CN112757971A - Calculation method for power supply capacity of AT power supply of heavy haul railway - Google Patents
Calculation method for power supply capacity of AT power supply of heavy haul railway Download PDFInfo
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Abstract
When the method is used, a through power supply scheme of the heavy haul railway traction cable powered by the AT is calculated, mathematical models of electrical parameters such as CTN equivalent impedance, current distribution, voltage loss, CTN terminal network voltage and the like are deduced, and a power supply distance calculation method is further obtained to judge the power supply capability of the scheme, so that the existing heavy haul railway AT power supply scheme is conveniently improved, the electric phase separation of the existing heavy haul railway AT power supply system is greatly reduced, the power supply capability is improved, the power supply distance is prolonged, the locomotive runs more safely and reliably, and the running speed is improved.
Description
Technical Field
The invention relates to a heavy haul railway, in particular to a calculation method for the power supply capacity of AT power supply of the heavy haul railway.
Background
The system for supplying power by adopting the AT through the traction cable consists of a main substation MSS (MainSubStation) and a CTN.
The heavy-load railway traction load has high power and high transmission power requirement on a traction network, so that higher requirements are placed on the power supply capacity and the electric energy transmission reliability of a traction power supply system. The power supply modes of the heavy haul railway in China mainly comprise a direct power supply mode with a backflow line and an Autotransformer (AT) power supply mode 2. At present, a heavy haul railway traction power supply system still has certain defects, and is mainly embodied as follows: firstly, the net voltage of a contact net is low; secondly, the overload phenomenon of the traction power transformation is serious; thirdly, the power factor of a line traction substation which is not popularized and applied in a plurality of AC/DC type locomotives is low; speed drop and limited power supply capability of a traction power supply system during over-current phase separation of the train; the negative sequence problem caused by train operation; sixthly, the utilization rate of the regenerative braking energy of the train is lower.
AT present, no good calculation method is used for calculating the power supply capacity of the existing Autotransformer (AT) power supply mode, and the method is not beneficial to being modified to overcome the defects.
Disclosure of Invention
In view of the above situation, in order to overcome the defects of the prior art, the invention provides a method for calculating the power supply capacity of the AT of the heavy haul railway, which effectively solves the problems mentioned in the background of the prior art.
In order to achieve the purpose, the invention provides the following technical scheme: the invention comprises the calculation of the current and voltage distribution and the power supply capacity of the CTN, wherein the current and voltage distribution of the CTN is as follows: when power is supplied to a locomotive in an AT traction network, current directions of traction cables C1 and C2 in FIG. 1 are opposite and have the same magnitude, so that mutual impedance between the traction cables and an AT contact network can be ignored, leakage reactance of AT in CTN is ignored for convenience of analysis, and FIG. 1 is simplified into an equivalent circuit shown in FIG. 2, namely, a 2 × 27.5kV AT network is equivalent to a 27.5kV network and comprises a traction cable loop and a contact network equivalent loop, the simplified equivalent contact line, equivalent steel rail and equivalent negative feeder are respectively T ', R ' and F ', wherein F ' is an ideal lead, in the figure, E ' is MSS traction bus voltage, and equivalent impedances zAA and zBB of unit lengths of T ' and R ' are respectively:
zAA＝(zT+zF-2zTF)/4 (1)
zBB＝(zT+2zR+zTF-zRF-3zTR)/2 (2)
wherein z isT、zRAnd zFThe self-impedance of the contact line, the steel rail and the negative feeder line respectively; z is a radical ofTF、zRFAnd zTRThe mutual impedance between the contact line and the negative feeder line, between the steel rail and the negative feeder line, and between the contact line and the steel rail respectively;
