(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.
of a hydro pump storage plant in Germany. For that application of the GIL, a tunnel was built in the mountain for a 420-kV overhead line. To this day, a total of more than 100 km of GILs have been built worldwide at high-voltage levels ranging from 135 to 550 kV. The GILs have also been built at power plants having adverse environmental conditions. Thus, in situations for which overhead lines are not feasible, the GIL may be an acceptable alternative since it provides a solution for a line without reducing transmission capacity under any kinds of climate conditions. This is because the GIL transmission sys- tem is independent of environmental conditions since it is completely sealed inside a metallic enclosure.
Applications of GIL include connecting high-voltage transformers with high-voltage switchgear within power plants, connecting high-voltage transformers inside the cavern power plants to over- head lines on the outside, connecting gas-insulated switchgear (GIS) with overhead lines, and serv- ing as a bus duct within GIS.
At the beginning, the GIL system was only used in special applications owing to its high cost.
Today, the second-generation GIL system is used for high-power transmission over long distances owing to the substantial reduction in its cost. This is accomplished not only by its much lower cost but also by the use of an N2–SF6 gas mixture for electrical insulation. The advantages of the GIL system include low losses, low magnetic field emissions, greater reliability with high transmission capacity, no negative impacts on the environment or the landscape, and underground laying with a transmission capacity that is equal to an overhead transmission line.
EXAMPLE 3.19
A power utility company is required to build a 500 kV line to serve a nearby town. There are two possible routes for the construction of the necessary power line. Route A is 80 mi long and goes around a lake. It has been estimated that the required overhead transmission line will cost $1 mil- lion per mile to build and $500 per mile per year to maintain. Its salvage value will be $2000 per mile at the end of 40 years.
On the other hand, route B is 50 mi long and is an underwater (submarine) line that goes across the lake. It has been estimated that the required underwater line using submarine power cables will cost $4 million to build per mile and $1500 per mile per ear to maintain. Its salvage value will be $6000 per mile at the end of 40 years.
It is also possible to use GIL in route C, which goes across the lake. Route C is 30 mi in length.
It has been estimated that the required GIL transmission will cost $7.6 million per mile to build and
$200 per mile to maintain. Its salvage value will be $1000 per mile at the end of 40 years. It has also been estimated that if the GIL alternative is elected, the relative savings in power losses will be $17.5 × 106 per year in comparison with the other two alternatives.
Assume that the fixed charge rate is 10% and that the annual ad valorem (property) taxes are 3% of the first costs of each alternative. The cost of energy is $0.10 per kWh. Use any engineering economy interest tables* and determine the economically preferable alternative.
Solution
OVERHEAD TRANSMISSION:
The first cost of the 500 kV overhead transmission line is
P = ($1,000,000/mi)(80 mi) = $80,000,000
and its estimated salvage value is
F = ($2000/mi)(80 mi) = $160,000
*For example, see Engineering Economy for Engineering Managers, T. Gönen, Wiley, 1990.
The annual equivalent cost of capital invested in the line is A1 80 000 000A P1040 100 000A F1040
80
= −
=
$ , , ( ) $ , ( )
$
% %
/ /
,, , ( . ) $ , ( . )
$ , ,
000 000 0 10226 100 000 0 00226 8 180 800
−
= −−$266=$ ,8 180 534,
The annual equivalent cost of the tax and maintenance is
A2 = ($80,000,000)(0.03) + ($500/mi)(80 mi) = $2,440,000 The total annual equivalent cost of the overhead transmission line is
A A A= + = +
=
1 2 8 180 534 2 440 000 10 620 534
$ , , $ , ,
$ , ,
SUBMARINE TRANSMISSION:
The first cost of the 500 kV submarine power transmission line is P = ($4,000,000/mi)(50 mi) $200,000,000
and its estimated salvage value is
F = ($6000/mi)(50 mi) $300,000
The annual equivalent cost of capital invested in the line is
A1 A P1040 A F
1040
200 000 000 300 000 2
= −
=
$ , , ( ) $ , ( )
$
% %
/ /
000 000 000 0 10296 300 000 0 00296 20 591
, , ( . ) $ , ( . )
$ , ,
−
= 1112
The annual equivalent cost of tax and maintenance is
A2 = ($200,000,000)(0.03) + ($1500/mi)(50 mi) = $6,075,000 The total annual equivalent cost of the overhead transmission line is
A A A= + = +
=
1 2 20 591112 6 075 000 26 666 112
$ , , $ , ,
$ , ,
GIL TRANSMISSION:
The first cost of the 500 kV GIL transmission line is
P = ($7,600,000/mi)(30 mi) $228,000,000
and its estimated salvage value is
F = ($1000/mi)(30 mi) $30,000
The annual equivalent cost of capital invested in the GIL line is A1 228 000 000A P1040 30 000A F1040
22
= −
=
$ , , ( ) $ , ( )
$
% %
/ /
88 000 000 0 10226 30 000 0 00226 23 315 28
, , ( . ) $ , ( . )
$ , ,
−
= 00−$ .67 5=$ ,23 315 212 5, .
