Baseline Electricity Analysis

Homework

  1. Name three advantages of baseline analysis.
  1. Why were electricity costs regulated?
  1. Under deregulation, what must be separated and why?
  1. What is ‘real-time’ billing?
  1. In non-real rate structures, what are the four components of most electricity bills?
  1. Why do utilities typically charge for the peak rate of electricity use?
  1. What is a seasonal demand charge?
  1. Why do utilities typically charge for low power factor?
  1. Who owns the transformer in a primary rate?
  1. If the electrical demand is 1,600 kW and the power factor is 0.88, calculate the reactive power (kVAr) and supplied power (kVA).
  1. If the supplied power is 1,000 kVA and electrical demand is 930 kW, calculate power factor and reactive power (kVAr).
  1. 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

  1. 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

  1. 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-monthfor first 4,000 kVA:

$14 /kVA-month for all additional kVA

  1. Calculate the annual cost savings if a customer purchases the transformer and switches from a secondary to primary rate ifE = 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
  1. 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)

  1. 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

  1. 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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