How to Calculate BM/OM Ratio? — Theoretical Basis and Its Application —

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Methodologies Guidebook. CDM. Released!. How to Calculate BM/OM Ratio? — Theoretical Basis and Its Application —. Climate Experts Ltd. Naoki Matsuo n_matsuo@climate-experts.info. What information is needed?. Conventional Methodologies - PowerPoint PPT Presentation

Transcript of How to Calculate BM/OM Ratio? — Theoretical Basis and Its Application —

How to Calculate BM/OM Ratio?— Theoretical Basis and Its Application —

Climate Experts Ltd.Naoki Matsuo

n_matsuo@climate-experts.info

Released!

What information is needed?

Conventional MethodologiesOnly calculation method of EF (emission factor) is providedLittle consideration of “What is the Baseline Scenario”

Important Information:Which measures (plants) would be used to provide the electricity to meet the demand? Capacity/Fuel/Operation/When? (for BM) How? (for OM)

Shall the Methodology provide the steps “How to identify the Baseline Scenario” ?

A Simple Example I

Scenarios:PJS: NG: 100 MW (EFNG)BLS: Coal: 60 MW (EFCoal) identification needed!

• This means that – the project replaces the installation plan of

60 MW coal fired power plant, or – postpone the power development plan,

which may be by coal fired power plant(s) with 60% of the project capacity (for a while).

A Simple Example II

Build Margin Component:60 MW / 100 MW = 0.6

Operating Margin Component:(100 MW – 60 MW) / 100 MW = 0.4

Emission Factor of the Displacement EffectEFDisplacement = 0.4·EFGrid(OM) + 0.6·(EFCoal – EFNG)

Emission Reductions = ElectricityNG· EFDisplacement

AssumptionOperation pattern of Coal and NG are identical

NG100 MW

Coal 60 MW

PJS BLS

OM

BM

Formula

wOM =CAPADD

PJ −CAPADDBL

CAPADDPJ

⎣ ⎢ ⎤

⎦ ⎥

wBM =CAPADD

BL

CAPADDPJ

⎣ ⎢ ⎤

⎦ ⎥

Assumption Operation pattern of PJ and BL plants are identical In case operation pattern is different, generated electricity

is used, instead of capacity. (not easy to obtain BL plant’s electricity generation because it depends on the grid operation method)

Theory I

Theory II

BE = dE0

EBAUBL

∫ EFBAU (E) + dE0

EADDBL

∫ EFADDBL(E)

PE = dE0

EBAUPJ

∫ EFBAU (E) + dE0

EADDPJ

∫ EFADDPJ (E)

ER = dEEBAU

PJ

EBAUBL

∫ EFBAU (E) − dEEADD

BL

EADDPJ

∫ EFADDPJ(E)[ ] + dE

0

EADDBL

∫ ΔEFADD(E)

dE EF (E)E0

E0+δE

∫ ≅ δE EF (E0 )

EROM ≅ (EADDPJ − EADD

BL) EFBAU(EBAUPJ )− EFADD

PJ(EADDPJ)[ ]

= EADDPJ

•(EADD

PJ − EADDBL)

EADDPJ

⎢ ⎢ ⎢

⎥ ⎥ ⎥• EFBAU(EBAU

PJ )− EFADDPJ(EADD

PJ)[ ]

= EADDPJ

• wOM • EFOM

ERBM = EADDPJ

•1

EADDPJ

⎢ ⎢ ⎢

⎥ ⎥ ⎥• dE ΔEFADD(E)

0

EADDBL

≅ EADDPJ

•EADD

BL

EADDPJ

⎢ ⎢ ⎢

⎥ ⎥ ⎥• EFBM

= EADDPJ

• wBM • EFBM

CD-ROM (E-Card) available !