P Pl tPower PlantsA1M15ENY

Lecture No 4Lecture No. 4Jan Špetlík

spetlij@fel cvut cz - subject in e-mail ENY”spetlij@fel.cvut.cz subject in e mail „ENY

Department of Power Engineering Faculty of Eelectrical Engineering CTU Technická 2Department of Power Engineering, Faculty of Eelectrical Engineering CTU, Technická 2, 166 27 Praha 6

Calculation of the parameters:

2 2 2 3 2 2 22 2 2 3 2 2 2

2 2

3. .1 10 3.1,529 .185 127,5. . 2,3 2 3. 2 3.185

k k ks r

k

U I PX X

I− −′≅ = = = Ω

20 200

234,5 234,5 80. . 0,178 234,5 234,5 20s s sR R Rϑ

ϑ+ +

= = = Ω+ +

12 32 2

127,5 .10 0,178 1,064 3. 3.185

kr s

k

PR RI

′ = − = − = Ω

( )

2

22

..

r

SM r

R UMs RX X R

′= =

Ω ′⎛ ⎞′+ + +⎜ ⎟( )21,064

rs r sX X R

sU

+ + +⎜ ⎟⎝ ⎠

( )2

2

, ..100. 1,0642,3 2,3 0,178

ss

π= =

⎛ ⎞+ + +⎜ ⎟⎝ ⎠⎝ ⎠

3 2

23,386.10 .

1 064UM

s

−

=⎛ ⎞1,0645,308 0,178

ss

⎛ ⎞+ +⎜ ⎟⎝ ⎠

Short Circuit Currents in AuxiliaryShort circuit currents are calculated according to EN 60 909-0Force and thermal effects are calculated according to EN 60 909-1 and EN 60 865-1 (IEC standard 909, 865)( , )

Basic classification:Basic classification:

Three phase initial symmetrical short-circuit current3kI ′′

Single phase initial symmetrical short-circuit current

Three phase transient short circuit current

1kI ′′

I ′ Three phase transient short-circuit current

Three phase steady-state short-circuit current

3kI

3kI

Peak short-circuit currentpi

Thermal equivalent short-circuit currentkeI

Short Circuit Current Types

32. kI ′′ 21 .kt

keI i dt= ∫

pi 32. kI ′ 2 I

0ke

kt∫

p 3k32. kI

Short Circuit Current TypesCalculation of peak short circuit current acc. to EN 60 909-0:

. 2.p ki Iκ ′′= 3. /1,02 0,98. R Xeκ −= +

Short Circuit Current TypesCorrection factor for impedance, overview from ČSN 33 3020:κ

Short Circuit Current TypesCalculation of thermal equivalent short circuit current acc. to EN 60 909-0:

F t f th l ff t f DC t d AC t.ke e kI k I ′′= ek m n= +

Factor for thermal effects of DC component m and AC component n are present in graphs in standard EN 60 909-0

Short Circuit Current TypesFactor from ČSN 33 3020:ek

Thermal EffectsCross section dimensioning:

.ke kI tS ( ) 2020 .

lF V F KcK

ϑ ϑ ϑ+ +ke kSKϑ

≥( ) 20

20 1

.lnF V F K

F

Kϑϑ ϑ

ρ ϑ ϑ+

=+

Att ti ! I dditi t th t ti di i i t tAttention! In addition to that, cross section dimensioning must respect voltage increments, operation currents and/or economy of losses

Earthing and Touch VoltageDimensioning of system earting must correspond the short circuit:

I. Calculation of earthing resistance:

a) Earth rod b) Circular earth electrode

4..lnEE

LR ρ= 2

2. ..lnEE

DR ρ π=.ln

2. .ERL dπ 2 .ln

.ERD dπ

Earthing and Touch Voltagec) Earthing mesh

Acc to EN 50522:

2 4E E

ERD A

ρ ρ π= =

Acc. to EN 50522:

2. 4D A

(Dwight’s formula)

A formula acc. to std. IEEE respecting earthing conductor length:(Laurent-Niemahn’s formula):

( )1 1 1 1. .4 4 2 . .

