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A line that meets one of these two requirements is a safer line.
Talking about the Grounding Fault Protection Measures of Street Lamps
The zero-resistance resistance RNP at the end of the street lamp line is equal to the phase-preserving resistance RφP at normal temperature, and the TN-S system can judge whether the ground fault can be cut off in time according to the measured RNP value of the zero-resistance resistance. The networked TT system of the street lamp determines the safety of the fault point voltage according to the ratio of the two grounding resistors. The error of the simplified calculation of the end single-phase short-circuit current is generally less than 3.4%. The phase-resistance resistance Rφp of the cable is the phase line resistance and the protection line resistance. The line has been built, and the grounding fault can be cut off in time with the segmentation protection.
First, check and judge
(1) TN-S zero protection system:
At the end of the street lamp line, measure the resistance RNP between the zero (N) line and the ground line (PE), referred to as zero-resistance.
(2) Networking TT system for street lamps:
In the street light substation, measure the grounding resistance RNE of the zero (N) row and the grounding resistance RPE of the ground (PE) row,
1. If 6 × RPE <RNE, the fault point to ground voltage is less than 33 volts in a humid environment safe voltage.
2. If RPE <0.5 ohm, the ground can be disconnected automatically after the root fuse is grounded.
Second, the reasons for judgment:
(1) TN-S zero protection system:
As shown in Figure 1, the resistance between the N-line and the PE line is measured at the end of the street lamp line and is recorded as RNP. The resistance between the N row and the PE row is measured at the beginning, which is recorded as the beginning of the RN (the N row is disconnected from the PE row, and the terminal N is connected to the PE), as shown in FIG.
Use RN to indicate the resistance of the neutral line from the beginning to the end, and RP to represent the equivalent resistance between the PE row and the end of the PE line. RNP=RN + RP
And RN starts with P = RP + RN
So RNP=RN starts at P...(1)
The cross-section cable has the same cross-section as the phase line. The resistance between the phase line and the PE line is measured at the beginning (the end phase line is connected to the PE line), which is recorded as RφP often, see Figure 3.
When the street light is not shining, there is
RφP often = RN starts from P... (2)
Substituting (1),
Get RNP=RφP often...(3)
That is to say, the TN-S system measures the resistance RNP between N and PE at the end, which is the phase-preserving resistance RφP at the normal temperature when the terminal is single-phase grounded. Since the current during the short circuit is large, the wire will heat up and the value of RNP is increased by 50%, so the phase-resistance resistance RφP.L of the single-phase short-circuit current at the end is calculated.
There are many PE lines on the PE row, and there are many grounding poles between the PE row and the PE wire end, so the equivalent resistance RP between the PE wire end and the PE row is smaller than the PE wire resistance. If the measured zero-resistance RNP is larger than the calculated resistance of the N-line plus PE line, the PE line may have a breakpoint. There should be no breakpoints on the PE line. This method can be used to check for the presence or absence of breakpoints.
(2) Networking TT system for street lamps:
1. If you do 6×RPE < RNE
As shown in Figure 4, when a ground fault occurs, the phase line is connected to the PE line, and the power supply voltage U=220 volts is added between the PE line (point P in the figure) and the neutral point N of the transformer (ignoring the phase line resistance), ie Add in two series resistor RPE
And on the RNE. The sum of the PE line-to-ground voltage UPE and the transformer neutral-to-ground voltage U zero is equal to the supply voltage of 220 volts.
This result shows that the PE line to ground voltage UPE at ground fault is less than the safe voltage of 33 volts in a humid environment.
2. If you do RpE < 0.5 Euro
Set the transformer grounding resistance RNE < 4 ohms, the phase line resistance is 1 ohm, and the total loop resistance will be less than 0.5 + 4 + 1 = 5.5 ohms. The short-circuit current will be greater than 220 ÷ 5.5 ohms = 44 amps, and the rated current of the fuse at the root of the pole is less than 10 amps, which can be blown in 5 seconds.
