# Thermodynamics: Otto Cycle

1. Mar 9, 2016

### jdawg

1. The problem statement, all variables and given/known data
A four cylinder, four stroke internal combustion engine has a bore of 3.7 in and a stroke of 3.4 in. The clearance volume is 16% of the cylinder volume at bottom dead center and the crankshaft rotates at 2400 RPM. The processes within each cylinder are modeled as an air standard Otto cycle with a pressure of 14.5 lbf/in2 and a temperature of 60 °F at the beginning of compression. The maximum temperature in the cycle is 5200 °R. Based on this model, calculate the net work per cycle in Btu and the power developed by the engine in horse power.

2. Relevant equations

3. The attempt at a solution
So p1=14.5 lbf/in2 and T1=520°R. The max temperature is at T3 so T3=5200°R. V3=V2. V4=V1.

And they tell you in the problem that the clearance volume ( V2) is 16% of the volume at bottom dead center (V1). I would think that meant V2=(0.16)*(V1), but the solution I'm looking at says V2=(0.16)*(V1+V2).

What exactly is the stroke? All my book says about it is that its the distance the piston moves in one direction, which kind of makes me think its the total volume? The solution I have kind of treats it as a height because they multiply the stroke by the area they calculated with the bore:
ΔV1-2=(π(3.7/2)2)(3.4)=0.02116 ft3

I feel like the terminology is the main thing that's tripping me up. If someone could clarify I would really appreciate it! :)

2. Mar 9, 2016

### SteamKing

Staff Emeritus
The stroke is the distance the piston moves up or down during one revolution of the crankshaft. The clearance volume is that small space which remains when the piston is at the top of its stroke. The swept volume is the product of the area of the cylinder and the piston stroke.

3. Mar 9, 2016

### jdawg

Ok thanks! Now I'm trying to find vr2 and its coming out wrong. I looked up the value for vr1 to be 158.58 and calculated V1=0.026 ft3 and V2=0.004716 ft3.

So now using (vr2/vr1)=(V2/V1) and found vr2=28.76... But it should be 25.373.

They did (1/6.25)*(158.58)=vr2
How did they find the compression ratio to be 6.25? It seems like they found the compression ratio without using V1 and V2. I know 1/6.25=0.16, is it a coincidence that the inverse of the compression ratio is the percentage of the clearance volume at bottom dead center?

4. Mar 9, 2016

### SteamKing

Staff Emeritus
The compression ratio is defined using the geometric properties of the cylinder:

$CR = \frac{SV+CV}{CV}$

5. Mar 9, 2016

### jdawg

I don't understand why it doesn't work when I just plug in my values to (vr2)=(V2/V1)*(vr1)

Are the values for V1 and V2 incorrect?

6. Mar 9, 2016

### SteamKing

Staff Emeritus
Your calculation for SV = 0.02116 ft3 looks OK. IDK if you have calculated the CV correctly, though.

I would work in cubic inches for these calculations and convert to cubic feet later. 1 ft3 = 1728 in3

7. Mar 9, 2016

### jdawg

Thanks, I'll try that!

8. Dec 14, 2016

### jiasd

Can you show the complete solution I wanna compare with mine