E E ERA L A N N lπ πρ ρ

⎡ ⎤⎡ ⎤ ⎢ ⎥= + ≅ +⎢ ⎥ ⎢ ⎥+ Δ⎣ ⎦ ⎣ ⎦

(Laurent Niemahn s formula):

( )2. .N N l+ Δ⎣ ⎦ ⎣ ⎦A formula respecting depth of the mesh:

1 1 1 0 165 2l hπ π⎡ ⎤⎛ ⎞Δ ⎛ ⎞1 1 1 0,165. . 2.. ln 1 1,128.4 2. . 2. 2.E E

FeZn

l hRA N l S Aπ πρ

π

⎡ ⎤⎛ ⎞Δ ⎛ ⎞= + −⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟Δ ⎝ ⎠⎢ ⎥⎝ ⎠⎣ ⎦

Earthing and Touch Voltaged) Combination mesh + rod

Detailed formula:Detailed formula:

2

1 1..1 0 9EE

R nη=2

1 2

.1 0,9E

E E

nR R

η+

Earthing and Touch VoltageII. Current dimensioning of earthing conductors and electrodesa) Current duration less than 5 s

min .ln

ke k

K F

I tSK ϑ ϑ=

+

Materiál β K

Al 228 148

Cu 234,5 226

1 Fϑ ϑ+ Fe 202 78

Rem. Acc. to EN 50522 is the initial temperature 20°C and final 300°CExample of current dimensioning of FeZn conductors for final temperature 300°C:

Earthing and Touch Voltageb) Current duration more than 5 s- From the graphs in EN 50522, attach. CContinuous permitted current for FeZn conductors (temp. 300°C)

Re-computation to other final temperature:

Earthing and Touch VoltageIII. Touch voltage and earth potential rise

Earth potential rise UE means voltage between selected point andEarth potential rise UE means voltage between selected point andreference (remote) earth within corresponding short circuit level and location.

where IE is current to earthTT k i h l i d l hi

.E E EU R I=

I r I ′′=TT network with low impedance neutral earthingIT network with resonant impedance earthingIT(r) network with resistance neutral earthing

1.E kI r I=

I I

2 2Re.E s LI r I I= +

IT(r) network with resistance neutral earthingR.EI r I=

Cresidual earth fault curent, maximum 10% I

Reduction factor r is respecting partial outlet of short circuit current or earth fault current through ground cables, cable shieldings or t f t l R d ti f t i th d fi d Itransformer neutral. Reduction factor is thus defined as

03.EIrI

=

Earthing and Touch VoltageTouch voltage UTp is potential between conductive parts, which can be touched by human or animal. Real touch voltage level is significantlytouched by human or animal. Real touch voltage level is significantly influenced by human (animal) body impedance.

F b t ti b

2U U≤

For substations above1 kV AC:

2.E TpU U≤

4.E TU U≤Or:

4.E TpU U≤+ implementation of special measures acc. to EN 50522(non-conductive asphalt belt, equipotential grading etc.)

In this case it is necessary to isolate earthing system of substation (power plant) from public earthing systems (min. 20 m)!

Force Effects – Rigid ConductorsFor rectangular shaped conductors:(IEC 865-1 / EN 60865-1)

For correct dimensioning isdecisive :decisive :a) Bending stress of a conductorb) Bending forces on supports

Force on central conductor is:

203 3

3. . .2. 2m p

m

lF id

μπ

=m

effective distancedd =mdk

Force Effects – Rigid Conductorsa) Bending stress on a conductor:

3.o mM F lV Vσ β= =Dyn. and stat. force ratio factor: Vσ. . .