respectively defining a long loop (formed by a power supply cable and a contact network connected by MSS) and a short loop (formed by a power supply cable and a contact network between traction substations on two sides of a locomotive station), and reducing an equivalent circuit in the figure 2 to a 27.5kV traction side to obtain CTN and the likeValue loop 1, as shown in FIG. 3(a), in which IcpAnd l'cpThe current flowing through the long loop cable and the short loop cable when the locomotive is positioned between the p-th seat and the p + 1-th traction substation is respectively; the Ip1 and the Ip2 are currents passing through the long loop and the short loop AT when the locomotive is positioned AT the pth seat and the pth +1 seat traction substation respectively; i is locomotive current, voltage drops of a loop 1 and a loop 2 in the graph 3(a) are respectively analyzed, and the voltage drops are equivalent to an equivalent loop 2 shown in the graph 3(b) due to neglect of leakage reactance of AT;
FIG. 3(b) is further simplified to obtain an equivalent circuit shown in FIG. 4. In the figure, ZTTThe leakage reactance of the traction transformer is obtained; e is equivalent voltage after the reduction to the 27.5kV side; lpDistance of the locomotive from MSS; lp-DpAnd DpThe lengths of a long loop and a short loop of the locomotive between the p-th seat and the p + 1-th traction substation are respectively set; x is the number ofpThe equivalent impedance z of the unit length of the traction cable loop in the figure 4 is reduced to 27.5kV traction side for the distance between the time when the locomotive is positioned between the p-th traction substation and the p + 1-th traction substation and the time when the locomotive is positioned between the previous substations1Equivalent impedance z of contact net unit length2See formula (3) and formula (4), respectively;
z1＝[RC1+RC2-2RC1C2+jω(LC1+LC2-2LC1C2)]/16 (3)
z2＝zAA+zBB (4)
wherein R isC1And RC2Resistance per unit length of the traction cables C1 and C2, respectively; l isC1And LC2Respectively a traction cable C1And C2The inductance per unit length of (1); rC1C2And LC1C2The mutual resistance and mutual inductance of the cables in unit length are respectively;
writing kirchhoff current law and kirchhoff voltage law equations for the circuit in which the locomotive is located in fig. 4 can obtain:
solving according to the formula (5) to obtain Ip1And Ip2Respectively as follows:
wherein k isp1＝[(z1+z2)Dp+ZTT]/[(z1+z2)Dp+2ZTT]；kp2＝z2/[(z1+z2)Dp+2ZTT]；
In the CTN, the long loop voltage drop and the short loop voltage drop form a voltage drop delta U from the head end of the CTN to the position of the locomotive1Comprises the following steps:
ΔU1＝Icpz1(lp-xp)+Ip1(ZTT+z2xp) (8)
wherein, Icp＝I；
By substituting and simplifying formula (6) for formula (8): delta U1＝[z1(lp-xp)+(kp1-kp2xp)(ZTT+z2xp)]I(9)；
Calculating the power supply capacity: the voltage drop Δ Un at the end locomotive is:
wherein k ism1＝[(z1+z2)Dm+ZTT]/[(z1+z2)Dm+2ZTT]；km2＝z2/[(z1+z2)Dm+2ZTT]；InAnd ImThe current of the nth locomotive and the current of the mth locomotive respectively; lmIs the m-th machineDistance of the vehicle from MSS; dmThe length of a short loop at the position of the mth locomotive; x is the number ofmThe distance between the position of the mth locomotive and the previous substation;
let ZAn＝RAn+jXAn＝[lm-xm+(ZTT/z2+xm)km2Dm]z1，ZBn＝RBn+jXBn＝(ZTT/z2+xn)kn2[ZTT+z2(Dn-xn)]Then, the voltage loss Δ U at the end locomotive is obtained from the equation (10)nComprises the following steps:
wherein phi ismAnd phinPower factor angles for the mth and nth locomotives, respectively; i ismAnd InThe current values of the mth locomotive and the nth locomotive are respectively;
in the same way, get the f (f)<n) voltage drop Δ U at a locomotivefComprises the following steps:
wherein the content of the first and second substances,
kf1＝[(z1+z2)Df+ZTT]/[(z1+z2)Df+2ZTT]；kf2＝z2/[(z1+z2)Df+2ZTT]；xfthe distance between the position of the f-th locomotive and the previous power substation; i isfCurrent of the f-th locomotive; lfThe distance between the f-th locomotive and the MSS;
let ZAf＝RAf+jXAf＝(lf-xf)z1，ZBf＝RBf+jXBf＝(kf1-kf2xf)(ZTT+z2xf)z1ZCf ═ RCf + jXCf ═ (1-kf1+ kf2xf) Dmz1, then the voltage loss Δ U at the f-th locomotive isfComprises the following steps:
wherein, Z'Af＝RAfcosφm+XAfsinφm；Z′Bf＝RBfcosφf+XBf×sinφf，φfIs the power factor angle of the f-th locomotive; z'Cf＝RCfcosφm+XCfsinφm；IfThe current value of the f locomotive is obtained;
setting a limit value epsilon by taking the formula (11) and the formula (13) as a judgment basis of the farthest power supply distance, recording a first locomotive close to a traction substation as m is 1, determining the position of the mth locomotive, calculating locomotive current or actual line locomotive current by utilizing locomotive power, taking the maximum value of the locomotive current as the current of the locomotive, and calculating the voltage loss delta U of the corresponding locomotive at the corresponding position at the calculated position m,1,2, n, if max (Δ U)m,)And E is less than or equal to epsilon, the calculation is finished, the number of the locomotives is n, and the available power supply distance is l.
