本标准提供了三相双回路平面排列的单芯电缆的金属套涡流损耗计算方法。当金属套单点或交叉互连时金属套中没有明显的环流。如果金属套两端互连时，金属套的显著环流将导致载流量降低。对于双回路环流损耗计算方法在考虑中。 JB/T 10181.2-2000 电缆载流量计算 第1部分: 载流量公式(100%负荷因数)和损耗计算第2节: 双回路平面排列电缆金属套涡流 JB/T10181.2-2000 标准下载解压密码：www.bzxz.net

Machinery Industry Standard of the People's Republic of China
Cable Ampacity Calculation
Part 1: Ampacity Formula (100% Load Factor) and Loss Calculation
Section 2: Double-circuit planar arrangement cables
Metal sheath eddy current loss factor
Calculation of the current rating of electric cablesPart l: Current rating equations(100% foad factor) and calculation of lossesSection 2: Sheath eddy current loss factors for two circuits in flat formation1 scope
JB/T10181.2-2000
idtIEC60287-1-2:1993
This standard provides the eddy current losses of metal-sheathed single-core cables with three-phase double-circuit planar arrangement Calculation method. There is no significant circulation in the metal sleeve when the metal sleeve is interconnected at a single point or across. If the two ends of the metal sleeve are interconnected, significant circulation of the metal sleeve will lead to a reduction in the ampacity. The calculation method for double loop circulation loss is under consideration. This method provides correction of the loss factor of the metal sheath of separately laid three-phase loop cables. For the cable parameter m (m=W/107·R) less than 0.1, corresponding to the longitudinal resistance of the metal sheath being greater than 314μ2/m at the system frequency of 50Hz, the correction coefficient can be ignored. Therefore, this method is applicable to most specifications of aluminum-sheathed cables, while lead-sheathed cables do not need to be taken into account unless their cross-sections are particularly large. These factors are tabulated and calculated based on the basic formula for sheath loss. Calculating these values ​​requires specialized calculation programs that are not readily available on the general market. The simplified formula for deriving the coefficients of the table is in the test. 2 Referenced standards
The provisions contained in the following standards constitute provisions of this standard by being quoted in this standard. The editions shown were valid at the time of publication of the standard. All standards are subject to revision and parties using this standard should explore the possibility of using the latest version of the standard listed below. JB/T10181.1-2000 Calculation of cable ampacity Part 1: Ampacity formula (100% load factor) and loss calculation Section 1: General provisions
3 symbols
Symbols used in this standard and its parameters are given in the table below. The symbols used in other parts of this standard may indicate different quantities. A,B,C,D are used to calculate the coefficients Ds
the outer diameter of the metal sleeve
that is exactly tangent to the inner surface of the trough of the corrugated metal sleeve. The diameter of an imaginary concentric cylinder whose diameter is exactly tangent to the wave crest of the corrugated metal sleeve. The loss coefficient caused by the eddy current in the metal sleeve caused by the cable conductor current. Approved by the National Machinery Industry Bureau on 2000-04-24mm
2000-10 -01 Implementation
JB/T10181.2-2000
The resistance of the conductor's AC resistance at its maximum operating temperature
S, T, U, V are used to obtain Coefficient in J interpolation c
Spacing between cable centers in adjacent circuits (see Figure 1) Average diameter of the metal sheath or shield
System frequency
Due to the currents in adjacent cables The loss coefficient caused by eddy currents in the metal sheath of the cable × 10 -7
Distance from the center of the cable in the same loop
Thickness of the metal sheath
Coefficient used in 6.5
Under single circuit, the loss factor of high-resistance metal sleeve is under single circuit, the loss factor of low-resistance metal sleeve is under double circuit, the loss factor of low-resistance metal sleeve is under operating temperature, the resistivity of metal sleeve material at operating temperature, the angular frequency of the power supply system ( 2πf)
4 Method Description
4.1 Overview
This method is similar to the calculation method for single loop in JB/T10181.1. The formula for the loss factor of the metal sleeve suitable for the longitudinal resistance of the metal sleeve is m0.1 (R=314μα/m at 50Hz). The empirical formula for calculating the correction coefficient of the low-resistance metal sleeve is also given.
