Adiabatic process

  • Thread starter menco
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Homework Statement


In a diesel engine, the piston compresses the air-fuel mixture from an initial volume of 630 cm^3 to a final volume of 30cm^3. If the initial temperature of the air-fuel mixture is 45 degrees C and the process is occurring adiabatically, determine the final temperature. Comment on the significance of the result.


Homework Equations


V(initial) = 6.3x10^-4 m^3
V(final = 3.0x10^-5 m^3
T(initial) = 318.15 K
T(final) = ?
y = 1.4

T(final) = T(initial)*(V(initial)/V(final))^y-1

The Attempt at a Solution



T(final) = 318.15(6.3x10^-4/3.0x10^-5)^0.4
T(final) = 1075.28 K or 802.13 degrees C


Is that on the right track? I found the equation online but I am a little confused how the equation is actually formed from T(f) = P(f)*V(f) / nR which I have in my textbook.
 

Answers and Replies

  • #2
ehild
Homework Helper
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During an adiabatic process, no heat exchange occurs, so the change of internal energy is do to the work alone: dU=-PdV. PV=nRT is valid but the temperature and pressure changes during the process: [itex]P=nRT/V[/itex]. For an ideal gas, the internal energy is [itex]U=nC_vT[/itex], [itex]dU=nC_vdT[/itex], so [itex]nC_vdT=-(nRT/V) dV[/itex]. n cancels. Collect the like terms and integrate

[tex]\int{\frac{dT}{T}}=\int{-\frac{R}{C_v}\frac{dV}{V}}[/tex]

[tex] \ln(T)=-\frac{R}{C_v} \ln(V) +const[/tex].

R/Cv can be written in terms of γ=Cp/Cv: [itex]\frac{R}{C_v}=\gamma -1[/itex].

You can rewrite the equation as [tex] \ln(TV^{\gamma-1})=const [/tex],

that is [tex]TV^{\gamma-1}=const[/tex] .

ehild
 

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