52
Electric Power Distribution Engineering
53
Load Characteristics
Example 2.7 3500
The average load factor of a substation is 0.65. Determine the average loss factor of its feeders, if the substation services
) 3000 W k ( s d n a m e d m2000 u m i x a m y l h t 1141 n o 1000 M
a. An urban area b. A rural area Solution a. For the urban area, FLS = 0.3FLD + 0.7(F LD) 2
65) + 0. 7(0. 6 5) = 0.3(0 . 65
2
= 0.49
d a o l k a e p l a u n n A
Unsold energy
Annual avg. load
b. For the rural area, FLS = 0.16FLD + 0.84(F LD) 2
0
( . 65 65) + 0. 840 ( . 65 65) = 0.160
r t r r r e l y y r y r y h i l u s b e o b e b e b e g u a u a a r c A p r M a J u n J u m m a F e b r M A u p t e m O c t o v e e c e J a n D N S e
2
= 0.53
Time (months)
FIGURE 2.11 A monthly load curve.
Example 2.8 or, from Equation 2.8,
Assume that the Riverside distribution substation of the NL&NP Company supplying Ghost Town, which is a small city, city, experiences an annual peak load of 3500 kW. The total annual energy supplied to the primary feeder circuits is 10,000,000 kWh. The peak demand occurs in July or August and is due to air-conditioning air-conditioning load.
Annualload factor =
a. Find the annual average power demand. b. Find the annual load factor.
=
Solution Assume a monthly load curve as shown in Figure 2.11.
The unsold energy, as shown in Figure 2.11, 2.11, is a measure of capacity and investment cost. Ideally, it should be kept at a minimum.
Totalannualenergy Year
107 kWh/year 8760 h/year
Example 2.9 Use the data given in Example 2.8 and suppose that a new load of 100 kW with 100% annual load factor is to be supplied from the Riverside substation. The investment cost, or capacity cost, of the power system upstream, that is, toward the generator, from this substation is $18.00/kW per month. Assume that the energy delivered to these primary feeders costs the supplier supplier,, that is, NL&NP,, $0.06/kWh. NL&NP
1141 kW kW
b. From Equation 2.6, the annual load factor is F LD LD
107 kW Wh/year h/year 3500 kW × 8760 3500
= 0.326
a. The annual average power demand is Annual P av
Totalannual energ energyy Annualpeakload Annu alpeakload× 876 8760 0
Annualaverageload Annual peak demand
1141kW 3500 350 0 kW
a. Find nd the new annual load factor on the substation. b. Find Find the total annual cost to NL&NP to serve this load. Solution Figure 2.12 shows the new load curve after the addition of the new load of 100 kW with 100% load.
0.326
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Electric Power Distribution Engineering
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Load Characteristics
Example 2.10 3600
) 3000 W k ( d n a m e d m2000 u m i x a m y l h t 1241 n o M1000
Assume that the annual peak-load input to a primary feeder is 2000 kW. A computer program that calculates voltage drops and I2R losses shows that the total copper loss at the time of peak W. The total annual energy supplied to the sending end of the feeder is load is I 2R = 100 k W. 5.61 × 106 kWh.
∑
New load curve
d a o l k a e p l a u n n a w e N
Old load curve
a. By using Equation 2.40, determine the annual loss factor. b. Calculate Calculate the total annual copper loss loss energy and and its its value value at $0.06/kWh. Solution a. From Equation 2.40, the annual loss factor is FLS = 0.3FLD + 0.7F L2D
New annual average load
where
0
r r e l y e r e r r y r y h r i l y u s t m b e o b e u a u a a r c A p M a J u n J u u g b o b a F e b r M A e p t e e O c t o v e m e c e m J a n S D N
F LD =
Time (months)
5.61×106 kWh 2000kW ×8760h/year
= 0.32
FIGURE 2.12 The new load curve after the new load addition.
