Saut, Bonifacio R. BSME-5 5. The fixed element invested in a 100,000-kw power plant is $15,000,000. Find its average annual cost per kw, based on straight line depreciation. Estimated life, 15 yrs; salvage value, 20%; r,5%.
Solution: Annual Depreciation ¿
( $ 15,000,000(1−0.20) ) 15
¿ $ 800,000
$ 800,000 Rate ¿ 100,000 kw
(
)
¿ $ 8 per kw
7. Ten years ago a small steam plant of 2000-kw capacity costing $125 per kw was erected. The life was estimated at 15 years and the salvage value at 5%. At present, abandoning the old plant in favor of a new Diesel plant is being considered. A market has been found for the old engine and boiler equipment at $15,000. The remainder of the old plant can be utilized in the new and is given a valuation of $8500. Depreciation has been figured on the straight line basis. What is the difference between the depreciated book value of the old plant and its sale value? How would the difference be taken care of in the reconstruction?
Capital Cost=$ 125 /kW ( 2000 kW )=$ 250000 Salvage Value=( 0.05 ) 250000=$ 12500 D 10=10
( 25000−12500 )=$ 158 333.3 15
Book Value ,C 10=250000−158333.3=$ 91666.67 Sale Value=15000+85000=235000
Book Value−SaleValue=91666.67−23500=$ 68166.67
8. The load duration curve and fuel characteristic for a 45,000 kw plant are given in Prob. 20, Chapter 2. Other data for the same station are: Initial cost, $135 per kw; estimated life, 20 years; salvage value, 8% of first cost; interest rate, 5%; taxes and insurance, 3 ½ %; labor, 45 men; average salary, $175 per month; maintenance and repairs, $65,000 per annum; oil, waste and supplies, $20,000 per annum; coal, $8.25 per tonne. Find production cost per kw hour.
Solution:
I nitial C ost =$ 135 ( 45000 )=$ 6,075,00 0
Annual depreciation reserve=
6075000(1−0.08) =$ 279,450 20
Interest +insurance=( 0.05+0.035 ) $ 6,075,000=$ 516,375 Labor=$ 175 ( 45 ) ( 12 )=$ 94,500 Maintenance+ repairs=$ 65,000
Oil, waste∧supplies=$ 20,000 Average Cost of coal per yr .=
1tonne $ 8.25 24 hrs 365 days ( 25.39lbhr )( 2000 lb )( 1tonne )( 1 day )( 1 yr ) average
¿ $ 917.4275
Annual production cost=¿ $976,242.4275
Assumption :
Annual load factor=50 Annual production=( 45,000 kW ) ( 0.5 ) ( 8760 hrs )=19,710,000 kWh
Rate=
976,242.4275 =$ 0.0495 /kWh 19,710,000
9. The annual costs expected by a utility system in supplying a certain residential suburb of 45,000 customers are; Fixed element, $345,000; energy element, $180,000; customer element, $300,000; return on investment, $200,000. 17, 050,00 kw hr will be registered on customers’ meters during the year and their maximum demand on the power plant will be of the order of 5500 kw. Diversity factor from Table 2-2.
(a) Form a straight line meter rate. (b) Form a three charge rate, putting ¾ of the profit in the energy element, ¼ in fixed element. (c) Form a room rate in which the customer element is a fixed 4 kw hr per room. (Assume average home, 6 rooms.) Energy element is uniformly distributed.
