JPH037843Y2 - - Google Patents

Description

[Detailed explanation of the idea]

[Purpose of the invention] Industrial field of application The present invention relates to the structure of a three-phase power cable, and is particularly concerned with reducing impedance, which causes voltage drop, and reducing unbalance thereof. Conventional technology Conventionally, as shown in Figure 5, low impedance cables include a 12-core type (a), a three-phase coaxial type (b),
A three-phase flat type (c) in which the conductors of each phase R, S, and T are arranged in parallel in a flat plate shape is known. All of these have a smaller average impedance than the general-purpose three-core twisted structure, but for example, in the three-phase coaxial type (b) and three-phase flat type (c), the impedance of each phase becomes unbalanced, resulting in a voltage drop. This causes fluctuations and voltage unbalance, and in induction motors in particular, there are problems such as insufficient rotational torque and temperature rise, resulting in poor efficiency. Furthermore, each of the types (a) to (c) does not have a sufficient impedance reduction effect, and there has been a demand for a cable with even lower impedance. Problems to be Solved The present invention has been made to solve the above problems, and an object of the present invention is to provide a power cable having a structure that can reduce impedance and impedance unbalance between phases. [Structure of the invention] Means for solving the problem The impedance of a power cable consists of conductor resistance and inductance. Conductor resistance is determined primarily by the cross-sectional area of the conductor (insulated wire core), and inductance is determined by the geometric dimensions (arrangement) of the conductor. Therefore, reducing the impedance of a power cable depends on how to reduce the inductance by devising the geometrical dimensions of the conductor. The present invention was developed with attention to the above points, and its feature is that three insulated wire cores with three-phase identification markings or a multiple of these insulated wire cores are arranged in parallel. A flat cable is constructed by applying a sheath all at once, and this flat cable is
The cables are arranged in parallel or multiples thereof, and the phase identification display between each flat cable is sequentially changed. Embodiments Hereinafter, the present invention will be specifically described with reference to drawings showing embodiments thereof. FIG. 1 is a sectional view of the cable of the present invention. In the figure, 1 is a flat cable, in which three insulated wire cores 4 made of a conductor 2 and an insulating coating 3 are arranged in parallel.
It is molded into a flat shape with a sheath 5 applied all at once. The insulated wire core 4 is provided with three-phase identification marks R ₁ , S ₁ , and T ₁ in order from the left. Flat cables 1' and 1'' having a similar structure are arranged in parallel on the lower surface of this flat cable 1, and are connected together to form a three-phase power cable A. Flat cables 1', 1 The phase identification display for the three insulated wire cores is as follows:
As shown in the figure, S ₂ , T ₂ , R ₂ and T ₃ , R ₃ , S ₃ are sequentially shifted by one phase. In addition, at both ends of each cable 1 to 1'', connect insulated wire cores of the same phase as R ₁ - R ₂ - R _3. In the above structure, calculate the impedance of power cable A and compare it with the conventional product. The result is as follows: In the embodiment of the present invention shown in Fig. 1, first the insulated wire core is
Calculate the inductance of R ₁ . The number of times that the magnetic flux generated by the current Ir ₁ flowing through the wire core R ₁ interlinks with Ir ₁ itself is Ψ _R1R1 = (1/2 + 2logS/r) Ir ₁ ² ×10 ^-7 per 1 m of cable length. Awb〕(1) However, r: Conductor radius (m) S: Vertical distance between any point P and the wire center (However,
Point P is very far away. ) On the other hand, the number of times that the magnetic flux generated by the current flowing through the other insulated wire cores interlinks with Ir ₁ is as follows:
The results are as follows. However, a: Distance between conductor centers of insulated wire cores arranged in parallel, b: Distance between conductor centers of flat cable, c=√ ² + ² d=√ ² + (2) ² e=√ (2) ^{2 +2} . Ψ _R1R2 = (2logS/e) I _R1 I _R2 ×10 ^-7 [Awb] (2) ₁ Ψ _R1R3 = (2logS/d) Ir ₁ Ir ₃ ×10 ^-7 [Awb] (2) ₂ Ψr ₁ s ₁ = (2logS/a) Ir ₁ Is ₁ ×10 ^-7 [Awb] (2) ₂ Ψ _R1S1 = (2logS/a) I _R1 I _S1 ×10 ^-7 [Awb] (2) ₃ Ψ _R1S2 = (2logS/ _{b ) I R1} ^I _{S2 ​} ^​ _{​ ​ ​} I _T1 ×10 ^-7 [Awb] (2) ₆ Ψ _R1T2 = (2logS/c) I _R1 I _R2 ×10 ^-7 [Awb] (2) ₇ Ψ _R1T3 = (2logS/2b) I _R1 I _T3 ×10 ^-7 [Awb] (2) ₈ I The total magnetic flux linkage Ψ _R1 with _R1 is expressed by formula (1) and formula (2) ₁ ~ (2) ₈
Therefore, Ψ _R1 = Ψ _R1R1 + Ψ _R1R2 +……+Ψ _R1T3 …(3). Here, to simplify the calculated values, we assume that I _R1 = I _R2 = I _R3 = I _R I _S1 = I _S2 = I _S3 = I _S I _T1 = I _T2 = I _T3 = I _T ...(4) do. Although conductors of the same phase are actually connected at both ends of the cable, the current in each core should be slightly different. However, the positions of each wire center are geometrically balanced, and the difference is presumed to be small. For this reason, calculations were performed here using the above assumptions. From equations (2) to (4), Ψ _R1 = (1/2+2logS/r)I _R2 ×10 ^-7 +2I _R {I _R (logS/e+logS/d) +I _S (logS/a+logS/b+logS/2c) +I _R (logS/2a+logS/c+logS/2b)} ×10 ^-7 [Awb] ...(5) _1Here , assuming a load such as a three-phase induction motor,
It is assumed that the following symmetrical three-phase alternating currents flow in each of the R, S, and T phases. I _R = I I _S = α ² I I _R = αI ...(5) ₂ However, α = -1/2 + j√3/2), α ² = -1/2 - j√3/2) ...(5 ) ₃ (5) ₂ From equations (5) ₃ , equation (5) ₁ becomes as follows. Ψ _R1 (1/2+2logS/r)I ² ×10 ^-7 +2I ² {logS ₂ /
