Unit5ppts

POWER PLANT ENGINEERING
UNIT 5
ECONOMICS , SELECTION CRITERIAS
& ENVIRONMENTAL ISSUES
S.PALANIVEL ASSOCIATE PROF./MECH. ENGG
KAMARAJ COLLEGE OF ENGG. & TECH. VIRUDHUNAGAR(NEAR)

LOAD CURVE
TIME - hours
L
O
A
D
In
M
w

• BASE LOAD : LOAD BELOW WHICH DEMAND
NEVER FALLS. LOAD REQUIRED TO BE
SUPPLIED 100 %
• PEAK LOAD : MAX. LOAD SUPPLIED – IT MAY
BE 15 – 20 % OF TIME
• INTERMEDIATE LOAD : LOAD SUPPLIED
BETWEEN BASE AND PEAK.

LOAD CURVE & LOAD DURATION CURVE

Load duration curve

700
• RATED CAPACITY OF UNIT : Installed Capacity
• PEAK LOAD : MAXIMUM LOAD PRODUCED
• AVERAGE LOAD : GENERATED POWER /HOURS
• INSTALLED CAPCITY = 1000 MW
• TOTAL PLANNED SHUT DOWN = 720 Hrs.
• FORCED OUTAGE = 50 Hrs.
• UNIT IN SERVICE = 8760-720-50 =7990 Hrs.
0-6 6-8 8-9 9-17 18-20 20-22 22-24
600 700 800 900 750 700 650
Daily loading pattern of a Power Station (in MWs)

• MAX LOAD = 900 MW
• MINI LOAD = 600 MW
• AVERAGE LOAD = SUM OF LOAD/TOTAL HOURS
= 6X600+2X700+800+8X900+3X750+2X700+2X650
24
= 17950/24 =747.91 MW
LOAD FACTOR = AVERAGE LOAD/PEAK LOAD
= 747.91/900 = 0.831
CAPACITY FACTOR = AVERAGE LOAD/RATED CAPACITY
= 747.91/1000 = 0.74791
RESERVE FACTOR =LOAD FACTOR/CAPACITY FACTOR
=0.831/0.74791= 1.111
USE FACTOR = AV.LOADX8760/INST.CAP.X HOURS RUN
= 747.91X8760/1000X7990 =0. , Say 0.819S.PALANIVEL ASSOCIATE PROF./MECH. ENGG

DEMAND & DIVERSITY FACTORs
IF LIGHTING LOAD = 35 KW , MACHIENRIES LOAD = 450 KW,
MISC. LOAD = 15 KW are connected
200 Nos of 60 W BULB IN SERVICE FOR 16 Hours. 5 nos. of
M/c of 1500 W and 10 nos. of M/c OF 750 W IN SERVICE for
2 Shifts, 7 hours/Shift.
CONNECTED LOAD : SUM OF RATED CAPACITY OF ALL
EQUIPMENTS AND OUTLETS ON CONSUMER’S CIRCUIT
CONNECTED LOAD =35+ 450+15=500 KW
L.L = 16X60X 200 =19200 W = 19.2 KW
M.L.= 2X7X5X1500+ 2X7X10X750=210000 W = 210 KW
MISC.L= 5 KW
DEMAND FACTOR = ACTUAL MAX. CONSUMPTION
TOTAL CONNECTED LOAD
= 19.2 + 210 +5/500 =234.2/500 =0.4684

DIVERSITY FACTOR
= TOTAL LOAD CONNECTED/MAX. CONSUMPTION
DIV. FACTOR (SYSTEM)= 500/234.2 = 2.1349
DIV. FACTOR (L.L.) = 35/19.2 =1.8229
DIV. FACTOR (M.L.) = 450/210 = 2.1428
DIV. FACTOR (Misc. L.) = 15/5 = 3

• HEAT RATE =Heat Input (I)/Net kW output(L)
• Plant Net Efficiency = Net output(L)/Heat Input (I)
• INCREMENTAL HEAT RATE :Amount of energy
needed to generate additional output at any given
load
• When plant output decreases, heat rate increases,
where as the incremental heat rate decreases.

COST/PER K.W
• DETERMINED BY
• FIXED COST + O & M COST+ FUEL COST
• COST IS BASED ON POWER EXPORTED YEARLY
• FIXED COST : INTEREST, DEPRECIATION, INSURANCE,
TAXES ON CAPITAL INVESTED.
• CAPITAL COST = CONSTRUCTION COST+ LAND COST
• CONSTRUCTION COST = DESIGN+ PRICE OF EQPTS+
TRANSPORT+ ERECTION + COMMISSIONING

Total Annual Cost ( Ct )
Cc = Construction cost = 1430 Cr.
I = INTEREST RATE = 12 % = 0.12
D= DEPRECIATION = 100/15 =6.67 % = 0.0667
T= TAXES = 14 % = 0.14
W = TOTAL ANNUAL WAGES = 126 Cr.
M = MTCE. CHARGES = 30 Cr.
Cf = Fuel Cost = Cost of fuel including its trans.
Ct= I + D + T x Cc +(W+M)+ Cf
100

