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ideal gas and engine cylinder

ideal gas and engine cylinder

ideal gas and engine cylinder

(OP)
I am trying to understand what would happen in a certain scenario regarding an internal combustion engine.  I am trying to determine why might a cylinder fill with more atmosphere when comparing two atmosphere's that have the same density, but one has higher pressure.

Lets say we have two cylinders that are the same volume and are both sealed under vacuum with a valve.  Cylinder A is placed in an atmosphere of an ideal gas that is at 70F and 29inHg.  Cylinder B is placed in another atmosphere that is 80F and 29.5inHg.  So both atmosphere have the same gas DENSITY.

So now if you open both valves for the same amount of time, do both containers fill with the same amount of air?

RE: ideal gas and engine cylinder

PV=nRT

RE: ideal gas and engine cylinder

(OP)
PV=nRT Ideal gas law

Can you please explain in more detail how to apply the ideal gas law to this example?  

Don't both atmosphere weigh the same per volume and doesn't that weight cause the atmosphere to fill the cylinders?

RE: ideal gas and engine cylinder

One equation.  One unknown.  Solve for the unknown.

RE: ideal gas and engine cylinder

(OP)
Cyl A:

70F -> 294.26111K
29inHg -> 98205.2565PA

Density 1.16272  

Volume=2.99638e-3 x R


Cyl B:

80F -> 299.816666K
29.5475-> 100059.304Pa

Density 1.162721

Volume=2.99638e-3 x R


Based on these calc's the volume of air in each cyl is the same....  

I was told that the air with the higher pressure would yield more volume in the cylinder.  But my reasoning could not verify this and it appears that these calculations agree with me.   

Do my calc's appear corrent?

 

RE: ideal gas and engine cylinder

You're confused.  Volume of a cylinder is a mechanical construct given by bore and stroke.  Not sure what you're giving as fact and what you're giving as supposition, but the two gases must be at differing densities if the volume is the same, but T/P is different.  That's the whole point of the ideal gas law.

TTFN

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RE: ideal gas and engine cylinder

(OP)
Your right I am confused, but you see my calculations, so what am I doing wrong?

How is t/p different?

RE: ideal gas and engine cylinder

(OP)
If my original post is not clear, let me try to restate my question.  Which cylinder A or B will have more molecules of gas after the valve is opened?

RE: ideal gas and engine cylinder

(OP)
I wish I was still in school! Why do you ask?

RE: ideal gas and engine cylinder

Perhaps my original answer was unclear.  So I'll write it again.

PV=nRT

RE: ideal gas and engine cylinder

(OP)
Thanks.

40 moles vs 34 moles.

I forgot what the definition of a mole was!!!

RE: ideal gas and engine cylinder

Look, if the density is the same then how do you get a difference in the number of moles.
The gas law stated should prove this, so you have a math error or your original statement about the equality of densities does not square with your P,T values.

RE: ideal gas and engine cylinder

On second blush, you might have a more interesting problem, since in each case the atmosphere pushes the ideal gas into the cylinder with an amount of work=PV, so the internal energy in the cylinder is increased by PV; thus the new    internal energy is equal to the enthalpy of an equal weight of atmospheric gas.

Now you will end up with a new T in each case to which you can apply the ideal gas law.


 

RE: ideal gas and engine cylinder

(OP)


I thought about this some more last night and got confused again because I didn't understand how i could have a different number of moles if the density was the same?

Zekeman comments may be more relevant to this problem.  

So I assume now i have to base it on a new T that is created after the "work' has been done through the gas transfered to each cylinder.  (Zekeman is this what you mean?)

RE: ideal gas and engine cylinder

Yes

RE: ideal gas and engine cylinder

There should be no confusion about your P and T values.  If they are correct, then n = PV/RT.  Again, you are still unclear about what is fact/measurement, and what is supposition.  

Your original statement:

"Lets say we have two cylinders that are the same volume and are both sealed under vacuum with a valve.  Cylinder A is placed in an atmosphere of an ideal gas that is at 70F and 29inHg.  Cylinder B is placed in another atmosphere that is 80F and 29.5inHg.  So both atmosphere have the same gas DENSITY."

Your last sentence is incorrect, based on the first two sentences.  Density, as measured by n/V is equal to P/RT, so doing the math shows that condition A has a higher gas density than condition B.

TTFN

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RE: ideal gas and engine cylinder

The density at that combination of pressure and temperature is the same and so the number of moles in the two cases MUST be the same, as the volume V1 at set condition (P1,T1) is the same as volume V2 at set condition (P2,T2)


V1 =P1/(n1*T1)
V2 = P2/(n2*T2)

V1 = V2 implies P1/(n1*T1) = P2/(n2*T2)


But with your values
P1/T1 = P2/T2

So n1 = n2
 

RE: ideal gas and engine cylinder

(OP)
What is looks like to me, at this point, is that either when the valves are open each cylinder will end up with the same number of molecules based on PV=nRT or Zekeman is on the right path with looking at the problem based on the amount of Work=PV.

