Decrease in pressure due to adiabatic expansion derivation?

[monnapomona]

Homework Statement

Show that $\frac{dp}{p}$ =$\frac{\gamma}{\gamma-1}$$\frac{dT}{T}$ if the decrease in pressure is due to an adiabatic expansion.

Homework Equations

Poisson equations:
Pv$^{\gamma}$
Tv$^{\gamma - 1}$

Ideal Gas Law:
Pv=R$_{d}$T, where R$_{d}$ is the dry air gas constant.

Hydrostatic Equation:
$\frac{dp}{dz}$ = - ρg

The Attempt at a Solution

I tried making those two equations equal to each other since they are equivalent (i think) and differentiating on both sides but i ended up with $\frac{\gamma - 1}{\gamma}$ in the final result...

EDIT: I was looking up adiabatic atmosphere and found this. I'm wondering where the formula p$^{\gamma - 1}$T$^{\gamma}$ = constant comes from...

 

[ehild]

Homework Statement

Show that $\frac{dp}{p}$ =$\frac{\gamma}{\gamma-1}$$\frac{dT}{T}$ if the decrease in pressure is due to an adiabatic expansion.

Homework Equations

Poisson equations:
Pv$^{\gamma}$ =??
Tv$^{\gamma - 1}$=??

Ideal Gas Law:
Pv=R$_{d}$T, where R$_{d}$ is the dry air gas constant.

Hydrostatic Equation:
$\frac{dp}{dz}$ = - ρg

The Attempt at a Solution

I tried making those two equations equal to each other since they are equivalent (i think) and differentiating on both sides but i ended up with $\frac{\gamma - 1}{\gamma}$ in the final result...

EDIT: I was looking up adiabatic atmosphere and found this. I'm wondering where the formula p$^{\gamma - 1}$T$^{\gamma}$ = constant comes from...

How is γ related to Cp and Cv?

You can consider air as ideal gas. Use the Ideal Gas Law to eliminate V. ehild

 

[monnapomona]

How is γ related to Cp and Cv?

You can consider air as ideal gas. Use the Ideal Gas Law to eliminate V.


ehild

I think $\gamma$ = Cp/Cv and R = Cp - Cv.
I used an entropy equation and made it equal to 0:
0 = Cp*ln(T2/T1) - R*ln(p2/p1) and got (R/Cp)*ln(p2/p1) = ln(T2/T1). So I solved for R/Cv = $\gamma$ -1 / $\gamma$... and the final result was
(P2/P1)^(($\gamma$ -1) / $\gamma$) = T2/T1

Am I on the right track with this?

 

[ehild]

You are on the right track, but write the equation with the differentials.

ehild

 

[monnapomona]

You are on the right track, but write the equation with the differentials.

ehild

This is where I'm stuck... could i say P2/P1 = P and T2/T1 = T, then take the ln of the equation to bring the exponent down?

 

[ehild]

Assume that P1=Po and T1=To are fixed, and P2=Po+dP, T2=To+dT, dT and dP very small. How are dP and dT related? Have you studied differentials? ehild

 

[monnapomona]

Assume that P1=Po and T1=To are fixed, and P2=Po+dP, T2=To+dT, dT and dP very small. How are dP and dT related? Have you studied differentials?


ehild

Is dt and dp directly proportional?

And I have calculus background from a few years ago so I'm a bit rusty on some concepts... like differentials.

 

[ehild]

Try to refresh your Calculus knowledge.

You have the equation (P2/P1)^((γ -1) / γ) = T2/T1.

Write P2/P1=(Po+dP)/Po=1+dP/Po and T2/T1=1+dT/To. dP/Po << and dT/To <<1. The original equation becomes
(1+dP/Po)^(γ-1)/γ=1+dT/To.
Use the approximation (1+δ)ⁿ ≈1+nδ,valid for δ<<1.

Spoiler

You can check it, what is (1+0.001)³, for example?

What do you get?


ehild

 

[monnapomona]

Try to refresh your Calculus knowledge.

You have the equation (P2/P1)^((γ -1) / γ) = T2/T1.

Write P2/P1=(Po+dP)/Po=1+dP/Po and T2/T1=1+dT/To. dP/Po << and dT/To <<1. The original equation becomes
(1+dP/Po)^(γ-1)/γ=1+dT/To.
Use the approximation (1+δ)ⁿ ≈1+nδ,valid for δ<<1.

Spoiler

You can check it, what is (1+0.001)³, for example?

What do you get?


ehild

You would get 1.003003 so its approximately 1. Curious question, why does this approximation matter to the proof or derivation of the original question?

 

[ehild]

1. Curious question, why does this approximation matter to the proof or derivation of the original question?

You get the desired formula with that approximation.


ehild