PIPE-TYPE AMPACITY EXAMPLE

This appendix contains a sample calculation of pipe-type cable ampacity using the procedures outlined in this chapter and detailed in the references.

The cable configuration is as shown in Figure A3.1-1.

Figure A3.1-1 Cable configuration for pipe-type ampacity example.

tis:=0.005 Thickness of i nsulation shield tapes, inches ρ_shield:=0.7 ohm -meters Electrical resistivity of stainl ess steel shield tapes Skid W ire Size/Type 0.1in. x 0.2in., 2 x 3in. lay

ρ_skidwire:=0.7 ohm -meters Electrical resistivity of stainl ess steel skidwire

f:=60 Hz

LF=0.62 Dail y (24-hour) loss factor, per unit (entered below)

n:=3 Num ber of cables wi thin pipe or conduit

N:=2 Pipe Data:

Pipe is HPFF

OD_pipe:=10.75 Pipe outside diameter, inches

ID_pipe:=10.25 Pipe inner diameter, inches

t_coating:=0.07 Pipe coating (Som astic) thickness, inches

ρ_coating:=3.5 Thermal resistivity of the pipe coating, C°-m/w

Pipe-Type Cable Ampacity Worked Example

Cables are 345kV HPFF cables with 2500kcmil segmental copper conductors, 905 mils kraft paper i nsulation, 0.1x0.2, 2x3in. lay stainless steel skid wires in an 10-inch cable pipe, 2 circuits, 0.62 loss factor.

Cable Data:

A:=2500 Conductor area, CI

ρ_conductor:=0.017241 ohm -meters Electrical resistivity, copper conductor

ks:=0.39 Conductor ski n effect factor, in oil

kp:=0.46 Conductor proximitty effect factor, in oil, cradled

Dc:=1.824 Diam eter of the segm ental conductor, inches

Tc:=85 Maximum normal conductor operating tem perature, °C

tcs :=0.005 Thickness of conductor semiconducting shield, i nches

ti:=0.905 Insulation wall thi ckness, inches Values for

Paper Insulation

SIC:=3.5 Diel ectric constant of the insulation

tanδ:=0.0023 Dissipation factor of the insulation, numeric

ρ_insulation:=6.00 Thermal resistivity of the insulation, C°-m/Watt

Center_Depth_cbf:=39 Value for center-line depth of cbf is arbitrary Calculate Cable Geometry:

D_cs:=Dc+2tcs Diameter over the conductor shield, inches

D_insulation:=D_cs+2ti Diameter over the insulation, inches

D_is:=D_insulation+2tis Diameter over the insulation shield, inches

D_skidwire:=D_is+1.5×0.1 Diameter over 0.1" skid wires, inches

D_earth:=OD_pipe+2t_coating

Clearanc e ID_pipe

2 −1.366D_skidwire

(

ID_pipe −D_skidwire

)

2 1 D_skidwire

ID_pipe −D_skidwire

⎛⎜

⎝

⎞⎟

− ⎠

⎡⎢

⎣

⎤⎥

⎦

0.5 +

:=

Clearance in pipes, inches D_cs=1.834

D_insulation=3.644 D_is=3.654 D_skidwire=3.804 D_earth=10.89 Clearanc e=1.992 Installation Data:

x1:=−15 x2:=15 Horizontal location of pipe center, inches burial1:=42 burial2:=42 Burial depth to pipe center, inches

Ta:=25 Ambient earth temperature, °C

f:=60 Power frequency, Hz

E:=345000 1.05× System maximum line to line voltage, volts

ρ_native:=0.9 Thermal resistivity of the native earth, C°-m/w

ρ_backfill:=0.5 Thermal resistivity of the duct concrete, C°-m/w

Width_cbf:=53 Height_cbf:=29 Width and height of backfill, inches

x1:=x125.4 x2:=x225.4 x1=−381

Width_{cb f}:=Width_cbf25.4 Width_cbf=1.346×10³ mm

Height_cbf:=Height_cbf25.4 Height_cbf=736.6 mm

Center_Depth_cbf:=Center_Depth_{cb f}25.4 Center_Depth_cbf=990.6 mm Calculate the dielectric losses:

W_d

2πfSIC E 3

⎛⎜

⎝

⎞⎟

⎠

2

tanδ10⁻⁹

18 ln Dinsulatio n Dcond_shield

⎛⎜

⎝

⎞⎟

⎠

:= W_d=10.741 W /m/phase

Calculate Conductor Resistance:

R_dc20

ρ_conductor Area_conductor

:= R_dcT R_dc20 Tc−(−234.5)

20−(−234.5)

⎡⎢⎣ ⎤⎥⎦

:= R_dc20=1.361×10⁻⁵

R_dcT:=R_dcT1.025 Assume 2.5% Stranding of Conductor Ohms/meter

Xs 8πf ks( ) 10⁻⁷ R_dcT

:= Xp 8πf kp( ) 10⁻⁷

R_dcT :=

Y_cs Xs² 192+0.8Xs²

:= S:=D_skidwire Y_cs=0.056

Y_cp Xp²

192+0.8Xp²

⎛⎜⎜

⎝

⎞⎟⎟

⎠

Dco ndu ctor S

⎛⎜

⎝

⎞⎟

⎠

2

0.312 D_conductor S

⎛⎜

⎝

⎞⎟

⎠

2

1.18 Xp² 192+0.8Xp²

⎛⎜⎜

⎝

⎞⎟⎟

⎠

+0.27 +

...

⎡⎢

⎢⎢

⎢⎢

⎢⎣

⎤⎥

⎥⎥

⎥⎥

⎥⎦

:=

Y_cp =0.061 R_acc^:=R_dcT⎡⎣1⁺1.5 Y

(

_cs+Y_cp

)

_⎤⎦

Metric Conversion of Variables Area_conductor A

1.9735

:= Area_conductor=1.267× 10³ mm²

D_conductor:=Dc25.4 D_conductor=46.33 mm

Dcond_shield:=D_cs25.4 Dcond_shield=46.584 mm

D_insulation:=Din sulation25.4 D_insulation=92.558 mm

Dinsl_shield:=D_is25.4 Dinsl_ shield=92.812 mm

D_skidwire:=D_skidwire25.4 D_skidwire=96.622 mm

OD_pipe:=OD_pipe25.4 OD_pipe=273.05 mm

ID_pipe:=ID_pipe25.4 ID_pipe=260.35 mm

D_earth:=D_earth25.4 D_earth=276.606 mm

burial1:=burial125.4 burial2:=burial225.4 burial1=1.067×10³ mm

Ohms/meter

Calculate Mutual Reactance (assume cradled configuration) X_m 2 2πf

( )

10⁻⁷ln 2.3Dinsl_shield

D_skidwire

⎛⎜

⎝

⎞⎟

:= ⎠

X_m=5.977× 10⁻⁵ Ohms

Y_sc R_s R_acc

1

1 R_s X_m

⎛⎜

⎝

⎞⎟

⎠

2 +

:=

Y_sc=1.314× 10⁻³

R_acs^:=R_dcT⎡⎣1⁺ 1.5 Y

(

_cs+ Y_cp+ Y_sc

)

_⎤⎦

Calculate the pipe loss increments to AC resistance:

Y_p

0.0438D_skidwire+ 0.0226ID_pipe R_dcT10⁶

:= Y_p =0.578

R_acp^:=R_dcT⎡⎣1⁺ 1.5 Y

(

_cs+ Y_cp+ Y_sc

)

+^Yp⎤⎦

R_acc=2.06×10⁻⁵ R_acs=2.063× 10⁻⁵ R_acp=3.075×10⁻⁵ Ohms/meter

Qs R_acs R_acc

:= Qp

R_acp R_acc

:= Qs=1.002 Qp=1.493 AD/DC resistance ratios

Calculate the shield and skidwire loss increments:

Area of shield tape is width of tape times thi ckness. Resistance of shield tape is the area times the helical length times the resistivity divided by the area. Shield tape thickness is 0.005in stainless steel, with 1/8"

lapped, with typical tape width of 7/8". Assume there are 2 tapes.

widthshield _tape 7 825.4

:= thicknessshield _tape:=0.005× 25.4 lap_shield 1 825.4 :=

areashield _tape:=thicknessshield _tapewidthshield _tape lay_shield:=widthshield _tape−lap_shield

R_shield

ρ_shield areashield _tape

1

πDinsl_shield lay_shield

⎛⎜

⎝

⎞⎟

⎠

2 +

:= R_shield=3.804 Ohms/meter

Skid wire resistance

Area of elipse is pi * major_radius * minor_radius. Area of skid wire is half this. Skid wire is 0.1 x 0.2, 3-inch lay, 2 wires. Material is stainless steel.