Solution:
Fixed Element Energy Element Customer Element Profit Element Total
$ 345,000 $ 180,000 $ 300,000 $ 200,000 $ 1,025,000 $ 1,025,000 =$ 0.06 per kw hr 17,050,000 kw hr
a)
Straigh line meter rate=
b)
1 Demand Charge=$ 345,000+ ( $ 200,000 )=$ 395,000 4 5500 kW × 4=22,000 kw Unit demand charge=
$ 395,000 $ 17.95 = 22,000 kw kw
Therefore demand charge per month = $1.15 per kW of maximum demand
3 Energy Charge=$ 180,000+ ( $ 200,000 ) =$ 330,000 4 Unit energy charge=
$ 330,000 =$ 0.0194 per kw hr 17,050,000 kw hr
Service Charge=$ 300,000 300,000 45,000 cust service charge per month= =$ 0.55 12 $
11. The annual costs of operating an electric system are:
Coal, $300,000 Franchise and publicity, $70,000 Station depreciation, $190,000 Station supplies, $30,000 Primary line dep., $190,000 Interest, taxes and insurance: on plant and primary line, $50,000; on secondary lines, $85,000 Secondary lines dep., $120,000 Plant maintenance: Fixed $1000, variable $2000 Secondary lines maintenance, $50,000 Labor: Generation, $105,000; Distribution, $80,000; Accounting, $150,000 Superintendence and management, $50,000 Dividends to stockholders, $350,000 Annual station output 70 x 106 kw hr, 10% energy loss in transmission; peak load, 20,000 kw; diversity, 3.33; 100,000 customers. Compute a three charge rate in which one-half of the dividends are in the service charge, the other half in the demand charge. Solution: Fixed Element Energy Element Customer Element Profit Element Total
$ 481,000 $ 517,000 $ 475,000 $ 350,000 $ 1,823,000
1 Demand Charge=$ 481,000+ ( $ 350,000 ) =$ 656,000 2 20000 kw ×3.33=66600 kw
Unit demand charge=
$ 656,000 $ 9.85 = 66,600 kw kw
Therefore demand charge per month is $0.82 per kW of maximum demand Energy charge=
$ 517,000 =0.82¢ per kw hr 63 ×10 6 kw hr
1 Service Charge=$ 475,000+ ( $ 350,000 ) =$ 650,000 2 650,000 100,000 cust service charge per month= =$ 0.542 12 $
13. Annual costs in a certain power system are:
For fixed costs: Plant, 1,750,000; primary lines, $600,000; secondary lines, $1,250,000. For operating cost: Plant, $75,000 indirect and $950,000 direct. For distribution system, $500,000. Direct customer expense, $400,000; profit, 8% of fixed cost. Peak load on plant, 45,000 kw; diversity factor, 4; annual plant output, 1.2 x 108 kw hr. Assume 50,000 customers and 20% transmission loss. Find the straight line meter rate. Solution: Fixed Element Energy Element Customer Element Profit Element Total
$ 3,600,000 $ 1,025,000 $ 900,000 $ 288,000 $ 5,813,000
Annual Plant Output =( 1.2× 108 kw hr ) ×0.80=96,000,000 kw hr Straight Line Meter Rate=
$ 5,813,000 =$ 0.061 per kw hr 96,000,000 kw hr
14. Using data of Prob. 13, construct a Doherty rate, putting profit into the proper elements, in
proportion to the investment. Solution: Fixed Element
$ 3,600,000
Energy Element Customer Element Profit Element Total
$ 1,025,000 $ 900,000 $ 288,000 $ 5,813,000
Demand Charge=$ 3,600,000+288,000=$ 3,888,000 45000 kw × 4=180000 kw
Unit demand charge=
$ 3,888,000 $ 21.6 = 180,000 kw kw
Therefore demand charge per month is $ 1.80 per kw of maximum demand Energy charge=
$ 1,025,000 =$ 0.011 per kw hr 96,000,000 kw hr
Service Charge=$ 900,000 900,000 50,000 cust Service charge per month= =$ 1.50 12 $
15. An air preheater installation will cost $12,500. Its life is assumed to be 8 years. Salvage
value is nothing. Annual maintenance and repair is estimated to average $150. Use compound interest at 6% and find the annual cost of the preheater. Solution: 0.06 ( ) Sinking Fund Payment ¿ $ 12,500−0 ( 1+0.06 )8 −1
(
)
¿ $ 1,262.95 Annual Depreciation Reserve Annual Maintenance Annual Cost
1,262.95 150.00 $ 1412.95
17. A 30 mhp condensate pump motor has been burned beyond repair. The plant superintendent has two replacement alternatives. Manufacturer “A” offers to replace the original (which was an “A” motor) for $510. Manufacturer “B” offers a cheaper motor at $400 but can only guarantee 87% efficiency whereas the “A” motor is guaranteed for 89%. The installation operates 25% of the time at full load, and 75% of the time at half load where the two efficiencies become 85% and 84%
respectively. Assume a motor comparison of 5 years, interest rate 8%, equal maintenance cost. Electric energy is charged for at the rate of 1 ½
¢
per kw hr.