ed−1/2logS ³ /2abc−1/2logS ³ /4abc+j√3/
2(log1/2)}×10 ^-7 = I ² (1/2+2logS/r+2logS ² /ed−log8a ² b ² c ²
/S ⁶ +j√3log1/2)×10 ^-7 =I ² (1/2+log8a ² b
² c ² /r ² e ² d ² +j√3log1/2)×10 ^-7 [Awb] ...(5) _4Here , if a unit current of I = 1A flows,
Inductance Lr ₁ can be found. Therefore, L _R1 is Lr ₁ = (1/2 + log8a ² b ² c ² / r ² e ² d ² + j√3log1/
2) ×10 ^-7 [H/m] ...(6) becomes ₁ . Similarly, Ls ₁ to Lr ₃ are determined. Ls ₁ = log2a ² b ² /r ² cd+1/2+j√3logd/2c...(6
) ₂ Lt ₃ =loga ² b ² ed/r ² c ⁴ +1/2+j√3loge/d)...
…(6) ₉ Furthermore, the inductances LR, LS, and LT of each pair are 1/LR=1/LR ₁ +1/LR ₂ +1/LR ₃ (7) 1/LS=1/LS ₁ +1/LS ₂ +1/ LR ₃ (8) 1/LT=1/LT ₁ +1/LT ₂ +1/LT ₃ (9) Substitute specific values into these equations and compare with each type of impedance. Here, calculations are performed assuming that the total conductor size in each phase is 500 mm ² . In an embodiment of the invention, there are three conductors (e.g. R ₁ ,
R ₂ , R ₃ ) constitute one phase. Therefore,
The conductor cross-sectional area A of one insulated wire core is A=500/3
= 166mm ² , and the conductor radius r is r = 7.75mm
(However, this is for stranded wire). Also, if the insulation thickness of each insulated wire core is 2 mm, and the minimum insulation thickness when formed into a flat cable is 1.6 mm,
a=19.5mm, b=227mm. If the insulator has this thickness, it will have sufficient insulation performance for a 600V cable as long as normal materials (vinyl chloride, polyethylene, etc.) are used. Further, the values of c, d, and e are determined from the values of a and b. On the other hand, the conductor resistance (per equivalent) is determined by the conductor size, and when r = 7.75 mm, it is 0.0473 (Ω/Km)
It is. Substitute these conditions into equation (6) to find Lr ₁ . Similarly, by substituting L _R2 , Lr ₃ ......Lr ₃ into each equation and then substituting the results into equations (7) to (9) above, the inductances of each phase LR, LS,
LT is LR = 0.639 - j0.0274 [H/m] LS = 0.763 + j0.0133 [H/m] LT = 0.772 [H/m] (10) Therefore, the impedance of each phase is as follows. Become. ZR=0.0482−j0.0201 [Ω/Km] ZS=0.0469+j0.0240 [Ω/Km] ZT=0.0473+j0.0243 [Ω/Km] (11) The equivalent impedance of the line changes depending on the power factor of the power source. do. Now, change the power supply voltage as shown in Figure 6.
If Er, current is I, and power factor is cosθr, the voltage Es at the sending end is Es=Er+(R+jX)I...(12) where R+jX is the total impedance of the line, R: total resistance [Ω/Km ] X: Total reactance [Ω/Km] ES ² = (E _R・cosθr+R・I) ² + (Er・sinθr+X・I) ² ...(13) If the line is short, you can think of θr=θs. , Es=Er+(R・cosθr+X・sinθr)I (14) The equivalent impedance Z is Z=R・cosθr+X・sinθr...(15) Since R and X have already been found,
Cosθr, that is, the change in impedance Z with respect to the power factor can be calculated. The results calculated in this manner are shown in FIGS. 2 and 3. These figures also show impedances of five types of conventional products for comparison. These conventional products are those having the structures shown in FIGS. 5a to 5c, and 3×CV and CVT having a three-core twisted structure. In order to match the conditions, all cables are 600V, and the total conductor size for each phase is ^500mm2 . In addition, each dimension such as the insulator thickness was determined by referring to commercially available products and estimating the value for a product equivalent to 500 mm ² . First, in the case of the 12-core type shown in Figure 5a, four wires constitute one phase, so the cross-sectional area of each wire is 500/4.
From this, the conductor radius r of each covered wire is 6.75 mm. At this time, the outer diameter D of the covered wire is 17.4 mm. Calculate the inductance from the relationship between this geometric dimension, the current flowing through each conductor, and the resulting magnetic flux linkage, and then add the conductor resistance to find the total impedance Z of the line, Z = 0.0473+j0.0178 (Ω/Km). Next, in the case of the three-phase coaxial type shown in Figure 5b,
Use a type with a conductor in the center, radius a of the center conductor = 13.45 mm, inner radius b ₁ of the intermediate conductor = 19.45 mm,
Outer radius b ₂ = 24.45 mm, inner radius of outer conductor C ₁ = 30.05
mm, outer radius c ₂ = 33.55 mm. Since the impedance of this type differs depending on the phase order of the applied voltage, positive phase, I _R = I, I _S = α ² I, Zt = αI, reverse phase, I _R = I, I _S = αI, Zt = Calculations were performed in the same manner as above for both α ² I and . As a result, in the case of positive phase, Z _R = 0.0452−j0.0011 (Ω/Km) Z _S = 0.0644 + j0.0092 (Ω/Km) Z _T = 0.0272 + j0.0582 (Ω + Km) In the case of negative phase, Z _R = 0.0452−j0.0011(Ω/Km) Z _S =0.0242+j0.0092(Ω/Km) Z _T =0.0674+j0.0582(Ω/Km). Next, in the case of the three-phase flat type shown in Figure 5c,
The length a of the conductor is 60.0 mm, the width b is 12.0 mm, and the pitch D between the conductors is 22.6 mm. From this value, Z _R = Z _T = 0.0850 + j0.0431 (Ω/Km) Z _S = 0.0473 + j0.0213 (Ω/Km). Next, in the case of a CVT with a three-core twisted structure, the conductor radius r = 13.45 mm and the outer diameter of the covered wire D = 32.9 mm, and from these values, Z = 0.0473 + j0.0795 (Ω/Km). This value is the same for R, S, and T. Finally, in the case of 3×CV with a 3-core twisted structure, the conductor radius r = 13.45 mm, the outer diameter of the covered wire D = 37.1 mm, and for R, S, and T, Z = 0.0473 + j0.0719 (Ω/Km). Summer. From the above results, the diagrams shown in FIGS. 2 and 3 were obtained, and as is clear from these diagrams, the three-phase parallel flat structure according to the present invention is the best. In order to further experimentally verify this, the impedance (maximum value) at power factor cosθ = 0.5
I asked for This is shown in the table below.

