Well I'm doing a 1000 word paper on diesel trucks/engines and it has to be math related. Anything involving figures, ratios, etc... Heh... just anything with numbers suits. I've got some on compression ratios and a little on BTU's. Just kinda hoping if someone could give me some more facts/ideas that'd be worth adding to this paper. It's kind of short notice so I need some quick thinkin eh

Thanks in advance

 

How many miles to pay back the additional cost of a diesel engine vs. gas

Step 1: get the gallons of fuel we will use in x miles


For the gas car:
(gallons) x
(x miles)/(29 miles per gallon)= (x miles) x ---------- = --- gallons
(29 mpg) 29


For the diesel car:
(gallons) x
(x miles)/(32 miles per gallon)= (x miles) x ---------- = --- gallons
(32 miles) 32



Step 2: get the cost for that much fuel

gas: ((x/29) gallons)(1.16 dollars per gallon)= x(1.16/29) dollars

diesel: ((x/32) gallons)(1.04 dollars per gallon)= x(1.04/32) dollars

Now that we know how much it will cost in fuel to drive both x miles,
we need to make one final observation. When we defined x we said it
was the miles we had to drive so that we paid off the additional cost
of a diesel engine. Thus we know that the cost of driving the gas car
x miles is equal to the cost of driving the diesel car plus the cost
of the diesel engine. In a math equation this is:

x(1.16/29) = x(1.04/32) + 427.65

So all we need to do is solve for x.

x(0.04) = x(0.0325) + 427.65
x(0.04) - x(0.0325) = 427.65
x((0.04) - (0.0325)) = 427.65
x(0.0075) = 427.65
x = 427.65/0.0075
= 57,020 miles

 

I reccommend:

www.howstuffworks.com

There you can do a search for diesel engine operation and find out quite a bit.

Also try:

www.dieselpub.com

Which is the trade magazing called diesel progress, looks at things from an engineering perspective.

Learn your math! Very needed--wish I learned more of it!

 

Who did you tick off?

Story problems, YUCK!!

I was honored with a speeding ticket once. Went to court, the judge had me do an essay on E=MC squared to keep my record clean.

What did you do?

Mike

 

p=mv

Momentum = Mass x Velocity (both of which are readily available in large amounts in the CTD)

Newton's three laws of motion:

I. Every object in a state of uniform motion tends to remain in that state of motion unless an external force is applied to it.
-In our case, the external force is something called "BOMBING"

II. The relationship between an object's mass m, its acceleration a, and the applied force F is F = ma.
- Again, plenty of "m" and lots of "a" , especially if you use the force (BOMBING)

III. For every action there is an equal and opposite reaction.
- The happier you are with the trucks performance due to your BOMBING, the angrier your wife gets with all the money you spent to do it

 

To figure exhaust flow rate
CFM= Engine Displ (cu in) x Full Load RPM x Eff. x (Exh. Temp Deg F+ 460)
C x 941760
CFM= exhaust flow rate in cubic feet per minute.
Efficiency= .85 for naturally aspirated engines
Efficiency= 1.4 for turbo charged engines
Efficiency= 1.2 for engines with scavenging blower
C= 1 for two-cycle engine C= 2 for four-cycle engines

If Exhaust temperature is not available, use:
T= 1200 Deg F for gasoline engines
T= 900 Deg F for diesel engines

 

Horsepower = torque x rpm divided by 5252

That's on www.howstuffworks.com under "How horsepower works." Also, be sure to look at "How Diesel engines work" and "How torque works." Lots of nice numbers and explanations there.

 

I've read alot off of 'How Stuff Works' and you're right though, it really is quite helpful but I was hoping for a little bit more. And really I think a couple of you might have answered my problems... *cheers*... Pretty simple and straight forward formula's too I might add... gotta love math, lol

 

If you REALLY wanna get into numbers and confusing stuff, check out the delicate relationship between intake and exhaust turbo housing sizes! Do a search on here (the button is at the top of this page) for "turbo housing size". Think about it...larger intake housing = more volume of compressed air, but slower spool-up speed. Larger exhaust housing = lower EGT's, but slower turbine speed, and slower spool-up. Smaller intake housing = faster spool-up, but less air volume. Smaller exhaust housing = faster spool-up, but higher EGT's.

The guys at Ford think they have the answer to this whole problem of housing sizes. Their solution is the variable vane technology. Basically, the fins on the turbine (I'm not sure if which turbine, or if it's both) can actually change pitch just like a jet engine. As far as I know, they're still ironing out the details and problems w/ this, seeing as how they just threw the 6.0L together and put it in all their new trucks for the past 2 yrs.

You might be able to find some info on some turbo manufacturer's websites about how much air different turbo sizes will move and all that kinda stuff.

You could also get into twin turbo designs where one feeds into the other. You could figure out what size housings you'd want for that kinda setup.