The cycle in Fig. 20-31 represents the operation of a gasoline internal combustion engine. Volum

The cycle in Fig. 20-31 represents the operation of a...

Key Concepts

Adiabatic Process

An adiabatic process is one in which no heat is exchanged with the surroundings. For an ideal gas undergoing an adiabatic process, the relationship TV^(?-1) = constant holds, linking the temperature and volume changes. This concept is crucial when analyzing the compression and expansion strokes in an internal combustion engine cycle, where the processes are often modeled as adiabatic.

Constant Volume Process

A constant volume process occurs when the volume remains unchanged while the system undergoes heat transfer. In such a process, any added or removed heat causes a change in temperature and pressure. This process is key in engine cycle analysis, such as during the combustion (heat addition) step in a gasoline engine, where the volume is assumed constant.

Specific Heat Ratio (?)

The specific heat ratio, ?, is the ratio of the specific heat at constant pressure to the specific heat at constant volume. It is a critical parameter in determining how temperature and pressure change during adiabatic processes. In internal combustion engine cycles, ? influences the efficiency and the relationships between different state variables during compression and expansion.

Thermal Efficiency of Engine Cycles

Thermal efficiency is a measure of how effectively an engine converts the heat from fuel into useful work. In the context of ideal engine cycles, such as those approximating gasoline engines, thermal efficiency is derived from the relationships between temperatures and pressures at various points in the cycle. It is an important concept in evaluating the performance of combustion engines.

Ideal Gas Behavior

The assumption of ideal gas behavior simplifies the analysis of thermodynamic processes by allowing the use of well-known relations like the Ideal Gas Law (pV = nRT) and adiabatic equations. This concept is fundamental when analyzing the cyclic processes in internal combustion engines, where the working fluid (the air-fuel mixture) is often treated as an ideal gas.

Transcript

00:01 For this problem on the topic of entropy, we are shown a cycle in the figure representing the operation of a gasoline internal combustion engine.

00:08 We have the volume v3 to be four times the volume v1, and we want to assume that the intake mixture is an ideal gas with gamma equal to 1 .3.

00:19 We are asked to find the ratios t2 over t1, t3 over t1, t4 over t1, p3 over p1, p4 over p1, as well as the efficiency.

00:30 Of the engine.

00:33 Now the pressure at 2 p2 is equal to 3 times the pressure p1 and this is given in the problem statement.

00:44 The volume v2 is equal to the volume v1 which from the ideal gas equation is nr t 1 over p1 and so the temperature t2 is p2 v1.

01:03 2 over nr which is 3 p1 v1 over nr which is 3 p1 v1 over nr which is 3 t1 so the ratio we require t2 over t1 is then simply 3 now for part b the process 2 to 3 is adiabatic so we have t2 v 2 to the power gamma minus 1 is equal to t3 v3 to the power gamma minus 1.

01:49 And using the result from a, we have v3 is 4v1, v2 is equal to v1, and gamma is equal to 1 .3.

02:05 So we get the ratio of temperatures, t3 to t1, to be t3, divided by t2 over 3, which is 3 multiplied by v2 over v3 all to the power gamma minus 1.

02:31 And so this is 3 into 1 over 4 to the power 0 .3.

02:40 And so the ratio t3 by t1 is equal to 1 .98.

02:46 Now for part c, the process for what to 1 is adiabatic.

02:58 So we have t4 v4 to the power gamma minus 1 is equal to t1 v1 to the power gamma minus 1.

03:13 And since v4 is 4v1, we have t4 over t1 that is required equal to v1 over v4 over v4 all to the gamma minus 1.

03:28 And we can write this as 1 over 4 to the power 0 .3, which gives us t4 over t1 to be 0 .66.

03:46 For part td, the process 2 to 3 is adiabatic, which gives us p2 v2 to the power gamma equal to p3 v3 to the power gamma.

04:02 Or the pressure p3 is equal to v2 over v3 or to the power gamma times p2.

04:13 Now we know v3 is 4v1, v2 is equal to v1, p2 is 3 p1 and gamma is 1 .3.

04:20 So the required ratio p3 over p1 is 3 divided by 4 to the power 1 .3, which can, gives the ratio of pressures we require to be 0 .495.

04:46 For part e, the process 4 to 1 is adiabatic.

04:50 So again, we have p4 v4 to the power gamma is p1 v1 to the power gamma.

04:59 And so the ratio p4 to p1 is v1 over v4 to the power gamma...

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