# Estimating engine parameters

1. Jul 17, 2011

### dboozer

1. The problem statement, all variables and given/known data
Several velocities, time, and length scales are useful in understanding what goes on inside engines. Make estimates of the following quantities for a 1.6-liter displacement four-cylinder spark-ignition engine, operating at wide-open throttle at 2500 rev/min.

a. The mean piston speed and the maximum piston speed.
b. The maximum charge velocity in the intake port (the port area is about 20 percent of the piston area).
c. The time occupied by one engine operating cycle, the intake process, the compression process, the combustion process, the expansion process, and the exhaust process. (Note: The word process is used here not the word stroke.)
d. The average velocity with which the flame travels across the combustion chamber.
e. The length of the intake system (the intake port, the manifold runner, etc.) which is filled by one cylinder charge just before the intake valve opens and this charge enters the cylinder (i.e., how far back from the intake valve, in centimeters, one cylinder volume extends in the intake system).
f. The length of exhaust system filled by one cylinder charge after it exits the cylinder (assume an average exhaust gas temperature of 425ºC).
You will have to make several appropriate geometric assumptions. The calculations are straightforward, and only approximate answers are required.

2. Relevant equations
$\bar{S_p}=2LN$
$\frac{S_p}{\bar{S_p}}=\frac{\pi}{2}\sin\theta \left( 1+\frac{\cos \theta}{(R^2+\sin^2\theta)^{1/2}}\right)$

For flat topped pistons,
${V_d}=\frac{c \pi B^2 L}{4}$, where c is the number of cylinders.

Assuming a stroke length of .08 m, the bore diameter is .0799 m.

If the connecting rod length is 0.15m,
$R = \frac{2l}{L} = \frac {2 \bullet 0.15m}{0.1 m} = 3. 75$.

3. The attempt at a solution
Using L = 0.08 m and N = 2500 rpm, the mean piston speed is
$\bar{S_p} = 6.667$ m/s.

Using graphing software, the max speed occurs at $\theta$ = 1.33732 and is 10.79626 m/s.

I'm not sure how to begin the 2nd part of the question. I'm thinking it has something to do with

$\rho v A = k$, where k is a constant.

Last edited: Jul 17, 2011
2. Jul 17, 2011

### dboozer

It makes sense that the faster the piston moves down, the faster the air is drawn into the chamber.

Assuming a constant density,

$v_1=v_2 \frac{A_2}{A_1}= \frac{10.796 m/s}{20\%}=53.98 m/s$

3. Jul 17, 2011

### dboozer

I think I got most of it... by looking at a valve timing diagram, I should be able to use the rpm to get an idea of how each process is. This will answer parts c and d. I think I can use the diagram to answer e and f, but I'm not 100% sure on that.

http://www.crazyengineers.com/forum/mechanical-automobile-engineering/43144-engine-valve-timing-diagram.html [Broken]

Last edited by a moderator: May 5, 2017