CN103902815A - Method for building model of transmission line of medium voltage distribution network - Google Patents

Abstract

The invention provides a method for building a model of a transmission line of a medium voltage distribution network. The method comprises the following steps that (1) according to the transmission line part of circuitry, pi-type and T-type equivalent circuits of the transmission line can be obtained; (2) according to the pi-type and T-type equivalent circuits and parameter values capable of being measured on site, simplified pi-type and T-type equivalent circuits are obtained; (3) according to the simplified pi-type and T-type equivalent circuits, on-site actual requirements and computer calculation requirements, a front L-type equivalent circuit and a rear L-type equivalent circuit are obtained. The method for building the model of the transmission line of the medium voltage distribution network has the advantages that both the front L-type equivalent circuit and the rear L-type equivalent circuit of the transmission line can be applied to network parameter calculation of the power distribution network; the front L-type equivalent circuit and the rear L-type equivalent circuit can reduce the amount of computer calculation to a great degree; by independently adopting the front L-type equivalent circuit or the rear L-type equivalent circuit, different parameters can be shown from the reactive power input end in different reactive power transmission directions of the same transmission line.

Description

A kind of method for building up of medium voltage distribution network transmission line model

[technical field]

The present invention relates to a kind of medium-voltage distribution technical field, relate in particular to a kind of method for building up of medium voltage distribution network transmission line model.　

[background technology]

In medium voltage distribution network, because middle pressure power transmission line circuit length is shorter, the parameter higher pressure power transmission and transformation network of middle pressure transmission line is little a lot, be difficult for calculating, middle pressure transmission line sectional area of wire is less, and therefore power distribution network transmission line resistance and reactance size are substantially equal, because the electric pressure of medium voltage distribution network exceeds doubly a lot compared with low-voltage network, therefore susceptance can not be ignored, and the impact that in medium voltage distribution network, electricity is led need to determine whether in transmission line model, to ignore through demonstration;

Traditional transmission line model is π type and T-shaped model, this transmission line model bilateral impedance in high voltage power transmission and transforming network equates, apply more extensive, but in MV distribution systems, it is unknown pressing due to piezoelectricity in most of medium-voltage distribution room, adopt π type and T-shaped model to calculate comparatively complicated, need to simplify transmission line model.　

[summary of the invention]

The object of the invention is to solve in MV distribution systems, it is unknown pressing due to piezoelectricity in most of medium-voltage distribution room, adopt π type and T-shaped model to calculate comparatively complicated, the deficiency that need to simplify transmission line model and the method for building up of a kind of novel medium voltage distribution network transmission line model that provides.　

The present invention is achieved through the following technical solutions: a kind of method for building up of medium voltage distribution network transmission line model, comprises that step is as follows:

(1) according to circuitry transmission line portions, can known transmission line π type and T-shaped equivalent circuit;

(2) according to the parameter value going out of π type and T-shaped equivalent circuit and the actual measurement of on-the-spot energy, draw π type and the T-shaped equivalent circuit of simplification;

(3) according to the π type of simplifying and T-shaped equivalent circuit according to needs on-the-spot actual and computing machine calculating, draw front L-type equivalent circuit and rear L-type equivalent circuit.　

Further, described step (1) is according to circuitry transmission line portions, can known transmission line π type and T-shaped equivalent circuit in, the resistance of the resistance R transmission line in Z=R+jX, X is transmission line reactance, ohmic loss is directly proportional to transmission line current flowing size to reactance loss, and it is irrelevant with transmission-line voltage grade, be directly proportional to cable length simultaneously, resistance R can be calculated according to electric wire handbook, reactance X can calculate according to electric wire handbook, also can know according to circuitry, it is that the leakage power of cable and the electricity that produces are led that electricity in Y=G+jB is led G, B is cable earth capacitance electric current and susceptance that corresponding charge power produces, leakage power and charge power are directly proportional to transmission-line voltage square, with to flow through transmission line choke irrelevant, be directly proportional to cable length simultaneously, electricity is led G and can be calculated according to electric wire handbook, leakage power is made up of two parts: one is insulation resistance loss, one is dielectric loss, susceptance B can calculate according to electric wire handbook.　

Further, described step (2) is according to the parameter value going out of π type and T-shaped equivalent circuit and the actual measurement of on-the-spot energy, draw π type and the T-shaped equivalent circuit of simplification, suppose 185 square millimeters of cable model YJV-8.7-15kV of medium voltage distribution network cross-section of cable list core, circuit is single phase power supply;

Its parameter is conductor AC resistance 0.128 Ω/km, and reactance is 0.096 Ω/km, and electric capacity is 0.3 μ F/km, supposes that this single-phase cable length is 1km longer in actual conditions,

At this moment cable data is calculated as R=0.128 Ω, X=0.096 Ω, B=9.42x10-5S

The size of current flowing through on transmission line is:

I=250A　

Transmission-line voltage grade is:

$U=10000/\sqrt{3}V$

Suppose that the power transmitting on circuit is:

P=1400kW　

Q=351.19kvar　

Therefore power factor is:

cosφ=0.97　

The realistic ruuning situation of this power factor

So can draw up following formula:

${\mathrm{\&Delta;Q}}_B={\mathrm{<figure-callout\; id="2"\; label="BU"\; filenames="140310145823.png"\; state="\{\{state\}\}">BU</figure-callout>}}^2=9.42\&times;{10}^{-5}\&times;\frac{{\left({10}^4\right)}^2}{{\left(\sqrt{3}\right)}^2}=3140\mathrm{var}$

△P _R=I ²R=250 ²×0.128=8000W　

△Q _X=I2X=250 ²×0.096=6000var　

△P _G=△P _G1+△P _G2　

At this moment calculate △ P _gsize, wherein △ P _g=△ P _g1+ △ P _g2

1) △ P _g1insulation resistance loss, △ P _g2dielectric loss,

We first see insulation resistance

According to electric wire handbook (second edition 2008 enlarged editions) (Wang Chunjiang chief editor) P299 page formula (3-2-17)

Wherein insulation resistance

ρ _v---insulation resistance coefficient: $R_v={\mathrm{\&rho;}}_v\frac{\mathrm{\&delta;}}{A}=\frac{1}{{\mathrm{<figure-callout\; id="1"\; label="G\; G"\; filenames="140310145823.png"\; state="\{\{state\}\}">G</figure-callout>}}_G}$

According to electric wire handbook (second edition 2008 enlarged editions) (Wang Chunjiang chief editor) P299 page table 3-2-13

Crosslinked polyethylene (YJV) insulating resistance of cable coefficient is 1016～1017/ (Ω cm)

δ---insulation thickness

Suppose that insulation thickness is 0.5cm(and can reduces calculated resistance toward getting in little, thereby increase leakage power)

A---electrode area

185mm ²cable interior diameter is 2cm, girth 6.28cm, and 1km cable is 100000cm length,

Calculate by the poorest condition $R_v={\mathrm{\&rho;}}_v\frac{\mathrm{\&delta;}}{A}={10}^{16}\&times;\frac{0.5}{5\&times;{10}^5\&times;6.28}=7.962\&times;{10}^9\mathrm{\&Omega;}$

Therefore for 10kV circuit, insulation resistance Leakage Current has:

$I_{\mathrm{<figure-callout\; id="1"\; label="G"\; filenames="140310145823.png"\; state="\{\{state\}\}">G</figure-callout>}1}=U/R_v=10000/(\sqrt{3}\&times;7.962\&times;{10}^9)=7.251\&times;{10}^{-7}A$

Therefore insulation resistance loss power is:

△P _G1=UI _G1=0.0042W　

(2) according to electric wire handbook (second edition 2008 enlarged editions) (Wang Chunjiang chief editor) P304 page formula (3-2-24)

△P _G2=W _d=U ²ωCtgδ　

Wherein tg δ calculates according to crosslinked polyethylene maximal value 0.0025 in table 3-2.22:

△P _G2=W _d=U ²ωCtgδ=3140×0.0025=7.85W　

△P _G=△P _G1+△P _G2=0.0042+7.85=7.854W　

Therefore total electricity is led power

1) therefore total conduction current is: ${\mathrm{\&Delta;I}}_G=\frac{P_G}{U}=\frac{7.854}{\frac{10000}{\sqrt{3}}}=1.36\&times;{10}^{-3}A$

And load current is 250A, current mutual inducer for line outlet end rated current is 300A, and current transformer full accuracy is 0.1%, and the minimum current that can differentiate is 0.3A, and electricity lead loss electric current only for current transformer can be differentiated 1/219th of electric current,

Simultaneously two kinds of electricity are led active loss and are all directly proportional to line voltage distribution grade square, and flow through line current size and have nothing to do, and electric pressure substantially constant, and therefore to lead loss be an individual constant to electricity substantially, and variation range is very little,

In sum, in power distribution network transmission line model, electricity is led and can be ignored,

2) by ${\mathrm{\&Delta;Q}}_B={\mathrm{<figure-callout\; id="2"\; label="BU"\; filenames="140310145823.png"\; state="\{\{state\}\}">BU</figure-callout>}}^2=9.42\&times;{10}^{-5}\&times;\frac{{\left({10}^4\right)}^2}{{\left(\sqrt{3}\right)}^2}=3140\mathrm{var}$

The susceptance reactive power compensation power that cable earth capacitance forms is directly proportional to line voltage distribution grade square, line current size is irrelevant with flowing through, and electric pressure substantially constant, therefore to lead loss be constant to electricity substantially, and variation range is very little, and the power of susceptance B is 3140var, and substantially constant, the electric current calculating is 0.54A, can be resolved the current transformer identification of rate 0.3A,

Therefore susceptance parameter can not be ignored, and need to calculate susceptance power back-off;

3) by △ P _r=I ²r=250 ²× 0.128=8000W

Ohmic loss is directly proportional to current squaring, irrelevant with line voltage distribution size, circuit voltage swing is basicly stable simultaneously, and ohmic loss is very large with load variations variation range, and it is 1.39A that full load calculates size of current, can be resolved the current transformer identification of rate 0.3A;

