Adiabatic Engine Ratios

  • Thread starter Varaia
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This one had three bullets by it so it's a pretty tough one.

The cycle in the picture (see attachment) represents the operation of a gasoline internal combustion engine. Assume the gasoline air intake mixture is an ideal gas with (gamma)=1.30. What are the ratios a) T2/T1 b)T3/T1 c)T4/T1 d)p3/p1 e)p4/p1 also f) what is the engine efficiency?

Work and rambling thoughts so far... please tell me if I'm wrong.
This would be a change in entropy problem for the first couple of ratios right?
(delta)S=Sfinal-Sinitial=nRln(Vf/Vi)+nCvln(Tf/Ti)
yet it is an adiabatic operation so
pV^(1.3)=a constant during the adiabatic process
and since pV=nRT
TV^(1.3-1)= a constant
or TiVi^(1.3-1)=TfVf^(1.3-1)

now in the picture there are two adiabatic processes one when the gas is expanding 2 to 3 so temperature is decreasing and one from 4 to 1 where the gas is being compressed and temperature is increasing so the ratio of
a) T2/T1 since there is no volume change= 3.00 (duh)


other ratio's I can see T3/T4=3.00,

P1V1=T1 3P1V1=T2 T1=1/3(T2) Still just not seeing how they get the other ratios

Question: What formula do I use to find out b,c,d, and e?
then I can do f.) on my own.

Numerical answers b)1.98 c)0.660 d)0.495 e)0.165 f)34%
 

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  • #2
Andrew Mason
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Varaia said:
This one had three bullets by it so it's a pretty tough one.
Question: What formula do I use to find out b,c,d, and e?
then I can do f.) on my own.

Numerical answers b)1.98 c)0.660 d)0.495 e)0.165 f)34%
I am assuming that P2 = 3.00P1, although your graph is a little confusing, So of course T2/T1 = 3.00

For the adiabatic expansion from 2 to 3,

[tex]P_2V_2^\gamma = P_3V_3^\gamma[/tex]

[tex]P_3 = \frac{P_2V_2^\gamma}{V_3^\gamma}[/tex]

Since V3 = 4V2 = 4V1 and P2 = 3P1, you can work out P3 in terms of P1. And the ideal gas equation will give you T3.

Getting P4 is similar to finding P3.

[tex]P_4 = \frac{P_1V_1^\gamma}{(4V_1)^\gamma}[/tex]

The efficiency part is a little complicated, I think.

Edit: To find the efficiency, I thought one would have to determine the work done in the cycle by integrating under the curves.
[itex]\eta = \frac{W}{Q_H} = \frac{Q_H-Q_C}{Q_H} = 1 - \frac{Q_C}{Q_H}[/tex]

But looking at this again, I see that is not necessary since there is no heat change from 2 to 3 or 4 to 1 (adiabatic). The heat is gained and lost at constant volume from 1 to 2 and 3 to 4. So:

[tex]Q_C = C_v\Delta T_{3-4}[/tex] and [tex]Q_H = C_v\Delta T_{1-2}[/tex]

the efficiency is easily determined from the temperatures at 1, 2, 3 and 4.

AM
 
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