Has the advantages that: when the method is used, the through power supply scheme of the AT-powered heavy haul railway traction cable is calculated, mathematical models of electrical parameters such as CTN equivalent impedance, current distribution, voltage loss, CTN terminal network voltage and the like are deduced, and a power supply distance calculation method is further obtained to judge the power supply capacity of the scheme, so that the existing AT power supply scheme of the heavy haul railway is conveniently transformed, the electric phase of the existing AT power supply system of the heavy haul railway is greatly reduced, the power supply capacity is improved, the power supply distance is prolonged, the locomotive runs more safely and reliably, and the running speed is improved.
Drawings
The accompanying drawings, which are included to provide a further understanding of the invention and are incorporated in and constitute a part of this specification, illustrate embodiments of the invention and together with the description serve to explain the principles of the invention and not to limit the invention. In the drawings:
FIG. 1 is a schematic diagram of the power supply system of the present invention;
FIG. 2 is a CTN equivalent circuit diagram of the present invention;
FIG. 3 is a CTN equivalent circuit diagram of the present invention;
FIG. 4 is a single-vehicle equivalent circuit diagram of the traction cable through power supply system of the invention;
FIG. 5 is a multi-vehicle equivalent circuit diagram of the traction cable through power supply system of the invention.
Detailed Description
The following description of the embodiments of the present invention will be made in detail with reference to the accompanying drawings 1 to 5.
The embodiment is given by fig. 1 to 5, and the invention provides a method for calculating the power supply capacity of the AT power supply of the heavy haul railway, which includes calculating the current and voltage distribution and the power supply capacity of the CTN, where the current and voltage distribution of the CTN is: when power is supplied to a locomotive in an AT traction network, current directions of traction cables C1 and C2 in FIG. 1 are opposite and have the same magnitude, so that mutual impedance between the traction cables and an AT contact network can be ignored, leakage reactance of AT in CTN is ignored for convenience of analysis, and FIG. 1 is simplified into an equivalent circuit shown in FIG. 2, namely, a 2 × 27.5kV AT network is equivalent to a 27.5kV network and comprises a traction cable loop and a contact network equivalent loop, the simplified equivalent contact line, equivalent steel rail and equivalent negative feeder are respectively T ', R ' and F ', wherein F ' is an ideal lead, in the figure, E ' is MSS traction bus voltage, and equivalent impedances zAA and zBB of unit lengths of T ' and R ' are respectively:
zAA＝(zT+zF-2zTF)/4 (1)
zBB＝(zT+2zR+zTF-zRF-3zTR)/2 (2)
wherein z isT、zRAnd zFThe self-impedance of the contact line, the steel rail and the negative feeder line respectively; z is a radical ofTF、zRFAnd zTRThe mutual impedance between the contact line and the negative feeder line, between the steel rail and the negative feeder line, and between the contact line and the steel rail respectively;
respectively defining a long loop (formed by a power supply cable and a contact network connected by an MSS) and a short loop (formed by a power supply cable and a contact network between traction substations on two sides of a locomotive location), and reducing the equivalent circuit in the figure 2 to a 27.5kV traction side to obtain a CTN equivalent loop 1, as shown in figure 3(a), wherein IcpAnd l'cpThe current flowing through the long loop cable and the short loop cable when the locomotive is positioned between the p-th seat and the p + 1-th traction substation is respectively; the Ip1 and the Ip2 are currents passing through the long loop and the short loop AT when the locomotive is positioned AT the pth seat and the pth +1 seat traction substation respectively; i is locomotive current, voltage drops of a loop 1 and a loop 2 in the graph 3(a) are respectively analyzed, and the voltage drops are equivalent to an equivalent loop 2 shown in the graph 3(b) due to neglect of leakage reactance of AT;