Therefore resistance parameter can not be ignored, and needs calculated resistance power attenuation;

4）△Q _X=I ²X=250 ²×0.096=6000var　

Reactance Power consumption and electric current square is directly proportional, irrelevant with line voltage distribution size, circuit voltage swing is basicly stable simultaneously, and reactance loss is very large with load variations variation range, and it is 1.04A that full load calculates size of current, can be resolved the current transformer identification of rate 0.3A;

Therefore reactance parameter can not be ignored, and need to calculate blind power loss;

Therefore judge that can the standard that parameter can be ignored be resolved the mutual inductor identification that rate is the highest;

The π type that therefore can be simplified and T-shaped equivalent circuit;

Further, described step (3) is according to the π type of simplifying and T-shaped equivalent circuit is actual according to scene and the needs of computing machine calculating, draw front L-type equivalent circuit and rear L-type equivalent circuit, described front L-type refers to that power back-off point is built up in reactive power transmitting terminal, the reactive power of line charging power and power supply, after transmitting terminal merges, enters energy receiving end after transmission line loss; Described rear L-type refers to that power back-off point is built up in reactive power receiving end, the reactive power of power supply is after transmission line loss, merge with line charging power, enter energy receiving end, cable data is calculated as R=0.128 Ω, X=0.096 Ω, the size of current flowing through on B=9.42x10-5S transmission line is:

I=250A　

Transmission-line voltage grade is:

$U=10000/\sqrt{3}V$

Suppose that the power transmitting on circuit is:

P=1400kW　

Q=351.19kvar　

Therefore power factor is:

cosφ=0.97　

The realistic ruuning situation of this power factor

So can draw up following formula:

△P _R=I ²R=250 ²×0.128=8000W　

△Q _X=I ²X=250 ²×0.096=6000var　

1) power direction for meritorious by A to B, idle by A to B, L-type equivalent circuit after adopting, it is as follows that process falls in calculating voltage:

$\mathrm{\&Delta;U}=\frac{\mathrm{PR}+\mathrm{QX}}{U}=\frac{1400\&times;0.128+354.33\&times;0.096}{\frac{10}{\sqrt{3}}}=36.93V$

$\mathrm{\&delta;U}=\frac{\mathrm{PX}-\mathrm{QR}}{U}=\frac{1400\&times;0.096-351.19\&times;0.128}{\frac{10}{\sqrt{3}}}=15.49V$

Suppose that A point voltage phase angle is benchmark phase angle, wire both end voltage is as follows:

Can be drawn by known conditions:

△P _R=I ²R=250 ²×0.128=8000W　

△Q _X=I ²X=250 ²×0.096=6000var　

2) power direction for meritorious by A to B, idle by A to B, L-type equivalent circuit before adopting, it is as follows that process falls in calculating voltage:

${\mathrm{\&Delta;Q}}_B={\mathrm{<figure-callout\; id="2"\; label="BU"\; filenames="140310145823.png"\; state="\{\{state\}\}">BU</figure-callout>}}^2=9.42\&times;{10}^{-5}\&times;\frac{{\left({10}^4\right)}^2}{{\left(\sqrt{3}\right)}^2}=3140\mathrm{var}$

${\mathrm{\&Delta;Q}}_X=I^{2}X=\frac{{\mathrm{<figure-callout\; id="2"\; label="P"\; filenames="140310145823.png"\; state="\{\{state\}\}">P</figure-callout>}}^2+Q^2}{U^2}X=\frac{{1400}^2+{354.33}^2}{\frac{{10}^2}{3}}\&times;0.096=6006.38\mathrm{var}$

Suppose that A point voltage phase angle is benchmark phase angle, wire both end voltage is as follows:

On transmission line, loss is:

$\mathrm{\&Delta;U}=\frac{\mathrm{PR}+\mathrm{QX}}{U}\frac{1400\&times;0.128+351.19\&times;}{\frac{10}{\sqrt{3}}}=36.88V$

$\mathrm{\&delta;U}=\frac{\mathrm{PX}-\mathrm{QR}}{U}=\frac{1400\&times;0.096-354.33\&times;0.128}{\frac{10}{\sqrt{3}}}=15.42V$

3) power direction for meritorious by A to B, idle by B to A, L-type equivalent circuit after adopting, it is as follows that process falls in calculating voltage:

$\mathrm{\&Delta;U}=\frac{\mathrm{PR}+\mathrm{QX}}{U}=\frac{1400\&times;0.128-354.33\&times;0.096}{\frac{10}{\sqrt{3}}}=25.15V$

$\mathrm{\&delta;U}=\frac{\mathrm{PX}-\mathrm{QR}}{U}=\frac{1400\&times;0.096+354.33\&times;0.128}{\frac{10}{\sqrt{3}}}=31.13V$

Suppose that A point voltage phase angle is benchmark phase angle, wire both end voltage is as follows:

On transmission line, loss is:

${\mathrm{\&Delta;P}}_R=I^{2}R=\frac{{\mathrm{<figure-callout\; id="2"\; label="P"\; filenames="140310145823.png"\; state="\{\{state\}\}">P</figure-callout>}}^2+Q^2}{U^2}R=\frac{{1400}^2+{354.33}^2}{\frac{{10}^2}{3}}\&times;0.128=8008.5W$

${\mathrm{\&Delta;Q}}_X=I^{2}X=\frac{{\mathrm{<figure-callout\; id="2"\; label="P"\; filenames="140310145823.png"\; state="\{\{state\}\}">P</figure-callout>}}^2+Q^2}{U^2}X=\frac{{1400}^2+{354.33}^2}{\frac{{10}^2}{3}}\&times;0.096=6006.38\mathrm{var}$

4) power direction for meritorious by A to B, idle by B to A, L-type equivalent circuit before adopting, it is as follows that process falls in calculating voltage:

$\mathrm{\&Delta;U}=\frac{\mathrm{PR}+\mathrm{QX}}{U}=\frac{1400\&times;0.128-351.19\&times;0.096}{\frac{10}{\sqrt{3}}}=25.20V$

$\mathrm{\&delta;U}=\frac{\mathrm{PX}-\mathrm{QR}}{U}=\frac{1400\&times;0.096+351.19\&times;0.128}{\frac{10}{\sqrt{3}}}=31.06V$

Suppose that A point voltage phase angle is benchmark phase angle, wire both end voltage is as follows:

Can be drawn by known conditions:

△P _R=I ²R=250 ²×0.128=8000W

△Q _X=I ²X=250 ²×0.096=6000var

5) power direction for meritorious by B to A, idle by A to B, L-type equivalent circuit after adopting, it is as follows that process falls in calculating voltage:

$\mathrm{\&Delta;U}=\frac{\mathrm{PR}+\mathrm{QX}}{U}=\frac{-1400\&times;0.128-351.19\&times;0.096}{\frac{10}{\sqrt{3}}}=-25.20V$

$\mathrm{\&delta;U}=\frac{\mathrm{PX}-\mathrm{QR}}{U}=\frac{-1400\&times;0.096-351.19\&times;0.128}{\frac{10}{\sqrt{3}}}=-31.06V$

Suppose that A point voltage phase angle is benchmark phase angle, wire both end voltage is as follows:

Can be drawn by known conditions:

△P _R=I ²R=250 ²×0.128=8000W

△Q _X=I ²X=250 ²×0.096=6000var

6) power direction for meritorious by B to A, idle by A to B, L-type equivalent circuit before adopting, it is as follows that process falls in calculating voltage:

$\mathrm{\&Delta;U}=\frac{\mathrm{PR}+\mathrm{QX}}{U}=\frac{-1400\&times;0.128+354.33\&times;0.096}{\frac{10}{\sqrt{3}}}=-25.15V$

$\mathrm{\&delta;U}=\frac{\mathrm{PX}-\mathrm{QR}}{U}=\frac{-1400\&times;0.096-354.33\&times;0.128}{\frac{10}{\sqrt{3}}}=-31.13V$

Suppose that A point voltage phase angle is benchmark phase angle, wire both end voltage is as follows:

On transmission line, loss is:

${\mathrm{\&Delta;P}}_R=I^{2}R=\frac{{\mathrm{<figure-callout\; id="2"\; label="P"\; filenames="140310145823.png"\; state="\{\{state\}\}">P</figure-callout>}}^2+Q^2}{U^2}R=\frac{{1400}^2+{354.33}^2}{\frac{{10}^2}{3}}\&times;0.128=8008.5W$

${\mathrm{\&Delta;Q}}_X=I^{2}X=\frac{{\mathrm{<figure-callout\; id="2"\; label="P"\; filenames="140310145823.png"\; state="\{\{state\}\}">P</figure-callout>}}^2+Q^2}{U^2}X=\frac{{1400}^2+{354.33}^2}{\frac{{10}^2}{3}}\&times;0.096=6006.38\mathrm{var}$

7) power direction for meritorious by B to A, idle by B to A, L-type equivalent circuit after adopting, it is as follows that process falls in calculating voltage:

$\mathrm{\&Delta;U}=\frac{\mathrm{PR}+\mathrm{QX}}{U}=\frac{-1400\&times;0.128-354.33\&times;0.096}{\frac{10}{\sqrt{3}}}=-36.93V$

$\mathrm{\&delta;U}=\frac{\mathrm{PX}-\mathrm{QR}}{U}=\frac{-1400\&times;0.096+354.33\&times;0.128}{\frac{10}{\sqrt{3}}}=-15.42V$

Suppose that A point voltage phase angle is benchmark phase angle, wire both end voltage is as follows:

On transmission line, loss is:

${\mathrm{\&Delta;P}}_R=I^{2}R=\frac{{\mathrm{<figure-callout\; id="2"\; label="P"\; filenames="140310145823.png"\; state="\{\{state\}\}">P</figure-callout>}}^2+Q^2}{U^2}R=\frac{{1400}^2+{354.33}^2}{\frac{{10}^2}{3}}\&times;0.128=8008.5W$

