Tariffs And Energy Management Example

Tariffs And Energy Management

1a) Discuss briefly the factors governing tariff structures.

The tariff structure of energy is governed by the factors:

Demand

The demand level of energy will determine the tariff structure of energy, when the demand is high the tariff is likely to be low and when the demand is low the tariff is likely to be high.

Rate

The rate charge of unit consumption of energy will also determine the structure of energy tariffs; the rate is the unit charge of one unit of energy consumed.

Distribution voltage

The distribution voltage will also determine the structure of energy tariff when the distribution voltage is high then the tariff is likely to be high.

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User application:

The other factor that determines the structure ids the user application, this is whether the consumption is for domestic application or for industrial or commercial purposes.

Accounting methods

The other determining factor is the accounting method used in determining the consumption level of electricity, the most used application of determining the consumption level is the use of electric meters that determine the watt consumption level over a period of time.

Social and economic factors:

The structure of energy tariff will also be determined through considering the social and economic factors in a region, the adoption of a certain structure will take into consideration other factors of a region.

1b) Describe a typical tariff for:-

1. i) domestic consumers

The domestic consumer will pay for the number of units consumed per month, these charges may include a monthly standing charge, for example the rate will be charged per KWH, for

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example a domestic consumer may be required to pay a twenty dollar standing charge and in addition to this pay 40 cents per KWH. The tariff will also change in terms of peak and off peak consumption, peak units will be charged higher than off peak.

1. ii) industrial consumers

An industrial consumer will still pay a standing monthly charge if applicable, he will also be required to pay for the number of units consumed per month, for industrial consumption of energy the consumer will pay less due to bulk consumption than the domestic consumer, peak charges will be higher than the off peak charges and summer tariffs will also be different from the winter charges where winter charges will be higher than the summer tariffs.

2) A factory with a foundry and a machine shop has a maximum demand of 750 kW and a load factor of 0.2. The tariff is 20 Pound/kW of maximum demand + 4 Pence/kWh of energy consumed.

1. a) Calculate the annual cost of electricity.

Load factor = (KWH/ total KWH) X 100

0.2 = (750/ Y)

Y= 3750

Total consumption = 3750

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Cost = 3750 X .04 = 150

Cost for maximum demand 750 X 20 = 15000

Total cost = 15000 +150 = 15150 pounds

1. b) Discuss briefly one way in which the energy cost of the factory may be reduced without reducing the number of kWh consumed.

Because most of tariffs are charged higher during peak hours then the firm may reduce the cost of energy by utilising the off peak hours when the tariffs are lower. Lower energy costs can also be achieved by reducing the load factor of the firm, if the firm achieves a lower load factors then the cost of energy will be lower which can be achieved without reducing the number of KWH consumed.

3a) State one method of raising the overall power factor of an industrial load.

Use of an power corrector either passive or active (PFC)

3b) Discuss briefly the advantages and disadvantages of applying power factor correction at:

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1. i) the main bus bars

Advantage:

– It is a useful method in power factor correction

– The use of a filter can reduce the harmonic current

Disadvantage:

– The rectifier used in this case is non linear and therefore there is energy at the harmonics of the frequency of the input current.

– The above problem also posses a problem to the energy providing company as they cannot rectify this problem, whereby they cannot use capacitors to rectify the harmonic current

1. ii) individual loads

Advantages:

– It is a useful method in power factor correction

– It is easy to adjust the correction level because the correction acts as a variable capacitor

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Disadvantages:

– It is not cost effective as it requires the installation on each load or machine

4) A factory has maximum demand of 400 kW at a lagging power factor such that 600 kVAr of re-active power has to be supplied.

The tariff in operation includes a maximum demand charge of 12 Pound/kVA.

Consider the effect of adding capacitor banks to the system in SIX stages, each drawing 100 kVAr leading. The capital charges on each 100 kVAr bank are 400 Pound per annum.

1. a) Calculate the net saving and overall power factor at each stage.

Power factor = real power (p) watts/ apparent power (s) volt amperes

S2 =P2 + Q2

Where Q is the reactive power measured in VAR

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initial

stage 1

stage 2

stage 3

stage 4

stage 5

stage 6

reactive  power Q

600

500

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Tariffs And Energy Management

400

300

200

100

0

Q2 (QXQ)

360000

250000

160000

90000

40000

10000

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0

P

400000

400000

400000

400000

400000

400000

400000

P2

160,000,000,000

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Tariffs And Energy Management

160,000,000,000

160,000,000,000

160,000,000,000

160,000,000,000

160,000,000,000

160,000,000,000

S2

160,000,360,000

160,000,250,000

160,000,160,000

160,000,090,000

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Tariffs And Energy Management

160,000,040,000

160,000,010,000

160,000,000,000

S

400000.45

400000.3125

400000.2

400000.1125

400000.05

400000.0125

400000

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power  factor

0.999998875

0.999999219

0.9999995

0.999999719

0.999999875

0.999999969

1

power  save

0

0.137499869

0.249999797

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0.337499763

0.39999975

0.437499747

0.449999747

savings

0

1.649998427

2.999997562

4.049997153

4.799997

5.249996965

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5.399996962

per  year savings

0

593.9994338

1079.999122

1457.998975

1727.99892

1889.998907

1943.998906

1. b) Plot a graph of net saving (vertically) against overall power factor (horizontally)

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1. c) Explain what your results and graph show.

From the above graph it is clear that as we increase the number of bank capacitors the reaction energy is reduced until it is zero, the firm makes a save on their cost of energy as shown above.

5) Energy-efficient lamps are miniature fluorescent lamp designed to replace ordinary GLS lamps.

1. a) Discuss briefly the advantages and disadvantages of energy-efficient lamps compared with other sources of illumination.

Advantages of miniature fluorescent

– They consume less energy than GLS lamps

– Longer life span

Disadvantages of miniature fluorescent

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– Costly to install and maintain

– Cannot be used at low voltages

– Do not lender desired colour

Advantages of GLS lamps

– Cheap to install and maintain

– They lender good colour

– Simple to operate

Disadvantages of GLS lamps

– Consume more energy

– Short life span

1. b) Obtain manufacture’s data and compare the total cost of using an energy-efficient lamp with that of an ordinary GLS lamp, assuming illumination is provided for 8000 hours.

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GLS cost in pounds

Florescent cost in pounds

initial cost

2.8

9.3

power consumption watt

100

7

voltage

240

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240

per hour

0.41666667

0.029166667

constant 10

4.16666667

0.291666667

8000 hours

33333.3333

2333.333333

total cost

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33336.1333

2342.633333

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References:

Electricity bulbs (2007) advantages and disadvantages of GLS and florescent, retrieved on 13th

October, available at

http://www.alertelectrical.com/Lamps-And-Tubes/GLS-Lamps/

The E Shops(2007)bulb prices, retrieved on 13th October, available at http://www.ebulbshop.c om/acatalog/Light_Bulbs_Fluorescent_tubes___Strip_lights_9.html

Lewis Evans (2006) Alternating Currents, Victoria University Press, New Zealand

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