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?
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
RE: ideal gas and engine cylinder
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
RE: ideal gas and engine cylinder
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
TTFN
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RE: ideal gas and engine cylinder
How is t/p different?
RE: ideal gas and engine cylinder
RE: ideal gas and engine cylinder
TTFN
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RE: ideal gas and engine cylinder
RE: ideal gas and engine cylinder
PV=nRT
RE: ideal gas and engine cylinder
TTFN
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RE: ideal gas and engine cylinder
40 moles vs 34 moles.
I forgot what the definition of a mole was!!!
RE: ideal gas and engine cylinder
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
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
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
RE: ideal gas and engine cylinder
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
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to calculate teh air density of both atmopheres
Cyl A:
70F -> 294.26111K
29inHg -> 98205.2565PA
Cyl B:
80F -> 299.816666K
29.5475-> 100059.304Pa
I get the same density 1.163kg.m^3 for each.
IRSTUFF - Please explain how this is wrong?
RE: ideal gas and engine cylinder
TTFN
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RE: ideal gas and engine cylinder
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
Any further insight on this would be appreciated.
RE: ideal gas and engine cylinder
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
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
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RE: ideal gas and engine cylinder
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RE: ideal gas and engine cylinder
RE: ideal gas and engine cylinder
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
Can this be calculated?
RE: ideal gas and engine cylinder
Yes, ostensibly, that's what the correction factor does for you already.
TTFN
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RE: ideal gas and engine cylinder
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
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
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
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
RE: ideal gas and engine cylinder
I am trying to show how I would determine the difference in density. For the example, with perfect gas, adiabatic conditions, etc.