${\mathrm{\&Delta;Q}}_X=I^{2}X=\frac{{\mathrm{<figure-callout\; id="2"\; label="P"\; filenames="140310145823.png"\; state="\{\{state\}\}">P</figure-callout>}}^2+Q^2}{U^2}X=\frac{{1400}^2+{354.33}^2}{\frac{{10}^2}{3}}\&times;0.096=6006.38\mathrm{var}$

8) power direction for meritorious by A to B, idle by A to B, L-type equivalent circuit before adopting, it is as follows that process falls in calculating voltage:

$\mathrm{\&Delta;U}=\frac{\mathrm{PR}+\mathrm{QX}}{U}=\frac{-1400\&times;0.128-351.19\&times;0.096}{\frac{10}{\sqrt{3}}}=-36.88V$

$\mathrm{\&delta;U}=\frac{\mathrm{PX}-\mathrm{QR}}{U}=\frac{-1400\&times;0.096+351.19\&times;0.128}{\frac{10}{\sqrt{3}}}=-15.49V$

Suppose that A point voltage phase angle is benchmark phase angle, wire both end voltage is as follows:

Can be drawn by known conditions:

△P _R=I ²R=250 ²×0.128=8000W

△Q _X=I ²X=250 ²×0.096=6000var

Can draw 1 from above calculating) 4) 5) 8) corresponding L-type equivalent circuit, compensation point is after reactive power is passed through transmission line;

2) 3) 6) 7) corresponding front L-type equivalent circuit, and compensation point is before reactive power is passed through transmission line;

Wherein 1) 8) equivalence, after the two difference is that active power and reactive power direction are oppositely, compensation point position is also reverse;

2) 7) equivalence, after the two difference is that active power and reactive power direction are oppositely, compensation point position is also reverse;

3) 6) equivalence, after the two difference is that active power and reactive power direction are oppositely, compensation point position is also reverse;

4) 5) equivalence, after the two difference is that active power and reactive power direction are oppositely, compensation point position is also reverse;

Can be drawn by above inference, transmission line model can be with 1) 5) represent L-type equivalent circuit after positive dirction; Transmission line model can be with 4) 8) represent L-type equivalent circuit after negative direction;

By drawing above following inference, transmission line model can be with 2) 6) represent L-type equivalent circuit before positive dirction, transmission line model can be with 3) 7) represent the front L-type equivalent circuit of negative direction;

Further, described according to the needs of on-the-spot reality and computing machine calculating, calculate the estimation of error of front L-type equivalent circuit and rear L-type equivalent circuit, the longer 1km of unit cable length in calculated examples, power approaches fully loaded 250A, power factor is 0.97, tally with the actual situation, in actual conditions, full load falls in transmission-line voltage can not be greater than 5%, according to this constraint condition, peak power and the length of transmission line that can transmit are all restricted, therefore transmission line is about as 5km most, corresponding maximum current 300A, therefore in the time that lowest high-current value reduces, length of transmission line is in the situation that sectional area is constant, can also increase to some extent,

Only having meritorious transmission, there is no idle transmission, power factor is 1, and voltage drop is only with meritorious relevant with cable resistance admittance parameter, but the determinative of decision tolerance variation is conductor length;

Meritorious transmission, only has idle transmission, and power factor is 0, and voltage drop is only in idle relevant with cable resistance admittance parameter, but the determinative of decision tolerance variation is conductor length;

On circuit, do not transmit meritorious load or burden without work, the voltage drop on cable is just fewer, and the front L that can measure and the tolerance of rear L-type equivalent circuit are just less, until surveying instrument loses resolution characteristic;

Circuit powers on source point while thering is no output power, only has charge power on circuit, and in example as cable length is got to limit radius of electricity supply 5km, charge power is 15700var;

Length of transmission line is shorter, and the power loss on transmission line is fewer, and the basic tolerance of front L and rear L-type equivalent circuit is less, because surveying instrument resolution is certain, the two is tending towards poor consistent, and in the time that conductor length is less than to a certain degree, surveying instrument loses resolution characteristic;

Circuit is longer, and front L-type and rear L-type equivalent circuit calculated tolerances are larger, but because cable voltage drop can not exceed 5%, therefore in general 10kV cable power supply radius is 5km, in economic current density;

Therefore only need to consider two kinds of situations below, just can judge whether whether front L-type and rear L-type equivalent circuit calculated value can be measured by voltage measurement element and current measurement device;

1, unit cable length is amplified to 10kV limit radius of electricity supply to 5km, under full-load power, the current/voltage calculated tolerances distance of front L-type and rear L-type;

1) power direction for meritorious by A to B, idle by A to B, L-type equivalent circuit after adopting, it is as follows that process falls in calculating voltage:

$\mathrm{\&Delta;U}=\frac{\mathrm{PR}+\mathrm{QX}}{U}=\frac{1400\&times;0.64+351.19\&times;0.48}{\frac{10}{\sqrt{3}}}=184.39V$

$\mathrm{\&delta;U}=\frac{\mathrm{PX}-\mathrm{QR}}{U}=\frac{1400\&times;0.48-351.19\&times;0.64}{\frac{10}{\sqrt{3}}}=77.46V$

Suppose that A point voltage phase angle is benchmark phase angle, wire both end voltage is as follows:

Can be drawn by known conditions:

△P _R=I ²R=250 ²×0.64=40000W　

△Q _X=I ²X=250 ²×0.48=30000var　

2) power direction for meritorious by A to B, idle by A to B, L-type equivalent circuit before adopting, it is as follows that process falls in calculating voltage:

$\mathrm{\&Delta;U}=\frac{\mathrm{PR}+\mathrm{QX}}{U}=\frac{1400\&times;0.64+366.9\&times;0.48}{\frac{10}{\sqrt{3}}}=185.70V$

$\mathrm{\&delta;U}=\frac{\mathrm{PX}-\mathrm{QR}}{U}=\frac{1400\&times;0.48-366.9\&times;0.64}{\frac{10}{\sqrt{3}}}=75.72V$

Suppose that A point voltage phase angle is benchmark phase angle, wire both end voltage is as follows:

On transmission line, loss is:

${\mathrm{\&Delta;P}}_R=I^{2}R=\frac{{\mathrm{<figure-callout\; id="2"\; label="P"\; filenames="140310145823.png"\; state="\{\{state\}\}">P</figure-callout>}}^2+Q^2}{U^2}R=\frac{{1400}^2+{366.9}^2}{\frac{{10}^2}{3}}\&times;0.64=40216.6W$

$\mathrm{\&Delta;}{Q}_X=I^{2}X=\frac{{\mathrm{<figure-callout\; id="2"\; label="P"\; filenames="140310145823.png"\; state="\{\{state\}\}">P</figure-callout>}}^2+Q^2}{U^2}X=\frac{{1400}^2+{366.9}^2}{\frac{{10}^2}{3}}\&times;0.48=30162.46\mathrm{var}$

Can see above the meritorious idle 162W of differing of 216W that differs of L-type circuit and rear L-type equivalent circuit result of calculation, reduction is 249.96A to mains side top electric current, lose 0.04A, and the recognition capability of current measurement device is 0.3A, front L-type and rear L-type equivalent circuit gained voltage drop difference are about 2V, and voltage measurement element highest resolution is 6V, therefore, front L-type equivalent circuit and rear L-type equivalent circuit result of calculation, in circuit electric parameter and through-put power, while reaching actual conditions allowable maximum, basically identical simultaneously;

2, current/voltage error while only having line charging power:

In the time only having charge power, 5 kilometers of line lengths, front L-type equivalent circuit calculates as follows:

$\mathrm{\&Delta;U}=\frac{\mathrm{PR}+\mathrm{QX}}{U}=\frac{15.7\&times;0.48}{\frac{10}{\sqrt{3}}}=1.31V$

$\mathrm{\&delta;U}=\frac{\mathrm{PX}-\mathrm{QR}}{U}=\frac{-15.7\&times;0.64}{\frac{10}{\sqrt{3}}}=1.74V$

On transmission line, loss is:

${\mathrm{\&Delta;P}}_R=I^{2}R=\frac{{\mathrm{<figure-callout\; id="2"\; label="P"\; filenames="140310145823.png"\; state="\{\{state\}\}">P</figure-callout>}}^2+Q^2}{U^2}R=\frac{{15.7}^2}{\frac{{10}^2}{3}}\&times;0.64=4.73W$

${\mathrm{\&Delta;Q}}_X=I^{2}X=\frac{{\mathrm{<figure-callout\; id="2"\; label="P"\; filenames="140310145823.png"\; state="\{\{state\}\}">P</figure-callout>}}^2+Q^2}{U^2}X=\frac{{15.7}^2}{\frac{{10}^2}{3}}\&times;0.48=3.55\mathrm{var}$

Rear L-type equivalent circuit calculates as follows:

$\mathrm{\&Delta;U}=\frac{\mathrm{PR}+\mathrm{QX}}{U}=\frac{0\&times;0.48}{\frac{10}{\sqrt{3}}}=0V$

$\mathrm{\&delta;U}=\frac{\mathrm{PX}-\mathrm{QR}}{U}=\frac{0\&times;0.64}{\frac{10}{\sqrt{3}}}=0V$

On transmission line, loss is:

${\mathrm{\&Delta;P}}_R=I^{2}R=\frac{{\mathrm{<figure-callout\; id="2"\; label="P"\; filenames="140310145823.png"\; state="\{\{state\}\}">P</figure-callout>}}^2+Q^2}{U^2}R=\frac{0^2}{\frac{{10}^2}{\sqrt{3}}}\&times;0.64=0W$

${\mathrm{\&Delta;Q}}_X=I^{2}X=\frac{{\mathrm{<figure-callout\; id="2"\; label="P"\; filenames="140310145823.png"\; state="\{\{state\}\}">P</figure-callout>}}^2+Q^2}{U^2}X=\frac{0^2}{\frac{{10}^2}{\sqrt{3}}}\&times;0.48=0\mathrm{var}$

Although voltage difference 1.31V+j1.74V, compare 5773.5 specified phase voltage and change and be about 0.03%, voltage measurement element cannot be measured, difference power is apart from not enough 10w also, current measurement device also cannot be measured;

Therefore can reach a conclusion: the front L-type equivalent circuit of transmission line and rear L-type equivalent circuit can be applied in power distribution network network calculation of parameter; Front L-type and rear L-type equivalent circuit can reduce computing machine calculating amount to a great extent; Before adopting separately, L or rear L-type equivalent circuit, can, at same transmission line under different idle transmission directions, look from reactive power input end, expresses different parameters.　

