CHAPTER XVII VARIABLE LOAD
1. A central station is supplying energy to a community through two sub-stations. One substation feeds four disturbing circuits; the other six. The maximum daily recorded demands are: POWER STATION Substation A Feeder 1 Feeder 2 Feeder 3 Feeder 4 Substation B Feeder 1 Feeder 2 Feeder 3 Feeder 4 Feeder 5 Feeder 6
12000 kW 6,000 kW 1,700 kW 1,800 kW 2,800 kW 600 kW 9,000kW 620 kW 1,500 kW 1,000 kW 2,900 kW 2,200 kW 3,000 kW
Calculate the diversity factor between substations, between feeders on substation A and between substations B. c. 1.28, 1.19, 1.32 a. 1.25, 1.15, 1.25 b. 1.36, 1.21, 1.35 d. 1.32, 1.17, 1.35 Solution:
The diversity factor between substations:
Thus;
The diversity factor between feeders on substation A:
Thus;
The diversity factor between feeders on substation B:
Thus:
2. A 75 MW power plant has an average load of 35,000 kW and a load factor of 65%. Find the reserve over peak. c. 25.38 MW a. 21.15 MW b. 23.41 MW d. 18.75 MW solution:
Solving for peak load:
Then;
3. The annual peak load on a 15,000 kW power plant is 10,500 kW. Two substations are supplied by this plant. Annual energy dispatched through substation A is 27,500,000 kW-hr wit h a peak load at 8,900 kW , 16,500,000 are sent through substation B with a peak load at 6,650 kW. Neglect line losses. Find the diversity factor between substations and capacity factor of the power plant. a. 1.48, 0.446 c. 1.75, 0.335 b. 1.48, 0.335 d. 1.75, 0.446 solution:
Thus;
⁄ ⁄ ( ⁄) Thus;
4. What is the daily average load in a certain power plant if the daily energy produced is 500,000 kw-hrs. a. 28.28 MW c. 19.61 MW b. 30.26 MW d. 20.83 MW Solution:
thus:
5. A distributing transformer supplies a group of general power customers having a connected load of 186 kW. Demand factor and diversity factor are 0.75 and 1.5 respectively. If the load factor for the group wil average 45% and the energy sells 3 ½ cents per kW-hr, what will be the monthly income (30-day) from energy delivered through this transformer? Assume average motor efficiency is 75% c. P 1,501.61 a. P 1,406.16 b. P 2,812.32 d. P 3,003.22
solution:
Solving for kW-hrs deliverd 30 days in one month:
With 75% motor efficiency, the max demand on disturbing transformer:
kW-hrs delivered 30 days in one month:
Then; Monthly income from energy delivered:
Thus;
6. What is the annual capacity factor of the plant if the annual energy produced in a 150 MW power plant is 500,000,000 kW-hrs? c. 56.785% a. 38.05% b. 44.04% d. 34.44% solution:
Thus;
7. A power plant is said to have /had a use factor of 48.5% and capacity factor of 42.4%. how many hrs. did it operate during the year? a. 6,600.32 hrs c. 8,600.32 hrs d. 5,658.23 hrs b. 7,658.23 hrs solution:
DERIVED FORMULA:
Thus;
8. A central station has an annual factor as follows: load factor 58.5%, capacity factor 40.9%, use factor 45.2%. the reversed carried over and above the peak load is 8,900 kW. Find the no. of hrs per year not in service. c. 783.33 hrs a. 833.37 hrs b. 733.38 hrs d. 873.33 hrs solution:
solving for the no. of hours operation per year:
Where:
Then;
From
Then; thus;
9. A 50, 000 kW steam plant delivers an annual output of 238,000,000 kW-hr with a peak load of 42,860 kW. What is the annual load factor and capacity factor? a. 0.634, 0.534 c. 0.634, 0.543 b. 0.643, 0.534 d. 0.643, 0.534 Solution: Load Factor
=
Solving for the Average Load; Ave. Load
Load Factor
Thus; Load Factor
= = 27,168.94 kW = =
= 0.634
=
Annual Capacity Factor =
Thus;
A nnual C apaci ty Factor
= 0. 543
10. Calculate the use factor of a power plant if the capacity factor is 35 and it operates 8000 hrs during the year? a. 38.325% c. 35.823% b. 33.825% d. 32.538% Solution:
= 8760
No. of hours Operation = 8760 800 Use Factor
= 0.038325
Thus;
Us e Factor = 38.325%
11.. Given a load factor, 0.48, installed capacity, 35,000 kW, reserve over peak, 3,000 kW, hours out of service per year 410. Find the capacity and the use factor. c. 43.89%, 46.04% a. 48.39%, 46.04% b. 43.89%, 44.06% d. 48.39%, 44.06% Solution: Capacity Factor
=
Solving for the annual Energy Produced: = Ave. Load x 8760 = Load Factor x Peak Load x 8760 = 0.48(32,000)(8760) = 133,553,600 kW-hrs Then; Capacity Factor
=
= 0.4389 = 43.89% Use Factor
= =
= 0.4604
Thus;
Us e Factor = 0.4604 or 46.04% C apacity Factor = 0.4389 or 48.39%
12. The system shown in the figure consists in part of transformer serving customers x,y,z. estimate the peak load on the transformer. X - store building with 5 kW lighting, 25 kW small motor power Y - store building with 18 kW lighting, 35 kW small motor power Z - office building with 55 kW lighting, 80 kW large motor. a. 75.08 kW c. 95.84 kW d. 105.76 kW b. 85.07 kW Solution:
Typical Demand Factors : For commercial lighting of stores and offices = 0.70 For general power service: 15 kW to 75 kW = 0.55 Over 75 kW = 0.50 Demand Factor Actual Max. Demand FOR CUSTOMER X: Actual Mac. Demand FOR CUSTOMER Y: Actual Max. Demand
FOR CUSTOMER Z: Actual Max. Demand
=
= Demand Factor x Connected Load = 5 (0.7)+ 25 (0.55) = 17.25 kW
= 18 (0.70) + 35 (0.55) = 31.85 kW
= 55 (0.70) + 80 (0.50) = 78.5 kW
THE TOTAL ACTUAL MAXIMUM DEMAND: Total Actual demand = 17.25 + 31.85 + 78.5 = 127.60 kW For Commercial Lighting and General Power Service: Diversity Factor = 1.5 THE PEAK LOAD ON THE TRANSFORMER: Peak load on the transformer
= =
Thus;
Peak load on the trans former = 85.07 kW
13. A daily load curve which exhibited a 15 minutes peak of 150,000 kW is drawn to scales of 1 cm = 3 hrs and 1 2 cm = 10,000 kW. The area under the curve is measured by a planimeter and found to be 60cm . What is the load factor based on the 15 minute peak? c. 0.55 a. 0.50 b. 0.75 d. 0.65 solution:
Solving for the average load:
Thus;
14. A distribution transformer supplies a group of general power customers having a connected load of 200 kW. If the demand and diversity factors are 0.75 and 1.5 respectively. Calculate the maximum simultaneous demands. a. 150 kW c. 100 kW b. 240 kW d. 200 kW solution:
Solving fort the sum of individual maximum demands or total actual
Thus;
15. If the rated capacity of the system is 50,000 kW and the actual maximum demand is 35,000 kW, what is the utilization factor? a. 0.60 c. 0.50 d. 0.80 b. 0.70 solution:
Thus;
