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APA
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Literature & Language
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English (U.S.)
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Topic:
Mechanical Engineering (Coursework Sample)
Instructions:
Baseline Electricity Analysis
Homework
1. Name three advantages of baseline analysis.
2. Why were electricity costs regulated?
3. Under deregulation, what must be separated and why?
4. What is ‘real-time' billing?
5. In non-real rate structures, what are the four components of most electricity bills?
6. Why do utilities typically charge for the peak rate of electricity use?
7. What is a seasonal demand charge?
8. Why do utilities typically charge for low power factor?
9. Who owns the transformer in a primary rate?
10. If the electrical demand is 1,600 kW and the power factor is 0.88, calculate the reactive power (kVAr) and supplied power (kVA).
11. If the supplied power is 1,000 kVA and electrical demand is 930 kW, calculate power factor and reactive power (kVAr).
12. If the electrical demand is 1,000 kW and electrical energy use is 250,000 kWh, calculate the adjustment to the avoided cost of demand for the following energy use block structure:
Energy: $0.04 /kWh for first 200 kWh/kW
$0.03 /kWh for next 100 kWh/kW
$0.02 /kWh for all additional kWh
13. Use the following rate structure to calculate the monthly service charge, energy charge, demand charge, power factor charge and total charge for a plant if E = 600,000 kWh, D = 900 kW, PF = 0.85. Determine the fraction of the total cost associated with each charge.
Service: $100 / month
Energy: $0.04 /kWh for first 200 kWh/kW
$0.03 /kWh for next 100 kWh/kW
$0.02 /kWh for all additional kWh
Demand: $12 /kW-month
Power factor: If PF < 0.90, additional demand charge of: P (kW) (0.90 – PF) / PF
14. Use the following rate structure to calculate the monthly service charge, energy charge, demand charge and total charge for a plant if E = 500,000 kWh, D = 1,000 kW, PF = 0.92. What would be the demand charge if the power factor was 1.00?
Service: $100 /month
Energy: $0.03 /kWh for first 250 kWh/kVA
$0.01 /kWh for all additional kWh
Demand: $18 /kVA-month for first 4,000 kVA:
$14 /kVA-month for all additional kVA
15. Calculate the annual cost savings if a customer purchases the transformer and switches from a secondary to primary rate if E = 500,000 kWh/month, D = 1,200 kW. If the transformer could be purchased for $20,000, determine the simple payback.
Primary Rate
Service: $95 /month
Demand: $13.80 /kW-month
Energy: $0.021 /kWh
Secondary Rate
Service: $16 /month
Demand: $14.10 /kW-month
Energy: $0.030 /kWh for first 125,000 kWh
$0.025 /kWh for over 125,000 kWh
16. Plant demand during each of three shifts per day and the on-peak and off-peak demand periods are shown below. Using the following rate structure, determine annual demand cost savings if: a) 500 kW is moved from first shift to third shift, and b) if 1,200 kW is moved from first shift to third shift.
Demand: $14 /kW-month
Greatest of: 100% of on-peak (weekdays: 7 am to 9 pm)
75% of off-peak (all other times)
17. Determine the annual cost savings from consolidating the following two electrical services into a single service with one meter if E1 = 500,000 kWh/month, E2 = 400,000 kWh/month and the typical demand profiles are shown in the graphs below.
Service: $16 /month
Demand: $14.10 /kW-month
Energy: $0.030 /kWh for first 125,000 kWh,
$0.025 /kWh for over 125,000 kWh
18. A plant's demand and power factor are 800 kW and 0.80 respectively for six months per year, and 700 kW and 0.85 respectively for the other six months per year. If demand costs $18 /kVA-month, determine the quantity of capacitance to maximize savings without over-correcting for power factor, and the annual savings from adding this quantity of capacitance.
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Content:
Base Line Analysis
By:
Institution:
Base line analysis homework
1 Importance of baseline analysis include:
* It aids in focusing efforts on only the most important areas.
* It helps in defining potential energy savings.
* The analysis helps in determining the accurate avoided energy costs used for calculating cost saving.
2 Electricity costs are regulated so as to ensure;
* Reliability of service with adequate supply of electricity
* Financial transparency and security
* Reasonable pricing
* Market is safeguarded against manipulation.
3 Historically electrical energy supply is regulated, and the government in many states is the sole provider of this energy. However, in the past few years this supply process is being opened up allowing competition from other utility providers. This gives the customers the power to choose whom they purchase this service from. Deregulation is important and offers the following advantages. Competition thus price protection, more services and products are made available, price stabilization and gives the customers freedom of choice.