8.m rV VZ Zσσ β= =

Section modulus:βFactor respecting an arrangement of support:

rVFactor respecting double C-O operation:

Section modulus:

b

IZ = Rectangular shape:3

Moment of inertia:

2

Ci l h

3.12a bI =

Requirement:

2.S

I r dS= ∫Moment of inertia: Circular shape:

4

4I rπ=

0,2.m qσ σ<

St i ld i t S 4Tubular shape:

( )4 4I π

Stress yield point:

0 2 120 180 MPaAlMgSiσ = − 0 2 80 MPaCuσ =Conventional, what causes elongation by 0,2%σ

( )4 41 24

I r r= −0,2 AlMgSi 0,2Cu

q – factor of plasticity.

Force Effects – Rigid ConductorsFactor of plasticity q

Force Effects – Rigid ConductorsIf there is more than one conductor per phase, it is necessary to add stressinduced by single phase conductors:

2⎛ ⎞ .. .

16.s s

s rF lV V

Zσσ = tot m sσ σ σ= +

R i t

230 . .

2.p s

sms

i lFn d

μπ

⎛ ⎞= ⎜ ⎟

⎝ ⎠Requirement:

0,2.tot qσ σ<

Force Effects – Rigid Conductorsb) Bending forces on supports

Force on support: Dyn. and stat. force ratio factor: FV0 8

pp

3. . .D F r mF V V Fα=

VFactor respecting double C-O operation:

0,20,8.min 1; ;max 2,7F

tot

Vσ

σ=

αFactor respecting an arrangement of support:rVFactor respecting double C-O operation:

Requirement:

H0,8. .D dHF F P

H T< =

+

Where P is mechanicalendurance of supportendurance of support

Force Effects – Rigid ConductorsFactors : from the fig:, ,F rV V VσorFor more accurate calculation itFor more accurate calculation itis possible to compute naturalfrequency of the arrangement:

2

..cE If

l mγ

=′

γ Factor (conductor plasticity)γEm′

Factor (conductor plasticity)

Young’s modulus

Mass per lengthm p g

Rem.: Conductor behavior will be the same, if a natural frequency is more than 10xf network.The worst results we obtain for circa 1-2xf network1 2xf networkl supporting arrangement should avoidresonance/

Force Effects – Rigid ConductorsFactors :, ,α β γ

0,50,5

A

B

αα

==

1β = 1,57γ =

0,6250,375

A

B

αα

==

0,5Aα =

0,73β =

0,5β =

2, 45γ =

3,56γ =0,5Bα =0,3751, 25

A

B

αα

==

,β

0,73β = 2, 45γ =

,γ

0, 41,1

A

B

αα

==

0,73β = 3,56γ =

Force Effects – Cable ConductorsForce effects on hanged cable conductors:

Force per length on cable conductor:2

0 3/

3. . .2. 4

k cl

m

I lFd l

μπ

′′=

Limitations:

Detailed computation in IEC 865-1

Limitations:

- Short circuit tensile force tF- Drop force after short circuit- Minimum conductors’ distance during swing out of the span

Horizontal displacement during swing out of the span

fFmina

b- Horizontal displacement during swing out of the span- Pinch force in the case of bundled conductors

hbpiF

Addenum to 4th LectureRigid conductor arrangement:Example:Conductors of three phase system 10,5 kV are rectangular, typeAl 63x10 mm (one per phase), span length between supports is 1m andphase distance is 0 5m Make a decision which type of support do youphase distance is 0,5m. Make a decision, which type of support do you select and whether a conductor use is suitable for conductor arrangement:1) Horizontal )2) VerticalMounted clamp means total height elongation by 2 cm

Input data:

Number of spans/supports: 4/5

3 25 kAkI ′′ = 1 7κ =

0,5 mmd = 0,2 120 MPaσ =

(without respecting double CO)3 25 kAkI 1,7κ (without respecting double CO)

Addenum to 4th LectureSelect a suitable support from following list

Minimum 50 Hz (wet) withstand

Leakage distance

Maximum working il l d

Section length H Diameter D

[ ]withstand [kV]

distance [mm] cantilever load

[kN]

length H[mm] [mm]

1. 75 174 5 130 60

75 187 10 130 722. 75 187 10 130 72

3. 75 195 16 130 90