How Does Thermodynamics Explain Power Generation in Motorcycle Engines?

[Kelvin]

In a motorcycle engine, after combustion occurs in the top of the cylinder, the piston is forced down as the mixture of gaseous products undergoes an adiabatic expansion. Find the average power involved in this expansion when the engine is running at 4000 rpm, assuming that the gauge pressure immediately after combustion is 15.0 atm, the initial volume is 50.0 cm[tex]^3[/tex], and the volume of the mixture at the bottom of the stroke is 250 cm[tex]^3[/tex]. Assume that the gases are diatomic and that the time involved in the expansion is one-half that of the total cycle.

I know work done in adiabatic process is
[tex]W=\frac{p_2 V_2 - p_1 V_1}{\gamma - 1}[/tex]

and for adiabatic process,
[tex]p_1 V_1^{\gamma} = p_2 V_2^{\gamma}[/tex]

so [tex]V_1 = 50.0 \hbox{ cm^3}[/tex], [tex]p_1 = 15.0 \hbox{ atm}[/tex],
[tex]\gamma = \frac{7}{5}[/tex]

but what is the final volume [tex]V_2[/tex], which I need to determine [tex]p_2[/tex]?

 

[Doc Al]

... the initial volume is 50.0 cm[tex]^3[/tex], and the volume of the mixture at the bottom of the stroke is 250 cm[tex]^3[/tex].

Isn't the final volume given?

 

[Kelvin]

actually, I don't even know what a motorcycle is, and how the cylinder looks like ...


what is meant by "the volume of the mixture at the bottom of the stroke" ?

 

[cronxeh]

the volume of the mixture at the bottom of the stroke is the final Volume

This is a motorcycle

 

[Kelvin]

sorry, I made typo mistakes
I mean I don't know how "motorcycle engine" works, and how does the "cyclinder" looks like...

 

[cronxeh]

Refer to this site for a visual:

http://auto.howstuffworks.com/engine3.htm
http://auto.howstuffworks.com/engine4.htm

The design in motorcycle is different (I really have no idea to be honest) - but the principle is the same

Edit: ah I found more on that site: http://auto.howstuffworks.com/inside-engines-roundup.htm

I think this is what you really looking for:
http://auto.howstuffworks.com/radial-engine.htm

 

[Kelvin]

so the assumption that "the expansion is one-half that of the total cycle" is wrong? it should be 1/4 ?

 

[Kelvin]

oh...I've found the answer :D
so...let me redo the problem

 

[Kelvin]

oh I got it correct
thanks a lot
my "picture of engine" is completely wrong ...