Therefore, the networked TT grounding protection system of the street lamp is safer if one of the above two conditions is fulfilled.
Third, the simplified calculation of the terminal short-circuit current
(1) Single-phase grounding short-circuit current calculation formula
If the street lamp line is single-phase grounded, the voltage of one phase is zero, the other two phases are not zero, and the three-phase voltage is asymmetrical. After analysis by the symmetrical component method, the low-voltage single-phase ground fault current Ik of the TN grounding system can be found in the design manual.
The subscript φ indicates the phase, P indicates the protection line, and φP indicates the phase guarantee.
(1) indicates positive sequence, (2) indicates negative sequence, (0) indicates zero sequence, S indicates system, T indicates transformer, m indicates bus, and L indicates low voltage line.
Calculate the short-circuit current according to the formula (7). Calculate the resistance and reactance of the positive sequence, negative sequence, and zero sequence. Calculate the phase-resistance and reactance of the cable, and calculate the phase-resistance and reactance of the transformer, bus, and system.
(2) Simplified calculation
The short-circuit current at the end of the street lamp line is calculated by using the minimum short-circuit current to verify the fuse or circuit breaker, and whether the ground fault can be disconnected within the specified time (for example, 5 seconds).
1. Simplify the calculation:
The simplified calculation formula is to calculate only the resistance Rφp.L of the low-voltage line of the street lamp, ignoring its reactance and other impedances, that is, calculating the single-phase grounding short-circuit current Ik†according to the formula (8).
2. Simplify the calculation error
The line resistance of the street lamp from the power supply to the end is relatively large, ignoring the reactance of the line and the impedance of the transformer, the busbar and the system, and the error generated is relatively small, and the analysis and judgment of the ground fault is much more convenient. The following is a representative street lamp distribution system to analyze the error caused by the simplified calculation.
Set the system short-circuit capacity 10MVA, transformer capacity 100kVA, the connection group is D, yn11, busbar for busbars, the specification is 4 (80×10), the length is 3 meters, the spacing is D=250mm, and the line is 25mm2 copper core cable. The distribution radius is 500 meters.
(1) Calculate the system impedance
Check the 10kV power distribution engineering design manual (the same below) Table 10-26, the system impedance of 10MVA is
Rφp.S=1.06 mΩ, Xφp.S=10.61 mΩ,
(2) Calculate the transformer impedance
Look at Table 10-28, the positive sequence impedance of 100kVA is
R(1).T=25.6 mΩ, X(1).T=58.7 mΩ,
In Note 1 of Table 10-28, there is a description, namely: ..., the zero sequence and phase-preserving impedance values ​​of the transformer connected to D, yn11 are equal to R(1) and X(1) in the table. and so
Rφp.T=25.6 mΩ, Xφp.T=58.7 mΩ,
(3) Calculate the bus impedance
Looking at Table 10-31, the impedance of the busbar is Rφp'=0.04 mΩ/m and Xφp'=0.361 mΩ/m.
The busbar impedance of 3 meters long is
Rφp.m=0.12 mΩ, Xφp.m=1.08 mΩ,
(4) Calculate cable impedance
Look at Table 10-32, the phase-preserving resistance of the 25mm2 cable with equal section is Rφp'=2.106 mΩ/m, and the phase-retaining reactance of the all-plastic cable is Xφp'=0.164 mΩ/m.
Rφp.L=500×2.106=1053 mΩ=1.053 Ω,
Xφp.T=500×0.164= 82 mΩ=0.082 Ω.
(5) Calculate the total impedance ZφP
Rφp=Rφp.s+ Rφp.T+ Rφp.m+ Rφp.L=1.06+0.12+25.6+1053=1079.78 mΩ=1.08 Ω ,
Xφp=Xφp.s+ Xφp.T+ Xφp.m+ Xφp.L=10.61+1.08+58.7+82 = 152.39 mΩ=0.152 Ω,
(6) Error analysis of simplified calculation
That is to say, the cable phase-preserving resistor Rφp.L accounts for 96.6% of the entire impedance Zφp.