Beneficial effect of the present invention is: the front L-type equivalent circuit of transmission line and rear L-type equivalent circuit can be applied in power distribution network network calculation of parameter; Front L-type and rear L-type equivalent circuit can reduce computing machine calculating amount to a great extent; Before adopting separately, L or rear L-type equivalent circuit, can, at same transmission line under different idle transmission directions, look from reactive power input end, expresses different parameters.　

[brief description of the drawings]

Fig. 1 is the derivation structural representation of medium voltage distribution network transmission line model of the present invention;

Fig. 2 is the variation proportion structure schematic diagram of the each parameter of the present invention with electric current and voltage;

Fig. 3 is the instance graph structural representation of medium voltage distribution network transmission line model of the present invention;

Fig. 4 is the instance graph structural representation after L-type Type Equivalent Circuit Model after substitution of the present invention;

Fig. 5 is the instance graph structural representation after L-type Type Equivalent Circuit Model before substitution of the present invention;

[embodiment]

Below in conjunction with the drawings and the specific embodiments, the present invention is described further:

As shown in Figure 1 and Figure 2, a kind of method for building up of medium voltage distribution network transmission line model, comprises that step is as follows:

(1) according to circuitry transmission line portions, can known transmission line π type and T-shaped equivalent circuit;

(2) according to the parameter value going out of π type and T-shaped equivalent circuit and the actual measurement of on-the-spot energy, draw π type and the T-shaped equivalent circuit of simplification;

(3) according to the π type of simplifying and T-shaped equivalent circuit according to needs on-the-spot actual and computing machine calculating, draw front L-type equivalent circuit and rear L-type equivalent circuit.　

Preferably, described step (1) is according to circuitry transmission line portions, can known transmission line π type and T-shaped equivalent circuit in, the resistance of the resistance R transmission line in Z=R+jX, X is transmission line reactance, ohmic loss is directly proportional to transmission line current flowing size to reactance loss, and it is irrelevant with transmission-line voltage grade, be directly proportional to cable length simultaneously, resistance R can be calculated according to electric wire handbook, reactance X can calculate according to electric wire handbook, also can know according to circuitry, it is that the leakage power of cable and the electricity that produces are led that electricity in Y=G+jB is led G, B is cable earth capacitance electric current and susceptance that corresponding charge power produces, leakage power and charge power are directly proportional to transmission-line voltage square, with to flow through transmission line choke irrelevant, be directly proportional to cable length simultaneously, electricity is led G and can be calculated according to electric wire handbook, leakage power is made up of two parts: one is insulation resistance loss, one is dielectric loss, susceptance B can calculate according to electric wire handbook.　

Preferably, described step (2) is according to the parameter value going out of π type and T-shaped equivalent circuit and the actual measurement of on-the-spot energy, draw π type and the T-shaped equivalent circuit of simplification, suppose 185 square millimeters of cable model YJV-8.7-15kV of medium voltage distribution network cross-section of cable list core, circuit is single phase power supply;

Its parameter is conductor AC resistance 0.128 Ω/km, and reactance is 0.096 Ω/km, and electric capacity is 0.3 μ F/km, supposes that this single-phase cable length is 1km longer in actual conditions;

At this moment cable data is calculated as R=0.128 Ω, X=0.096 Ω, B=9.42x10-5S

The size of current flowing through on transmission line is:

I=250A　

Transmission-line voltage grade is:

$U=10000/\sqrt{3}V$

Suppose that the power transmitting on circuit is:

P=1400kW　

Q=351.19kvar　

Therefore power factor is:

cosφ=0.97　

The realistic ruuning situation of this power factor

So can draw up following formula:

${\mathrm{\&Delta;Q}}_B={\mathrm{<figure-callout\; id="2"\; label="BU"\; filenames="140310145823.png"\; state="\{\{state\}\}">BU</figure-callout>}}^2=9.42\&times;{10}^{-5}\&times;\frac{{\left({10}^4\right)}^2}{{\left(\sqrt{3}\right)}^2}=3140\mathrm{var}$

△P _R=I ²R=250 ²×0.128=8000W　

△Q _X=I ²X=250 ²×0.096=6000var　

△P _G=△P _G1+△P _G2　

At this moment calculate △ P _gsize, wherein △ P _g=△ P _g1+ △ P _g2

1) △ P _g1be insulation resistance loss, △ Px2 is dielectric loss;

We first see insulation resistance

According to electric wire handbook (second edition 2008 enlarged editions) (Wang Chunjiang chief editor) P299 page formula (3-2-17)

Wherein insulation resistance

ρ _v---insulation resistance coefficient: $R_v={\mathrm{\&rho;}}_v\frac{\mathrm{\&delta;}}{A}=\frac{1}{{\mathrm{<figure-callout\; id="1"\; label="G\; G"\; filenames="140310145823.png"\; state="\{\{state\}\}">G</figure-callout>}}_G}$

According to electric wire handbook (second edition 2008 enlarged editions) (Wang Chunjiang chief editor) P299 page table 3-2-13

Crosslinked polyethylene (YJV) insulating resistance of cable coefficient is 1016～1017/ (Ω cm)

δ---insulation thickness

Suppose that insulation thickness is 0.5cm(and can reduces calculated resistance toward getting in little, thereby increase leakage power)

A---electrode area

185mm2 cable interior diameter is 2cm, girth 6.28cm, and 1km cable is 100000cm length;

Calculate by the poorest condition:

$R_v={\mathrm{\&rho;}}_v\frac{\mathrm{\&delta;}}{A}={10}^{16}\&times;\frac{0.5}{5\&times;{10}^5\&times;6.28}=7.962\&times;{10}^9\mathrm{\&Omega;}$

Therefore for 10kV circuit, insulation resistance Leakage Current has:

${\mathrm{<figure-callout\; id="1"\; label="I\; G"\; filenames="140310145823.png"\; state="\{\{state\}\}">I</figure-callout>}}_G=U/R_v=10000/(\sqrt{3}\&times;7.962\&times;{10}^9)=7.251\&times;{10}^{-7}A$

Therefore insulation resistance loss power is:

△P _G1=UI _G1=0.0042W　

(2) according to electric wire handbook (second edition 2008 enlarged editions) (Wang Chunjiang chief editor) P304 page formula (3-2-24)

△P _G2=W _d=U ²ωCtgδ　

Wherein tg δ calculates according to crosslinked polyethylene maximal value 0.0025 in table 3-2.22:

△P _G2=W _d=U ²ωCtgδ=3140×0.0025=7.85W　

△P _G=△P _G1+△P _G2=0.0042+7.85=7.854W　

Therefore total electricity is led power

1) therefore total conduction current is: ${\mathrm{\&Delta;I}}_G=\frac{P_G}{U}=\frac{7.854}{\frac{10000}{\sqrt{3}}}=1.36\&times;{10}^{-3}A$

And load current is 250A, current mutual inducer for line outlet end rated current is 300A, current transformer full accuracy is 0.1%, and the minimum current that can differentiate is 0.3A, and the electric current that electricity is led loss is only for current transformer can be differentiated 1/219th of electric current;

Simultaneously two kinds of electricity are led active loss and are all directly proportional to line voltage distribution grade square, and flow through line current size and have nothing to do, and electric pressure substantially constant, and therefore to lead loss be an individual constant to electricity substantially, and variation range is very little;

In sum, in power distribution network transmission line model, electricity is led and can be ignored;

2) by ${\mathrm{\&Delta;Q}}_B={\mathrm{<figure-callout\; id="2"\; label="BU"\; filenames="140310145823.png"\; state="\{\{state\}\}">BU</figure-callout>}}^2=9.42\&times;{10}^{-5}\&times;\frac{{\left({10}^4\right)}^2}{{\left(\sqrt{3}\right)}^2}=3140\mathrm{var}$

The susceptance reactive power compensation power that cable earth capacitance forms is directly proportional to line voltage distribution grade square, line current size is irrelevant with flowing through, and electric pressure substantially constant, therefore to lead loss be constant to electricity substantially, and variation range is very little, and the power of susceptance B is 3140var, and substantially constant, the electric current calculating is 0.54A, can be resolved the current transformer identification of rate 0.3A,

Therefore susceptance parameter can not be ignored, and need to calculate susceptance power back-off;

3) by △ P _r=I ²r=250 ²× 0.128=8000W

Ohmic loss is directly proportional to current squaring, irrelevant with line voltage distribution size, circuit voltage swing is basicly stable simultaneously, and ohmic loss is very large with load variations variation range, and it is 1.39A that full load calculates size of current, can be resolved the current transformer identification of rate 0.3A;

Therefore resistance parameter can not be ignored, and needs calculated resistance power attenuation;

4）△Q _X=I ²X=250 ²×0.096=6000var　

Reactance Power consumption and electric current square is directly proportional, irrelevant with line voltage distribution size, circuit voltage swing is basicly stable simultaneously, and reactance loss is very large with load variations variation range, and it is 1.04A that full load calculates size of current, can be resolved the current transformer identification of rate 0.3A;

Therefore reactance parameter can not be ignored, and need to calculate blind power loss;

Therefore judge that can the standard that parameter can be ignored be resolved the mutual inductor identification that rate is the highest;

The π type that therefore can be simplified and T-shaped equivalent circuit;