FIG. 3(b) is further simplified to obtain an equivalent circuit shown in FIG. 4. In the figure, ZTTThe leakage reactance of the traction transformer is obtained; e is equivalent voltage after the reduction to the 27.5kV side; lpDistance of the locomotive from MSS; lp-DpAnd DpThe lengths of a long loop and a short loop of the locomotive between the p-th seat and the p + 1-th traction substation are respectively set; x is the number ofpThe equivalent impedance z of the unit length of the traction cable loop in the figure 4 is reduced to 27.5kV traction side for the distance between the time when the locomotive is positioned between the p-th traction substation and the p + 1-th traction substation and the time when the locomotive is positioned between the previous substations1Equivalent impedance z of contact net unit length2See formula (3) and formula (4), respectively;
z1＝[RC1+RC2-2RC1C2+jω(LC1+LC2-2LC1C2)]/16 (3)
z2＝zAA+zBB (4)
wherein R isC1And RC2Resistance per unit length of the traction cables C1 and C2, respectively; l isC1And LC2Respectively a traction cable C1And C2The inductance per unit length of (1); rC1C2And LC1C2The mutual resistance and mutual inductance of the cables in unit length are respectively;
writing kirchhoff current law and kirchhoff voltage law equations for the circuit in which the locomotive is located in fig. 4 can obtain:
solving according to the formula (5) to obtain Ip1And Ip2Respectively as follows:
wherein k isp1＝[(z1+z2)Dp+ZTT]/[(z1+z2)Dp+2ZTT]；kp2＝z2/[(z1+z2)Dp+2ZTT]；
In the CTN, the long loop voltage drop and the short loop voltage drop form a voltage drop delta U from the head end of the CTN to the position of the locomotive1Comprises the following steps:
ΔU1＝Icpz1(lp-xp)+Ip1(ZTT+z2xp) (8)
wherein, Icp＝I；
By substituting and simplifying formula (6) for formula (8): delta U1＝[z1(lp-xp)+(kp1-kp2xp)(ZTT+z2xp)]I(9)；
Calculating the power supply capacity: the voltage drop Δ Un at the end locomotive is:
wherein k ism1＝[(z1+z2)Dm+ZTT]/[(z1+z2)Dm+2ZTT]；km2＝z2/[(z1+z2)Dm+2ZTT]；InAnd ImThe current of the nth locomotive and the current of the mth locomotive respectively; lmThe distance between the mth locomotive and the MSS; dmThe length of a short loop at the position of the mth locomotive; x is the number ofmThe distance between the position of the mth locomotive and the previous substation;
let ZAn＝RAn+jXAn＝[lm-xm+(ZTT/z2+xm)km2Dm]z1，ZBn＝RBn+jXBn＝(ZTT/z2+xn)kn2[ZTT+z2(Dn-xn)]Then, the voltage loss Δ U at the end locomotive is obtained from the equation (10)nComprises the following steps:
wherein phi ismAnd phinPower factor angles for the mth and nth locomotives, respectively; i ismAnd InThe current values of the mth locomotive and the nth locomotive are respectively;
in the same way, get the f (f)<n) voltage drop Δ U at a locomotivefComprises the following steps:
wherein the content of the first and second substances,
kf1＝[(z1+z2)Df+ZTT]/[(z1+z2)Df+2ZTT]；kf2＝z2/[(z1+z2)Df+2ZTT]；xfthe distance between the position of the f-th locomotive and the previous power substation; i isfCurrent of the f-th locomotive; lfThe distance between the f-th locomotive and the MSS;
let ZAf＝RAf+jXAf＝(lf-xf)z1，ZBf＝RBf+jXBf＝(kf1-kf2xf)(ZTT+z2xf)z1ZCf ═ RCf + jXCf ═ (1-kf1+ kf2xf) Dmz1, then the voltage loss Δ U at the f-th locomotive isfComprises the following steps:
wherein, Z'Af＝RAfcosφm+XAfsinφm；Z′Bf＝RBfcosφf+XBf×sinφf，φfIs the power factor angle of the f-th locomotive; z'Cf＝RCfcosφm+XCfsinφm；IfThe current value of the f locomotive is obtained;
setting a limit value epsilon by taking the formula (11) and the formula (13) as a judgment basis of the farthest power supply distance, recording a first locomotive close to a traction substation as m is 1, determining the position of the mth locomotive, calculating locomotive current or actual line locomotive current by utilizing locomotive power, taking the maximum value of the locomotive current as the current of the locomotive, and calculating the voltage loss delta U of the corresponding locomotive at the corresponding position at the calculated position m,1,2, n, if max (Δ U)m,)Less than or equal to epsilon, the calculation is finished, the number of the locomotives is n, and the available power supply distance is lm。