a. Which motor is the more economical buy? b. At what energy cost do they become equal alternatives? i= 8% n=5 years Motor A
Motor B
0.08 0.08 Annual depreciation reserve=510 =$Annual 86.93 depreciation reserve=400 =$ 68.18 5 ( 1+0.08 ) −1 (1+ 0.08 )5−1
(
Rate=
Annual Cost Annual prduction
0.015=
)
(
Rate=
)
Annual Cost Annual prduction
86.93 68.18 0.015= Annual production( ( 0.89∗0.25 ) +(0.85∗0.75)) Annual production ( ( 0.25∗0.87 ) + ( 0.84∗0.75 ) )
Annual production=6738.76 kWh
Annual production=5363.23 kWh
Motor A is more economical. 18. Make a comparative analysis of the production cost per kw hr of the two plants for which data are given. Annual production= 1 x 10^6 kw hr Diesel Plant Engine and generator $53,000 Swbd. And wiring $5,600 Miscellaneous $8,000 Building $10,500 Steam Plant Labor, per month $350 Turbogen. And condenser $24,000 Fixed charges Boiler and stoker 11% $20,000 Oil, per liter (0.621 Swbd.g/ml) and wiring $5,600 4.1 ¢ Miscellaneous $10,000 Fuel economy 0.49 lb/kw hr Building $12,000 Labor, per month $450 Fixed charges 12% Coal, per ton $3.5 19. Using Fuel economy 1.72 lb/kw hr data of the sample in Sec 3-10, form a block meter rate with the fixed element and one-half the customer element collected in the first 12 kw hr per month per customer. In the second block, reduce the rate to ¾ of the first block for the collection of the energy element and remainder of the customer element. Assume that ¾ of the customers
will average full consumption of the second block. Collect the profit element in the third and final block.
21. The load duration of a group of residential customers served by a substation is given by the following data: Percent of the year Kilowatts
0 95
10 50
20 40
30 30
40 29
50 29
60 28
70 28
80 22
90 12
100 8
Average efficiency of distribution is 95%. Customer’s rate is 8 ¢ per kw-hr for the first
25 kw-hr, 5¢ per kw hr for the next kw hr, 3¢ per kw hr for the next 50 kw hr, and 2¢ per kw hr for all remaining energy. What is the average monthly gross income from this group of customers if 20 of them average taking 300 kw hr per month, 50 of them take 130 kw hr per month, while the remainder average 70 kw hr per month each? How many customers are there in the whole group?
22. A customer’s meter reads 29,543 kw hr on May 1, and 29,598 kw hr on June 1. Find the
amount of his electric bill for May based on the following rates. a. 7¢ per kw hr b. 10¢ per kw hr for the first 35 kw hr; 5¢ per kw hr for the next 25 kw hr; 3¢ per kw hr for all in excess of 60 kw hr.
Solution: 29,598−29543=55 kw
a)
7¢ × 55 kw hr=385 ¢ kw hr
b) 10 ¢ × 35 kw hr=350 ¢ kw hr 5¢ × 20 kw hr=100 ¢ kw hr Total=450 ¢=$ 4.50
23. Assume that a customers’ maximum monthly demand was recorded as 120kw. His energy
consumption for the same period was 40,500 kw hr. His rate is: $2.40 per month per kw for the first 50 kw of maximum demand; $2.00 per month per kw for the excess of maximum demand over 50 kw; plus 5¢ per kw hr for the first 1000 kw hr per month; 3¢ per kw hr for the next 4000 kw hr per month; and 2¢ per kw hr for all energy in excess of 5000 kw hr. What is his bill for the month considered? Solution: $ 2.40 ×50 kw=$ 120 kw $ 2.00 ×70 kw=$ 170 kw 5¢ × 1000 kw hr=5000 ¢=$ 50 kw hr 3¢ × 4000 kw hr =12000¢=$ 120 kw hr 2¢ × 40000 kw hr =80000 ¢=$ 800 kw hr Total=$ 1230
24. A customer having a 7-room house used 55 kw hr during a certain month. What is his
electric bill for that month if his rate is that given as the example of “Room rate charge,” Sec 3-9, and there is a 5% discount for payment in 15 days? Solution: In one room ¿
55 kw =7.857 kw 7
10 ¢ × 3 kw hr ×7 rooms=210 ¢ kw hr 7¢ × 3 kw hr ×7 rooms=147 ¢ kw hr 4¢ × 1.857 kw hr ×7 rooms=51.9 ¢ kw hr Total=408.9 ¢
26. The rate for a commercial customer is $6.00 per kw per month for the first kw of maximum
demand, plus $5.00 per kw per month for the next 6 kw of maximum demand, plus $4.00 per kw per month for all of the maximum demand in excess of 7 kw, plus energy charge as follows: First 100 kw hr at 4¢ per kw hr. All remaining energy at 1¢ per kw hr. What type of rate is this? How much is the customer’s bill in a month when he registers 15-kw maximum demand and consumes 1850 kw hr? Solution: $ 6.00 ×1 kw=$ 6 kw $ 5.00 × 6 kw=$ 30 kw $ 4.00 ×8 kw=$ 32 kw 4¢ × 100 kw hr=400 ¢=$ 4 kw hr 1¢ × 1750 kw hr=1750 ¢=$ 17.5 kw hr Total=$ 89.5