Further, described step (3) is according to the π type of simplifying and T-shaped equivalent circuit is actual according to scene and the needs of computing machine calculating, draw front L-type equivalent circuit and rear L-type equivalent circuit, described front L-type refers to that power back-off point is built up in reactive power transmitting terminal, the reactive power of line charging power and power supply, after transmitting terminal merges, enters energy receiving end after transmission line loss; Described rear L-type refers to that power back-off point is built up in reactive power receiving end, the reactive power of power supply is after transmission line loss, merge with line charging power, enter energy receiving end, cable data is calculated as R=0.128 Ω, X=0.096 Ω, the size of current flowing through on B=9.42x10-5S transmission line is:

I=250A　

Transmission-line voltage grade is:

$U=10000/\sqrt{3V}$

Suppose that the power transmitting on circuit is:

P=1400kW　

Q=351.19kvar　

Therefore power factor is:

cosφ=0.97　

The realistic ruuning situation of this power factor

So can draw up following formula:

△P _R=I ²R=250 ²×0.128=8000W　

△Q _X=I ²X=250 ²×0.096=6000var　

1) power direction for meritorious by A to B, idle by A to B, L-type equivalent circuit after adopting, calculating voltage

${\mathrm{\&Delta;Q}}_B={\mathrm{<figure-callout\; id="2"\; label="BU"\; filenames="140310145823.png"\; state="\{\{state\}\}">BU</figure-callout>}}^2=9.42\&times;{10}^{-5}\&times;\frac{{\left({10}^4\right)}^2}{{\left(\sqrt{3}\right)}^2}=3140\mathrm{var}$

The process of falling is as follows:

$\begin{array}{c}\mathrm{\&Delta;U}=\frac{\mathrm{PR}+\mathrm{QX}}{U}=\frac{1400\&times;0.128+351.19\&times;0.096}{\frac{10}{\sqrt{3}}}=36.88v\\ \mathrm{\&delta;U}=\frac{\mathrm{PX}-\mathrm{QR}}{U}=\frac{1400\&times;0.096-351.19\&times;0.128}{\frac{10}{\sqrt{3}}}=15.49v\end{array}$

Suppose that A point voltage phase angle is benchmark phase angle, wire both end voltage is as follows:

Can be drawn by known conditions:

△P _R=I ²R=250 ²×0.128=8000W　

△Q _X=I ²X=250 ²×0.096=6000var　

2) power direction for meritorious by A to B, idle by A to B, L-type equivalent circuit before adopting, it is as follows that process falls in calculating voltage:

$\begin{array}{c}\mathrm{\&Delta;U}=\frac{\mathrm{PR}+\mathrm{QX}}{U}=\frac{1400\&times;0.128+354.33\&times;0.096}{\frac{10}{\sqrt{3}}}=36.93V\\ \mathrm{\&delta;U}=\frac{\mathrm{PX}-\mathrm{QR}}{U}=\frac{1400\&times;0.096-354.33\&times;}{\frac{10}{\sqrt{3}}}=15.42V\end{array}$

Suppose that A point voltage phase angle is benchmark phase angle, wire both end voltage is as follows:

On transmission line, loss is:

${\mathrm{\&Delta;P}}_R=I^{2}R=\frac{{\mathrm{<figure-callout\; id="2"\; label="P"\; filenames="140310145823.png"\; state="\{\{state\}\}">P</figure-callout>}}^2+Q^2}{U^2}R=\frac{{1400}^2+{354.33}^2}{\frac{{10}^2}{3}}\&times;0.128=8008.5W$

${\mathrm{\&Delta;Q}}_X=I^{2}X=\frac{{\mathrm{<figure-callout\; id="2"\; label="P"\; filenames="140310145823.png"\; state="\{\{state\}\}">P</figure-callout>}}^2+Q^2}{U^2}X=\frac{{1400}^2+{354.33}^2}{\frac{{10}^2}{3}}\&times;0.096=6006.38\mathrm{var}$

$\mathrm{\&delta;U}=\frac{\mathrm{PX}-\mathrm{QR}}{U}=\frac{1400\&times;0.096-351.19\&times;0.128}{\frac{10}{\sqrt{3}}}=15.49V$

3) power direction for meritorious by A to B, idle by B to A, L-type equivalent circuit after adopting, it is as follows that process falls in calculating voltage:

$\mathrm{\&Delta;U}=\frac{\mathrm{PR}+\mathrm{QX}}{U}=\frac{1400\&times;0.128+351.19\&times;0.096}{\frac{10}{\sqrt{3}}}=36.88V$

$\mathrm{\&Delta;U}=\frac{\mathrm{PR}+\mathrm{QX}}{U}=\frac{1400\&times;0.128-354.33\&times;0.096}{\frac{10}{\sqrt{3}}}=25.15V$

$\mathrm{\&delta;U}=\frac{\mathrm{PX}-\mathrm{QR}}{U}=\frac{1400\&times;0.096+354.33\&times;0.128}{\frac{10}{\sqrt{3}}}=31.13V$

Suppose that A point voltage phase angle is benchmark phase angle, wire both end voltage is as follows:

On transmission line, loss is:

${\mathrm{\&Delta;P}}_R=I^{2}R=\frac{{\mathrm{<figure-callout\; id="2"\; label="P"\; filenames="140310145823.png"\; state="\{\{state\}\}">P</figure-callout>}}^2+Q^2}{U^2}R=\frac{{1400}^2+{354.33}^2}{\frac{{10}^2}{3}}\&times;0.128=8008.5W$

${\mathrm{\&Delta;Q}}_X=I^{2}X=\frac{{\mathrm{<figure-callout\; id="2"\; label="P"\; filenames="140310145823.png"\; state="\{\{state\}\}">P</figure-callout>}}^2+Q^2}{U^2}X=\frac{{1400}^2+{354.33}^2}{\frac{{10}^2}{3}}\&times;0.096=6006.38\mathrm{Var}$

4) power direction for meritorious by A to B, idle by B to A, L-type equivalent circuit before adopting, it is as follows that process falls in calculating voltage:

$\mathrm{\&Delta;U}=\frac{\mathrm{PR}+\mathrm{QX}}{U}=\frac{1400\&times;0.128-351.19\&times;0.096}{\frac{10}{\sqrt{3}}}=25.20V$

$\mathrm{\&delta;U}=\frac{\mathrm{PX}-\mathrm{QR}}{U}=\frac{1400\&times;0.096+351.19\&times;0.128}{\frac{10}{\sqrt{3}}}=31.06V$

Suppose that A point voltage phase angle is benchmark phase angle, wire both end voltage is as follows:

Can be drawn by known conditions:

△P _R=I ²R=250 ²×0.128=8000W

△Q _X=I ²X=250 ²×0.096=6000var

5) power direction for meritorious by B to A, idle by A to B, L-type equivalent circuit after adopting, it is as follows that process falls in calculating voltage:

$\mathrm{\&Delta;U}=\frac{\mathrm{PR}+\mathrm{QX}}{U}=\frac{-1400\&times;0.128+351.19\&times;0.096}{\frac{10}{\sqrt{3}}}=-25.20V$

$\mathrm{\&delta;U}=\frac{\mathrm{PX}-\mathrm{QR}}{U}=\frac{-1400\&times;0.096-351.19\&times;0.128}{\frac{10}{\sqrt{3}}}=-31.06V$

Suppose that A point voltage phase angle is benchmark phase angle, wire both end voltage is as follows:

Can be drawn by known conditions:

△P _R=I ²R=250 ²×0.128=8000W

△Q _X=I ²X=250 ²×0.096=6000var

6) power direction for meritorious by B to A, idle by A to B, L-type equivalent circuit before adopting, it is as follows that process falls in calculating voltage:

$\mathrm{\&Delta;U}=\frac{\mathrm{PR}+\mathrm{QX}}{U}=\frac{1400\&times;0.128-354.33\&times;0.096}{\frac{10}{\sqrt{3}}}=-25.15V$

$\mathrm{\&delta;U}=\frac{\mathrm{PX}-\mathrm{QR}}{U}=\frac{1400\&times;0.096+354.33\&times;0.128}{\frac{10}{\sqrt{3}}}=-31.13V$

Suppose that A point voltage phase angle is benchmark phase angle, wire both end voltage is as follows:

On transmission line, loss is:

${\mathrm{\&Delta;P}}_R=I^{2}R=\frac{{\mathrm{<figure-callout\; id="2"\; label="P"\; filenames="140310145823.png"\; state="\{\{state\}\}">P</figure-callout>}}^2+Q^2}{U^2}R=\frac{{1400}^2+{354.33}^2}{\frac{{10}^2}{3}}\&times;0.128=8008.5W$

${\mathrm{\&Delta;Q}}_X=I^{2}X=\frac{{\mathrm{<figure-callout\; id="2"\; label="P"\; filenames="140310145823.png"\; state="\{\{state\}\}">P</figure-callout>}}^2+Q^2}{U^2}X=\frac{{1400}^2+{354.33}^2}{\frac{{10}^2}{3}}\&times;0.096=6006.38\mathrm{var}$

7) power direction for meritorious by B to A, idle by B to A, L-type equivalent circuit after adopting, it is as follows that process falls in calculating voltage:

$\mathrm{\&delta;U}=\frac{\mathrm{PX}-\mathrm{QR}}{U}=\frac{-1400\&times;0.096-354.33\&times;0.096}{\frac{10}{\sqrt{3}}}=-36.93V$

$\mathrm{\&delta;U}=\frac{\mathrm{PX}-\mathrm{QR}}{U}=\frac{1400\&times;0.096+354.33\&times;0.128}{\frac{10}{\sqrt{3}}}=-15.42V$

Suppose that A point voltage phase angle is benchmark phase angle, wire both end voltage is as follows:

${\mathrm{\&Delta;P}}_R=I^{2}R=\frac{{\mathrm{<figure-callout\; id="2"\; label="P"\; filenames="140310145823.png"\; state="\{\{state\}\}">P</figure-callout>}}^2+Q^2}{U^2}R=\frac{{1400}^2+{354.33}^2}{\frac{{10}^2}{3}}\&times;0.128=8008.5W$

