fundamental thermal machines for the thermodynamics of ...d. chiuchiu 1, g. gubbiotti2 1universit a...

Fundamental thermal machines for thethermodynamics of computation

J. Stat. Mech. 2016 053110

D. Chiuchiù1, G. Gubbiotti2

1Università degli Studi di Perugia

2Università Roma Tre

ICT-Energy conference 17/8/2016 Aalborg

D. Chiuchiù, G. Gubbiotti J. Stat. Mech. 2016 053110 ICT-Energy conference 1 / 18

Outline

1 Problem formulationGas enclosed by a pistonGas-Piston equationsDimensionless Gas-Piston Equations

2 Multiple scales method for the Gas-Piston equationsMultiple scales methodRelaxation to equilibriumIsothermal compression


Gas enclosed by a piston

x : piston position.

T : gas temperature.

T b: reservoir temperature.

F : external force.

Question: if the gas-piston system is at equilibrium and F or T b change,how do T and x evolve?

Standard thermodynamics:

• Describes the new final equilibrium.• No information on the dynamics.


Gas-Piston equations of Cerino et al. PRE 91, 032128

Working hypothesis

• Perfect gas.• Gas-Piston collisions are elastic.• Gas-Reservoir collisions sets the gas particle velocities according to

the Maxwell-Boltzmann distribution for the reservoir temperature T b.

• After each collision the gas temperature T =∑

i mv2 is computed.

• The gas distribution is always a Maxwellian for the temperature T .• Averaging Gas-Piston and Gas-Reservoir collisions we get equations

for x and T .


Gas-Piston equations of Cerino et al. PRE 91, 032128

ẍ +F

M− νNν + 1

1

xerfc

(√ m2T

ẋ)(ẋ

2+

T

m) +

νN

ν + 1exp

(− mẋ

2

2T

) ẋx

√2T

πm= 0,

Ṫ +2ẋ[mẋ

2+ T (1− 2ν)

]x(ν + 1)2

erfc(√ m

2Tẋ)+

√2T

πm

T − T bx

+2

x(ν + 1)2

√2mT

π

(2νT

m− ẋ2

)exp

(− mẋ

2

2T

)= 0,

M piston mass N gas particles numberm gas particle mass F external forceν mM T b reservoir temperature

Problems:

• termodynamic limits ν = m/M → 0, N →∞?• analytic solution?


Dimensionless Gas-Piston Equations

ẍ + F + erfc( εẋ√

2T

)ε2ẋ2 + Tx

+ exp(− ε

2ẋ2

2T

) ẋx

√2T

πε = 0,

Ṫ − 2 erfc( εẋ√

2T

) ẋx(ε2ẋ2 + T )− 2

√2T

πεẋ2

xexp

(− ε

2ẋ2

2T

)+

√2T

π

T − Tbεx

= 0.

• x , T , F , Tb are the dimensionless version of x , T , F , Tb• m/M → 0 and N →∞ are already taken;• ε2 = NmM is finite;• standard thermodynamics equilibrium conditions

xeq =TbF, ẋeq = 0, Teq = Tb.


Linear Dimensionless Gas-Piston Equations

• Not too far from the equilibrium• ε perturbation parameter

ẍ +F 2

Tb

(x − Tb

F

)+ 2

√2

πTbεF ẋ − F T − Tb

Tb= 0

Ṫ + 2F ẋ +

√2

πTb

F

ε(T − Tb) = 0

Treated as a third order equation for x coupled with a definition for T

TbF

...x +

[2

√Tb√2ε√

π− TbḞ

F 2+

ṪbF

+

√Tb√2√

πε

]ẍ +

[3F +

√2εṪb√π√Tb

+ 4F

π

]ẋ

+

[Ḟ +

√2F 2

√πε√Tb

]x −√2F√Tb√

πε= 0

T =TbF

ẍ + 2

√2

π

√Tbεẋ + Fx


Multiple scales method

ODE

(t, x , ẋ , ẍ , . . . ,

(n)x , ε

)= 0,

t independent variable, x = x(t) , ε small.

General idea:

1 Conjecture N behaviors with different natural time-scales.

2 Define N + 1 time functions (t0, . . . , tN), ti = ti (ε, t).Each ti isolates a single behaviour with a given time-scale.

3 Expand x and its time derivatives in ODE according to

x (t) =N∑i=0

εixi (t0, . . . , tN)+O(εN+1

), ẋ =

N∑i=0

∞∑j=0

εi∂xi∂tj

∂tj∂t

+O(εN+1

), . . .

4 Collect ε coefficients.

5 Transform the ODE in PDEs system.


Multiple scales method

6 Solve the ε0 equation for x0.

7 Subs x0 in the ε1 equation.

8 Set all the arbitrary functions in x0 to avoid secular terms in x1.

9 Solve the ε1 equation for x1.

10 Iterate for alle the remaining xi

At the end of the procedure:

• All the x0, . . . , xN are specified.• xap (t) =

∑N−1i=0 ε

ixi (t0, . . . , tN) +O(εN)

is asymptotic up totN(t, ε) = O (1).

