How to find the stroke and bore of a tractor engine

Question

Given that a 4-stroke and 4-cylinder diesel engine have a ratio between its stroke and its bore as 6:5. The engine operating at 2000 rpm uses 7.5 liters of fuel and 240 kilo liters per hour of air. Assuming that during each intake stroke, fresh charge of air enters the cylinder and it's equal to the displaced volume. Calculate the average volume in meters cube of the individual injections and stroke length and bore of each cylinder.

Honestly I don't know how to approach the question. I understand that the displaced volume is the piston displacement but how do I use it when values of stroke and length are not given? How do this 240 kiloliters per hour of air comes in? A help will be really appreciated.

Answer 1

This sounds suspiciously like homework but I'll give you an answer. "It is left as an exercise to the student" to figure out if my answer is worth anything.

You want liters per cylinder and you have 2000 RPM and 240,000 liters per hour with four cylinders.

2000 RPM times 60 minutes/hour is 120,000 revolutions per hour

Liters per revolution would be $\frac{240,000}{120,000} = 2 $

Since we have four cylinders and it takes two revolutions to cycle air through all cylinders, half the 2 liters would be distributed through the four cylinders so each cylinder is $\frac{1}{4}$ liter.

Four cylinders each of 0.25 makes for a 1000 CC engine. I did not know they made four cylinder diesel engines that small. News to me.

I'm still learning how to type in equations so I'll practice here.

We have two equations for stroke, S and bore, B:

$ratio=\frac{S}{B}=\frac{6}{5}$

and for a displacement of 0.25 liters, that would be 250 cc (keeping in mind the volume of an cylinder of radius r and height h is $\pi \times r^2 \times h$ or $\pi \times \frac{d^2}{4} \times h$ :

$250 cm^3 = \frac{\pi \times B^2}{4} \times S$

Combining the equations and reducing, we get:

$1000 = \pi \times B^3 \times \frac{6}{5}$

or

$B=\sqrt[3]{\frac{5000}{6 \pi}} $

If B = 6.5 cm, the stroke must be:

$S= 6.5cm \times \frac{6}{5} = 7.7 cm$

How about converting that to cubic meters? Sorry, I'm not doing that part.

Keep in mind this may or may not be right and your teacher is going to ask where you came up with any small part of it (and might read Engineering StackExchange).