[Effect of idea]

Since the present invention is as explained above, it has the following effects. (1) Impedance unbalance of each phase can be kept to a minimum. (2) The impedance can be lowered even more than conventional low impedance cables, and the voltage drop is also reduced. (3) Since the cable structure is flat, it has excellent flexibility. (4) Since flat cables are used closely together, it can be laid compactly and is easy to install.

[Brief explanation of drawings]

Figure 1 is a cross-sectional view showing one embodiment of the cable of the present invention, Figures 2 and 3 are graphs showing the relationship between the power factor and impedance of the cable of the present invention and a conventional product, respectively, and Figure 4 is a graph showing the relationship between the power factor and impedance of the cable of the present invention. 5A to 5C are sectional views of conventional cables, and FIG. 6 is a vector diagram. 1~1″,
6~6''...Flat cable, 4,4'...Insulated wire core, 5,5'...Sheath.

Claims (1)

[Scope of utility model registration request] A flat cable is constructed by arranging three insulated wire cores with three-phase identification markings or a multiple of these insulated wire cores in parallel and applying a collective sheath. Three-phase cables are arranged in multiples in parallel, the phase identification indicators between each flat cable are sequentially shifted by one phase, and the insulated wire cores of the same phase are connected to each other at both ends of each flat cable. Structure of power cable.