However, for double loops, an accurate empirical formula covering the entire range of coefficients would involve falsified polynomials and would therefore have little or no advantage over using an exact tabular interpolation method, whose advantage lies in the accuracy of the loss factor. Close to the original calculation and its calculation accuracy is better than 1%.
The empirical formula for the limit range of the coefficient is under consideration. In order to explain this method, a suitable manual calculation method is required. However, obviously to meet the requirement of providing loss factors for six cables, it is expected that the calculation is usually only feasible with the help of a computer. In this case, interpolation between tabular data (when necessary) is well justified.
However, in many cases the relevant parameter values ​​do not need to be interpolated, or sufficient accuracy can be achieved using a test method. The correction for the eddy current flowing within the metal sleeve can be derived using the same formula used in JB/T10181.1. 4.2 Summary of method
When the double loop is arranged in a plane (see Figure 1), the calculation of the cable metal sheath loss factor is given by the following formula: [4. H(α-3) N(~6) J(1-6)-g +G,]ara
JB/T10181.2-2000
Where: - Low in double circuit The loss factor of the resistor metal sleeve;. The loss factor of high-resistance metal sheath in a single circuit: H (1 ~ 3) - metal sheath resistance correction coefficient, the value obtained relative to cable 1, 2 or 3 in a single circuit: - the mutual influence between the circuits Coefficient, therefore depends on the corresponding phase sequence of cables 1 to 3 and 4 to 6; N (16) -
J (1 to 6) - - depends on each circuit cable (1 to 3) and ( 4~6) Coefficient of position; 8 - Coefficient of loss caused by eddy currents in the metal sheath of the cable caused by the current of the adjacent cable: G Coefficient of loss caused by eddy currents in the metal sheath caused by the current of the cable conductor: Coefficient The use of N and N is not directly related to any physical function, but is used to simplify the list, which is named arbitrarily. The values ​​of H, W and J are from Table 1 to Table 11, and are selected according to the following parameters as well as the cable position and the current phase sequence in the conductor × 10-7
where: f
—system Frequency (Hz);
R—metal sheath resistance at operating temperature (α/m). In the formula: S-the distance between the centers of cables in the same circuit (mm): d-the average diameter of the metal sleeve (mm).
In the formula: c
z=d/2s
The distance between the cable centers in an adjacent circuit (see Figure 1) +5
w=2πf|| tt | Shown: R
[% · H(1 ~ 3)·g, + G,]
4.3 Application criteria of formulas and coefficients
For metals with m value less than 0.1 Sheaths (including most lead-sheathed cables) can assume that the coefficients H, N, J, and g are 1, while G. is zero, in which case a double-circuit one can be used. No need to correct. When the m value is equal to or greater than 0.1, which is mostly the case except for small-section aluminum sheathed cables, the values ​​of H, N, J and gs should be calculated. The coefficient G, is important only when the value of m is equal to or greater than 1. 5Single circuit high resistance metal sleeve loss factor^. The formula for metal sleeve loss factor^. It is given by the following formula: A. C-
For the planar arrangement of three single-core cables, the coefficient C value is as follows: 2S
Center cable
Outside cable
Calculation of 6 series numbers H, N and J
JB/T10181.2—2000
6.1 For the allocation of coefficients for each cable, the timing and phase marking coefficients C
Special attention should be paid to the coefficients H, A and J depending Depending on the timing of the current flow and the actual position of the conductor. Press picture 1.Number the cables.
