Can some one show me where to begain?

[NAZ1998BMWM3]

Hay I was wondering if one of you guys could help me figure out this extra credit take home paper I got from school. I have some clue of where to begain but I would like to see if im correct. (Im not to good at math but im working on it.)

If someone could just show me how to begin I would appriciate it.

Question: A liquid cooled, four stroke 60 degree V-12 engine, running on gasoline. Its bore is 5.4 inches and the stroke is 6 inches, compression ratio is 6:1.
Show all work! or no credit, including units.
Hints, the engine is running under ideal conditions and for this problem only, assume volume=mass and 100 percent volume efficiency.

(1)Find the cubic inch displacement of the engine?


(2)Find how much air, this engine uses in one hour at 5500 rpm!


My question(s).
Is there a way to convert cubic inch displacement into liters?
Whats a normal compression ratio for most V-12 engines?
I multiply and divide to get my answers correct?
How the hell do I do number 2?


Thanks guys.
-Nick

[ken_vs_ryu]

http://www.csgnetwork.com/cubicinchcalc.html

[NAZ1998BMWM3]

wow sweet website. I just dont no what to type in for Bore/Stroke Ratio (B/S)?

[ohnoes]

wow sweet website. I just dont no what to type in for Bore/Stroke Ratio (B/S)?

Do you know what a ratio is? :confused

[NAZ1998BMWM3]

Yeah I know what the stoichiometric ratio is like 14.7 parts air to 1 part gas but I just need a little help with this paper.

[jettore]

Part 1 is just calculating cylinder volume. Pi x Height X radius_squared. That's easy you don't need a website to do that, just a calculator. This will be 1 cylinder, so multiply by 12 when done.

Why do you want to convert to liters and what does a regular v12 compression ratio have to do with the problem?

#2, I'm not 100% sure about. Might be just the displacement x 5500 rpm. Then that times 60 minutes to get volume of air in cubic inches.

[NAZ1998BMWM3]

Part 1 is just calculating cylinder volume. Pi x Height X radius_squared. That's easy you don't need a website to do that, just a calculator. This will be 1 cylinder, so multiply by 12 when done.

Why do you want to convert to liters and what does a regular v12 compression ratio have to do with the problem?

#2, I'm not 100% sure about. Might be just the displacement x 5500 rpm. Then that times 60 minutes to get volume of air in cubic inches.

I'm not sure but I think you can take cubic inches and figure out what the liters are in any engine.

I wanted to get some sort of idea of what a v12 ratio would be. Like 6.0 6.1 etc. I guess I answered that one for myself.

[Charlie]

Jesus dude, do you own homework.


I'm not sure but I think you can take cubic inches and figure out what the liters are in any engine.

.

Oh my god...

-Charlie

[NAZ1998BMWM3]

Jesus dude, do you own homework.



Oh my god...

-Charlie

Hay buddy, im just asking for help ok. Please understand that.

[Charlie]

Hay buddy, im just asking for help ok. Please understand that.

Go back and retake 8th grade math.

-Charlie

[hicks650i]

These are the most basic principles. If you can't figure this out, you are better off dropping the course.

'I'm not sure but I think you can take cubic inches and figure out what the liters are in any engine."

Even if you no NOTHING about motors, you should be able to figure out 1 inch = 2.54 cm. 1 cubic inch = 16.387064 cubic centimeters. 1000 cubic centimeters = 1 liter.

This is elementary school math. Seriously.

[jettore]

From my previous post for #2, it would not be 5500 hundred rpm. I forgot to factor in it's a 4 stroke engine.

What course is this for?

[NAZ1998BMWM3]

From my previous post for #2, it would not be 5500 hundred rpm. I forgot to factor in it's a 4 stroke engine.

What course is this for?

Automotive technology.


Like I said, math is my worst subject.


________________________________________

For number (2), it would be something like, 2750 RPMs X air mass or something? The air intake should be 14.7+1 part air per cylinder per complete cycle.

[xatlas0]

Automotive technology.


Like I said, math is my worst subject.


________________________________________

For number (2), it would be something like, 2750 RPMs X air mass or something? The air intake should be 14.7 per complete cycle.

This isn't a math question, this is a basic "do you understand how an engine works" question.

It looks like the others already helped you with the displacement. (0.5*bore)^2*pi*stroke*12=Displaced Volume (in^3)

Since it is a 4 stroke, you get full displacement every 2 revolutions. Assuming they want the volume of air displaced (since they didn't give enough information to find the true mass of air displaced) you would multiply the displacement by 0.5*RPM*60 to get the total displaced air over an hour. If it was a 2 stroke, then you would not multiply by 0.5, as the 2 stroke cycle takes place over 360 degrees, rather than 720.

If you are having trouble with this sort of thing, some extra books could help. I'd suggest Pulkrabeck's "Engineering Fundamentals of the Internal Combustion Engine".
In terms of a regular CR of most modern enignes, it is typically in the 10-11:1 range. This is largely due to the advent of knock sensors, the improvement in fuels, and improvements in fuel metering.

[pckl300]

... it would not be 5500 hundred rpm....
:rofl I hope not, that would be the highest revving V12 I ever heard of.

[MrBen]

I'm not sure but I think you can take cubic inches and figure out what the liters are in any engine.

Here just plug in your own numbers http://www.google.com/search?hl=en&q=350cubic+inches+to+litres&btnG=Search

[NAZ1998BMWM3]

For number (1) I came up with Pie/3.14x6.00x5.40=101.736 X 12=1220.832

Im still working on the second one. And thanks for your help with this guys.

I guess this is where im at with number (2) for now. 0.5x2750x60=82500??? What am I doing wrong?

[gekko16]

For number (1) I came up with Pie/3.14x6.00x5.40=101.736 X 12=1220.832

Im still working on the second one. And thanks for your help with this guys.

I guess this is where im at with number (2) for now. 0.5x2750x60=82500??? What am I doing wrong?

No.

You have to do Pi*(Bore/2)^2*Stroke*Cylinder Count.

[NAZ1998BMWM3]

No.

You have to do Pi*(Bore/2)^2*Stroke*Cylinder Count.

I came up with 101.736?

And I have no clue how to do the second one.

[LagunengrunE36]

For number (1) I came up with Pie/3.14x6.00x5.40=101.736 X 12=1220.832

Im still working on the second one. And thanks for your help with this guys.

I guess this is where im at with number (2) for now. 0.5x2750x60=82500??? What am I doing wrong?
Sounds delicious -- I love pie. :)

[bigdog68]

As interesting as this was to me and my analytical mind, you should post this in the STEP section, as this is right up those guys alley. however I did enjoy figuring this out, took me back to AP Physics Junior year of High School..... damn that was 9 years ago.