Provincial Electricity Authority, Thailand 1

Simple Thermal Modeling of Oil-

Immersed Distribution

Transformers in PEA. Thailand.Power Quality Control Section

Provincial Electricity Authority

Provincial Electricity Authority, Thailand 2

CONTENTS

OBJECTIVE INTRODUCTION SOPE OF WORK THEORY EXPERIMENTAL

WORKS CONCLUSION

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OBJECTIVE

To predict load current by measure top surface of transformer and ambient temperature.

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INTRODUCTION

Transformers are the largest part of investments of distribution system

It is advisable to know the insulation condition, operating life, and loading capacitance of a transformer to ensure its economical operation and secured service.

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INTRODUCTION

Rule of Thumb for PEA. “ Load current of transformers should not be greater than 80% of rated current.”

Measure load current of transformer. To ensure its operation.

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INTRODUCTION

Now PEA. use IR. Camera to do preventive maintenance for distribution system.

We found many electrical connection problems. Easy for us to identify problems and solve them.

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INTRODUCTION

For distribution transformer, there is no temperature base for each loading current to consider the operating temperature.

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SCOPE OF WORK

Consider 250 kVA (22kV/400-230V) Transformers in 22 kV distribution system, Chiangmai, Thailand.

Neglect heat that generate from harmonic current.

Assume load variation by time of transformers is very small, raw data may be for steady state.

Assume all transformers are healthy. No arc, corona, partial discharge inside transformer.

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THEORY

Losses in transformers

Heat transfer Equation

fitting Thermal-

Electrical analogy

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LOSSES IN TRANSFORMER

Core loss Copper loss

%Load

Ploss

Ploss = Pcore + PcuPloss = Pcore + Pcu

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1st LAW of THERMODYNAMIC

%Load

Ploss

%Load

Heat

PlossPloss

TR.TR.T surface

T amb

QQ

Electrical Electrical losses = HEAT losses = HEAT that transfer to that transfer to

surroundingsurrounding

Steady state , no accumulated energy

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EQUATION FITTING

%Load

Heat

%Load

ΔT

Q = UA.Q = UA. Δ ΔTT

T oil

QQ

T ambT surface

ΔΔT = Q / UAT = Q / UA ΔΔT = a(%L)T = a(%L)22+b(%L)+c+b(%L)+c

in consideration in consideration rangerange

UA is constantUA is constant

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THERMAL-ELECTRICAL ANALOGY

Q = UA.Q = UA. Δ ΔTT

T oil

QQ

T ambT surface

1/UA

Q = Q = ΔΔT/(1/UA)T/(1/UA)

QQ

T surfaceT surface T ambT amb

1/UA1/UAII

V1V1 V2V2

RR

I = I = ΔΔV/RV/R

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RESULT

Load Current Ambient

Temperature TR. Surface

max. Temperature

19 examples19 examples

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EQUATION FITTING

dTmax = dTmax = 0.00390.0039(%Load)(%Load)22 + +0.03230.0323 (%Load) (%Load) + 8.0857+ 8.0857

EQUATEQUATION:ION:

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EQUATION FITTING

n2

i ii 1

SDS (y Y)

sum of deviations squared (SDS.)sum of deviations squared (SDS.)

FOR 19 EXAMPLES , SDS = 69.59 CSDS = 69.59 C22

average percent absolute deviation (APD.)average percent absolute deviation (APD.)

2n

i i

i 1 i

100 Y yAPD

n Y

FOR 19 EXAMPLES , APD = 10.19%APD = 10.19%

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CONCLUSION

The top surface temperature of transformer is function of loading capacitance and ambient temperature.

dTmax = dTmax = 0.00390.0039(%Load)(%Load)22 + +0.03230.0323 (%Load) (%Load) + 8.0857+ 8.0857

Temperature rise at 80% of rated load is 35.63 C.35.63 C.

Provincial Electricity Authority, Thailand 18

Question