3 Steady-State Performance of Transmission Lines
3.13 EHV UNDERGROUND CABLE TRANSMISSION
As discussed in the previous sections, the inductive reactance of an overhead high-voltage ac line is much greater than its capacitive reactance. However, the capacitive reactance of an underground high-voltage ac cable is much greater than its inductive reactance because the three-phase conductors are located very close to each other in the same cable. The approximate values of the resultant vars (reac-tive power) that can be generated by ac cables operating at the phase-to-phase voltages of 132, 220, and 400 kV are 2000, 5000, and 15,000 kVA/mi, respectively. This var generation, due to the capacitive charging currents, sets a practical limit to the possible noninterrupted length of an underground ac cable.
This situation can be compensated for by installing appropriate inductive shunt reactors along the line. This “critical length” of the cable can be defined as the length of cable line that has a three-phase charging reactive power equal in magnitude to the thermal rating of the cable line. For example, the typical critical lengths of ac cables operating at the phase-to-phase voltages of 132, 200, and 400 kV can be given approximately as 40, 25 and 15 mi, respectively.
The study done by Schifreen and Marble [12] illustrated the limitations in the operation of high-voltage ac cable lines due to the charging current. For example, Figure 3.40 shows that the magnitude of the maximum permissible power output decreases as a result of an increase in cable length [13].
Figure 3.41 shows that increasing lengths of cable line can transmit full-rated current (1.0 pu) only if the load power factor is decreased to resolve lagging values. Note that, the critical length is used as the base length in the figures. Table 3.4 [14] gives characteristics of a 345-kV pipe-type cable.
Figure 3.42 shows the permissible variation in per-unit vars delivered to the electric system at each terminal of cable line for a given power transmission.
EXAMPLE 3.17
Consider a high-voltage open-circuit three-phase insulated power cable of length l shown in Figure 3.43. Assume that a fixed sending-end voltage is to be supplied; the receiving-end voltage floats,
and it is an overvoltage. Furthermore, assume that at some critical length (l = l0), the sending-end current IS is equal to the ampacity of the cable circuit, Il0. Therefore, if the cable length is l0, no load, whatever of 1.0 or the leading power factor can be supplied without overloading the sending end of the cable. Use the general long-transmission-line equations, which are valid for steady-state sinusoidal operation, and verify that the approximate critical length can be expressed as
l I V bls
0≅ 0
Solution
The long-transmission-line equations can be expressed as
VS = VRcosh γl + IRZcsinh γl (3.265)
Per-unit line length
Per-unit receiving-end power
0.98 lag PF of load
0 lead 0.4 lead 0.6 lead 0.8 lead
0.9 lead 0.98 lead
0.4 lag
nsmission
FIGURE 3.40 Power transmission limits of high-voltage ac cable lines. Curved lines: Sending-end current equal to rated or base current of cable. Horizontal lines: Receiving-end current equal to rated or base current of cable. (From Wiseman, R. T., Trans. Am. Inst. Electr. Eng., 26, 803 © 1956 IEEE.)
00 0.2
Per-unit line length
Per-unit receiving end power PF
loadof
FIGURE 3.41 Receiving-end current limits of high-voltage ac cable lines. Curved lines: Sending-end cur-rent equal to rated or base curcur-rent of cable. (From Wiseman, R. T., Trans. Am. Inst. Electr. Eng., 26, 803 © 1956 IEEE. Used with permission.)