Only the cable resistance is counted, and the simplified calculation of its reactance and other impedances is ignored, with an error of 3.4%.
The actual system short-circuit capacity is greater than 10MVA, the impedance is small, the calculation error will be less than 3.4%, the actual transformer capacity is greater than 100kVA, the impedance is small, the calculation error will be less than 3.4%, less than 100kVA, the impedance is larger, the error is Bigger.
The actual use of the street light box type substation busbar will not exceed 3 meters, the calculation error will be less than 3.4%, the street light cable is generally not larger than 25mm2, the resistance is larger, so the general calculation error will be 3.4% smaller. When the length of the street lamp cable is less than 500 meters, the resistance is small and the error will increase. It is not suitable to use the simplified calculation formula.
Fourth, the impedance of the cable ZφP.L
(1) Analysis of the composition of cable impedance
Figure 5 shows the low-voltage network where the neutral point is directly grounded. Where A, B, and C are phase lines, and N is a neutral line or a PE line.
When a three-phase short circuit occurs, there is no short-circuit current in the neutral line. Therefore, in the calculation of the three-phase short-circuit current, only the parameters of the phase line need to be considered. In general, the resistance r and the self-inductance L of the three phase lines are the same. In the cable, the mutual inductances Mab, Mbc, and Mca are also the same, which is taken as M.
That is, M=Mab=Mbc=Mca,
In the case of three-phase short circuit, the positive sequence voltage drops Ua1, Ub1, Uc1 and the positive sequence short-circuit currents Ia1, Ib1, Ic1 should satisfy the following relationship.
When a single-phase short circuit occurs, the short-circuit current will pass through the PE line. The resistance of the PE line is r0, the self-inductance is L0, and the mutual inductance between the PE line and each phase line is Ma0, Mb0, Mc0, respectively. In the cable, set Ma0= Mb0=Mc0=Mφp.
When a single phase is shorted, the zero sequence voltage of the network should satisfy the following relationship:
The first term in (13) is the fault phase self-impedance voltage drop, the second and third terms are the mutual inductance voltage drop of the other two relative fault phases, and the fourth term is the mutual inductance voltage drop of the PE line current (3I0) to the fault phase. Because the current of the PE line is opposite to the direction of the current in the fault phase, the item takes a negative sign. The fifth term is the self-impedance voltage drop of the PE line, and the sixth term is the mutual inductance voltage drop of the three phase lines to the PE line. Since the current direction in the phase line is opposite to the current direction of the PE line, the negative value is taken.
(2) Analysis results
(15) Description, when calculating the single-phase grounding short-circuit current, the resistance of the cable is the resistance of the phase line and the resistance of the PE line. The reactance of the cable is the self-reactance of the phase line, minus the mutual inductance between the phase line and the PE line. Times.
Considering the short circuit, the wire will heat up, and the resistance should be 1.5 times the resistance of 20 °C.
V. Segment protection
To cut off the ground fault in time, there must be a large enough short-circuit current. To increase the short-circuit current, it is mainly to reduce the distribution radius or increase the cross-section of the cable. It is of course more advantageous to increase the grounding pole and lower the grounding resistance. The line has been built, it is difficult to change the distribution radius and change the cable section. It is better to use segment protection to cut off the fault in time.
Most of the street lamp load is evenly distributed along the line. The current at the beginning is large, and the current is smaller at the end. The last rod has only one set of lamps, and the current is the smallest. Therefore, the rated current of the series fuse or circuit breaker can be reduced step by step along the line. The last rod uses the smallest fuse, and the middle uses a slightly larger fuse, so that the fault can be cut off in time, the whole line In other words, it has been done to cut off the fault in time.