4 Real time billing is being used by utility companies and their customers for real time payment of their services. The utility companies reduce their customer's cost by minimizing the time between electricity consumption and payment.
5 Components of electricity bills in non-real structures are:
* An energy charge
* A demand charge
* A service charge and,
* Power factor charge.
6 On load peak terms, there are three methods of pricing based on the time of use. Off-peak: this is when the electricity demand is lowest. The prices at this time are slightly lower. Mid-peak: this is when the electricity energy demand is moderate. These periods are generally during the day but not including the busiest times of the day. Prices are fixed moderately at this periods. On-peak: this is the period when the electricity demand is at its peak, highest. This includes the busiest times of the day and when households and most industries use electric power. Generally prices are directly proportional to demand.
7 Electricity cannot be stored unless, in batteries which are insufficient, utility energy generating companies continuously vary the amount of electricity produced to meet fluctuating demand in different seasons. Thus a utility company must purchase and produce enough energy to meet peak demand. They must maintain adequate generating, transmission and distribution to meet the load demand including a safety allowance especially on peak periods.
8 Low power factor is inefficient and expensive. Utility companies charge large amounts of cash to industrial customers who do not regulate the power factor to be as close to unity. Low power factor increases system losses by increasing current flows and affecting voltage drops thus reducing electrical system distribution capacity.
9 In general service primary rate, the customer owns the transformer. The transformer has fixed services with power factor charges. Due to the variable fuel cost the energy charge varies slightly on each month.
10
Demand = 1600kW,
Power factor = 0.88
Solution
KVAr = kW*
Hence,
kVAr = 1600 * tan 28.357
= 863.5 kVAR
Apparent Power = 1000 kVA
Demand = 930 kW
Solution
kVAr=
= 367.559 kVAr
=
Power factor
= 0.93
Demand = 1000 kW
Energy use = 250000kWh
Solution
Avoided Cost on Demand (ACD) is given by
ACD = B1 (kWh / kW) * (E1 $ / kWh – E2 $ / kWh) + B2 (kWh / kW) * (E2 $ / kWh – E3 $/ kWh)
ACD = 200 kWh / kW * ($ 0.04 / kWh - $ 0.03 / kWh) + 100 kWh / kW * ($ 0.03 / kWh - $ 0.02 / kWh)
= 2 + 0.06
= $ 2.06 / kW.
Energy = 600,000 kWh
Demand = 900 kW
Power factor = 0.85
Solution
Energy charge,
$ 0.04 / kWh for the first 200 kWh / kW * 900 kW
= 18000 kWh
$ 0.03 / kWh for the first 100 kWh / kW * 900
= 9000 kWh.
Demand charge,
Additional demand charge, demand penalty
= 52.94 kW
Power factor charge,
Power factor charge = demand penalty * $ 12 / kW – month
= 52.94 kW * $ 12 / kW
= $ 635.28 / month
Energy = 500,000 kWh
Demand = 1,200 kW
Power factor = 0.92
Solution.
Service charge is $ 100 / month
Energy charge,
$ 0.03 / kWh for the first 250 kWh / kW * 1000 kW
= 250000 kWh
$ 0.01 / kWh for all additional kWh.
Demand charge,
Calculating the demand penalty,
1000 * kW
= 86.9565 kW
Power factor charge = demand penalty * $ 100 / month
= 86.9565 * $ 100 / month
= $ 8695.65 / month
Solution
Primary rate,
Service = $ 95 * 12 = $ 1140 per year
Demand charges,
$ 13.80 / kW - month * 12000 kW * 12
= $198,720
Energy charges,
$ 0.021 / kW * 500,000 kWh / month * 12
= $ 126, 000
Total charges on primary rate under consideration is;
$ (1140 + 198,720 + 126,000)
= $ 325,860
On secondary rate;
Services = $ 16 * 12
= $ 192
Demand charges,
$ 14.14 / kW – month * 1200 kW * 12
= $ 203, 616
Energy charges,
[($ 0.03 / kWh * 125,000 kWh) + ($ 0.025 / kWh * 325,000)] * 12
= $ 11,875 * 12
= $ 142,500
Total charges on secondary rate,
$ (192 + 203,616 + 142,500)
= $ 346,308
Annual cost saving,
Primary rate + $ 20,000 (purchasing cost)
$ 325,860 + 20,000
= $ 345,860
Annual savings = 346,308 – 34...
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