The coefficients H (1, 2 and 3) in Table 1 are assigned based on the timing combined with the position of the cables so that the following single-circuit arrangements have the same timing:
Cable number
Coefficients used
Note: The letters R, S and T are used here for convenience and are the same as the commonly used symbols L, L, La, b, c or R, Y, B to distinguish timing and phase markings.
In the above example, cable 1 is always the outer conductor in the leading phase and is assigned the coefficient H. Cable 3 is the outer conductor in the lagging phase and is assigned the coefficient Hs.
It can be seen from this that the use of the symbols R, S and T to indicate the phase marking is not important but only the timing is significant. If there is an opposite timing for each circuit in a double circuit, the H values ​​must be assigned to the cables in opposite order. The values ​​of the assigned coefficients H depend on the timing within each circuit.
In a two-circuit arrangement, the use of symbols to indicate the phase makes sense for the following situations: the phase sign associated with the cable position in one circuit must be the same as the sequential phase sign of the other circuit, or the same as the reversed mirror phase sign of the other circuit. Table 2 gives two sets of coefficients N (1, 2, 3, 4, 5 and 6) corresponding to the sequential and reversed sequence. If the cable positions are marked in sequence and the phase sequence sign is consistent, the coefficients can be assigned in the same way as the coefficient H. Note that the values ​​of cables 4, 5 and 6 in the reversed sequence are reflections of the values ​​of cables 1, 2 and 3.
Many input parameters containing coefficients J (1, 2, 3, 4, 5 and 6) require the use of many tables. Tables 3 to 8 apply to each cable laid in the sequential order, and Tables 9 to 11 apply to each cable laid in the reversed sequence. The coefficients of cables 1 to 3 can also be used for cables 6 to 4 in this sequence, which are assigned the same pair of coefficients N.
Four examples of general avoidance situations are given below: Sequence
Cable number
Assign H
Assign N
Table 2 Sequence
JB/T 10181.2--2000
Table 3 to Table 8 Sequence
Assign J
Cable number
Assign H
Table 2 Sequence
Assign N
Table 3 to Table 8 Sequence
Assign J
Cable number
Assign H
Table 2 Reverse
Assign N
Table 9 to Table 11 Reverse
Assign J
Cable number
Table 2 Reverse
Assign N
Table 9 to Table 11 Reverse
Assign J
Cable number
Table 2 Reverse
Assign N
Table 9 to Table 11 Reverse|| tt||Distribution J
2 and 3) calculation
6.2 The coefficient H (1,
in Table 1) can be obtained from Table 1 using the parameters m and z and the position of each cable (see 6.1). When the m and z values ​​are interpolated between the data in Table 1, if you do not want to use the inspection method to obtain the interpolation, you can use the following method to obtain the H (a, b, c, d) value from the relevant part of Table 1, as shown in the following table: 2
Where mom, zo and z, are tabulated values, less than and greater than the m and z values. Tabulation:
(mm)
2- (zi-zo)
D-（HH）/M
C(H.-H)/z
D（H+HH,-H)/MZ..
