# How to Link between Stoichiometry, Energy Content of a Fuel, Combustion in a Cylinder-Piston, and Otto-Cycle, Compression Ratio, & Thermal Efficiency?

Picture 2, source:

I would like to do a real calculation of a piston. Below is data of my Toyota car engine 2NR-VE's geometry, 4 cylinder but in this case I take only one cylinder:

• Total cylinder capacity = 1496cc, so capacity of one cylinder = 374cc,
• Bore size = 7.25 cm,
• Compression Ratio (CR) = 11.5:1,
• Cylinder height (h) = 9.06 cm,

Calculation:

1. From the volume and the bore size we know that the 9.06cm is the total height of the cylinder, not the bore size as written in their website.
2. From the Bore size: Annulus (A) = 41.28cm^2,
3. From the compression ratio (CR) and Bore Size (A), --> the stroke = 7.79cm, and the clearance = 1.27cm,
4. The Displacement (D) = A * Stroke = 41.28cm^2*7.79cm = 321.57cc,
5. From the cylinder volume with assumption that Oxygen=21%, oxygen is 78.54cc = 0.07854L,
6. From internet (not official, but just consider it be so.) I got that the fuel's IUPAC is C10H24 and the Calorific Value is 44.2MJ/kg and its density = 0.723 (kg/L), then the calorific value = 31.96MJ/L,
7. From the stoichiometry, C10H24 + 16O2 --> 10CO2 + 12H2O. Calculating the required C10H24 at NTP which gas occupy 24.0548L volume, then moles of Oxygen at NTP = 0.07854L/(24.0548L/mole) = 0.003265 moles = 3.265mmoles.
8. Since the reaction requires 16 moles of O2 for every mole C10H24, then: Moles of C10H24 = Moles of Oxygen/16 = 0.003265moles/16 = 0.000204 moles = 0.204mmoles,
9. That 0.000204 moles C10H24 converted to liter at STP = 0.000204L*24.0548L/mole = 0.00491L = 4.91mL=4.91cc.
10. As in the point 6, the calorific value of that fuel = 31.96MJ/L, then calorie of every combustion = 156.91kJ.

My questions:

1. Where is the combustion occurs in the Otto P-V diagram? At point 2?
2. Is that calculations from step 1 to step 10 correct?
3. How that stoichiometry linked to the thermal efficiency of that engine? In the formula I read the thermal efficiency is not related to geometry.
4. Is it valid to use NTP for that calculation? NTP: 20 degree Celsius and 1 ATM, as fuel is not gas, it is liquid which boiling point is 204 degree Celsius, is.

Where is the combustion occurs in the Otto P-V diagram? At point 2?

Yes

Is that calculations from step 1 to step 10 correct?

Maybe, but what are you trying to accomplish? It looks fine on the surface.

How that stoichiometry linked to the thermal efficiency of that engine? In the formula I read the thermal efficiency is not related to geometry.

You read the wrong formula, theoretical thermal efficiency of the otto cycle is strictly related to the compression ratio.

Is it valid to use NTP for that calculation? NTP: 20 degree Celsius and 1 ATM, as fuel is not gas, it is liquid which boiling point is 204 degree Celsius, is.

Calculating combustion requires taking inlet conditions and compressing/heating up the combustion components from there. It still isn't clear what you're trying to accomplish. Note that a huge amount of your combustion energy goes out the exhaust.

• There is one more question that I forget to ask. As the requires fuel for single combustion per cylinder is 4.91cc, mean the 4-cylinder require 19.64cc. As my car stationary/idle at 750rpm where combustion occurs every 2 rpm the crankcase, means that with the car just idle it will burn 19.64cc*(750 rpm/2) = 7,365cc or around 7.4L. This is weird. It is just when idle. Nov 20, 2023 at 23:13