Otto Cycle - calculate the net work per cycle in Btu

In summary, the problem involves a four cylinder, four stroke internal combustion engine with specific values for bore, stroke, clearance volume, and rotation speed. The processes within each cylinder are modeled as an air standard Otto cycle and the task is to calculate the net work per cycle and the power developed by the engine. The stroke is the distance the piston moves during one revolution of the crankshaft and the clearance volume is the small space left at the top of the stroke. To find the compression ratio, the swept volume (area of the cylinder multiplied by the stroke) is added to the clearance volume and divided by the clearance volume. The values for V1 and V2 are used to find the compression ratio, which is then used to find the value
  • #1
jdawg
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Homework Statement


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.

Homework Equations

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! :)
 
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  • #2
jdawg said:

Homework Statement


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.

Homework Equations

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! :)
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.


svcv185w.jpg


​
 
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  • #3
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
jdawg said:
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?
The compression ratio is defined using the geometric properties of the cylinder:

##CR = \frac{SV+CV}{CV}##
 
  • #5
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
jdawg said:
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?
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
 
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  • #7
Thanks, I'll try that!
 
  • #8
Can you show the complete solution I want to compare with mine
 

1. What is the Otto Cycle and how does it work?

The Otto Cycle is a thermodynamic cycle that describes the processes involved in the operation of a typical spark-ignition engine. It consists of four processes: intake, compression, power, and exhaust. During the intake process, the fuel-air mixture is drawn into the engine cylinder. In the compression process, the mixture is compressed by the piston, increasing its temperature and pressure. The power process is where the spark plug ignites the mixture, causing it to expand and push the piston down, producing work. Finally, the exhaust process expels the remaining gases from the cylinder.

2. What is the net work per cycle in the Otto Cycle?

The net work per cycle in the Otto Cycle is the amount of work produced by the engine in one complete cycle. It is calculated by subtracting the work done during the intake and exhaust processes from the work done during the power process. This work output is measured in units of Btu (British thermal units) or Joules.

3. How do you calculate the net work per cycle in Btu for an Otto Cycle?

The net work per cycle in Btu for an Otto Cycle can be calculated using the following equation:

Net work per cycle (Btu) = (specific heat ratio x pressure ratio) / (specific heat ratio - 1) x (initial temperature - final temperature)

The specific heat ratio is the ratio of the specific heat at constant pressure to the specific heat at constant volume. The pressure ratio is the ratio of the maximum pressure during combustion to the minimum pressure during the exhaust process. The initial and final temperatures can be measured using a thermometer.

4. What factors affect the net work per cycle in the Otto Cycle?

The net work per cycle in the Otto Cycle is affected by several factors, including the compression ratio, fuel-air mixture, ignition timing, and engine speed. A higher compression ratio can result in a more efficient combustion process and therefore, a higher net work output. The fuel-air mixture must also be carefully controlled to achieve optimal combustion. Ignition timing, which determines when the spark plug ignites the mixture, can also impact the net work per cycle. Finally, the engine speed, or number of cycles per minute, can affect the amount of work produced.

5. How can the net work per cycle in the Otto Cycle be improved?

There are several ways to improve the net work per cycle in the Otto Cycle. One way is to increase the compression ratio, which can result in a more efficient combustion process. Another way is to optimize the fuel-air mixture to ensure complete and efficient combustion. Additionally, improvements in engine design, such as using a turbocharger or supercharger, can increase the net work output. Regular maintenance and tune-ups can also help maintain the engine's efficiency and improve the net work per cycle.

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