Add:
+B(mmo)
+(zz)
+D(mm).(z-zo)
Sum the coefficients
JB/T10181.2-2000
Repeat this procedure for the three cables in the loop to obtain H, H and Ha. 6.3 Calculation of coefficient N (1, 2, 3, 4, 5 and 6) in Table 2 The value of coefficient N can be obtained from Table 2 using the parameter y of each cable. The table has sequential and reverse values. Note that in the latter case, the coefficient values ​​of cables 4, 5 and 6 become mirror images of the coefficient values ​​of cables 1, 2 and 3. When interpolation is required Linear interpolation can meet the requirements. 6.4 Calculation of coefficients J (1, 2, 3, 4, 5 and 6) in Tables 3 to 11 The coefficient J value for each cable can be obtained from Tables 3 to 11 based on the current sequence and the parameters m, z and y. Tables 3 to 8 are applicable to 6 cables with sequential currents in the conductors. Tables 9 to 11 are applicable to cables 1 to 3 and 6 to 4 in reverse order. When all three parameters need to be interpolated, the following three-dimensional interpolation chart can be used. Each cable list is arranged into groups, one for each y value. Two groups can be selected, one with y values ​​less than the input value and the other with y values ​​greater than the input value. Each group requires J (a to の) values ​​and J (e to f) values ​​(similar to the interpolation method for H), as shown in the following chart: 20
Interpolate between the values ​​marked with * to obtain the required values ​​for each cable. The list calculation is as follows:
m\=mm
Calculation:
B(JJ) /M
C (JJ) /Z
D(JJ) /Y
S-［（J+J)-（J+J) J/M· y
F=[ (J+J) - (J+J) J/Z. y
=[(J+J) (J+J) 1/Mf. y
JB/T10181.2-2000
[ (J+J+J+J) - (J+J+J+J) J/M· Z. YThen add:
V. m' .z'y'
J=total
Use the same method to calculate the J value of the other 5 cables. 5 Calculation of coefficient G and g value
(B, ×t,)+
12×1012
N1o'×p
()17+ ​​×(β, × D, ×103 -1.6)8=1+ (
Where: P—Resistivity of metal sheath material at working temperature (2·m): D Outer diameter of metal sheath (mm).
Note: For corrugated metal sheath, replace D with the average outer diameter (2
+t):
Where: D—The diameter of an imaginary cylinder that is tangent to the outer surface of the corrugated metal sheath crest (mm); D.—The diameter of an imaginary cylinder that is tangent to the inner surface of the corrugated metal sheath trough (mm); t Thickness of metal sheath (mm).
7 Precautions for cable transposition
JB/T10181.2-2000||t t||Generally, the role of transposition is to shift all the conductors or metal sleeves, or both, gradually from one small section to another. If this transformation does not affect the phase sequence of the conductor current, as long as the transposition of each section of the line works in the same way relative to the phase sequence (that is, the timing and metal sleeve position requirements for each segment given in 6.1 remain the same), the transposition will not affect the use. The transposition can be carried out in the same direction as the phase or in the opposite direction. As long as the direction of each transposition of the two circuits is the same for the phase sequence, the direction of transposition will not affect the eddy current loss. It can be seen from this that if the two The conductor currents in the loops have reversed phase sequence, and the actual direction of transposition of one loop is opposite to the direction of transposition of the other loop. The value of the eddy current loss of the metal sheath depends only on the cable arrangement orientation, and once determined, it is suitable for any metal sheath in a certain position, regardless of the segmentation.
