海缆载流量计算

110 kV

300mm 2

1

1.1油道

D 油=19.5mm 1.2螺旋管D 螺=22.5mm 1.3导体D 导=

32.5mm 1.4导体屏蔽δ内屏=

0.5mm D 内屏=33.5mm 1.5油纸绝缘δ绝=

10.5mm D 绝=54.5mm 1.6绝缘屏蔽δ

外屏=0.6mm D 外屏=55.7mm 1.7铅护套δ铅=4mm D 铅=63.7mm 1.8衬垫层δ衬=0.75mm D 衬=65.2mm 1.9不锈钢带δ不=0.3mm D 不=65.8mm 1.10防水层δ防=0.8mm D 防=67.4mm 1.11PE 护套δ护=

3.5mm D 护=7

4.4mm 1.12麻+沥青δ

内麻= 2.5mm D 内麻=79.4mm 1.13铠装层δ铠=6mm D 铠=91.4mm 1.14沥青δ

沥=

0.5mm D 沥=92.4mm 1.15麻+沥青δ外麻= 2.5mm D 外麻=97.4mm 1.16PVC 带

δPVC =

0.2

mm

D PVC =

97.8

mm

2

2.1

海中段S=200m 陆上段

S=

300

mm

2.2

海中段L=1m 陆上段L=1m 2.3ρt =

1

K.m / W

2.42.5

土壤：θ0=25℃空气：θ0=

35

℃海水：

θ0=

20

℃

33.1

3.1.1在20℃时直流电阻

R 0=

0.0601

Ω / km =

6.01E-05Ω / m

充油海缆载流量及其它参数计算

电缆结构

敷设条件

敷设间距

埋设深度

两端接地

土壤热阻系数接地方式

环境最高温度

载流量计算

导体损耗计算

3.1.2在85℃时直流电阻R'=

7.5453E-05Ω / m

3.1.3

66单线外径：

2.49

mm

d i = 1.95cm

A= 3.213898cm 2d c '= 2.809727cm

k s =0.35894189

X s 2=

0.59780529Y s =X s 4/(192+0.8*X s 4)=0.00185854

3.1.4

k s =0.8d c =

3.25cm

X p 2= 1.33237229

海中段：

Y p =X p 4/(192+0.8*X p 4)*（d c /S)2{0.312(d c /S)2+1.18/[X p 4/(192+0.8X p 4)+0.27]} = 1.02437E-09

陆上段：

Y p =X p 4/(192+0.8*X p 4)*（d c /S)2{0.312(d c /S)2+1.18/[X p 4/(192+0.8X p 4)+0.27]} = 4.5528E-06