On transmission line, loss is:

${\mathrm{\&Delta;Q}}_X=I^{2}X=\frac{{\mathrm{<figure-callout\; id="2"\; label="P"\; filenames="140310145823.png"\; state="\{\{state\}\}">P</figure-callout>}}^2+Q^2}{U^2}X=\frac{{1400}^2+{354.33}^2}{\frac{{10}^2}{3}}\&times;0.096=6006.38\mathrm{var}$

8) power direction for meritorious by A to B, idle by A to B, L-type equivalent circuit before adopting, it is as follows that process falls in calculating voltage:

$\mathrm{\&Delta;U}=\frac{\mathrm{PR}+\mathrm{QX}}{U}=\frac{-1400\&times;0.128-351.19\&times;0.096}{\frac{10}{\sqrt{3}}}=-36.88V$

$\mathrm{\&delta;U}=\frac{\mathrm{PX}-\mathrm{QR}}{U}=\frac{-1400\&times;0.096+351.19\&times;0.128}{\frac{10}{\sqrt{3}}}=-15.49V$

Suppose that A point voltage phase angle is benchmark phase angle, wire both end voltage is as follows:

Can be drawn by known conditions:

△P _R=I ²R=250 ²×0.128=8000W

△Q _X=I ²X=250 ²×0.096=6000var

Can draw 1 from above calculating) 4) 5) 8) corresponding L-type equivalent circuit, compensation point is after reactive power is passed through transmission line;

2) 3) 6) 7) corresponding front L-type equivalent circuit, and compensation point is before reactive power is passed through transmission line;

Wherein 1) 8) equivalence, after the two difference is that active power and reactive power direction are oppositely, compensation point position is also reverse;

2) 7) equivalence, after the two difference is that active power and reactive power direction are oppositely, compensation point position is also reverse;

3) 6) equivalence, after the two difference is that active power and reactive power direction are oppositely, compensation point position is also reverse;

4) 5) equivalence, after the two difference is that active power and reactive power direction are oppositely, compensation point position is also reverse;

Can be drawn by above inference, transmission line model can be with 1) 5) represent L-type equivalent circuit after positive dirction.Transmission line model can be with 4) 8) represent L-type equivalent circuit after negative direction;

By drawing above following inference, transmission line model can be with 2) 6) represent L-type equivalent circuit before positive dirction, transmission line model can be with 3) 7) represent the front L-type equivalent circuit of negative direction;

Preferably, described according to the needs of on-the-spot reality and computing machine calculating, calculate the estimation of error of front L-type equivalent circuit and rear L-type equivalent circuit, the longer 1km of unit cable length in calculated examples, power approaches fully loaded 250A, power factor is 0.97, tally with the actual situation, in actual conditions, full load falls in transmission-line voltage can not be greater than 5%, according to this constraint condition, peak power and the length of transmission line that can transmit are all restricted, therefore transmission line is about as 5km most, corresponding maximum current 300A, therefore in the time that lowest high-current value reduces, length of transmission line is in the situation that sectional area is constant, can also increase to some extent,

Only having meritorious transmission, there is no idle transmission, power factor is 1, and voltage drop is only with meritorious relevant with cable resistance admittance parameter, but the determinative of decision tolerance variation is conductor length;

Meritorious transmission, only has idle transmission, and power factor is 0, and voltage drop is only in idle relevant with cable resistance admittance parameter, but the determinative of decision tolerance variation is conductor length;

On circuit, do not transmit meritorious load or burden without work, the voltage drop on cable is just fewer, and the front L that can measure and the tolerance of rear L-type equivalent circuit are just less, until surveying instrument loses resolution characteristic;

Circuit powers on source point while thering is no output power, only has charge power on circuit, and in example as cable length is got to limit radius of electricity supply 5km, charge power is 15700var;

Length of transmission line is shorter, and the power loss on transmission line is fewer, and the basic tolerance of front L and rear L-type equivalent circuit is less, because surveying instrument resolution is certain, the two is tending towards poor consistent, and in the time that conductor length is less than to a certain degree, surveying instrument loses resolution characteristic;

Circuit is longer, and front L-type and rear L-type equivalent circuit calculated tolerances are larger, but because cable voltage drop can not exceed 5%, therefore in general 10kV cable power supply radius is 5km, in economic current density;

Therefore only need to consider two kinds of situations below, just can judge whether whether front L-type and rear L-type equivalent circuit calculated value can be measured by voltage measurement element and current measurement device;

1, unit cable length is amplified to 10kV limit radius of electricity supply to 5km, under full-load power, the current/voltage calculated tolerances distance of front L-type and rear L-type

1) power direction for meritorious by A to B, idle by A to B, L-type equivalent circuit after adopting, it is as follows that process falls in calculating voltage:

$\mathrm{\&Delta;U}=\frac{\mathrm{PR}+\mathrm{QX}}{U}=\frac{1400\&times;0.64+351.19\&times;0.48}{\frac{10}{\sqrt{3}}}=184.39V$

$\mathrm{\&delta;U}=\frac{\mathrm{PX}-\mathrm{QR}}{U}=\frac{1400\&times;0.48-351.19\&times;0.64}{\frac{10}{\sqrt{3}}}=77.46V$

Suppose that A point voltage phase angle is benchmark phase angle, wire both end voltage is as follows:

Can be drawn by known conditions:

△P _R=I ²R=250 ²×0.64=40000W　

△Q _X=I ²X=250 ²×0.48=30000var　

2) power direction for meritorious by A to B, idle by A to B, L-type equivalent circuit before adopting, it is as follows that process falls in calculating voltage:

$\mathrm{\&Delta;U}=\frac{\mathrm{PR}+\mathrm{QX}}{U}=\frac{1400\&times;0.64+366.9\&times;0.48}{\frac{10}{\sqrt{3}}}=185.70V$

$\mathrm{\&delta;U}=\frac{\mathrm{PX}-\mathrm{QR}}{U}=\frac{1400\&times;0.48-366.9\&times;0.64}{\frac{10}{\sqrt{3}}}=75.72V$

Suppose that A point voltage phase angle is benchmark phase angle, wire both end voltage is as follows:

On transmission line, loss is:

${\mathrm{\&Delta;P}}_R=I^{2}R=\frac{{\mathrm{<figure-callout\; id="2"\; label="P"\; filenames="140310145823.png"\; state="\{\{state\}\}">P</figure-callout>}}^2+Q^2}{U^2}R=\frac{{1400}^2+{366.9}^2}{\frac{{10}^2}{3}}\&times;0.64=40216.6W$

${\mathrm{\&Delta;Q}}_X=I^{2}X=\frac{{\mathrm{<figure-callout\; id="2"\; label="P"\; filenames="140310145823.png"\; state="\{\{state\}\}">P</figure-callout>}}^2+Q^2}{U^2}X=\frac{{1400}^2+{366.9}^2}{\frac{{10}^2}{3}}\&times;0.48=30162.46\mathrm{var}$

Can see above the meritorious idle 162W of differing of 216W that differs of L-type circuit and rear L-type equivalent circuit result of calculation, reduction is 249.96A to mains side top electric current, lose 0.04A, and the recognition capability of current measurement device is 0.3A, front L-type and rear L-type equivalent circuit gained voltage drop difference are about 2V, and voltage measurement element highest resolution is 6V, therefore, front L-type equivalent circuit and rear L-type equivalent circuit result of calculation, in circuit electric parameter and through-put power, while reaching actual conditions allowable maximum, basically identical simultaneously;

2, current/voltage error while only having line charging power:

In the time only having charge power, 5 kilometers of line lengths, front L-type equivalent circuit calculates as follows:

$\mathrm{\&Delta;U}=\frac{\mathrm{PR}+\mathrm{QX}}{U}=\frac{15.7\&times;0.48}{\frac{10}{\sqrt{3}}}=1.31V$

$\mathrm{\&delta;U}=\frac{\mathrm{PX}-\mathrm{QR}}{U}=\frac{-15.7\&times;0.64}{\frac{10}{\sqrt{3}}}=1.74V$

On transmission line, loss is:

${\mathrm{\&Delta;P}}_R=I^{2}R=\frac{{\mathrm{<figure-callout\; id="2"\; label="P"\; filenames="140310145823.png"\; state="\{\{state\}\}">P</figure-callout>}}^2+Q^2}{U^2}R=\frac{{15.7}^2}{\frac{{10}^2}{3}}\&times;0.64=4.73W$

${\mathrm{\&Delta;Q}}_X=I^{2}X=\frac{{\mathrm{<figure-callout\; id="2"\; label="P"\; filenames="140310145823.png"\; state="\{\{state\}\}">P</figure-callout>}}^2+Q^2}{U^2}X=\frac{{15.7}^2}{\frac{{10}^2}{3}}\&times;0.48=3.55\mathrm{var}$

Rear L-type equivalent circuit calculates as follows:

$\mathrm{\&Delta;U}=\frac{\mathrm{PR}+\mathrm{QW}}{U}=\frac{0\&times;0.48}{\frac{10}{\sqrt{3}}}=0V$

$\mathrm{\&delta;U}=\frac{\mathrm{PX}-\mathrm{QR}}{U}=\frac{0\&times;0.64}{\frac{10}{\sqrt{3}}}=0V$

On transmission line, loss is:

${\mathrm{\&Delta;P}}_R=I^{2}R=\frac{{\mathrm{<figure-callout\; id="2"\; label="P"\; filenames="140310145823.png"\; state="\{\{state\}\}">P</figure-callout>}}^2+Q^2}{U^2}R=\frac{0^2}{\frac{{10}^2}{\sqrt{3}}}\&times;0.64=0W$

${\mathrm{\&Delta;Q}}_X=I^{2}X=\frac{{\mathrm{<figure-callout\; id="2"\; label="P"\; filenames="140310145823.png"\; state="\{\{state\}\}">P</figure-callout>}}^2+Q^2}{U^2}X=\frac{0^2}{\frac{{10}^2}{\sqrt{3}}}\&times;0.48=0\mathrm{var}$

