Temperature Modelling and Control of the Biomass ...npcw17.imm.dtu.dk/Proceedings/Session 2...

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Temperature Modelling and Control of the Biomass Pretreatment Process Remus Mihail Prunescu NPCW-17 26 January 2012

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Agenda πœ• πœŒπ‘π‘‡

πœ•π‘‘+ 𝛻𝑇 πœŒπ‘π’–π‘‡ = 𝛻𝑇 Γ𝑐𝛻𝑇 + 𝑆𝑇

Temperature

Modelling

Questions?

Classical and Advanced Control*

Project Description

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𝑒𝑅𝑛 = Δ𝑑[ 1 βˆ’ πœƒ 𝑣𝑇

𝑛+1 + πœƒπ‘£π‘‡π‘›] The Inbicon Demonstration Plant

Process Description I

Bio-Ethanol (144 kg) C5 Molasses (371 kg) Bio-Pellets (435 kg)

End Products

Pretreatment

Thermal Reactor Steam

Wheat Straw (1 ton)

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The Thermal Reactor Process Description II

𝐴 = πœ‹π‘Ÿ2

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Preliminary Analysis Temperature Modelling I

𝑓 π‘₯ = π‘Ž0 + π‘Žπ‘› cosπ‘›πœ‹π‘₯

𝐿+ 𝑏𝑛 sin

π‘›πœ‹π‘₯

𝐿

∞

𝑛=1

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Computational Fluid Dynamics i Temperature Modelling II

Heat Convection-Diffusion Equation:

πœ•(πœŒπ‘π‘‡)

πœ•π‘‘+ 𝛻𝑇 Γ𝑐𝑒𝑇 = 𝛻𝑇 Γ𝑐𝛻𝑇 + 𝑆𝑇

πœŒπ‘πœ•π‘‡

πœ•π‘‘+ πœŒπ‘π‘’π‘₯

πœ•π‘‡

πœ•π‘₯=

πœ•

πœ•π‘₯πœ…πœ•π‘‡

πœ•π‘₯+

πœ•

πœ•π‘¦πœ…πœ•π‘‡

πœ•π‘¦+ 𝑆𝑇

: : ∫

πœƒ

π‘Žπ‘ƒπ‘›+1𝑇𝑃

𝑛+1 = π‘Žπ‘ƒπ‘›π‘‡π‘ƒ

𝑛 + π‘ŽπΈπ‘›π‘‡πΈ

𝑛 + π‘Žπ‘Šπ‘› π‘‡π‘Š

𝑛 +

+π‘Žπ‘†π‘›π‘‡π‘†

𝑛 + π‘Žπ‘π‘›π‘‡π‘

𝑛 + π‘ŽπΈπ‘›+1𝑇𝐸

𝑛+1 + π‘Žπ‘Šπ‘›+1π‘‡π‘Š

𝑛+1 +

+π‘Žπ‘†π‘›+1𝑇𝑆

𝑛+1 + π‘Žπ‘π‘›+1𝑇𝑁

𝑛+1 + 𝑆𝑒

π‘Žπ‘ƒ,𝑁,𝑆,𝐸,π‘Š = 𝑓(πœƒ, 𝐷, 𝐹)

π‘Ž2 + 𝑏2 = 𝑐2

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Computational Fluid Dynamics ii Temperature Modelling II

π‘Žπ‘ƒπ‘›+1 + π‘Žπ‘

𝑛+1

π‘Žπ‘ƒπ‘›+1

𝑇𝑃𝑛+1 = π‘Žπ‘ƒ

𝑛 βˆ’ π‘Žπ‘π‘›

π‘Žπ‘ƒπ‘›

𝑇𝑃𝑛 +

+π‘ŽπΈπ‘›π‘‡πΈ

𝑛 + π‘Žπ‘†π‘›π‘‡π‘†

𝑛 + π‘Žπ‘Šπ‘› π‘‡π‘Š

𝑛 + π‘ŽπΈπ‘›+1𝑇𝐸

𝑛+1 +

+π‘Žπ‘†π‘›+1𝑇𝑆

𝑛+1 + π‘Žπ‘Šπ‘›+1π‘‡π‘Š

𝑛+1 +

+𝑆𝑒 + 2𝑇𝑛 π‘Žπ‘π‘› + π‘Žπ‘

𝑛+1

𝑆𝑒

Boundary Conditions:

𝐄𝐓 π‘₯ 𝑇𝑛+1 = 𝐀𝐓π‘₯ 𝑇

𝑛 + 𝐁𝐓𝑒𝑇𝑛

State Space Model:

π‘₯ + π‘Ž 𝑛 = 𝑛

π‘˜π‘₯π‘˜π‘Žπ‘›βˆ’π‘˜

𝑛

π‘˜=0

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Simulation Results Temperature Modelling III 𝑒π‘₯ = 1 +

π‘₯

1!+

π‘₯2

2!+

π‘₯3

3!+ β‹―

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PID Control (Results) Classical and Advanced Control I

𝐢 𝑠 = 𝐾𝑃 +𝐾𝐼

𝑠+ 𝐾𝐷𝑠

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LQRi (Results) Classical and Advanced Control II

𝑒 = βˆ’πΎπ‘₯

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MPC (Results) Classical and Advanced Control III

minΞ”π‘ˆπ‘˜

Ξ¦ =1

2 𝑦 π‘˜+1+𝑗|π‘˜ βˆ’ π‘Ÿπ‘˜+1+𝑗|π‘˜ 𝑄

2π‘βˆ’1

𝑗=0

+1

2 Ξ”π‘’π‘˜+𝑗|π‘˜ 𝑆

2π‘βˆ’1

𝑗=0

𝑒min < π‘’π‘˜+𝑗|π‘˜ < 𝑒max

Δ𝑒min < Ξ”π‘’π‘˜+𝑗|π‘˜ < Δ𝑒max

1 + 2 = 3

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:-??!??-:

Questions? Something grey and funny that fits on a line :-?