The working principle is as follows: when the method is used, the through power supply scheme of the AT-powered heavy haul railway traction cable is calculated, mathematical models of electrical parameters such as CTN equivalent impedance, current distribution, voltage loss, CTN terminal network voltage and the like are deduced, and a power supply distance calculation method is further obtained to judge the power supply capacity of the scheme, so that the existing AT power supply scheme of the heavy haul railway is conveniently transformed, the electric phase of the existing AT power supply system of the heavy haul railway is greatly reduced, the power supply capacity is improved, the power supply distance is prolonged, the locomotive runs more safely and reliably, and the running speed is improved.
Finally, it should be noted that: although the present invention has been described in detail with reference to the foregoing embodiments, it will be apparent to those skilled in the art that changes may be made in the embodiments and/or equivalents thereof without departing from the spirit and scope of the invention. Any modification, equivalent replacement, or improvement made within the spirit and principle of the present invention should be included in the protection scope of the present invention.
Claims (1)
1. A method for calculating the power supply capacity of the AT power supply of a heavy haul railway comprises the calculation of the current and voltage distribution and the power supply capacity of a CTN, and is characterized in that: the CTN current and voltage distributions: when power is supplied to a locomotive in an AT traction network, wherein current directions of a traction cable C1 and a current direction of a traction cable C2 are opposite and the current magnitudes are the same, so that mutual impedance between the traction cable and an AT contact network can be ignored, for convenience of analysis, leakage reactance of AT in the CTN is ignored, the AT network of 2 multiplied by 27.5kV is equivalent to a 27.5kV network, the equivalent network comprises a traction cable loop and a contact network equivalent loop, the simplified equivalent contact line, the equivalent steel rail and the equivalent negative feeder are respectively T ', R ' and F ', wherein F ' is an ideal lead, E ' is MSS traction bus voltage, and equivalent impedances zAA and zBB of unit lengths of T ' and R ' are respectively:
zAA＝(zT+zF-2zTF)/4 (1)
zBB＝(zT+2zR+zTF-zRF-3zTR)/2 (2)
wherein z isT、zRAnd zFThe self-impedance of the contact line, the steel rail and the negative feeder line respectively; z is a radical ofTF、zRFAnd zTRThe mutual impedance between the contact line and the negative feeder line, between the steel rail and the negative feeder line, and between the contact line and the steel rail respectively;
define long loops separately: the power supply cable and the contact network connected by the MSS form a short loop: by traction from both sides of the location of the locomotiveThe equivalent circuit is reduced to 27.5kV traction side to obtain a CTN equivalent loop 1, IcpAnd l'cpThe current flowing through the long loop cable and the short loop cable when the locomotive is positioned between the p-th seat and the p + 1-th traction substation is respectively; the Ip1 and the Ip2 are currents passing through the long loop and the short loop AT when the locomotive is positioned AT the pth seat and the pth +1 seat traction substation respectively; i is locomotive current, voltage drops of the loop 1 and the loop 2 are analyzed respectively, and the voltage drops can be equivalent to an equivalent loop 2 due to neglecting the leakage reactance of AT;