Any further insight on this would be appreciated.



 

RE: ideal gas and engine cylinder

"PV=nRT or Zekeman"

What's the difference? Both approaches are using the same equation, so the results must be the same.

TTFN

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RE: ideal gas and engine cylinder

(OP)
Here is why I asked the question to begin with.  For internal combustion engines they use a dyno correction factor (cf) to correct for horsepower in different atmopheres.

IF you calculate the cf using the values shown above the cf  will be less(more power) for the atmophere with the higher pressure.  I can not reason why this should be, hence why I asked the orignal question.

Here is the equation using to calc cf:

cf=1.18[(990/Pa)((Tc+273)/298)^0.5]-0.18

where:   cf = the dyno correction factor
          Pd = the pressure of the dry air, mb
          Tc = ambient temperature, deg C
 

RE: ideal gas and engine cylinder

That's because there are other factors involved besides just the cylinder, as the equation hints.  The actual combustion process is affected by the pressure and temperature of the reactants.  The back pressure on the exhaust may be a factor.  The efficiency of the cooling system is affected by temperature.

TTFN

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RE: ideal gas and engine cylinder

(OP)
Sure there are many factor to make horsepower.  But the atmoshere in which we are discussing has exactly the same amount of oxygen in it.  So unless I can figure out how atmosphere B causes more oxygen to fill in the cylinder I cannot understand why it woudl create more power or from where that power would come from?

RE: ideal gas and engine cylinder

Just because the amount of reactant is the same, it does not mean that the results would be the same.  A crude and not quite germane example was shown on Effing Science a while ago.  4 gallons of gasoline in a trough slow vaporizes and burns continuous for several minutes.  The same 4 gallons dispersed over about 10 cubic meters produces an instant explosion.

A 10°F temperature difference means that colder mixture must steal 10°F of joule heating to get to the same combustion temperature.  That takes away from the available power.

TTFN

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RE: ideal gas and engine cylinder

(OP)
So your saying that the potential energy stored in B is greater then in A?

Can this be calculated?

RE: ideal gas and engine cylinder

(OP)
Irstuff ->  Is this better?  The difference in the temperature of the charge, even though at the same density, can effect the combustion process additionally to create power.

Do you feel this effect is independent of pressure?

Might it be reasoanble that if we had a similar scenario with more water vapor in the charge that the vapor might increase the joules needed to reach combustion temp as well?



  

RE: ideal gas and engine cylinder

"create" is, I think, a misleading term; "result in more available" might be more accurate.  Obviously, the "power" is related to the total available energy, which for a given stoichiometry, is fixed by the chemistry/physics of the combustion.  How much of the resultant energy actually goes into turning the wheels is a different matter altogether.

Given that the correction factor includes pressure, the answer would appear to be no.

TTFN

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RE: ideal gas and engine cylinder

(OP)
I agree with your explanation on how the higher temperature
requires less energy to reach combustion.  But I can't figure out the role of the pressure.  

RE: ideal gas and engine cylinder


Lets say we have two cylinders that are the same volume and are both sealed under vacuum with a valve
Assuming fixed volume and no heat transfer. Do an energy balance.
With perfect gas constant specific heats.
Change of internal energy in volume =Ho* amount of air flowing in.   Where Ho is specific enthalpy of surrounding air.

dU/dt=wHo      dU/dt  rate of energy increase
               w is flow rate.
Subscripts  f and i    for final and initial conditosn
MfCvTf-MiCvTi=Ho(Mi-Mf)=CpToi(Mf-Mi)
neglecting iniial mass (vacuum conditions)
MfCvTf=CpToi Mf
Tf=gamma*Toi       where gamma =Cp/Cv

     
 

RE: ideal gas and engine cylinder

Continuing from my previous post.
Using PV=nRT   where Final pressure  p  is surrounging ambient    and T=gamma*Toi
PV=nRgamma*Toi
(Pcase1/Pcase2)*(Toicase2/Toicase1)=(ncase1/ncase2)

RE: ideal gas and engine cylinder

(OP)
sailday28-What point are you making?  Are you saying that one air sample has more enthalpy/

RE: ideal gas and engine cylinder

sailday28-What point are you making?  Are you saying that one air sample has more enthalpy/
I am trying to show how I would determine the difference in density.  For the example, with perfect gas, adiabatic conditions, etc.

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