Does the Euler method give a better insight into adiabatic processess?

[chuakoktong]

TL;DR Summary

Sorry this is very long but does it give a better insight to adiabatic process and free expansion process for a diatomic gas? I propose this sequence before and verified it is also an Euler method, using U=2.5PV, internal energy of diatomic gas and $\frac{dU}{dv}=-p$

I verified with others the equation below is an Euler method as well with $a$ can be any value such that it give the same $\frac{dE}{dv}=-1.4\frac{p}{v}$ but with $a$ other than one, it have no meaning in physics. For anyone that already understand Euler method can omit the part i have underline.

So from the numerical integration(Euler method)
the first step is interpret as

The internal energy of a diatomic gas with internal energy U=2.5PV undergo an adiabatic expansion with a very small increase in volume from $v_0$ to $v_0+h$ and the condition for it to happen, it have to lost an amount which is equivalent to $p_0h$ where the value of p is define as instantaneous pressure when the gas with initial pressure $p_0$ just about to expand, which is p_0 in this case. In other word, if a gas with initial pressure $p_0$ just about to expand and not able to lose this amount of energy $p_0h$ then it cannot define as adiabatic expansion? Same goes if the gas with same condition being compress and not able to increase internal energy of $p_0h$ when it just about to compress, then it cannot consider as adiabatic compression?

$2.5p_0v_0-p_0(h)=2.5p_1(v_0+h)$

and compare $p_1$ from this equation $p_0v_0^{1.4}=p_1(v_0+h)^{1.4}$

Both equation yield $p_1=p_0-1.4h\frac{p_0}{v}$ by using binomial approximation if h is very small value

Same goes for compression

$2.5p_0v_0-p_0(-h)=2.5p_1(v_0-h)$

and compare $p_1$ from this equation $p_0v_0^{1.4}=p_1(v_0-h)^{1.4}$

$p_1$ from both equation give $p_1=p_0+1.4h\frac{p_0}{v_0}$ when using binomial approximation if h is very small

For a free expansion, we can consider each step the lose in internal energy is 0 or $\frac{dU}{dv}=0$

so this the numerical integration which result in PV=constant

$2.5p_0v_0+0(h)=2.5p_1(v_0+h)$
$2.5p_1(v_0+h)+0(h)=2.5p_2(v_0+2h)$
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.
.
$2.5p_{n-1}(v_0+(n-1)h)+0(h)=2.5p_n(v_0+nh)$

whole sequence simplify to

$2.5p_0v_0=2.5p_n(v_0+nh)$_____________________________________________________________________________________________________________________
Consider the sequence below

$2.5p_0v_0-p_0(\frac{h}{n})=2.5p_1(v_0+\frac{h}{n})$

$2.5p_1(v_0+\frac{h}{n})-p_1(\frac{h}{n})=2.5p_2(v_0+2\frac{h}{n})$

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.

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$2.5p_{n-1}(v_0+(n-1)\frac{h}{n})-p_{n-1}(\frac{h}{n})=2.5p_n(v_0+h)$

and second sequence

$p_0v_0^{1.08687}-0.31313p_0v_0^{0.08687}(\frac{h}{n})=p_1(v_0+\frac{h}{n})^{1.08687}$

$p_1(v_0+\frac{h}{n})^{1.08687}-0.31313p_{1}(v_0+\frac{h}{n})^{0.08687}(\frac{h}{n})=p_2(v_0+2\frac{h}{n})^{1.08687}$

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.

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$p_{n-1}(v_0+(n-1)\frac{h}{n})^{1.08687}-0.31313p_{n-1}(v_0+(n-1)\frac{h}{n})^{0.08687}(\frac{h}{n})=p_n(v_0+h)^{1.08687}$

Does this make sense to say both sequence is the numerical integration for DE $\frac{dp}{dv}=-1.4\frac{p}{v}$,when n tend to infinity where the exact solution of this DE is $p_0v_0^{1.4}$=constant consider $p(v_0)=p_0$ which is actually the adiabatic process in diatomic gas.

I treat the first sequence $E=u(p,v)=2.5pv$ and $\frac{dE}{dv}=g(p,v)=-p$

using

$u_p\frac{dp}{dv}+u_v$

$2.5v\frac{dp}{dv}+2.5p=-p$

$\frac{dp}{dv}=-1.4\frac{p}{v}$

and second sequence $E=u_1(p,v)=pv^{1.08687}$, $\frac{dE}{dv}=g_1(p,v)=-0.31313p$

did the same as above which yield $\frac{dp}{dv}=-1.4\frac{p}{v}$

Attached also printscreen of python of the sequence but I afraid practically is not possible to show exactly the same so theoretically if n tend to infinity, is $p_n$ the same for both sequence? Consider $p(v_0)=p_0$ and h is constant.first sequence
[1]: https://i.stack.imgur.com/l6lyV.jpg
second sequence
[2]: https://i.stack.imgur.com/tiok9.jpg

Where the algorithm calculate $p_n=p(v)=p(51)$ using $n=50000,h=50,p(1)=1$ while the exact solution using $pv^{1.4}$=constant is $p(51)=\frac{1}{51^{1.4}}=0.00406819$

This is the answer verified by other people
Yes, the explicit Euler method is convergent with order 1. As the differential equations are equivalent and all are regular for $v\in[1,51]$ with smooth right sides, the method is applicable, converges for these ODE, and in the limit will converge to equivalent solutions.