3.1.5海中段：

R=7.5593E-05Ω / m 陆上段：R=7.5593E-05Ω / m

3.1.6要求的载流量

P=100MW I n =P/(30.5U)=

524.8638811A 3.1.7每相导体损耗W c =I n 2R=

20.824551W / m

3.23.2.1电缆电容C

ε= 3.6

C=ε/[18*ln(D 外屏/D 导）]*10-9= 4.1097E-10F / m

3.2.2每相介质损耗

tg δ=0.0033W d =ωCU 02tg δ=

1.718446962W / m

3.2.3单回路充电容量

P c =30.5

ωCU 02

=901.9507419VA / m 3.3

便，取陆上段的R 值

介质损耗因素R’（1+Y s +Y p ）=

海中段和陆上段的R 误差很小，为计算方

单线根数：π/4*单线外径2*单线根数 =

R 0*[1+α

20*（85-20}]=

计算集肤效应因素

8πf/R'*10-7*k p =

(4/π*A+d i 2)^0.5 =(d c '-d i )/(d c '+d i )*[(d c '+2*d i )/(d c '+d i )]2 =

8πf/R'*10-7*k s =

计算邻近效应因素

导体在85℃时交流电阻

R’（1+Y s +Y p ）=金属护套、钢丝铠装层损耗

3.3.1总损耗

a)金属护套、加强层并联电阻铅护套电阻

A s1=7.502117cm 2

ρs1=0.0000214Ω . cm

θ

s1=

70℃

α

s1=

0.004

R s1= 3.42E-06Ω / cm

=0.000342303Ω / m

加强层电阻

A s2=0.617322cm 2ρ

s2=

0.00007Ω . cm

R s2=0.000227Ω / cm

=0.022678586Ω / m

并联电阻

R s =0.00033721Ω / m

b)铠装层电阻钢丝电阻

钢丝：36

根铜丝：7根

θa1=65℃

ρ

a1= 1.38E-05Ω . cm

A a1=

10.17875cm 2R a1= 1.63E-06Ω / cm

=0.000163031Ω / m

隔离铜丝电阻

θa2=65℃

ρ

a2=

1.72E-06Ω . cm A a2=

1.979202cm 2

R a1= 1.03E-06Ω / cm =0.000102516Ω / m

并联电阻

R a =7.0742E-05Ω / m

c)每相回路护套产生电感海中段：

d=D 铅-δ铅=59.7mm

H s =

1.76E-08H /cm = 1.76E-06H / m 陆上段：H s = 4.61E-09H /cm = 4.61E-07H / m d)钢丝产生的电感分量

μe =400μ1=

43d f =

0.6

cm d A =

8.54cm

μt =

1γ=0.785398

钢丝节径比

h=

13

0.46(lg(2S/d)*10-8=

0.46(lg(2S/d)*10-8=

ρa1/A a1[1+αa 1(θa1-20)]=

π/4*铜丝直径2*铜丝根数 =

ρa1/A a1[1+αa 1(θa1-20)]=1.4R a1*R a2/(1.4R a1+Ra 2)=R s1*R s2/(R s1+Rs 2) =

铠装层结构：π/4*钢丝直径2*钢丝根数 =

π/4*（D 铅2-D 外屏2） =

ρs1/A s1[1+αs1(θs1-20)]=π/4*（D 不2-D 衬2） =

2ρs2/A s2=

β=arctg(h/π）= 1.33368162P=

h*d A =111.02cm H 1= 1.4095E-08H / cm = 1.40949E-06H / m H 2= 1.4095E-08

H / cm = 1.40949E-06H / m

H 3=0H / cm =

H / m

e)海中段总损耗：

B 1=0.00099572

Ω / m B 2=0.0004428Ω / m R e =

5.8475E-05

Ω / m

W (s+A)=15.728879W / m

d)陆上段总损耗：

B 1=0.00058763Ω / m

B 2=0.0004428Ω / m R e = 5.8475E-05

Ω / m

W (s+A)=15.3173587W / m

3.3.2铅护套损耗λ1和铠装层损耗λ

2

海中段：λ1=λ2=

λ1’=λ2’=

0.377652陆上段：

λ1=λ2=λ1’=λ2’=0.367772

3.43.

4.1绝缘热阻T 1：

ρt1=5

K.m / W

T 1=

0.42871608K.m / W

3.4.2热阻T 2：

a)包带和内护层热阻T 2’

ρ't2'=6K.m / W

ρ't2''=

3.5K.m / W

T'2=

0.108958K.m / W

b)内油麻热阻T 2''

ρ''t2=6K.m / W

T''2=

0.06211099K.m / W c)热阻T 2：

T 2=T'2+T''2=

0.171068714K.m / W

3.4.4外油麻热阻T 3:

ρt3=6K.m / W

T 3=

0.06462883K.m / W

热阻计算ρt1/2/π*ln(D 外屏/D 导）=

ρ't2'/2/π*ln(D 防/D 铅）+ρ't2''/2/π*ln(D 护/D 防）=

ρt1/2/π*ln(D 内麻/D 护）=

ρt3/2/π*ln(D 电缆/D 铠）=

ωH 2 =R s *R a /(R s +R a ) =

I 2{R e *(B 22+B 12+R e B 2)/[(R e +B 2)2+B 12]} =

W （s+A)/2/W c =W （s+A)/2/W c =

0.314μe(μ1*d f 2/P/d A )sin βcos γ*10-8 =0.314μe(μ1*d f 2/P/d A )sin βsin γ*10-8 =I 2{R e *(B 22+B 12+R e B 2)/[(R e +B 2)2+B 12]} =

