The Stochastic KiBaM ... or how charging probably keeps batteries - - PowerPoint PPT Presentation

The Stochastic KiBaM

... or how charging probably keeps batteries alive Holger Hermanns, Jan Krčál, Gilles Nies

Saarland University

May 5, 2015 Alpine Verification Meeting 2015

What 3 items would you take to a deserted island?

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What items up to 1 kg & 1 liter would you take?

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What items up to 1 kg & 1 liter would you take?

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Fly Your Satellite! educational program

A Cube Satellite Cube satellites for educational or scientific use

◮ Limits: 1 kg & 1 liter ◮ Mission time: up to 4 years

What do we have to squeeze in there?

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Fly Your Satellite! educational program

A Cube Satellite Cube satellites for educational or scientific use

◮ Limits: 1 kg & 1 liter ◮ Mission time: up to 4 years

What do we have to squeeze in there? We will focus on the battery!

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The kinetic battery model (KiBaM)

The two-wells illustration

p

b(t) 1−c a(t) c

I 1 − c c b(t) a(t)

Parameters

◮ c – Width of available charge tank ◮ p – Diffusion rate between tanks

KiBaM ODE System

˙ a(t) = −I + p b(t) 1 − c − a(t) c

• ˙

b(t) = p a(t) c − b(t) 1 − c

• 4 / 21

KiBaM (ctd.)

Example (Unbounded KiBaM)

1500 5000 9000

• 600

400 10 40 55 available bound load

The model supports:

◮ Discharging:

Load is positive (I > 0)

◮ Charging:

Load is negative (I > 0)

◮ Depletion:

Available charge reaches 0

(a(t) ≤ 0)

Analysis hard if load is not piecewise constant...

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Why is the KiBaM a good model?

The KiBaM captures some realistic effects: Recovery effect

2500 5000

• 1

5 10 15 20 25 30 35 available bound load

Rate-capacity effect

2500 5000 500 5 10 15 20 25 available bound load 2500 5000 700 5 10 15 20 25 available bound load 6 / 21

Solution of KiBaM ODEs

Solution of ODE system Kt,I

• a0

b0

• =
• qa(t)

ra(t) sa(t) qb(t) rb(t) sb(t)

• ·

  a0 b0 I   ⇒ Linear in a0, b0 and I

Coefficients

qa(t) = (1 − c)e−kt + c qb(t) = −(1 − c)e−kt + (1 − c) ra(t) = −c · e−kt + c rb(t) = c · e−kt + (1 − c) sa(t) = (1 − c)(e−kt − 1) k − t · c sb(t) = (1 − c)(1 − e−kt) k − t · (1 − c) ⇒ Not linear in t

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What can be added to the KiBaM?

1500 5000 9000

• 600

400 10 40 55 available bound load 1500 5000 9000

• 600

400 10 40 55 available bound load

• 5

5 10 15

• 5

5 10 15 available

• 5

5 10 15 0.02 0.04 0.06 0.08 0.1 density available 0.0622 0.0308 0.0622 0.0308 0.0622 0.0308 0.02 0.04 0.06 0.08 0.1 density

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Capacity bounds

Example (Bounded KiBaM)

1500 5000 9000

• 600

400 10 40 55 available bound load

◮ Switching ODE systems

˙ b(t) = p amax c − b(t) 1 − c

• ( ... can be solved)

◮ if current high enough

btresh(I) = bmax+I · 1 − c p

◮ But when?

t = −W u v · e− w

v

• − w

v

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Capacity bounds

Example (Underapproximated charging current)

1500 5000 9000

• 600

400 10 40 55 available bound load 1500 5000 9000

• 600

400 10 40 55 available bound load

Underapproximate charging load such that capacity bound is hit when load changes

¯ I(a0, b0) = −qa sa · a0 − ra sa · b0 + amax sa .

10 / 21

Random SoC and load

Example (Random initial SoC with random load) Random load

+ Random SoC

• 5

5 10 15

• 5

5 10 15 available bound 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 density

= t = 0 t = 20 t = 60

• 5

5 10 15

• 5

5 10 15

• 5

5 10 15

• 5

5 10 15

• 5

5 10 15 0.02 0.04 0.06 0.08 0.1 density

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How do we handle ?

