LT 和 HT 电机的电缆尺寸计算
LT &HT 电机的电缆尺寸如何计算?
为电机选择合适的电缆尺寸 无论是在安装和调试期间还是在运行条件下,都是行业的重要参数。这是安全、成本最小化和减少意外损失的一个非常重要的方面。尺寸过小的电缆可能会在电机运行期间燃烧,从而对人的生命、机器、基础设施、生产损失和更换成本造成风险。
而过大的导体不仅会为长期电缆产生不必要的成本,而且还会为与它们一起使用的电缆端接材料,即接线片、密封套、连接套件(以防出现任何故障)产生不必要的成本将来发生)和超大电缆桥架。与相应的较小尺寸电缆相比,铺设较大尺寸电缆的人工成本也将更高。考虑到所有这些要点,为我们的电机进行正确的电缆尺寸计算非常重要 .
- 必读:如何计算不同负载和家庭布线的电缆尺寸(美国和欧盟解决的示例)
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在详细介绍之前,让我们先弄清楚 LT 和 HT 电机之间的主要区别。
LT 和 HT 电机有什么区别?
好吧,正如LT(Low Tension,即低压)和HT(High Tension,即高压)或低扭矩和高扭矩这两个词分别描述了整个故事本身。
它还取决于电源电压的可用性,即在美国和欧盟,
LT 电机系列 =230V – 415V
HT 电机系列 =3.3 kV、6.6kV – 11kV
同时请记住,LT 电机 需要更新 HT 电机。
在其他地区,他们将LT电机归为1kV以下和HT电机超过1kV。
现在我们要讨论的主题是如何计算电机的电缆尺寸?
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125 Kw LT电机电缆尺寸计算
电机KW =125
Pf =0.8,效率 =94%
系统电压,V1 =415
电缆长度 =200 m
负载电流 =P / (1.732 x V x Pf x Eff) —> (P =√3 x Vx I CosΦ =三相电路)
主页>
=125000 / (1.732 x 415 x 0.8 x 0.94)
~ 230 A
这是理想条件下需要满足的满载电流电缆。但在实际情况下,有几个降额因素需要考虑。
电缆的电流额定值是针对 40* C 的环境温度定义的。如果环境温度高于此值,则电缆的载流能力会降低。
假设我们的电缆在空气中铺设在电缆桥架上,
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气温(摄氏度) | 20° | 25° | 30° | 35° | 40° | 45° | 50° | 55° | |
普通PVC | 1.32 | 1.25 | 1.16 | 1.09 | 1.00 | 0.90 | 0.80 | 0.80 | |
降级因素 | HR PVC | 1.22 | 1.17 | 1.12 | 1.06 | 1.00 | 0.94 | 0.87 | 0.80 |
XLPE | 1.20 | 1.16 | 1.11 | 1.06 | 1.00 | 0.95 | 0.88 | 0.82 |
与环境空气温度变化相关的评级因素
温度校正系数,电缆在空中时的 K1 =0.88(对于 50* Amb temp &XLPE 电缆)
电缆分组也会降低电缆的载流能力。如果将多条电缆组合在一起,它们都会变热。热量将无法正确消散,因此它会使电缆本身和与之接触的电缆变热。这将进一步提高温度。因此,我们必须根据分组因素降低电缆的载流能力。
让我们来看看最坏的情况,即 3 个相互平行的托盘有 9 根电缆相互接触。
没有。机架数 | 没有。每个机架的电缆数 | 没有。每个机架的电缆数 | ||||||||
1 | 2 | 3 | 6 | 9 | 1 | 2 | 3 | 6 | 9 | |
1 | 1.00 | 0.98 | 0.96 | 0.93 | 0.92 | 1.00 | 0.84 | 0.80 | 0.75 | 0.73 |
2 | 1.00 | 0.95 | 0.93 | 0.90 | 0.89 | 1.00 | 0.80 | 0.76 | 0.71 | 0.69 |
3 | 1.00 | 0.94 | 0.92 | 0.89 | 0.88 | 1.00 | 0.78 | 0.74 | 0.70 | 0.68 |
6 | 1.00 | 0.93 | 0.90 | 0.87 | 0.86 | 1.00 | 0.76 | 0.72 | 0.65 | 0.66 |
表:电缆分组系数(托盘数量系数),K2 =0.68(对于 3 个托盘,每个托盘有 9 根电缆)
总降额系数 =K1 x K2
= 0.88×0.68 =0.5984
我们选择 1.1 KV、3 芯、240 Sq.mm、铝、XLPE、铠装电缆用于单条运行
点击放大表格
1.1 KV、3 芯、铝/铜导体、XLPE 绝缘铠装电缆的技术细节
240 Sq.mm XLPE 铠装铝电缆在空气中的电流容量为 402 安培
240 Sq.mm 电缆的总降额电流 =402×0.5984 = 240.55 Amp
电阻 = 0.162 Ω / Km 和
电抗 = 0.072 Ω / Km
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用于交流系统的 PVC/XLPE 铝电缆的估计电压降 | ||||
(电压降 - 伏特/公里/安培) | ||||
导体标称面积(平方毫米) | P.V.C.电缆 | XLPE 电缆 | ||
单相 | 三相 | 单相 | 三制 | |
1.5 | 43.44 | 37.62 | 46.34 | 40.13 |
2.5 | 29.04 | 25.15 | 30.98 | 26.83 |
4 | 17.78 | 15.40 | 18.98 | 16.44 |
6 | 11.06 | 9.58 | 11.80 | 10.22 |
10 | 7.40 | 6.41 | 7.88 | 6.82 |
16 | 4.58 | 3.97 | 4.90 | 4.24 |
25 | 2.89 | 2.50 | 3.08 | 2.67 |
35 | 2.10 | 1.80 | 2.23 | 1.94 |
50 | 1.55 | 1.30 | 1.65 | 1.44 |
70 | 1.10 | 0.94 | 1.15 | 1.00 |
95 | 0.79 | 0.68 | 0.83 | 0.70 |
120 | 0.63 | 0.55 | 0.66 | 0.56 |
150 | 0.52 | 0.46 | 0.55 | 0.48 |
185 | 0.42 | 0.37 | 0.44 | 0.40 |
240 | 0.34 | 0.30 | 0.35 | 0.30 |
300 | 0.28 | 0.26 | 0.30 | 0.26 |
400 | 0.24 | 0.22 | 0.24 | 0.22 |
500 | 0.23 | 0.20 | 0.23 | 0.20 |
630 | 0.20 | 0.18 | 0.21 | 0.18 |
800 | 0.19 | – | 0.20 | – |
1000 | 0.18 | – | 0.18 | – |
Voltage drop, V2 =0.3 Volts/Km/Amp (as per Havell’s brochure)