0.4(μt -1)*d f /d A cos 2β*10-8 =

ω(H s +H 1+H 3) = ωH 2 =R s *R a /(R s +R a ) =

ω(H s +H 1+H 3) =

3.4.5土壤热阻T 4:

a)海中段：

ρt4=

0.8K.m / W D e =97.8mm =0.0978m

u=

2L / D e =

20.44989775

T 4=

0.472439

K.m / W

b)陆上段：

ρt4=

1K.m / W D e =97.8mm =0.0978m

u=

2L / D e =

20.44989775

S1=

0.3

m

T 4= 1.19796302K.m / W

c)空气段（受光照）：

Z=0.21E= 3.94

g=

0.6

h=Z/D e g +E =

4.787258592W / m 2(K)6/4

n=

1K A =0.656698Δθd =

-0.00616

H=1000

W / m 2

Δθ=

50

℃

ζ=

0.8Δθ

ds =

=34.93168391

(Δθd )1/4=

2.3(Δθd )1/4= 2.4117015(Δθd )1/4= 2.39439884(Δθd )1/4= 2.39703857(Δθd )1/4= 2.3966349(Δθd )1/4= 2.39669661(Δθd )1/4= 2.39668718(Δθd )1/4

= 2.39668862(Δθd )1/4=

2.3966884

T 4=0.28366985K.W / m

3.53.5.1海中段:

I={[Δθ-W d (0.5T 1+T 2+T 3+T 4)]/R/[T 1+(1+λ1)T 2+(1+λ1+λ2)(T 3+T 4)]}1/2 =727.0995519A

3.5.2陆上段:

I={[Δθ-W d (0.5T 1+T 2+T 3+T 4)]/R/[T 1+(1+λ1)T 2+(1+λ1+λ2)(T 3+T 4)]}1/2 =514.7657435A

3.5.3空气段（受光照）：

Wd{[1/(1+λ1+λ2)-0.5]T1-n λ2T2/(1+λ1+λ2)}=ρt4/2/π*ln[u+(u 2-1）0.5]=ρt4/2/π*{ln[u+(u 2-1）0.5]+ln[1+(2L/S 1)2]}=

π*D e *h/(1+λ1+λ2)[T 1/n+T 2(1+λ1)+T 3(1+λ1+λ2)]=ζD e H/(1+λ1+λ2）[T 1/n+T 2(1+λ1）+T 3（1+λ1+λ2）]{(Δθ+Δθd +Δθds )/[1+K A (Δ

θs )1/4]1/4 ={(Δθ+Δθd +Δθds )/[1+K A (Δθs )1/4]1/4

={(Δθ+Δθd +Δθds )/[1+K A (Δθs )1/4]1/4 ={(Δθ+Δθd +Δθ

ds )/[1+K A (Δ

θs )1/4]1/4 =

1/[πD e h(Δθs )1/4]=

载流量{(Δθ+Δθd +Δθds )/[1+K A (Δθs )1/4]1/4 ={(Δθ+Δθd +Δθds )/[1+K A (Δ

θs )1/4]1/4 ={(Δθ+Δθd +Δθds )/[1+K A (Δθs )1/4]1/4 ={(Δθ+Δθd +Δθds )/[1+K A (Δθs )1/4]1/4

=

I={[Δθ-W d (0.5T 1+T 2+T 3+T 4)-ζD e HT 4]/R/[T 1+(1+λ1)T 2+(1+λ1+λ2)(T 3+T 4)]}1/2 =589.3503313A