• Crosscheck on the ansatz on the number of timescales N.


Back to the Linear Dimensionless Gas-Piston Equations

Additional working hypothesis:

• Tb is constant (w.l.o.g. Tb = 1)

• F is slowly varying over time(F = F (εt)

).

[..........Long expression for xap..........]

[..........Long expression for Tap.........]


Relaxation to equilibrium

• F = 1• x(t = 0) = 1 + x0, ẋ(t = 0) = ẋ0, T (t = 0) = 1 + T0

xap(t) =1 + K3ε2 exp

(√2π

t

ε(πε2 − 1)

)+ exp

(−εt(π + 2)√

2π

)·

·[C1 sin(t) + C2 cos(t) +

(− tε

3

2(εtθ2 − 2η)C1 + θtC2ε2 + C3ε

)sin(t)

+(− θ

2t2

2C2ε

4 + (C2ηt − C3tθ)ε3 + (C5 − C1tθ)ε2)cos(t)

]+O

(ε3)

withη = −

√2π4

(π + 4), θ = − (π2−4π−4)

4π

C1 = ẋ0, C2 = x0, C3 =(πT0+πx0+2x0)√

2π, C5 = −πT02 , K3 =

πT02.


Relaxation to equilibrium - xap goodness

• Simulate x numerically• Compute ∆x = maxt∈[0,∞[(|x − xap|)• Study ∆x as function of ε, x0, ẋ0 and T0

−10 −5 0 5 1010

−3

10−2

10−1

−2

0

2

4

−10 −5 0 5 1010

−3

10−2

10−1

−2

0

2

4

−10 −5 0 5 1010

−3

10−2

10−1

−2

0

2

4


Relaxation to equilibrium - thermodynamics

Analytic expression for the heat exchanged with the reservoir betweent = 0 and t = t

Qrelap (0, t) =x(0)− x(t) +ẋ2(0)−ẋ2(t)

2 +T (0)−T (t)

2

=(1 + x0 − xap(t)) +ẋ20−ẋ2ap(t)

2 +1+T0−Tap(t)

2 +O (ε) ,

• New non-equilibrium result.• Analytic curve of the heat exchanged with the reservoir.


Isothermal compression

• x(t = 0) = 1, ẋ(t = 0) = 0, T (t = 0) = 1• Gas compressed in a finite time, then relaxes to equilibrium

F =

1 for t < 0

1 + faεt for 0 ≤ t ≤ 1aε

1 + f for t >1

aε.

• f is F increment. 1aε is the compression duration.

xap(t) =

1faεt+1 + ε2(− 2a

2f 2

(faεt+1)5 +√

2π

af (π+2)(faεt+1)3

)+O

(ε4)

if t ∈ [0, 1aε ]

as in relaxation case if t > 1aε .


Isothermal compression - thermodynamics

Analytic formula for the net heat exchanged with the reservoir during thecompression and the following relaxation

Q linap (a) =x(0)− 1 +ẋ2(0)

2 +T (0)−1

2 + f a ε

∫ 1aε

0x(t)dt

= ln(1 + f )−2(faε)2

[(1+f )2−12

][(1+f )+

12

](1+f )4

+2faε2

√2π

[(π+

32 )(1+f )

2−π+24]

(1+f )2+O

(ε3)

• New result: analytic expectation value for the heat produced dutingan isothermal compression lasting 1aε .

• Consistency: the second principle is satisfied by this formula.


Isothermal compression - thermodynamics

Validity constrains of Q linap (a):

• f > 0• af . 1• f − ln(1 + f )� ε2

Quality test for Q linap (a)

• Simulate numerically x• Compute

∆Q = |Q lin(a)− Q linnum(a)|• Study ∆Q as function of ε, a ed f• ∆Q = O

(ε3)

101

100

101

0.25

0.5

0.75

1

1.25

1.5

1.75

2

2.25

2.5

2.75

3

1.5

2

2.5

3

3.5

4

4.5

100

102

102

101

100

1.4

1.6

1.8

2

2.2

2.4

2.6

2.8

3

3.2


Conclusions

• Dimensionless version of the eq. from PRE 91, 032128.• Application of the multiple scales method to the Linear Gas-Piston

Equations.

• Analytic formulas on the thermodynamic of non-equilibrium processes.• Thermodynamic cycles with the “slow expansion” approach.



Problem formulationGas enclosed by a pistonGas-Piston equationsDimensionless Gas-Piston Equations

Multiple scales method for the Gas-Piston equationsMultiple scales methodRelaxation to equilibriumIsothermal compression

fundamental thermal machines for the thermodynamics of ...d. chiuchiu 1, g. gubbiotti2 1universit a...

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