The annual equivalent cost of the tax and maintenance is
A2 = ($228,000,000)(0.03) + ($200/mi)(30 mi) = $6,846,000 The total annual equivalent cost of the GIL transmission line is
A A A= + = +
=
1 2 23 315 212 5 6 846 000 30 221 212 5
$ , , . $ , ,
$ , , .
Since the relative savings in power losses is $17,500,000, then the total net annual equivalent cost of the GIL transmission is
Anet= −
=
$ , , . $ , ,
$ , , .
30 221 212 5 17 500 000 12 721 212 5
The results show that the use of overhead transmission for this application is the best choice.
The next best alternative is the GIL transmission. However, the above example is only a rough and very simplistic estimate. In real applications, there are many other cost factors that need to be included in such comparisons.
EXAMPLE 3.20
Consider transmitting 2100-MVA electric power across 30 km by using an overhead transmission line (OH) versus by using a GIL. The resulting power losses at peak load are 820 and 254 kW/km for the overhead transmission and the GIL, respectively. Assume that the annual load factor and the annual power loss factor are the same and are equal to 0.7 for both alternatives. Also, assume that the cost of electric energy is $0.10 per kWh. Determine the following:
(a) The power loss of the overhead line at peak load (b) The power loss of the GIL
(c) The total annual energy loss of the overhead transmission line at peak load (d) The total annual energy loss of the gas insulated transmission line at peak load
(e) The average energy loss of the overhead transmission line (f) The average energy loss of the GIL at peak load
(g) The average annual cost of losses of the overhead transmission line (h) The average annual cost of losses of the GIL
(i) The annual resultant savings in losses using the GIL
(j) Find the breakeven (or payback) period when the gas-insulated line alternative is selected, if the investment cost of the gas-insulated line is $200,000,000
Solution
(a) The power loss of the overhead transmission line at peak load is (Power loss)OH line = (829 kW/km)30 km = 24, 870 kW (b) The power loss of the GIL transmission line at peak load is
(Power loss)GIL line = (254 kW/km)30 km = 7620 kW
(c) The total annual energy loss of the overhead transmission line at peak load is (Total annual energy loss)at peak=(24,870 kW 87)( 660 h/yr
21,786 10 kWh/yr4 )
= ×
(d) The total annual energy loss of the gas-insulated line at peak load is (Total annual energy loss)at peak=(7620 kW 8760)( hh/yr 6675.2 10 kWh/yr4 )
= ×
(e) The total annual energy loss of the overhead transmission line at peak load is (Average annual energy loss)OH line=0.7(21,786×110 kWh/yr)
15,250.2 10 kWh/yr
4
= × 4
(f) The average energy loss of the gas-insulated line at peak load is (Average annual energy loss)GIL line=0.7(6675kWhh/yr)
672.64 10 kWh/yr4
=4 ×
(g) The average annual cost of losses of the overhead transmission line is (Average annual cost of losses)OH line=($0.10/kWWh 15,250.2 10 kWh/yr
1525.02 10 /yr
4
3
)( )
$
×
= ×
(h) The average annual cost of losses of the GIL is
(Average annual cost of losses)GIL line=($0.10/kkWh 4672.64 10 kWh/yr 467.264 10 /yr
4
3
)( )
$
×
= ×
(i) The annual resultant savings in power losses using the GIL is
Annual savings in losses=(Annual cost of losses)) ( )
$
OH line− Annual cost of lossesGIL line
= 1525×× − ×
= ×
10 467 264 10 1057 736 10
3 3
3
$ .
$ . /yr
(j) If the GIL alternative is selected
Breakeven period Total investment cost Savings p
= eer year
years
= $ , ,× ≅
$ .
200 000 000 1057 736 103 189