Lecture T7: Heat Engines: The Otto Cycle

The lecture focused on applying the thermodynamic tools we have been discussing to estimating the performance (work and efficiency) of heat engines. To do this, we represent the workings of the engine in terms of thermodynamic processes. We then apply the first law of thermodynamics to estimate the heat and work transfers arriving at expressions for cycle work and cycle efficiency. As a final step, it is typical to then re-write these expressions in terms of typical design parameters which are more convenient for the engineer than the thermodynamic variables. We will only model ideal performance in this class (no friction, no unrestrained expansion, no heat loss where it isn't desired, etc.). In 16.050 you will develop higher fidelity models that take account of these effects. We did two PRS questions. The first question was designed to emphasize the importance of identifying the heat and work balances for each leg of a cycle. This should be your first step in analyzing a cycle. The second PRS problem was designed to show that different engines will be represented by different thermodynamic processes. This will affect the work output and the efficiency.

Responses to 'Muddiest Part of the Lecture Cards'

(27 respondents out of 67 students in class)

1) Why isn't the last leg of the PRS question constant vol.? (1 student) It is constant volume. 4-1 not clear about the cooling (1 student) The 4-1 process is the same as for the Otto cycle.

2) About the compression ratio, where did you get the V's from?? How does V1=V4 and V2=V3? (1 student) These are the volumes taken from the thermodynamic diagram.

3)When we say something is adiabatic, Q=0, but that doesn't mean delta T = 0 does it? (1 student) No it doesn't. For example, if I have an adiabatic process and I do work on the system, the temperature will rise.

4) In the refrigerator problem, we saw that removing heat from something takes work. why is this not so on leg 4 of the Otto cycle? (1 student) All that is required is that the first law be satisfied. So on leg 4, heat is removed, but there is no work. Where did the energy come from? The system. For the refrigerator or for any heat engine, we convert work to heat (refrigerator) or heat to work (engine) by arranging several processes in a loop so that the system comes back to its original thermodynamic state.

5)What is step 1'-1''? (2 students) This is the intake and exhaust stroke. On the intake stroke, the piston pulls a fresh volume of reactants into the cylinder, on the exhaust stroke it pushes the exhaust products out. This takes very little work (it is like breathing for example--in, out). It ends up as two lines on the p-v diagram that fall on top of one-another -- so no net work.

6) Is there such a thing as an "ideal" heat engine cycle (one that maximizes efficiency), or does the efficiency always depend on the design parameters of the engine? (1 student). The efficiency does always depend on the design parameters. However, for a given high temperature at which heat is added (Qin) and low temperature at which heat is extracted (Qout) there is one particuolar set of thermodynamic processes that maximizes the efficiency. It is called the Carnot cycle. We will not discuss this cycle in Unified, but you can read about it in any thermo text book, or wait until 16.050.

7) How to use efficiency when dealing with engines. (1 student) Efficiency is one of the most important performance metrics for engines. It tells us how much work we get out of the fuel.

8) 2-3 and 4-1: How is that constant volume? (1 student) The process 2-3 is approximately constant volume because the time it takes for the chemical reaction to occur (and the heat to be released is much less than the time it takes for th piston to move. For the process 4-1, it isn't strictly constant volume as described above, since the piston does move in and out. But the net work of this exhaust/intake stroke is very nearly zero, so we neglect it and assume that the air in the cylinder is cooled at constant volume.

9) Does delta u = q-w=0 for the cycle necessarily mean that deltaT=0 (isothermal)? (1 student) It doesn't mean that the cycle is isothermal. It just means that if I come around to the same point on the cycle again it is at the same temperature.

10) No mud (17 students). Very good.