If $q=pv^a$, then $\frac{dq}{dv}=v^a\frac{dp}{dv}+apv^{a-1}=(-1.4+a)pv^{a-1}=(-1.4+a)\frac{q}v$ and the Euler method for this equation reads
$$
q_{n+1}=q_n+\Delta v\,(-1.4+a)\frac{q_n}{v_n}\iff p_{n+1}v_{n+1}^a=p_{n}v_{n}^a-\Delta v\,(1.4-a)p_{n}v_{n}^{a-1}
$$
In your case, $a=1$ or $a=1.08687$ (with $1.4-a=0.31313$).

 

[Chestermiller]

I'm having trouble understanding what this is about. Are you trying to say that solving the differential equation numerically using Euler's method somehow gives you added insight into what is happening physically in an adiabatic reversible expansion?

 

[chuakoktong]

I'm having trouble understanding what this is about. Are you trying to say that solving the differential equation numerically using Euler's method somehow gives you added insight into what is happening physically in an adiabatic reversible expansion?

Hi Chestermiller thank for your feedback. That what i meant.

Just like solving a differential equation regarding draining of water in a cylindrical tank through a small hole at bottom of the tank.

We consider pressure of water at bottom of tank with water of height h as
p=$\rho gh$

and pressure also define as

p=$0.5\rho v^2$

Hence velocity of water coming out of the small hole is $v=\surd 2gh$
and velocity time area of the small hole,A we get change of volume= $A\surd 2gh$

since the volume is decreasing we define $\frac{dV}{dt}=-A\surd 2gh$
but $V=\pi r^2h$ where r is constant hence $\frac{dV}{dt}=\pi r^2\frac{dh}{dt}$
so we can derive what is $\frac{dh}{dt}$ and solve this differential equation to get the relationship of the height h with time.

If using Euler method will consider the initial condition $h(t_0)=h_0$
Initial volume of water in the tank=$\pi r^2h_0$
Since the rate flow of water from the small hole is dependency of h, consider it just about to flow out from the tank with height $h_0$ and after passing a very short amount of time,t it lose a volume equal to $A\surd 2gh_0t$ and attain a new height namely $h_1$, then the next step we repeat the same thing, consider now the height is $h_1$ and water just about to flow out from the hole and after passing a very short amount of time t,it lose a volume equal to $A\surd 2gh_1t$ and attain a new height $h_2$ and so on.

This is what the equation represent
$\pi r^2h_0-A\surd 2gh_0t=\pi r^2h_1$
$\pi r^2h_1-A\surd 2gh_1t=\pi r^2h_2$
and so on

ie r=1,$h_0=1$,t=0.01 then we can calculate $h_1,h_2$ and so on. Definitely we will not use this equation for calculation as it is very inefficient but does it give a better insight of the process flow? Adiabatic process should be the same consideration by taking internal energy,U=$\frac{1}{\gamma-1}PV$ and $\frac{dU}{dv}=-p$

Does this make sense?

 

[Chestermiller]

To me, the numerical approximation to the solution to the differential equation makes no more sense than the analytic solution to the same equation. But that's just me.

 

[atyy]

I think the Euler method forms part of the intuition one has for differential equations, so this could be helpful.

 

[chuakoktong]

To me, the numerical approximation to the solution to the differential equation makes no more sense than the analytic solution to the same equation. But that's just me.

Hi Chestermiller, how about the analytical solution and approximation solution below for an adiabatic process?
Consider an adiabatic expansion with initial condition $p(v_0)=p_0$ undergo an infinitesimal small expansion from $v_0$ to $v_0+h$

The analytical solution can be obtain from
$p_0v_0^{\gamma}=p_1(v_0+h)^{\gamma}$
$p_1=p_0(\frac{v_0}{v_0+h})^{\gamma}$

using binomial approximation,
$p_1=p_0(\frac{1}{1+\frac{h}{v_0}})^{\gamma}=p_0(1+\frac{h}{v_0})^{-\gamma}\approx p_0-h\gamma \frac{p_0}{v_0}$

While $p_1$ obtain from the numerical approximation

$\frac{1}{\gamma-1}p_0v_0-p_0h=\frac{1}{\gamma-1}p_1(v_0+h)$
$p_1=p_0\frac{1-\frac{h(\gamma-1)}{v_0}}{1+\frac{h}{v_0}}=p_0(1-\frac{h(\gamma-1)}{v_0})(1+\frac{h}{v_0})^{-1}\approx p_0-h\gamma\frac{p_0}{v_0}$

It hold also for adiabatic compression
$p_0v_0^{\gamma}=p_1(v_0-h)^{\gamma}$
$p_1=p_0(\frac{v_0}{v_0-h})^{\gamma}=p_0(\frac{1}{1-\frac{h}{v_0}})^{\gamma}=p_0(1-\frac{h}{v_0})^{-\gamma}\approx p_0+h\gamma\frac{p_0}{v_0}$

While $p_1$ obtain from the numerical approximation

$\frac{1}{\gamma-1}p_0v_0-p_0(-h)=\frac{1}{\gamma-1}p_1(v_0-h)$
$p_1=p_0\frac{1-\frac{h(\gamma-1)}{v_0}}{1+\frac{h}{v_0}}=p_0(1+\frac{h(\gamma-1)}{v_0})(1-\frac{h}{v_0})^{-1}\approx p_0+h\gamma\frac{h}{v_0}$

 

[scottdave]

I can see where numerical methods may help you understand what's going on with the differential equation.

When I first learned derivatives, we took a tiny Δx change, then find the corresponding new y value, then find Δy from that and find the slope.

Then learning about integrals, we approximated area by adding up tiny rectangles. The numerical solutions to differential equations are using that - finding the approximation of an integral.