Definition (Transformation Law of Random Variables) For fX-distributed vector X, injective and continuously differentiable function

g : Rd → Rd, express density of Y := g(X) as fY(y) = fX

• g−1(y)
• ·
• det
• Jg−1(y)
• ◮ Transform density of SoC conditioned on I = i:

fT(a, b | i) = f0

• K−1

T,i[a; b]

• ·
• ekT
• ◮ Integrate information of the load afterwards

fT(a, b) = ∞

−∞

f0

• K−1

T,i[a; b]

• · ekT · g(i) di.

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Bounded random SoC and load +

Example (Evolution over time)

t = 0 t = 20 t = 60

• 5

5 10 15

• 5

5 10 15

• 5

5 10 15

• 5

5 10 15

• 5

5 10 15 0.02 0.04 0.06 0.08 0.1 density

Imposing bounds 0 and 10

=⇒

• 0.0622

0.0308 0.0622 0.0308 0.0622 0.0308 0.02 0.04 0.06 0.08 0.1 density

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What can happen at the capacity bound?

The upper bound scenario . . . . . . bound charge available charge a = b

• (amax, b)
• (amax,¯

b)

• (amax, b′)
• ¯

KT

• K

t,i

∀ ≤ t ≤ T

KT,i

• (a, b)
• (ai, bi)
• (amax,B(a, b))

KT,i

K

T ,I ( a , b )

KT,0 K

T , 1 ◮ Moving within the bounds ◮ Sliding along the bound ◮ Moving from the capacity bound back within the bounds.

I(a, b) = (amaxe−kt − rba − qbb)/(rasb − rbsa), B(a, b) = −qba + qab + (qbsa − qasb) · I(a, b).

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GOMX–1 Cubesat

◮ 2-Unit Cube Satellite ◮ launched 21.11.2013 ◮ tracking airplanes using their ADS-B signal ◮ Logging plenty of internal (battery) data

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Satellite Model

Markov Task Process

Middle

190 mA 90 min.

Low start

90 mA 90 min.

Transfer

400 mA 5 min.

High

250 mA 90 min.

3 5 1 8 2 5 3 5 2 5 1 4 3 5 1 8 2 5 1 2

◮ Orbit Time: 99 min.

(1/3 in eclipse)

◮ Communication: when

close to Aalborg, DK

◮ Battery: 5000 mAh, 7.2

V, Li-Ion

◮ Solar charge: 400 mA

Additional Randomness:

◮ SoC uniformly distributed between 70% and 90% full (battery in equilibrium)

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Satellite Model

Markov Task Process

Middle

190 mA +N(0,5) 90 min.

Low start

90 mA +N(0,5) 90 min.

Transfer

400 mA +N(0,5) 5 min.

High

250 mA +N(0,5) 90 min.

3 5 1 8 2 5 3 5 2 5 1 4 3 5 1 8 2 5 1 2

◮ Orbit Time: 99 min.

(1/3 in eclipse)

◮ Communication: when

close to Aalborg, DK

◮ Battery: 5000 mAh, 7.2

V, Li-Ion

◮ Solar charge: 400 mA

Additional Randomness:

◮ SoC uniformly distributed between 70% and 90% full (battery in equilibrium) ◮ White noise in the workload model

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Computation

◮ Iterative approach stacks integrals:

fT(a, b) = ∞

−∞

f0

• K−1

T,i[a; b]

• · ekT · g(i) di.

◮ We discretize the battery SoC, we keep continuous time

17 / 21

Computation

◮ Iterative approach stacks integrals:

fT(a, b) = ∞

−∞

f0

• K−1

T,i[a; b]

• · ekT · g(i) di.

◮ We discretize the battery SoC, we keep continuous time

We tried other SHS tools

◮ SiSat ◮ Faust2

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Computation

◮ Iterative approach stacks integrals:

fT(a, b) = ∞

−∞

f0

• K−1

T,i[a; b]

• · ekT · g(i) di.

