Factor in power system math for EEE job preparation

The important factor in power system math includes load factor, demand factor, diversity factor, plant capacity factor, plant use factor, plant utilization factor, etc mathematical problem, and solution.

To know about the Theory related to these factors in the power system, visit this page at a glance, Click Here. This will help you to understand the basic terms and laws in detail.

Objective:

In this study, we learn about this type of factor in the power system mathematical solution in order to compete for the EEE job preparation in Bangladesh.

Contents in this page:

  • Basic laws and terms
  • Explanation of EEE job math solution
  • Suggestions
  • Reference books

Basic laws and terms for Factor in power system math

  1. Demand Factor=\frac{Maximum Demand}{Connected Load}
  2. Load \ Factor = \frac{Average \ Load}{Maximum \ Demand}
  3. Diversity \ Factor = \frac{Sum \ of \ individual \ maximum \ demand}{Maximum \ demand \ on \ power \ station}
  4. Plant \ Capacity \ Factor = \frac{actual \ energy \ produced}{maximum \ energy \ that \ could \ have \ been \ produced} = \frac{average \ demand}{plant \ capacity} 
  5. Annual \ plant \ capacity \ factor = \frac{annual \ kWh \ output}{plant \ capacity \times 8760}
  6. Plant capacity = reserve capacity + maximum demand
  7. Plant \ use \ factor = \frac{station \ output \ in \ kWh}{plant \ capacity \times hours \ of \ use}
  8. Units generated per annum = average load in kW × hours in a year
  9. Plant \ Utilization \ factor = \frac{Maximum \ demand}{plant \ capacity}

Explanation of EEE job math solutions

Q1. [TGTDCL-18(BUET) & PGCBL-17]

A power station has a maximum demand of 15,000kw. The annual load factor is 50% and the plant capacity factor is 40%. Find the Reserve capacity of the plant.

Solution:

We know that,

 Load \ Factor = \frac{Average \ Load}{Maximum \ Demand}

So Average load = (0.5 × 15,000) KW = 7500 KW

And

Plant \ Capacity \ Factor = \frac{actual \ energy \ produced}{maximum \ energy \ that \ could \ have \ been \ produced} = \frac{average \ demand}{plant \ capacity}

 

 

So Plant capacity = (7500 ÷ 0.4) KW = 18750 KW

Reserve capacity = plant capacity – maximum demand

= (18750 – 15000) KW

= 3750 KW (Answer)


 

Q2. [PGCB-18(BUET)]

The load duration curve in a power plant varies from 4GW to 12GW. Find the load factor. If transmission loss is 3% and reserve power is 25%, find the plant capacity.

Solution:

We see that the load varies from 4GW to 12GW which means the maximum demand is 12GW and so 

the average load= 12GW – 4GW is = 8GW.

Load \ Factor = \frac{Average \ Load}{Maximum \ Demand}

= 8 GW ÷ 12 GW

= 0.6667

= 66.67% (Answer)

Now the transmission loss 3% means by losing 3% of the sending end power of the plant it became 12GW receiving end power. Reserve power 25% means 20% of sending end power.

Maximum sending end Power = 12GW + (12GW × 3%) = 12GW + 0.36GW = 12.36 GW and

reserve Power = 12.36GW × 20% = 12.36× 0.2 = 2.472 GW

Plant capacity is the total of sending power and reserve power.

Now plant capacity = (12.36 GW + 2.472 GW) = 14.84 GW (Answer)


Factor in power system math for EEE job preparation

Q3. [RPGCL-17(BUET)]

A generating station has a maximum demand of 25 MW, a load factor of 60%, a plant capacity factor of 50%, and a plant use factor of 72%.

Find

(1) The Reserve capacity of the plant

(2) The daily energy produced and

(3) The maximum energy that could be produced daily if the plant while running as per schedule, were fully loaded.

Solution:

Load \ Factor = \frac{Average \ Load}{Maximum \ Demand}

So Average demand = (0.6 × 25) MW = 15 MW

Plant \ Capacity \ Factor = \frac{actual \ energy \ produced}{maximum \ energy \ that \ could \ have \ been \ produced} = \frac{average \ demand}{plant \ capacity}

So plant capacity = (15 ÷ 0.5) MW = 30 MW

  1. Reserve capacity of the plant = plant capacity – maximum demand = (30 – 25) MW = 5 MW (Answer)
  2. Daily energy produced = average demand × 24 hours= (15 MW × 24 h) = 360 MWh (Answer)
  3. we know that plant use factor = actual energy produced in a day divided by maximum energy that could be produced. So maximum energy that could be produced = (360 ÷ 0.72) = 500 MWh per day (Answer)

Q4. [NWPGCL-17 (BUET)]

A thermal station has the following data:

Maximum demand = 2000 KW, load factor = 40%,

Boiler Efficiency= 85%, turbine efficiency= 90%,

Coal consumption= 0.9 Kg/KWh, cost of 1 ton of coal = Rs300.

Determine (1) Thermal efficiency and (2) Coal bill per annum.

Solution:

(1) Thermal efficiency = ηboiler × ηturbine = 0.85 × 0.9 = 0.765 = 76.5%

(2) We know that,

Units generated per annum = average load × hours in a year

= load factor × maximum demand × hours in a year

= (0.4 × 20000 × 8760) KWh

= 7008 ×104 KWh

Coal consumption per annum = (0.9 × 7008 ×104) kg

= 63072 tons. [ 1 Ton = 1000 kg]

So coal bill per annum = Rs300 × 63072 = Rs1892160


 

Q5. [Dhaka WASA-17]

At the end of a power distribution system a certain feeder supplies three distribution transformers each supplying a group of customers whose connected loads are as under:

Transformer No 1:  Load 10 KW, demand factor 0.65 and diversity of group 1.5

Transformer No 2:  Load 12 KW, demand factor 0.6 and diversity of group 3.5

Transformer No 3: Load 15 KW, demand factor 0.7 and diversity of group 1.5

If the diversity factor among the Transformers is 1.3. Find the maximum load on the feeder.

