ideal gas and engine cylinder

[fullcircle69]

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?

 

[ione]

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

 

[fullcircle69]

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.

 

[IRstuff]

"PV=nRT or Zekeman"

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

TTFN

FAQ731-376

 

[fullcircle69]

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

 

[IRstuff]

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

FAQ731-376

 

[ione]

For better under standing

http://www.land-and-sea.com/dyno-tech-talk/corrected-horsepower.htm

 

[fullcircle69]

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?

 

[IRstuff]

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

FAQ731-376

 

[fullcircle69]

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

Can this be calculated?

 

[IRstuff]

Not what I said.

Yes, ostensibly, that's what the correction factor does for you already.

TTFN

FAQ731-376

 

[fullcircle69]

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?

 

[IRstuff]

"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

FAQ731-376

 

[fullcircle69]

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.

 

[sailoday28]

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

 

[sailoday28]

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)

 

[fullcircle69]

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

 

[sailoday28]

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.