# Car engine piston

1. Sep 24, 2014

### gezibash

Recently I have been studying thermodynamics and I wanted to analyze a fictitious car piston of my own design. Indeed I have quite some questions, but the most important is this one, and maybe someone could give me an opinion on my progress.

I started off by selecting the fuel, which according to some internet source (I forgot, can't cite) suggested that I should use Octane, so I wrote the chemical equation,

$$C_8H_{18} + 12.5(O_2+3.76N_2) \rightarrow 8CO_2 + 9H_2O + 23.5N_2$$
and I used the AFR ultimately computing the ratio to be,

$$i = \frac{m_o}{m_f} = 15.0279$$
After this, I computed the densities of air and octane,

$$\rho_A = 1.2754 \frac{\text{kg}}{\text{m}^3}$$
and

$$\rho_F = 703 \frac{\text{kg}}{\text{m}^3}$$
I randomly selected a Vp = 1.8L volume, so I went ahead and wrote these two equations,

$$m_A = i \cdot m_F \\ \frac{1}{\rho_A}m_A + \frac{1}{\rho_F}m_F = V_p$$
From which, when you solve it, you get a total mass of

$$m_T = m_A + m_F = 2.448 \; \text{grams}$$
My question is, is this feasible? Does the entire mass of the fuel and air mixture amount to about 2.448 grams in a single cylinder? If not, where did I go wrong?

Also, I would be most grateful if someone could point to a book or anything of the educational nature on this subject?

My next steps from here would be to try and figure the heat of combustion and then I will start to compute an Otto Cycle, perhaps later even use a more realistic intake/exhaust cycle.

2. Sep 24, 2014

### billy_joule

2.5grams sounds OK.
Real engines do not operate at stoichiometric air fuel ratios, they have a massive excess of air.

3. Sep 25, 2014

### gezibash

Thanks a lot. Yeah, I suppose I should fix that.

4. Sep 30, 2014

### Staff: Mentor

Typical air fuel ratios are indeed 15. Show us how you calculated the densities of the fuel and the air. I hope you took into account the mole fractions of the gases in the mixture.

Chet

5. Sep 30, 2014

### gezibash

Hi Chet, thanks for the reply. Sadly I did not manually calculate the densities, I used Tables from the internet to get them at 20°C, 1 bar - since I figured those would be intake conditions.

6. Sep 30, 2014

### Staff: Mentor

For those conditions, I got 2.28 gm for an afr of 15.

I got mole fraction of air = partial pressure of air (atm) = 0.9833
mole fraction of fuel = partial pressure of fuel (atm) = 0.01667
mass of air = 2.136 gm
mass of fuel = 0.1424 gm

Chet

7. Sep 30, 2014

### gezibash

Yes, that seems pretty good. I was just making sure that the answer itself is logical, so anything around 2-3 grams would be on the right track here.

8. Sep 30, 2014

### Staff: Mentor

A way of checking this is to figure out the fuel consumption rate. You can multiply by the number of cylinders and the engine speed to get the fuel consumption rate. Assume that, at cruising speed, the engine speed is about 1500 rpm. Does this make sense in terms of gas mileage?

Chet

9. Sep 30, 2014

### gezibash

Yeah, but I would have to time the cylinder stroke. Check my train of thought here,

$$n_1\;m_1\;N_c = \dot{m}$$
Where $n_1 = 1500\;\text{rpm}$, $m_1 = 2.448\;\text{gm}$ and $N_c = 8$ (number of cylinders).

10. Sep 30, 2014

### gezibash

Lol I could not have been wronger than my previous post. I used the combined mass of both the fuel and air. Clearly that is not a fuel consumption rate. I need to use $m_1\approx 0.15 \text{gm}$

11. Sep 30, 2014

### billy_joule

You also need to account for how often intakes occur. Presumably it's a four stroke engine...Not a mythical half stroke engine as your equation implies..

The average V8 will cruise close to 1500rpm at highway speed (100km/hr)
With that information you can go ahead and convert your fuel mass flow rate to litres/100km (or mpg if you're that way inclined) so you can see how your values compare with real world fuel consumption data.