Max PSI Before a Certain Temperature is Reached

In summary, the conversation discusses the problem of determining the maximum pressure that can be fed to a combustion engine before the fuel reaches its auto ignition point. The ideal gas law is considered, but it is noted that the gas inside the combustion chamber heats up as it is compressed. The conversation also explores the timing aspect of the engine and the importance of the spark in achieving maximum power output. The concept of the Otto Cycle engine design is mentioned, and the need for further research into cylinder pressure for a specific engine is acknowledged.
  • #1
imadagron89
7
0
Greetings all, this is my first post. I'm attempting to solve a problem regarding different octane fuels for a combustion engine and how pressure from a turbo can influence the Auto Ignition Point (Temperature where a fuel can ignite before reaching its proper cylinder position). The items of interest here are going to be pressure, temperature, and volume. My first thought was to turn to the ideal gas law but no matter which way I turn I can't seem to find an answer to properly convey that certain octane fuels have a certain max amount of psi you can feed pre-ignition before it can reach the auto ignition point. Math is below.
PV=nRT

What I know:

Auto Ignition Temp of 93 Octane = 280C or 553.15K
Auto Ignition Temp of Ethanol = 365C or 638.15K

My work:

First I want to find how many moles can fit inside of a single cylinder of a 2.5 liter engine 4 cylinder engine. 2.5L/4 = .625L.

Assuming when the pressure is in the downward position the atmospheres = 1

Temperature for this example when the cylinder is downward = 311.15K

So plugging information into the equation we get
1*.625L = n* .082(L atm K−1 mol−1) * 311.15K = 0.0244960669114967 mole

Now when the cylinder is completely in the up position the volume decreases from .625L to 0.05 L. I want to figure out what make pressure we can have given the moles we just calculated for since it won't change.

P(atm) *.05L=0.0244960669114967 * .082(L atm K−1 mol−1) * 553.15K = 22.2219 atm
22.2219 atm = 326.57 psi.

Since the volume decreased 12.5 times, we can assume the pressure inside the cylinder if starting at 1 atm would increase to 12.5 atm or 183.75psi.

326.57 psi - 183.75psi = 142.82 psi. So what I get is that basically this fuel will never achieve the ability to detonate until ~143 psi which is not the case. Usually 22psi is the limit you can go before you start entering dangerous territory. What am I doing wrong? I know this is calculated for a perfect world but its not anywhere near what I know is the right answer. What am I doing wrong?
 
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  • #2
You are ignoring the fact that the gas inside the combustion chamber heats up as it is compressed. Wikipedia has a fairly relevant entry.
 
  • #3
MrAnchovy said:
You are ignoring the fact that the gas inside the combustion chamber heats up as it is compressed. Wikipedia has a fairly relevant entry.
You are correct however since we are looking for the max pressure we can acquire before the auto ignition temp of when 93 octane can ignite without spark (553.15K), we input this into the equation to find what pressure would allow this to be heated to this temperature. Good input but the final temperature is a constant in the regard of this testing. Anyone else have any input to help me out?
 
  • #4
I think where I'm having trouble is the timing aspect of an engine. Essential you can adjust when the spark actually happens during the compression cycle to maximize the pressure/force against the piston it achieve greater power output. If you spark the fuel/oxygen mixture earlier it allows it a larger time to expand the controlled explosion in the cylinder increasing pressure. Taking that out I'm assuming that the spark is only happening when the cylinder is in the upright position. I think I may have found where my experiment stops because I have no idea how to even go about finding out information past the ideal gas law.
 
  • #5
That sounds like a question whose answer should be found in textbooks or tutorialsa out Otto Cycle engine design. Have you searched for those?
 
  • #6
imadagron89 said:
the final temperature is a constant in the regard of this testing
That's funny because I've never heard of the law of conservation of temperature. The law of conservation of energy on the other hand says that when you do work against a gas to compress it, the temperature of the gas goes up. The pressure at TDC is therefore much greater than the intake pressure times the compression ratio.
 
  • #7
MrAnchovy said:
That's funny because I've never heard of the law of conservation of temperature. The law of conservation of energy on the other hand says that when you do work against a gas to compress it, the temperature of the gas goes up. The pressure at TDC is therefore much greater than the intake pressure times the compression ratio.
You're not getting what I'm saying, at the temperature of 280C is when the Auto Ignition Temperature of 93 Octane is achieved, anything greater than that temperature doesn't matter because it will still be achieved. Once the temperature gets to 280C the relevant pressure at that temperature is the absolute max threshold for this experiment. I'm trying to find that threshold to say that (ex. As long as you stay below x psi turbo output on 93 Octane fuel, it cannot come within reach of the auto ignition temperature.
 
  • #8
anorlunda said:
That sounds like a question whose answer should be found in textbooks or tutorialsa out Otto Cycle engine design. Have you searched for those?
This is what I'm looking into now but I'm not sure if I have the proper schooling to use this type of physics/mathematics. From what I'm finding cylinder pressure on a naturally aspirated car can reach upwards of 1500psi at peak and 300psi at idle so I need to sit back, think, and figure out where I need to go from here in order to calculate this properly. Ultimately I need the cylinder pressure from my specific engine the Subaru EJ25 series but I'm unsure how that would be possible.
 

What is the importance of knowing the max PSI before a certain temperature is reached?

The maximum PSI (pounds per square inch) a material can withstand before a certain temperature is reached is crucial information for engineers and designers. It helps them determine the appropriate materials to use for a specific application and ensure the safety and durability of their products.

How is the max PSI before a certain temperature is reached determined?

The max PSI before a certain temperature is reached is determined through various testing methods, such as tensile strength testing and thermal expansion testing. These tests involve subjecting the material to different temperatures and measuring its ability to withstand pressure before it reaches its breaking point.

Why is the max PSI before a certain temperature is reached different for different materials?

Different materials have different physical and chemical properties, which affect their ability to withstand pressure and temperature. For example, metals have a higher max PSI before a certain temperature is reached compared to plastics because they have stronger intermolecular forces holding them together.

What happens if a material exceeds its max PSI before a certain temperature is reached?

If a material exceeds its max PSI before a certain temperature is reached, it can lead to mechanical failure or structural damage. The material may deform, crack, or break, depending on its properties and the amount of pressure and temperature it's exposed to. This can result in safety hazards and costly repairs or replacements.

How can the knowledge of max PSI before a certain temperature is reached be applied in real-life situations?

The knowledge of max PSI before a certain temperature is reached is crucial in various industries, such as aerospace, automotive, and construction. It helps engineers and designers select the right materials for specific applications and ensure the safety and reliability of their products. It is also essential for maintenance and troubleshooting to prevent failures and accidents.

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