4E max =7.791047 kV / mm E min =

4.788992 kV / mm

55.1已知条件

S=

300

mm 2

θf =160℃K=226θi =85℃X=0.45Y=

0.14

β=

234.5

5.2

t sc =1

ε=[1+X （t/S)1/2+Y(t/s)]1/2=

1.01313742

5.3计算I AD

I AD =

31133.56

A = 31.13356

kA

5.4计算I SC

I SC =

εI AD =

31.54256979

kA

66.1已知条件

S=750.211692mm 2θf =160℃F=1K=41θi =75℃β=

230

ζ2=2000000

J / K.m ζ3=1700000

J / K.m

ρ2=5K.m / W ρ3=6K.m / W ζ1=

1450000

J / K.m

δ =

4

mm

M=0.1004096.2

ε=

1.06055828

6.3计算I AD

I AD =

15250.69A = 15.25069

kA

6.4计算I SC

I SC =

εI AD =

16.17424612

kA

77.1

lnGMRc=ln(D 导/2)-(D 螺/2)4/[(D 导/2)2-(D 螺/2)2]2*ln(D 导/D 螺)

电场强度

U 0/[D 内屏/2*ln(D 绝/D 内屏）=U 0/[D 绝/2*ln(D 绝/D 内屏）=

导体短路电流计算

正、负序阻抗计算

非绝热因素非绝热因素线芯几何平均半径GMRc

铅护套短路电流计算

[(ζ2/ρ2）1/2+（ζ3/ρ3）1/2] /（2ζ1δ*0.001）*F =1+0.61M*t 1/2-0.069(M*t 1/2)2+0.0043(M*t 1/2)3 ={K 2S 2ln[（θf +β)/(θi +β)]/t}1/2=

{K 2S 2ln[（θf +β)/(θi +β)]/t}1/2=

+[3(D 螺/2)2-(D 导/2)2]/{4[(D 导/2)2-(D 螺/2)2]} = 2.6867703

GMRc=e lnGMRc =14.68417378mm 7.2

lnGMRs=ln(D 铅/2)-(D 外屏/2)4/[(D 铅/2)2-(D 外屏/2)2]2*ln(D 铅/D 外屏)

+[3(D 外屏/2)2-(D 铅/2)2]/{4[(D 铅/2)2-(D 外屏/2)2]} = 3.419247207

GMRs=e lnGMRs =30.54641123mm

7.3

n=1m=2S=200000

mm

Xm=Xs=

j

0.56660914Ω / km

7.4Z 1=Z 2=R c +Xm 2R s /(Xm 2+R s 2)+ j 2ω*10-4ln[(nm)1/3S/GMRc]- j Xm 3/(Xm 2+Rs 2)

=0.32637064 + j 0.19752409Ω / km

8

8.1X c =j

0.01534079Ω / km

8.2

ρ

大地=

30Ω .m D e =

513763.8mm

X m =0.23363096Ω / km

8.3

R g =π2f*10-4=0.049347939Ω / km R c +3X c =

0.07559312

+ j 0.04602238Ω / km 0.050675899 + j 0.239918Ω / km 0.490347199

+ j

0.700893Ω / km 0.26378013 + j 0.11223962Ω / km

Z 0=

0.339373

+ j

0.158262Ω / km

9

9.1电缆内感L i

Li=μ0/(8*π)=

0.00000005

H / m 9.2电缆电感L

L=

1.9336E-06

H / m

求零序阻抗Z0金属护套和大地组成的感抗Xm 93.8*ρ

大地

1/2

*1000=j 2ω*10-4*ln[D e /(S 2/3(nm)2/9*GMRs 1/3]= j 铅护套几何平均半径GMRs

j 2ω*10-4*ln(GMRs/GMRc)1/3 =j 2ω*10-4*ln[(nm)1/3S/GMRs]=求正负序阻抗Z 1、Z 2

求Xm 、Xs(海中段)因陆上段和空气段的长度与海中段相比可忽略不计

求零序阻抗Z 0

短路电流以金属护套为回路的自感抗Xc 求电缆电感L

Li+[2ln(2S/D c )]*10-7=

3R s (R g +X m ) =R s +3R g +3X m =3R s (R g +X m )/(R s +3R g +3X m ) =

R c +3X c +3R s (R g +X m )/(R s +3R g +3X m )=