Air/fuel ratio question ( again)

  • Thread starter morry
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In summary, the person is seeking help with finding the volumetric air/fuel ratio and equivalence ratio for a given combustion chemical equation. They are also questioning the validity of the equation and thanking the person named Gokul for their help with the exam.
  • #1
morry
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air/fuel ratio question (urgent again)

Ok, I have an exam on monday and both my lecturers and tutor have gone walkabouts. So you guys are my last resort. Thanks. :)
I get given a combustion chemical equation.
Looks like this : aC6H18 + bO2 + 3.76bN2 --> 9.5CO + 9.5CO2 + 9.5H2O + 71.5N2
I solve for the coefficients.

I can find the gravimetric a/f ratio, but I have no idea what to do to find the volumetric a/f ratio.

Also, for the equivalence ratio, I have the actual ratio, so to find the stoich. ratio, do I just have to rewrite the equation so it looks like --> xCO2 + yH2O +zN2??

Thanks again guys.
 
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  • #2
morry said:
Ok, I have an exam on monday and both my lecturers and tutor have gone walkabouts. So you guys are my last resort. Thanks. :)
I get given a combustion chemical equation.
Looks like this : aC6H18 + bO2 + 3.76bN2 --> 9.5CO + 9.5CO2 + 9.5H2O + 71.5N2
I solve for the coefficients.
Looks like there's an error in the equation. For instance, solving for 'a' from Carbon gives 6a = 19, but solving from Hydrogen gives 18a = 19. This is clearly a contradiction. So, the given equation appears to be flawed...or is there something I'm missing ?

I can find the gravimetric a/f ratio, but I have no idea what to do to find the volumetric a/f ratio.
Assuming you can correctly find a and b, you can calculate the molar a/f ratio (using the fact that 1 mole of O2 is found in about 4.76 moles of air). The molar ratio is the same as the volumetric ratio, because a mole of any gas occupies the same volume under the same conditions (temperature, pressure). Finding the gravimetric ratio requires multiplying by molar masses (molecular weights).
Also, for the equivalence ratio...
How is the equivalence ratio defined ?
 
  • #3
Im not sure if that equation is right, I am pretty sure, but not 100%.

I understand the vol. ratio now. Thanks :)

As for the equivalence ratio, its defined as actual ratio/stoich ratio.

But don't worry about trying to explain everything to me, I had the exam this morning, I think it went well. :)

Thanks for the help Gokul!
 

Related to Air/fuel ratio question ( again)

What is air/fuel ratio?

Air/fuel ratio is the ratio of air to fuel in a combustion process. It is typically expressed as a numerical value, such as 14:1, where 14 parts of air are mixed with 1 part of fuel.

Why is air/fuel ratio important?

Air/fuel ratio is important because it directly affects the efficiency and performance of a combustion process. An incorrect ratio can lead to incomplete combustion, which can result in decreased power and increased emissions.

What is the ideal air/fuel ratio for gasoline engines?

The ideal air/fuel ratio for gasoline engines is typically between 14:1 and 15:1, depending on the specific engine and operating conditions. This range allows for complete combustion and optimal performance.

How do you measure air/fuel ratio?

Air/fuel ratio can be measured using a variety of tools, such as a wideband oxygen sensor or a handheld exhaust gas analyzer. These tools measure the amount of oxygen in the exhaust gases and can calculate the air/fuel ratio based on that data.

What factors can affect air/fuel ratio?

Several factors can affect air/fuel ratio, including engine load, engine speed, and ambient temperature. Other factors such as fuel quality, engine condition, and exhaust system design can also impact the ratio.

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