Engine (Otto) Cycle Thermodynamics

" ..... This is in fact enough information to fully describe a simple system from a thermodynamic standpoint. The diagrams are useful when one wants to calculate the work done by the system, the integral of the pressure with respect to volume. One can often quickly calculate this using the PV diagram as it is simply the area enclosed by the cycle."

http://en.wikipedia.org/wiki/Pressure_volume_diagram

Guest in the quoted text is correct. Work or energy is the product of force by distance, and as it turns out it also the product of pressure by volume.

The net work out is the area inside the loop in the diagram you showed, or more particularly it is the area below the top part of the curve less the area below the bottom part of the curve. These two work quantities are the output work (due to combustion) and the input work (in the compression cycle.

In the old days, the area within an engine indicator diagram would be measured with a planimeter. Today one might take closely spaced pressure readings and do an approximate integral using the trapezoid rule in, say, a spreadsheet.

I can't imagine why or how somebody could understand allthose calculations and at same time ask what you are asking.Is that possible?.-

Well true, but I thought there was a "simpler" or maybe another way or making the calculations witht eh internal energy for example or any other values I had in these calculations.

Working from what I remember....and using your diagram numbers. The work input between 1 and 2 is due to compression work and must be equal to the change of internal energy in the working fluid as per rules re conservation of energy (given that the compression is adiabatic). The internal energy change 2 to 3 is the result of combustion (i.e. added energy through chemical reaction), the change 3 to 4 due to working fluid expansion and so to work out, and the change 4 to 1 energy loss through exhaust.

On the basis of the above then surely the work out is also equal to the difference in internal energy change 3 to 4 less the change 1 to 2. Yes ? !

I found still very larger number for the calculations steps shown here.

The closer I could get was with this relation:

W=R*T1*{(V1/V2)^(k-1)-1}/(k-1)

Coming from:

T2/T1 = (V1/V2)^(k-1) = (P2/P1)^((k-1)/k))

q - w = delta_u = Cv*(T2-T1)

Cp-Cv=R

Cp/Cv=k

Cv=R/(k-1)

With this I get more approximate values but still 150% higher than what I get from the experimental settings... It would be more acceptable to be accurate at + or minus 15%

Well now, practice is different from the calculation.

You have made a theoretical assessment without considering reality. In a real combustion engine we have losses, heat losses, friction losses, flow losses.

Generally there are three efficiencies in an ICE, these are

1. Thermal efficiency (W in - W out) / W in, or also (T max - T exhaust) / T max

2. The deviation from from the theoretical process that includes above mentioned heat losses such as through cylinder walls and flow losses of the working fluid, rest gases in cylinder from previous cycle, etc. etc. and

3. Mechanical losses due to friction.

You have only considered the first point.

The thermal efficiency is only dependent on the CR. is is around 58 - 60 % in spark ignited engine and more, near ~ 68 % in compression ignition engine. The formula for that you can find in any book on ICE's.

The efficiency of point 2 is the most variable depending on the engine layout, mostly influenced by the ratio of stroke to piston diameter. I ranges for 40 % to 75%.

The friction losses of point 3 are the lowest. Efficiencies are near 80 to 90 % as the auxiliary equipment needed for the engine to run is calculated in. That includes things like the alternator, oil pump, water pump, etc.

If you really want to assess overall efficiency of an engine you better pay attention to the deviation of the real process from the idealized process, which is the only point that you have considered.

Getting the work output from here, you just need to take the integral of PdV. this would require you to come up with a function from 1-2 and 3-4 but you can probably get a pretty good estimate by just taking (P3-P2)(V4-V2). This will give a good estimate of the power stroke. (1 every 2 cycles of the cylinder) Your drawing neglected the pump work of bringing the air into the engine though. 1-5 and 5-1 are not constant pressure and thus will have a negative work that takes away from the overall work.