TABLE 3.4 Characteristics of a 345-kV Pipe-Type Cable Maximum Electric Stress, 300 V/milMaximum Electric Stress, 350 V/mil Characteristics
Power Factor (%)
Power Factor (%) Conductor size, kcmil1,0001,2501,5002,0001,0001,2301,3002,000 Insulation thickness, mils ER kV, 200 kV
1,2501,1731,1101,035980915885835 IT, A0.55856386807300.3576623636688 0.36537217808600.3637724776847 Rated three-phase mVA
0.53503814064360.3344372392413 0.33904314665160.3393432463508 Z, Ω/mi0.403∠78.00.381∠78.00.363∠80.00.338∠80.00.377∠77.20.355∠78.50.367∠79.40.319∠80.0 Y S, mi ×10-41.08∠89.71.20∠89.71.32∠89.71.53∠89.71.26∠89.71.41∠89.71.54∠89.71.78∠89.7 A numeric mi ×10−86.6∠283.96.8∠84.36.9∠84.97.19∠84.86.9∠83.57.1∠84.17.5∠84.67.54∠84.8 Z0, Ω61.2∠5.956.4∠5.552.5∠4.967.1∠6.954.7∠6.350.1∠3.648.9∠3.242.3∠4.9 Ic A ∠mi21.624.126.530.525.128.130.83.37
3Φ charging, KVA/mi12,90014,40013,80013,80015,00016,80018,00021,300 S4, mi0.527.126.525.724.00.522.922.221.319.4 0.330.330.029.328.20.326.225.725.223.9 Nominal pipe size, in, 108/ in. Earth resistivity, thermal Ω cm8080808080808080 Conductor temperature, ºC
7070707070707070 3Φ dielectric loss, W/ft0.512.213.714.917.30.514.215.817.320.1 0.37.3819.010.30.38.59.510.412.0 3Φ total loss, W/ft0.529.030.832.234.30.530.532.133.535.2 0.30.329.731.533.034.8 Ratio watts dielectric loss, Φ total loss
0.542.244.546.050.50.546.349.352.637.5 0.338.836.935.330.50.335.033.131.734.5 Source:Wiseman, R. T., Trans. Am. Inst. Electr. Eng., 26, 803 © 1956 IEEE.
and
IS = IRcosh γl + VRYcsinh γl (3.266) Since at critical length, l = l0 and
IR = 0 and
IS=Il0
from Equation 3.264, the sending-end current can be expressed as
Il0=V YR csinhγl0 (3.267)
or
Il V YR c
l l
e e
0= −
γ0 γ0
2 (3.268)
0
0
0 0.2
0.2
0.4 0.4
0.6 0.6
0.8 0.8
1.0 1.0
1.2 1.4 1.6 1.8 2.0 Per-unit line length
Per-unit receiving-end power
Reduced power-transmission levels
Permissible per-unit variation in vars
delivered for a given power transmission
Maxim um
po wer 1.0
±0.9 ±0.8 ±0.7±0.6
±0.5 ±0.4 ±0.3 ±0.2 ±0.1
FIGURE 3.42 Permissible variations in per-unit vars delivered to electric system at each terminal of ac cable line for given power transmission. (From Wiseman, R. T., Trans. Am. Inst. Electr. Eng., 26, 803 © 1956 IEEE.)
IS
VR IR = 0
VS = VS 0˚
l
Open-circuit cable (no load)
FIGURE 3.43 Cable system for Example 3.17.
Il V YR c l l l l
Similarly, from Equation 3.263, the sending-end voltage for the critical length can be expressed as
Neglecting higher powers of γl0,
VS≅VR + l
Substituting Equation 3.273 into Equation 3.268,
Neglecting the second term,
Il0≅V YS cγl0 (3.279)
Therefore, the critical length can be expressed as
l l Therefore, the critical length can be expressed as
l l
Since, for cables, g ≪ b,
y ≅ b∠90°
and assuming
Il0≅ ∠ °Il0 90
from Equation 5.231, the critical length can be expressed as
l b
l S
0≅ 0
× I
V (3.285)
EXAMPLE 3.18
Figure 3.44a shows an open-circuit high-voltage insulated ac underground cable circuit. The criti-cal length of uncompensated cable is l0 for which IS = I0 is equal to cable ampacity rating. Note that, Q0 = 3VSI0, where the sending-end voltage VS is regulated and the receiving-end voltage VR
floats. Here, |VR| differs little from |VS| because of the low series inductive reactance of cables.
On the basis of the given information, investigate the performances with IR = 0 (i.e., zero load).
IS = I0
QS = Q0(3Φ)
VR IR = 0
VS
VS VR
VS VR
l = l0
(a)
(b)
(c) l1
Q0
2Q0 2Q0
l2
l1 l2 l3
FIGURE 3.44 Insulated HV underground cable circuit for Example 3.18.
(a) Assume that one shunt inductive reactor sized to absorb Q0 magnetizing vars is to be pur-chased and installed as shown in Figure 3.44b. Locate the reactor by specifying l1 and l2 in terms of l0. Place arrowheads on the four short lines, indicated by a solid line, to show the directions of magnetizing var flows. Also show on each line the amounts of var flow, expressed in terms of Q0.
(b) Assume that one reactor size 2Q0 can be afforded and repeat part (a) on a new diagram.
(c) Assume that two shunt reactors, each of size 2Q0, are to be installed, as shown in Fig-ure 3.44c, hoping, as usual, to extend the feasible length of cable. Repeat part (a).
Solution
The answers for parts (a), (b), and (c) are given in Figure 3.45a, b and c, respectively.