Although voltage difference 1.31V+j1.74V, compare 5773.5 specified phase voltage and change and be about 0.03%, voltage measurement element cannot be measured, difference power is apart from not enough 10w also, current measurement device also cannot be measured;

Therefore can reach a conclusion: the front L-type equivalent circuit of transmission line and rear L-type equivalent circuit can be applied in power distribution network network calculation of parameter; Front L-type and rear L-type equivalent circuit can reduce computing machine calculating amount to a great extent; Before adopting separately, L or rear L-type equivalent circuit, can, at same transmission line under different idle transmission directions, look from reactive power input end, expresses different parameters;

Embodiment 1;

Shown in Fig. 3, in the yi word pattern loop of five intermediate points, Z1 is source point, and Z2, Z3, Z4, Z5 are intermediate point, and wherein Z5 is end intermediate point.Wherein gain merit into idle direction and size as shown in the figure;

(Z1-Z2), (Z2-Z3), (Z3-Z4), (Z4-Z5) be the transmission line that connects Z1, Z2, Z3, Z4, Z5;

P(Z1-Z2), P(Z2-Z3), P(Z3-Z4) and, P(Z4-Z5) be the active power on transmission line;

Q(Z1-Z2), Q(Z2-Z3), Q(Z3-Z4) and, Q(Z4-Z5) be the reactive power on transmission line;

Shown in Fig. 4, rear L-type equivalent circuit: wherein power direction and meritorious reactive power size are known, therefore in the idle porch that enters intermediate point, to the idle rechargeable energy of the circuit of whole transmission line, access compensation, the original reactive power sum of the energy after compensation and transmission line flows into the intermediate point at the idle rechargeable energy compensation point of circuit place;

Transmission line (Z1-Z2) power direction for meritorious by A to B, idle by B to A, L-type equivalent circuit after adopting, line charging reactive-load compensation point is built up in A point;

Transmission line (Z2-Z3) power direction for meritorious by A to B, idle by A to B, L-type equivalent circuit after adopting, line charging reactive-load compensation point is built up in D point;

Transmission line (Z3-Z4) power direction for meritorious by A to B, idle by B to A, L-type equivalent circuit after adopting, line charging reactive-load compensation point is built up in E point;

Transmission line (Z4-Z5) power direction for meritorious by A to B, idle by B to A, L-type equivalent circuit after adopting, line charging reactive-load compensation point is built up in G point;

After setting up model, calculate respectively the line loss on every transmission lines;

Shown in Fig. 5, front L-type equivalent circuit, wherein power direction and meritorious reactive power size are known, therefore in the exit of idle outflow intermediate point, to the idle rechargeable energy of the circuit of whole transmission line, access compensation, the original reactive power sum of the energy after compensation and transmission line, after transmission line impedance loss, enters the intermediate point of the compensation point other end;

Transmission line (Z1-Z2) power direction for meritorious by A to B, idle by B to A, L-type equivalent circuit before adopting, line charging reactive-load compensation point is built up in B point;

Transmission line (Z2-Z3) power direction for meritorious by A to B, idle by A to B, L-type equivalent circuit before adopting, line charging reactive-load compensation point is built up in C point;

Transmission line (Z3-Z4) power direction for meritorious by A to B, idle by B to A, L-type equivalent circuit before adopting, line charging reactive-load compensation point is built up in F point;

Transmission line (Z4-Z5) power direction for meritorious by A to B, idle by B to A, L-type equivalent circuit before adopting, line charging reactive-load compensation point is built up in H point.　

The announcement of book and instruction according to the above description, those skilled in the art in the invention can also carry out suitable change and amendment to above-mentioned embodiment.Therefore, the present invention is not limited to embodiment disclosed and described above, also should fall in the protection domain of claim of the present invention modifications and changes more of the present invention.In addition,, although used some specific terms in this instructions, these terms just for convenience of description, do not form any restriction to the present invention.　

Claims (5)

1. a method for building up for medium voltage distribution network transmission line model, is characterized in that, comprises following simplification step:

(1) according to circuitry transmission line portions, can known transmission line π type and T-shaped equivalent circuit;

(2) according to the parameter value going out of π type and T-shaped equivalent circuit and the actual measurement of on-the-spot energy, draw π type and the T-shaped equivalent circuit of simplification;

(3) according to the π type of simplifying and T-shaped equivalent circuit according to needs on-the-spot actual and computing machine calculating, draw front L-type equivalent circuit and rear L-type equivalent circuit.　

2. the method for medium voltage distribution network transmission line model according to claim 1, it is characterized in that: described step (1) is according to circuitry transmission line portions, can known transmission line π type and T-shaped equivalent circuit in, the resistance of the resistance R transmission line in Z=R+jX, X is transmission line reactance, ohmic loss is directly proportional to transmission line current flowing size to reactance loss, and it is irrelevant with transmission-line voltage grade, be directly proportional to cable length simultaneously, resistance R can be calculated according to electric wire handbook, reactance X can calculate according to electric wire handbook, also can know according to circuitry, it is that the leakage power of cable and the electricity that produces are led that electricity in Y=G+jB is led G, B is cable earth capacitance electric current and susceptance that corresponding charge power produces, leakage power and charge power are directly proportional to transmission-line voltage square, with to flow through transmission line choke irrelevant, be directly proportional to cable length simultaneously, electricity is led G and can be calculated according to electric wire handbook, leakage power is made up of two parts: one is insulation resistance loss, one is dielectric loss, susceptance B can calculate according to electric wire handbook.　

3. the method for medium voltage distribution network transmission line model according to claim 1, is characterized in that: described step (2), according to the parameter value going out of π type and T-shaped equivalent circuit and the actual measurement of on-the-spot energy, draws π type and the T-shaped equivalent circuit of simplification; Suppose 185 square millimeters of cable model YJV-8.7-15kV of medium voltage distribution network cross-section of cable list core, circuit is single phase power supply;

Its parameter is conductor AC resistance 0.128 Ω/km, and reactance is 0.096 Ω/km, and electric capacity is 0.3 μ F/km, supposes that this single-phase cable length is 1km longer in actual conditions;

At this moment cable data is calculated as R=0.128 Ω, X=0.096 Ω, B=9.42x10-5S

The size of current flowing through on transmission line is:

I=250A　

Transmission-line voltage grade is:

Suppose that the power transmitting on circuit is:

P=1400kW　

Q=351.19kvar　

Therefore power factor is:

cosφ=0.97　

The realistic ruuning situation of this power factor

So can draw up following formula:

△P _R=I ²R=250 ²×0.128=8000W　

△Q _X=I ²X=250 ²×0.096=6000var　

△P _G=△P _G1+△P _G2　

At this moment calculate △ P _gsize, wherein △ P _g=△ P _g1+ △ P _g2

1) △ P _g1insulation resistance loss, △ P _g2it is dielectric loss;

We first see insulation resistance

According to electric wire handbook, wherein insulation resistance

ρ _v---insulation resistance coefficient:

Be 10 according to electric wire handbook crosslinked polyethylene (YJV) insulating resistance of cable coefficient ¹⁶～10 ¹⁷/ (Ω cm)

δ---insulation thickness

Suppose that insulation thickness is 0.5cm(and can reduces calculated resistance toward getting in little, thereby increase leakage power)

A---electrode area

185mm ²cable interior diameter is 2cm, girth 6.28cm, and 1km cable is 100000cm length;

Calculate by the poorest condition

Therefore for 10kV circuit, insulation resistance Leakage Current has:

　

Therefore insulation resistance loss power is:

△P _G1=UI _G1=0.0042W　

2) according to the formula of electric wire handbook (3-2-24)

△P _G2=W _d=U ²ωCtgδ　

Wherein tg δ calculates according to crosslinked polyethylene maximal value 0.0025 in table 3-2.22

△P _G2=W _d=U ²ωCtgδ=3140×0.0025=7.85W

△P _G=△P _G1+△P _G2=0.0042+7.85=7.854W　

Therefore total electricity is led power

1) therefore total conduction current is:

And load current is 250A, current mutual inducer for line outlet end rated current is 300A, current transformer full accuracy is 0.1%, and the minimum current that can differentiate is 0.3A, and the electric current that electricity is led loss is only for current transformer can be differentiated 1/219th of electric current;

Simultaneously two kinds of electricity are led active loss and are all directly proportional to line voltage distribution grade square, and flow through line current size and have nothing to do, and electric pressure substantially constant, and therefore to lead loss be an individual constant to electricity substantially, and variation range is very little;

In sum, in power distribution network transmission line model, electricity is led and can be ignored;

2) by

The susceptance reactive power compensation power that cable earth capacitance forms is directly proportional to line voltage distribution grade square, line current size is irrelevant with flowing through, and electric pressure substantially constant, therefore to lead loss be constant to electricity substantially, and variation range is very little, and the power of susceptance B is 3140var, and substantially constant, the electric current calculating is 0.54A, can be resolved the current transformer identification of rate 0.3A,

Therefore susceptance parameter can not be ignored, and need to calculate susceptance power back-off;

3) by △ P _r=I ²r=250 ²× 0.128=8000W

Ohmic loss is directly proportional to current squaring, irrelevant with line voltage distribution size, circuit voltage swing is basicly stable simultaneously, and ohmic loss is very large with load variations variation range, and it is 1.39A that full load calculates size of current, can be resolved the current transformer identification of rate 0.3A;

Therefore resistance parameter can not be ignored, and needs calculated resistance power attenuation;

4）△Q _X=I ²X=250 ²×0.096=6000var　

Reactance Power consumption and electric current square is directly proportional, irrelevant with line voltage distribution size, circuit voltage swing is basicly stable simultaneously, and reactance loss is very large with load variations variation range, and it is 1.04A that full load calculates size of current, can be resolved the current transformer identification of rate 0.3A;

Therefore reactance parameter can not be ignored, and need to calculate blind power loss;

Therefore judge that can the standard that parameter can be ignored be resolved the mutual inductor identification that rate is the highest, the π type that therefore can be simplified and T-shaped equivalent circuit.　