Solution:

For Transformer No 1,

Sum of the maximum demand = connected load × demand factor = 10 Kw × 0.65 = 6.5 Kw

Diversity \ Factor = \frac{Sum \ of \ individual \ maximum \ demand}{Maximum \ demand \ on \ power \ station}

So maximum demand on Transformer 1 = 6.5 Kw ÷ 1.5 = 4.33 KW

Maximum demand on Transformer 2 = (12 × 0.6) ÷ 3.5 = 2.057 KW

Maximum demand on Transformer 3 = (15 × 0.7) ÷ 1.5 = 7 KW

As the diversity factor among Transformers is 1.3,

Maximum demand on feeder = (4.3 + 2.057 + 7) ÷ 1.3 = 10.3 KW


 

Q6. [NPPCBL-17]

A generating station is to supply four regions of load whose peak loads are 10 MW, 5 MW, 8 MW, and 7 MW. The diversity factor at the station is 1.5 and the average annual load factor is 60%. Calculate: (1) the maximum demand on the station and (2) annual energy supplied by the station.

Solution:

  • Maximum demand on station = Sum of maximum demands of the regions ÷ Diversity factor

= (10 + 5 + 8 + 7) MW ÷ 1.5

= 20 MW

  • Units Generated per any = Maximum demand × Load factor × Hours in a year

= (20 × 103 × 0.6 × 8760) Wh

= 105.12 ×106 KWh


 

Q7. [PGCL-17]

A power plant has a maximum demand of 1500 KW, load factor 70%, plant capacity, and plant use factor 54.5%. Find (1) The energy produced daily (2) The reserve capacity of the plant and (3) the maximum energy that could have been produced daily, if the plant runs all the time.

Solution:

  • Average demand = L.F. × Maximum demand = (0.7 × 1500) KW = 1050 KW

Energy produced daily = average demand × 24 hours

= (1050 × 24) kWh

= 25200 KWh

  • Plant capacity factor = average demand ÷ plant capacity

∴ Plant capacity = 1050 ÷ 0.545 = 1926.61 KW

Reserve capacity = plant capacity – maximum demand

= (1926.61 – 1500) KW

= 426.61 KW

  • Plant use factor is equal to actual energy produced in a day divided by maximum energy that could be produced

So maximum energy that could be produced in a day = 25200 ÷ 0.545 = 46.2 KWh


 

Q8. [EGCBL-17]

The daily based load in a system is 3500 MW. The load changes to 4000 MW at 6:00 p.m. only and persists till 11:00 p.m. find the load factor.

Solution:

Load factor = average load ÷ maximum demand

The maximum load 4000 MW which is varying from 6.00 p.m. to 11.00 p.m. that is only 5 hours a day. The load in this 5 hours becomes (4000 KW × 5) and the rest (24-5)=19 hours become (3500 KW × 19). So the average load must be the total loads in KW divided by Hours in a day.

\fn_jvn =\frac{(3500 \times 19) + (4000 \times 5)}{24 \times 4000}

= 90.1 %


{The important factor in power system math for EEE job preparation in Bangladesh.}

Q9. [CBPCL-16]

A customer has a maximum demand of 200 KW at a load factor of 40%. If the tariff is Rs100 per KW of maximum demand plus 10 paise per KWh, find the overall cost per kWh.

Solution:

Units consumed per year = maximum demand load factor × hours in a year = (200 × 0.4 × 8760) = 700800 kWh

Annual charges = annual maximum demand charges + annual energy charges

= Rs[(100 × 200) + (0.1 × 700800)] [ As 10 paise = 0.1 Rs]

= Rs90080

Overall cost per KWh = Rs(90080 ÷ 700800)

= Rs0.1285

= 12.85 paise


 

Q10. [APSCL-16]

A power station of plant capacity 100 MW, average load 90 MW, and maximum load 95 MW find the load factor, plant capacity factor, and utility plant utilization factor.

Solution:

Load factor = average load ÷ maximum demand

= 90 MW ÷ 95 MW = 94.74%

Plant capacity factor = average load ÷ plant capacity

= 90 MW ÷ 100 MW = 90%

Plant utilization factor = maximum demand ÷ plant capacity

= 95 MW ÷ 100 MW = 95%


 

Q11. [BOF-16]

Find the diversity factor; a power station has the following data:

Lighting load: 150 KW for 10:00 a.m. to 7:00 p.m.

Residential load: 50 KW for 7:00 p.m. to 11:00 p.m.

Pumping load: 55 KW for 3:00 p.m. to 10:00 p.m.

Solution:

We know that,

Diversity factor = sum of individual maximum demand ÷ maximum demand of the circuit at the time of system place

= (150 + 50 + 55) Kw ÷ 150 Kw

= 1.7


The important factor in power system math for EEE job preparation in Bangladesh.

Analyze Yourself:

Suggestions + Reference Books

V.K. Mehta- Power System Book- Chapter-3;

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Factor in power system math for EEE job preparation

This Post Has One Comment

  1. Rehman Zaman

    It was very much useful!

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