8 Calculation examples of eddy current losses
8.1 IntroductionbzxZ.net
The cable sizes in the following examples are arbitrary and do not represent any special type of cable. In many cases, interpolation is not required, or the parameter value can be obtained by verification and interpolation using a partial table. However, when the data intervals in the table are too large to be suitable for verification or calculation with a calculator When calculating, it is useful to use an interpolation routine, which is not difficult to calculate manually or by computer programming. 8.2 Example 1
The numerical parameters in this example are consistent with the data in the table and do not require interpolation. Settings:
Metal sheath average diameter
Aluminum sheath thickness
Lead sheath resistance
Conductor resistance
Metal sheath resistivity
(See JB/T10181.1-2000 Table 1). Cable axis spacing in each circuit
Distance between circuits
c=90mm
t=3.18 mm
R=62.9×10-6 Q /m
R=11.3×10-6 2/m
P=2.8264×10-8 /m
s=150mm
c=375mm
314×10-7
62.9 ×150
(1+0.53)300
62.9×10-6
=Cx0.0180
11.Correction of the original value of 3×10-6
:
Assume that the conductors are connected in reverse order
(m0. 5 z=0. 3)
(m=0. 5 z=0. 3 y=0. 4)
Substituting into (1) we get:
The formula for No. 1 cable is:
JB/T10181.22000
4π×314
V2.8264 × 10-8 ×107
g=1+ （ 3.18
×(118.2×93.18×10 -3-1.6)=1.0261.74
G= (118.2 × 3.18)*
12×1012
Aiα\=5.57[(0. 0270×1.2200×1. 0605×1.0100×1.026)+0.0017]=0. 211
In this example, the parameter values ​​are chosen arbitrarily, so it is necessary to interpolate the data in the table. Settings:
Metal sheath average diameter
Aluminum sheath thickness
Aluminum sheath resistance
Conductor resistance
Aluminum sheath resistivity
(See JB/T10181.1--2000 Table 1)
Cable axis spacing in each circuit
Distance between circuits
d100mm
t,=2. 6 mm
R=35×10-6 α /m
R-9×10~6 2/m
p=2.8264×10-82/m
s=150 mm
c=400mm
Take cable 1 and sequential current:
C=1.5, enter 1.5×0.0495=0.0743
a) Interpolation of H:
Take the white coefficient H table:
JB/T10181.2-2000
314×10-7
35×10-6
2×150
35×10-6
9×10-6
2×150
4 yuan×314
2.8264×10-8×107
=C×0.0495
)1.74 ×(118.2 ×102.6×10-3 -1.6) =1.018g=1+(
(118.2×2.6)*
12×1012
Fm—mg=0.500
Z=(z,- z) =0. 050
mm0.897-0.500=0.397||tt ||zzF0. 333—0. 300=0. 033
B=(1.347-1.220)/0.5=
C=(1.309-1.220)/0.05=
D-(1. 503+1. 220—1. 309—1. 347) / (0. 5×0. 05)=2. 680 Add:
B·(m-mo)=0.254x0.397=
C. (z—zo)=1.780×0. 033=
JB/T10181.2-2000
D.(mm)·(zz,)=2.68×0.397×0.033=0.0351E1.4146
b) Interpolation of A
From the coefficient table of N:
y- y=0. 375-0. 3=0. 075
c) Interpolation of J
N0. 9432+(0.9238-0.9432)
(0.4- 0.3)
(0.9920.995)/0.5=
(=(0.991-0.995)/0.1=
D=(0. 991-0. 995) /0.2=
×0.075=0.929
Q=[(0.995+0.984)- (0.992+0.991)]/ (0.5×0.1)=R-[(0.995+0. 982)-(0. 991+0. 991)]/(0. 1X0.2)=S=[ (0. 995+0. 983) - (0. 992+0. 991)J/ (0. 5 X0. 2) =7=[(0.992+0.991+0.991+0.964)-(0.995+0.984+0.983+0.982)J/ (0.5×0.1×0.2)=Addition:
=—0.006X0.397=
=-0.04X0.033=
—0.02×0.175=075
c) Interpolation for J
N0. 9432+(0.9238-0.9432)
(0.4- 0.3)
(0.9920.995)/0.5=
(=(0.991-0.995)/0.1=
D=(0. 991-0. 995) /0.2=
(0. 991+0. 991)]/(0. 1X0.2)=S=[(0.995+0.983) - (0.992+0.991)J/(0.5X0.2)=7=[(0.992+0.991+0.991+0.964) -(0.995+0.984+0.983+0.982)J/ (0.5×0.1×0.2)=Addition:
=—0.006X0.397=
=- 0.04X0.033=
—0.02×0.175=075
c) Interpolation for J
N0. 9432+(0.9238-0.9432)
(0.4- 0.3)
(0.9920.995)/0.5=
(=(0.991-0.995)/0.1=
D=(0. 991-0. 995) /0.2=
(0. 991+0. 991)]/(0. 1X0.2)=S=[(0.995+0.983) - (0.992+0.991)J/(0.5X0.2)=7=[(0.992+0.991+0.991+0.964) -(0.995+0.984+0.983+0.982)J/ (0.5×0.1×0.2)=Addition:
=—0.006X0.397=
=- 0.04X0.033=
—0.02×0.175=
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