4. the method for medium voltage distribution network transmission line model according to claim 1, it is characterized in that: described step (3) is according to the π type of simplifying and T-shaped equivalent circuit is actual according to scene and the needs of computing machine calculating, draw front L-type equivalent circuit and rear L-type equivalent circuit, π type and T-shaped equivalent circuit have following shortcoming: circuit structure more complicated, being not suitable for carrying out multi head linear equation sets up vertical, can produce polynary polynomial equation group, increase computational complexity, convergence is unstable; π type and T-shaped equivalent circuit bilateral impedance equate, be not inconsistent with actual conditions, due to field condition complexity, the two-way resistance reactance of wire is unequal, the high-order infinitesimal that certain unequal part is respectively parameters, can ignore in most cases, still for the accuracy of result, adopt L-type equivalent circuit to have the inconsistent advantage of abundant performance wire two ends parameter;

Therefore circuit reduction is become to front L-type and rear L-type equivalent circuit;

Described front L-type refers to that power back-off point is built up in reactive power transmitting terminal, and the reactive power of line charging power and power supply, after transmitting terminal merges, enters energy receiving end after transmission line loss; Described rear L-type refers to that power back-off point is built up in reactive power receiving end, and the reactive power of power supply, after transmission line loss, merges with line charging power, enter energy receiving end, suppose that cable data is calculated as R=0.128 Ω, X=0.096 Ω, B=9.42x10-5S

The size of current flowing through on transmission line is:

I=250A　

Transmission-line voltage grade is:

Suppose that the power transmitting on circuit is:

P=1400kW　

Q=351.19kvar　

Therefore power factor is:

cosφ=0.97　

The realistic ruuning situation of this power factor

So can draw up following formula:

△P _R=I ²R=250 ²×0.128=8000W　

△Q _X=I ²X=250 ²×0.096=6000var　

1) power direction for meritorious by A to B, idle by A to B, L-type equivalent circuit after adopting, it is as follows that process falls in calculating voltage:

Suppose that A point voltage phase angle is benchmark phase angle, wire both end voltage is as follows:

Can be drawn by known conditions:

△P _R=I ²R=250 ²×0.128=8000W　

△Q _X=I ²X=250 ²×0.096=6000var　

2) power direction for meritorious by A to B, idle by A to B, L-type equivalent circuit before adopting, it is as follows that process falls in calculating voltage:

Suppose that A point voltage phase angle is benchmark phase angle, wire both end voltage is as follows:

On transmission line, loss is:

3) power direction for meritorious by A to B, idle by B to A, L-type equivalent circuit after adopting, it is as follows that process falls in calculating voltage:

Suppose that A point voltage phase angle is benchmark phase angle, wire both end voltage is as follows:

On transmission line, loss is:

4) power direction for meritorious by A to B, idle by B to A, L-type equivalent circuit before adopting, it is as follows that process falls in calculating voltage:

Suppose that A point voltage phase angle is benchmark phase angle, wire both end voltage is as follows:

Can be drawn by known conditions:

△P _R=I ²R=250 ²×0.128=8000W

△Q _X=I ²X=250 ²×0.096=6000var

5) power direction for meritorious by B to A, idle by A to B, L-type equivalent circuit after adopting, it is as follows that process falls in calculating voltage:

Suppose that A point voltage phase angle is benchmark phase angle, wire both end voltage is as follows:

Can be drawn by known conditions:

△P _R=I ²R=250 ²×0.128=8000W

△Q _X=I ²X=250 ²×0.096=6000var

6) power direction for meritorious by B to A, idle by A to B, L-type equivalent circuit before adopting, it is as follows that process falls in calculating voltage:

Suppose that A point voltage phase angle is benchmark phase angle, wire both end voltage is as follows:

On transmission line, loss is:

7) power direction for meritorious by B to A, idle by B to A, L-type equivalent circuit after adopting, it is as follows that process falls in calculating voltage:

Suppose that A point voltage phase angle is benchmark phase angle, wire both end voltage is as follows:

On transmission line, loss is:

8) power direction for meritorious by A to B, idle by A to B, L-type equivalent circuit before adopting, it is as follows that process falls in calculating voltage:

Suppose that A point voltage phase angle is benchmark phase angle, wire both end voltage is as follows:

Can be drawn by known conditions:

△P _R=I ²R=250 ²×0.128=8000W

△Q _X=I ²X=250 ²×0.096=6000var

Can draw 1 from above calculating) 4) 5) 8) corresponding L-type equivalent circuit, compensation point is after reactive power is passed through transmission line;

2) 3) 6) 7) corresponding front L-type equivalent circuit, and compensation point is before reactive power is passed through transmission line;

Wherein 1) 8) equivalence, after the two difference is that active power and reactive power direction are oppositely, compensation point position is also reverse;

2) 7) equivalence, after the two difference is that active power and reactive power direction are oppositely, compensation point position is also reverse;

3) 6) equivalence, after the two difference is that active power and reactive power direction are oppositely, compensation point position is also reverse;

4) 5) equivalence, after the two difference is that active power and reactive power direction are oppositely, compensation point position is also reverse;

Can be drawn by above inference, transmission line model can be with 1) 5) represent L-type equivalent circuit after positive dirction.Transmission line model can be with 4) 8) represent L-type equivalent circuit after negative direction;

By drawing above following inference, transmission line model can be with 2) 6) represent L-type equivalent circuit before positive dirction, transmission line model can be with 3) 7) represent the front L-type equivalent circuit of negative direction.　

5. the method for medium voltage distribution network transmission line model according to claim 4, it is characterized in that: described according to the needs of on-the-spot reality and computing machine calculating, calculate the estimation of error of front L-type equivalent circuit and rear L-type equivalent circuit, the longer 1km of unit cable length in calculated examples, power approaches fully loaded 250A, power factor is 0.97, tally with the actual situation, in actual conditions, full load falls in transmission-line voltage can not be greater than 5%, according to this constraint condition, peak power and the length of transmission line that can transmit are all restricted, therefore transmission line is about as 5km most, corresponding maximum current 300A, therefore in the time that lowest high-current value reduces, length of transmission line is in the situation that sectional area is constant, can also increase to some extent,

Only having meritorious transmission, there is no idle transmission, power factor is 1, and voltage drop is only with meritorious relevant with cable resistance admittance parameter, but the determinative of decision tolerance variation is conductor length;

Meritorious transmission, only has idle transmission, and power factor is 0, and voltage drop is only in idle relevant with cable resistance admittance parameter, but the determinative of decision tolerance variation is conductor length;

On circuit, do not transmit meritorious load or burden without work, the voltage drop on cable is just fewer, and the front L that can measure and the tolerance of rear L-type equivalent circuit are just less, until surveying instrument loses resolution characteristic;

Circuit powers on source point while thering is no output power, only has charge power on circuit, and in example as cable length is got to limit radius of electricity supply 5km, charge power is 15700var;

Length of transmission line is shorter, and the power loss on transmission line is fewer, and the basic tolerance of front L and rear L-type equivalent circuit is less, because surveying instrument resolution is certain, the two is tending towards poor consistent, and in the time that conductor length is less than to a certain degree, surveying instrument loses resolution characteristic;

Circuit is longer, and front L-type and rear L-type equivalent circuit calculated tolerances are larger, but because cable voltage drop can not exceed 5%, therefore in general 10kV cable power supply radius is 5km, in economic current density;

Therefore only need to consider two kinds of situations below, just can judge whether whether front L-type and rear L-type equivalent circuit calculated value can be measured by voltage measurement element and current measurement device;

(1), unit cable length is amplified to 10kV limit radius of electricity supply to 5km, under full-load power, the current/voltage calculated tolerances distance of front L-type and rear L-type;

1) power direction for meritorious by A to B, idle by A to B, L-type equivalent circuit after adopting, it is as follows that process falls in calculating voltage:

Suppose that A point voltage phase angle is benchmark phase angle, wire both end voltage is as follows:

Can be drawn by known conditions:

△P _R=I ²R=250 ²×0.64=40000W　

△Q _X=I ²X=250 ²×0.48=30000var　

2) power direction for meritorious by A to B, idle by A to B, L-type equivalent circuit before adopting, it is as follows that process falls in calculating voltage:

Suppose that A point voltage phase angle is benchmark phase angle, wire both end voltage is as follows:

On transmission line, loss is:

Can see above the meritorious idle 162W of differing of 216W that differs of L-type circuit and rear L-type equivalent circuit result of calculation, reduction is 249.96A to mains side top electric current, lose 0.04A, and the recognition capability of current measurement device is 0.3A, front L-type and rear L-type equivalent circuit gained voltage drop difference are about 2V, and voltage measurement element highest resolution is 6V, therefore, front L-type equivalent circuit and rear L-type equivalent circuit result of calculation, in circuit electric parameter and through-put power, while reaching actual conditions allowable maximum, basically identical simultaneously;

(2) current/voltage error while, only having line charging power:

In the time only having charge power, 5 kilometers of line lengths, front L-type equivalent circuit calculates as follows

On transmission line, loss is:

Rear L-type equivalent circuit calculates as follows:

On transmission line, loss is:

Although voltage difference 1.31V+j1.74V, compare 5773.5 specified phase voltage and change and be about 0.03%, voltage measurement element cannot be measured, difference power is apart from not enough 10w also, current measurement device also cannot be measured;

Therefore can reach a conclusion: the front L-type equivalent circuit of transmission line and rear L-type equivalent circuit can be applied in power distribution network network calculation of parameter; Front L-type and rear L-type equivalent circuit can reduce computing machine calculating amount to a great extent; Before adopting separately, L or rear L-type equivalent circuit, can, at same transmission line under different idle transmission directions, look from reactive power input end, expresses different parameters.

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