minor_radius:=0.1×25.4 major_radius 0.225.4 2

:= area_skidwire π

2minor_radius major_radius :=

lay_skidwire:=3× 25.4

Length of a hel ix is (1 + (pi*D/Lay)^2)^.5

R_skidwire

ρ_skidwire area_skidwire 1

πD_skidwire lay_skidwire

⎛⎜

⎝

⎞⎟

⎠

2 +

:= R_skidwire=0.284 Ohms/meter

R_s 1

1 R_skidwire

1 R_skidwire

+ 1

R_shield

+ 1

R_shield +

:= R_s=0.132

D_b=985.973

:= Thermal diffusivity of native soil mm²

hour α_soil=1.834× 10³

D_x:=1.02 α_soil24 Diam eter beyond which 24 hour average losses apply, mm D_x=213.978 Earth thermal resistance for AC losses at loss factor

2burial1+ 4burial1²−D_earth² D_x dielectric losses at 100% loss factor

2burial1 + 4burial1²−D_earth² D_x Calculate Cable Thermal Resistances

R_i

ρinsulatio n

2π ln Dinsl_shield D_{cond uctor}

Tm_oil=66.827 Estimate of m ean temperature in duct

R_oil n U

Calculate External Thermal Resistances

Calculate geometric correction factor for backfill envelope:

x:=Height_cbf x=short dimension of backfill y:=Width_cbf y=long dim ension of backfil l

D_b e

ΔT_d W_d R_i

2 +R_oil+ Rpip e_coating+R_earth'+ R_mutual'+Rcorrection '

⎛⎜

⎝

⎞⎟

:= ⎠ ΔT_d=20.896

ΔT:=Tc−ΔT_d−Ta ΔT 39.104= Adjust "iterated" values

below until they equal

"calculated".

I_rated ΔT

∑RacRth

:= I_rated=940.2

I_total:=2I_rated I_total=1880.4

Tc_calc I_rated²_⎡⎣R_accR_i+R_acsR_oil⁺ R_acp

(

Rpip e_coating+ R_earth+ R_{mu tual}+Rcor rection

)

_⎤⎦

ΔT_d+Ta +

...

:=

Tc_calc=85

Tshield_calc Tc_calc−I_rated²R_accR_i W_dR_i 2

−

:= Tshield_calc=69.357 T_shield≡69.357

Toil_calc^:=Tshield_calc⁻⎛⎝I_rated²R_acs+W_d⎞⎠^R₂^{oil Toilcalc=66.827 Tmoil≡66.827}

Tearth_interface Toil_calc ⎛I_rated²R_acs+ W_d

⎝ ⎞

⎠^R₂^oil

− ⎛I_rated²R_acp +W_d

⎝ ⎞

⎠^Rpip e_coating

− :=

Tearth_interface=63.477 Mutual thermal resistance between two pipes (just for normal ampacities)

d'₁₂^:=⎡⎣(x1−x2)²+ (burial1+ burial2)²⎤⎦^.5 Distance between one cable pipe and image of other cable pipe

d₁₂^:=⎡⎣(x1−x2)²+ (burial1−burial2)²⎤⎦^.5 Distance between one cable and the other F₁₂

d'₁₂ d₁₂

:= R_mutual

ρ_backfill

2π n LFln F

( )

₁₂

:= R_{mu tual}=0.161

R_mutual'

ρ_backfill 2π n ln F

( )

₁₂

:= R_{mu tual'}=0.26

Thermal resistance correction for native soil for AC losses

Rcor rection LF

ρ_native−ρ_backfill

( )

2π n N G_b

:=

Rcor rection=0.313 Thermal

resistance correction for native soil for dielectric losses Rcor rection '

ρ_native−ρ_backfill

( )

2π n N G_b

:=

Rcor rection'=0.505

Calculate Normal Ampacity on primary cable

∑RacRth^:=R_accR_i+R_acsR_oil⁺ R_acp

(

Rpipe_coating+ R_earth+ R_mutual+R_correction

)

_LF≡0.62

APPENDIX 3.2 EXTRUDED AMPACITY

In document Increased Power Flow Guidebook (Page 192-199)