the change in the heat transfer. And that heat transfer, which we’re told in the question is equal to 0.35 of the change in the internal energy. So we can substitute that into here, and we can say delta U times 1 minus 0.35. OK? And this term here is going to be equal to the change in the internal energy per unit mass times mass flow that we calculated here.
Skip to 7 minutes and 38 seconds And so we can express this now as a delta little u multiplied by m dot times 1 minus 0.35. And so that will be equal to 33 times 10 to 6 times this term here, 0.001422– so that’s the first bit– times this bit here, which is 0.65. So our final answer comes out to be 30,500 watts as the total output from this engine. And if we bear in mind that one brake horse power is around 746 watts, then this is about 40 brake horsepower. And that compares the maximum power output of, say, a Ford Focus, which is 83 brake horsepower. So this seems like a reasonable answer.
Combustion in cylinder (worked example)
A vehicle is powered by a 4-cylinder 1600cc engine. On a level road at a constant speed of 60km/hr the vehicle has a fuel consumption of 12L/100km. The fuel has a density of 711 kg/m\(^3\) 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 the engine.
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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