Therefore,
a. The new annual load factor on the substation is F LD LD = =
2 F LS LS = 0.3 × 0.3 2 + 0.7 × 0. 3 2
Annualaverageload Annual peak demand
≅
b. From Equation 2.25,
1141+ 100 3500 +100
345 345 = 0.3
F LS
b. The total annual and additional additional cost to NL&NP to serve the additional additional 100 kW load has two cost components, namely, (1) annual capacity cost and (2) annual energy cost. Therefore,
Average power loss Power loss at peak load
or
Annual Ann ual addi additio tional nal cap capaci acity ty cos costt = $18/kW /kW/mo /month nth × 12 mon month/ th/yea yearr × 100 kW
Averag Ave ragee pow power er los losss = 0.1677 × 100 kW
, = $21600 and
01677
. = 1677
kW
Therefore,
Annua Ann uall en ener ergy gy co cost st = 100 kW × 8,76 8,76 0 h/year h/year × $0.06/ 6/kW kWh h Total annual coppe copperr loss = 16.77 kW ×8760 h/year
= $52, 560
= 146, 905
kWh
Therefore, Total annual additi additional onal costs costs = Annual capacit capacityy cost +Annual eenergy nergy cost = $2 1, 600 + $ 52, 560 = $74,160
and Costt of tot Cos total al ann annual ual cop copper per los losss = 146,9 05 kWh kWh × $0.06/k 6/kWh Wh 814 = $8, 814
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Electric Power Distribution Engineering
57
Load Characteristics
Example 2 .11
Thus, the diversi�ed reactive power ( Q) is
Assume that one of the DTs of the Riverside substation supplies three primary feeders. The 30 min annual maximum demands per feeder are listed in the following table, together with the power factor (PF) at the time of annual peak load.
∑
3
P i × tanθ
i =1
F D
1800 × tan 18.2° + 2000 × tan 31. 79 ° + 2200 × tan 25. 84 ° = 1.15
Demand kW
PF
1
1800
0.95
2
2000
0.85
3
2200
0.90
Feeder
Q=
.8 kvar = 2518
Therefore, D g = (P 2 + Q 2)1 / 2 = S
Assume a diversity factor of 1.15 among the three feeders for both real power ( P ) and reactive power (Q).
2
= (5 21 7 + 2 51 8.8
a. Calculate the 30 min annual maximum demand on the substation transformer in kilowatts and in kilovoltamperes. b. Find the load diversity in kilowatts. c. Select a suitable substation transformer size if zero load growth is expected and if company policy permits as much as 25% short-time overloads on the distribution substation transformers. Among the standard three-phase (3 ϕ) transformer sizes available are the following:
2 1 / 2
)
= 5 79 3. 6 0 kV A
b. From Equation 2.17, the load diversity is 3
LD =
∑D − D i
g
i =1
= 6000 − 5217 = 783kW
2500/3125 kVA self-cooled/forced-air-cooled 3750/4687 kVA self-cooled/forced-air-cooled 5000/6250 kVA self-cooled/forced-air-cooled 7500/9375 kVA self-cooled/forced-air-cooled
c. From the given transformer list, it is appropriate to choose the transformer with the 3750/4687-kVA rating since with the 25% short-time overload, it has a capacity of 4687 × 1.25 = 5858.8 kVA
d. Now assume that the substation load will increase at a constant percentage rate per year and will double in 10 years. If the 7500/9375 kVA-rated transformer is installed, in how many years will it be loaded to its fans-on rating?
which is larger than the maximum demand of 5793.60 kVA as found in part (a). d. Note that the term fans-on rating means the forced-air-cooled rating. To �nd the increase ( g ) per year,
Solution a. From Equation 2.10,
(1+ g )10 = 2 F D =
1800 + 2000 + 2200 D g
= 1.15
hence, 1 + g = 1.07175
Therefore,
or D g
6000 1.15
5217 kW
g = 7.175%/year
P
Thus,
To �nd power in kilovoltamperes, �nd the PF angles. Therefore, (1.07175)n × 5793. 60 = 9375 kVA PF1 = cos θ1 = 0.95 → θ 1 = 18.2° PF2 = cos θ2 = 0.85 → θ 2 = 31.79 ° PF3 = cos θ 3 = 0.90 → θ 3 = 25.84 °
or (1.07175)n
1. 6182