◮ We discretize the battery SoC, we keep continuous time

We tried other SHS tools

◮ SiSat ◮ Faust2

=⇒ Cannot handle the KiBaM system, cannot compare with our accuracy

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Results – SoC Distribution after 1 year

SoC distribution for decreasing battery size 5000 mAh:

0.5167 1.7 · 10−63 0.5167 1.7 · 10−63 0.5167 1.7 · 10−63

• 350
• 300
• 250
• 200
• 150
• 100
• 50

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Results – SoC Distribution after 1 year

SoC distribution for decreasing battery size 5000 mAh:

0.5167 1.7 · 10−63 0.5167 1.7 · 10−63 0.5167 1.7 · 10−63

• 350
• 300
• 250
• 200
• 150
• 100
• 50

2500 mAh:

0.5167 6.6 · 10−31 0.5167 6.6 · 10−31 0.5167 6.6 · 10−31

• 200
• 150
• 100
• 50

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Results – SoC Distribution after 1 year

SoC distribution for decreasing battery size 5000 mAh:

0.5167 1.7 · 10−63 0.5167 1.7 · 10−63 0.5167 1.7 · 10−63

• 350
• 300
• 250
• 200
• 150
• 100
• 50

2500 mAh:

0.5167 6.6 · 10−31 0.5167 6.6 · 10−31 0.5167 6.6 · 10−31

• 200
• 150
• 100
• 50

1250 mAh:

0.5167 1.7 · 10−10 0.5167 1.7 · 10−10 0.5167 1.7 · 10−10

• 160
• 140
• 120
• 100
• 80
• 60
• 40
• 20

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Results – SoC Distribution after 1 year

SoC distribution for decreasing battery size 5000 mAh:

0.5167 1.7 · 10−63 0.5167 1.7 · 10−63 0.5167 1.7 · 10−63

• 350
• 300
• 250
• 200
• 150
• 100
• 50

2500 mAh:

0.5167 6.6 · 10−31 0.5167 6.6 · 10−31 0.5167 6.6 · 10−31

• 200
• 150
• 100
• 50

1250 mAh:

0.5167 1.7 · 10−10 0.5167 1.7 · 10−10 0.5167 1.7 · 10−10

• 160
• 140
• 120
• 100
• 80
• 60
• 40
• 20

625 mAh:

0.4978 0.0365 0.4978 0.0365 0.4978 0.0365

• 100
• 80
• 60
• 40
• 20

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Results – SoC Distribution after 1 year

Effect of noisy loads (1250 mAh battery)

◮ without noise:

0.5167 1.7 · 10−10 0.5167 1.7 · 10−10 0.5167 1.7 · 10−10

• 160
• 140
• 120
• 100
• 80
• 60
• 40
• 20

◮ with noise:

0.563 2.2 · 10−10 0.563 2.2 · 10−10 0.563 2.2 · 10−10

• 180
• 160
• 140
• 120
• 100
• 80
• 60
• 40
• 20

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Results – SoC Distribution after 1 year

Could a 1 unit satellite survive with a 5000 mAh battery?

◮ 9 solar panels:

0.5167 1.7 · 10−63 0.5167 1.7 · 10−63 0.5167 1.7 · 10−63

• 350
• 300
• 250
• 200
• 150
• 100
• 50

◮ 6 solar panels:

5 · 10−138 0.99999 5 · 10−138 0.99999 5 · 10−138 0.99999

• 480
• 460
• 440
• 420
• 400
• 380
• 360
• 340
• 320

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The last slide!

Summary

◮ We extended

p

b(t) 1−c a(t) c

I 1 − c c b(t) a(t)

with and

◮ We get

0.5167 1.7 · 10−63 0.5167 1.7 · 10−63 0.5167 1.7 · 10−63

• 350
• 300
• 250
• 200
• 150
• 100
• 50

for models.

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The last slide!

Summary

◮ We extended

p

b(t) 1−c a(t) c

I 1 − c c b(t) a(t)

with and

◮ We get

0.5167 1.7 · 10−63 0.5167 1.7 · 10−63 0.5167 1.7 · 10−63

• 350
• 300
• 250
• 200
• 150
• 100
• 50

for models. Future Work

◮ Battery wear ◮ Randomized capacity bounds ◮ Temperature dependency ◮ Energy optimal scheduling (GOMX–3!)

21 / 21

The last slide!

Summary

◮ We extended

p

b(t) 1−c a(t) c

I 1 − c c b(t) a(t)

with and

◮ We get

0.5167 1.7 · 10−63 0.5167 1.7 · 10−63 0.5167 1.7 · 10−63

• 350
• 300
• 250
• 200
• 150
• 100
• 50

for models. Future Work

◮ Battery wear ◮ Randomized capacity bounds ◮ Temperature dependency ◮ Energy optimal scheduling (GOMX–3!)

Questions?

21 / 21