ZTTthe leakage reactance of the traction transformer is obtained; e is equivalent voltage after the reduction to the 27.5kV side; lpDistance of the locomotive from MSS; lp-DpAnd DpThe lengths of a long loop and a short loop of the locomotive between the p-th seat and the p + 1-th traction substation are respectively set; x is the number ofpThe unit length equivalent impedance z of a traction cable loop is reduced to 27.5kV traction side for the distance between the time when the locomotive is positioned between the p-th traction substation and the p + 1-th traction substation and the previous substation1Equivalent impedance z of contact net unit length2See formula (3) and formula (4), respectively;
z1＝[RC1+RC2-2RC1C2+jω(LC1+LC2-2LC1C2)]/16 (3)
z2＝zAA+zBB (4)
wherein R isC1And RC2Resistance per unit length of the traction cables C1 and C2, respectively; l isC1And LC2Respectively a traction cable C1And C2The inductance per unit length of (1); rC1C2And LC1C2The mutual resistance and mutual inductance of the cables in unit length are respectively;
the loop where the locomotive is located is written with kirchhoff current law and kirchhoff voltage law equations, and the following can be obtained:
solving according to the formula (5) to obtain Ip1And Ip2Respectively as follows:
wherein k isp1＝[(z1+z2)Dp+ZTT]/[(z1+z2)Dp+2ZTT]；kp2＝z2/[(z1+z2)Dp+2ZTT]；
In the CTN, the long loop voltage drop and the short loop voltage drop form a voltage drop delta U from the head end of the CTN to the position of the locomotive1Comprises the following steps:
ΔU1＝Icpz1(lp-xp)+Ip1(ZTT+z2xp) (8)
wherein, Icp＝I；
By substituting and simplifying formula (6) for formula (8): delta U1＝[z1(lp-xp)+(kp1-kp2xp)(ZTT+z2xp)]I (9)；
Calculating the power supply capacity: the voltage drop Δ Un at the end locomotive is:
wherein k ism1＝[(z1+z2)Dm+ZTT]/[(z1+z2)Dm+2ZTT]；km2＝z2/[(z1+z2)Dm+2ZTT]；InAnd ImThe current of the nth locomotive and the current of the mth locomotive respectively; lmDistance MSS for m-th locomotiveSeparating; dmThe length of a short loop at the position of the mth locomotive; x is the number ofmThe distance between the position of the mth locomotive and the previous substation;
let ZAn＝RAn+jXAn＝[lm-xm+(ZTT/z2+xm)km2Dm]z1，ZBn＝RBn+jXBn＝(ZTT/z2+xn)kn2[ZTT+z2(Dn-xn)]Then, the voltage loss Δ U at the end locomotive is obtained from the equation (10)nComprises the following steps:
wherein phi ismAnd phinPower factor angles for the mth and nth locomotives, respectively; i ismAnd InThe current values of the mth locomotive and the nth locomotive are respectively;
in the same way, get the f (f)<n) voltage drop Δ U at a locomotivefComprises the following steps:
wherein k isf1＝[(z1+z2)Df+ZTT]/[(z1+z2)Df+2ZTT]；kf2＝z2/[(z1+z2)Df+2ZTT]；xfThe distance between the position of the f-th locomotive and the previous power substation; i isfCurrent of the f-th locomotive; lfThe distance between the f-th locomotive and the MSS;
let ZAf＝RAf+jXAf＝(lf-xf)z1，ZBf＝RBf+jXBf＝(kf1-kf2xf)(ZTT+z2xf)z1，ZCf＝RCf+jXCf＝(1-kf1+ kf2xf) Dmz1, the voltage loss at the f-th locomotive is Δ UfComprises the following steps:
wherein, Z'Af＝RAfcosφm+XAfsinφm；Z′Bf＝RBfcosφf+XBf×sinφf，φfIs the power factor angle of the f-th locomotive; z'Cf＝RCfcosφm+XCfsinφm；IfThe current value of the f locomotive is obtained;
setting a limit value epsilon by taking the formula (11) and the formula (13) as a judgment basis of the farthest power supply distance, recording a first locomotive close to a traction substation as m is 1, determining the position of the mth locomotive, calculating locomotive current or actual line locomotive current by utilizing locomotive power, taking the maximum value of the locomotive current as the current of the locomotive, and calculating the voltage loss delta U of the corresponding locomotive at the corresponding position at the calculated positionm,1,2, n, if max (Δ U)m,)Less than or equal to epsilon, the calculation is finished, the number of the locomotives is n, and the available power supply distance is lm。
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| FR2254898A1 (en) * | 1973-12-14 | 1975-07-11 | Secheron Atel | |
| CN105109362A (en) * | 2015-09-14 | 2015-12-02 | 西南交通大学 | Electrified railway traction power supply system |
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