=0.3 x 230 x (200 / 1000)
=13 V
Terminal voltage at Motor, V2 =415 -13 =402 V
% Drop =(V2 – V1) / (V1)
=(415 – 402) x 100 / (415)
=3.13%
To decide 240 Sq.mm cable, cable selection condition should be checked
- Cable derating Amp (240.55 Amp)is higher than full load current of load (230 Amp) = OK
- Cable voltage Drop (3.13%)is less than defined voltage drop (10%) = OK
- Cable short circuit capacity (22.56 KA) is higher than system short circuit capacity at that point ( X KA) = OK
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240 Sq.mm cable satisfied all three condition, so it is advisable to use 3 Core 240 Sq.mm cable.
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Cable Size Calculation for 350 KW HT Motor
In case of LV system cable can be selected on the basis of its current carrying capacity and voltage drop but in case of MV/HV system cable short circuit capacity is an important/deciding factor. So in case of HT motor, the cable short circuit capacity alone is enough to determine the cable size as rest two parameters will automatically follow.
Consider the below example:
Motor KW =350
Pf =0.8, Efficiency =94%
System Voltage, V1 =6.6 KV
Cable length =200 m
Load Current =P / (1.732 x V x Pf x Eff)
=350000 / (1.732 x 6600 x 0.8 x 0.94)
=41 A
Suppose Short circuit level/Fault level for H.T. system, Ish (for duration t=1sec) =26.2 KA
With Aluminum conductor, XLPE insulated cable =
=278.72 Sq.mm
Hence nearest higher size 300 sq mm is required.
We can see from the below table also that the short circuit capacity of 300sqmm cable is 28 KA which is more than our fault level.
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Click image to enlarge
(6.6KV UNEARTHED / 11KV EARTHED GRADE)
Technical Details For 6.6 KV, 3 Core, Aluminum/Copper Conductor, XLPE Insulated, Armored Cables
We can see that this will automatically satisfy other two conditions also.
Let’s select 6.6 KV, 3 core, 300 Sq.mm, Aluminum, XLPE, Armored cable for single run
Temperature Correction Factor, K1 when cable is in the Air =0.88 (for 50* Amb temp &XLPE cable)
Cable Grouping Factor (No of Tray Factor), K2 =0.68 (for 3 trays having 9 cable each)
Total derating factor =K1 x K2 =0.88 x 0.68 =0.5984
Current capacity of 300 Sq.mm XLPE Armored aluminum cable in Air is 450 Amp
Total derating current of 300 Sq.mm Cable =450×0.5984 = 269.28 Amp
Resistance = 0.130 Ω / Km and
Reactance = 0.0999 Ω / Km
Voltage drop =0.26 Volts / Km / Amp (as per Havell’s brochure)
=0.26 x 200 x 41 / 1000
=2.132 V
Terminal voltage at Motor, V2 =6600 – 2.132 =6597.868 V
% Drop =(V1 – V2) / (V1)
=(6600 – 6597.8) x 100 / (6600)
=0.032%
To decide 300 Sq.mm cable, cable selection condition should be checked
- Cable derating Amp (269.28 Amp ) is higher than full load current of load (41 Amp ) = OK
- Cable voltage Drop (0.032% )is less than defined voltage drop (5% ) = OK
- Cable short circuit capacity (28.20 KA ) is higher than system short circuit capacity at that point (26.2 KA ) = OK
300 Sq.mm cable satisfied all three condition , so it is advisable to use 3 Core 300 Sq.mm cable.
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