# Name three advantages of baseline analysis

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```							                      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

1
\$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                      Secondary Rate
Service: \$95 /month               Service: \$16 /month
Demand: \$13.80 /kW-month          Demand: \$14.10 /kW-month
Energy: \$0.021 /kWh               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)

2
on-peak                       off-peak
3,000

Demand (kW)
2,000

1,000

0
AM

PM

:0 M
PM

AM
11 0 P
0

0

0

0
:0

:0

:0

:0
07

03

09

07
Time

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
900

800

700
Electrical Demand (kW)

600

500

400

300

200

100

0
1             2    3    4   5   6   7   8   9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
Hour of day

Meter 1   Meter 2

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.

3

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