• Fuel cost : Total fuel required (T) x Cost/T
• Total fuel reqd.= Total Power generated/Fuel Kg/Kwh
• C.F.= 85% Installed = 420 MW
• C.F.= Average load/Installed capacity
• Ave.load = 0.85x420 =357 MW
• Total power generated =0.85x420x103
x8760
=312.732X 107
KWH
Fuel reqd./KWh = 1 Kg. (Net Heat rate = 1Kg/1KWh)
•

Annual Fuel reqd.=312.732X10 7
Kg.=3127320T
Fuel Cost =3127320X1600 =Rs. 5003712000 =500.03712X10 7
Total cost = (0.12+0.067+0.14)x1450x107
+(120+30) 107
+
fuel cost
= 474.15 X 107
+150 X 107
+500.3712 X 107
= 1124.5212 Cr.
Cost/Unit = Total Cost/Units Generated = 1124.5212/312.732
=3.5958
Cost of Generation = Rs. 3.60/kW
Profit margin = 43 p/unit : Selling price = 3.6 +0.43 =4.03
Auxiliary consumed = 8 %:
Power exported = (1-0.08)x units generated
=0.92x312.732 x107
Units: Revenue=4.03x0.92x312.732 Cr.
= 1159.485163 Cr.

Determine the generating cost per unit of 80 MW power station with the
following data
Capital cost = Rs. 160 x 107
; Annual cost of fuel = Rs. 32 x 106
;
Annual wages and taxes = Rs. 36 x 106
; Interest and depreciation = 10 %
of capital cost : Annual load factor = 45 %
Assume, Maximum load = 80 MW
Load factor = 0.45 = Average load /Maximum load
Average load = 0.45 x 80 = 36 MW
Total energy generated = Average load x 8760 = 36 x 8760 =
315.360 x106
kWH (units)
Generation cost = Fixed cost + Running Cost
Fixed cost = % of (interest + depreciation ) on Capital cost
Fixed cost = 0.1 x 160 x 107
= Rs.160 x 106

Running cost = Annual fuel cost + Annual wages and taxes
= 32 x 106
+ 36 x 106
= 68 x 106
Generation Cost = 160 x 106
+ 68 x106
= Rs. 228 x 106
Cost of generation of 1 unit
= Total generation cost/Total energy generated (units)
= 228 x106
/315.360 x106
= Rs.0. 7229 = 72.3 paise.

Find the cost of generation per kWh from the following data :
Capacity of the plant = 150 MW ; capital cost = Rs. 20,000 per kW
installed. Interest and depreciation = 10 % on capital ;
Fuel consumption = 1.2 kg/kWh; Fuel cost = Rs. 400 per tone; salaries,
wages, repair and maintenance=Rs.150x106
per year; Max. demand =
120 MW ; load factor = 50 %
Load factor = Average load/max. demand = 0.5
Average load = 0.5 x 120 MW = 60 MW
Total energy produced=Average loadx8760 = 60 x103
x 8760
=525.6 x 106
kWh(units)
Capital cost = 20000 x 150 x 103
=Rs. 300 x 107
Total cost of generation = Fixed cost + Running cost
Fixed cost = % (Interest + depreciation) x Capital cost
= 0.1 x 300 x 107
= Rs. 300 x 106

Fuel consumption / unit = 1.2 kg/kWh
Fuel requirement for annual generation = 1.2 x 525. 6 x 106
= 630.72x106
kg. = 630.72 x103
T.
Annual fuel cost = 630.72 x 103
x 400 = Rs. 252.288 x 106
Cost on salaries, wages, repair and maintenance = Rs. 150 x 106
Running cost =Rs. 252.288 x 106
+ Rs. 150 x 106
= Rs.402.288 x 106
Total generation cost = Fixed cost + Running cost
= Rs.300x106
+ Rs. 402.288 x 106
= Rs. 702.288 x 106
Cost of generation/unit
= Total generation cost/Total energy produced
= Rs. 702.288 x 106
/ 525.6 x 106
= Rs. 1.34

DEPRECIATION
• STRAIGHT LINE METHOD :
• DEPRECIATION OCCURS UNIFORMLY
• MONEY SAVED NEGLECTS ANY INTEREST
• IF ‘A’ = CAPITAL , G= SALVAGE VALUE,
N= YEARS OF USEFUL LIFE
THEN DPRECIATION “D”=(A-G)/N

DEPRECIATION
• SINKING FUND METHOD :
• SUM OF MONEY IS SET ASIDE EVERY YEAR & INVESTED TO
EARN COMPOUND INTEREST
• IF “P” IS SUM INVESTED EVERY YEAR FOR ‘N’ YEARS TO
HAVE DPRECIATION AMOUNT ‘S’ AND THE RATE OF
INTEREST IS ‘I’
• VALUE OF MONEY AT END OF 1 YEAR = P
• VALUE AT THE END OF 2 YEAR = (P+Pi) + P
• VALUE AT 3RD
YEAR = (P+Pi)Xi +Pi+P
• VALUE AT N-1 YEAR = P(1+i) (N-1-1)
• VALUE AT N YEAR = P(1+i) (N-1)
• SUM OF INVESTMENT in N YEARS = P+P(1+i)….+P (1+i) (N-1)
• S = P (1+i) (N)
-1) = A – G =A’ : P= A’ i
i (1+i) (N)
-1)

PRESENT WORTH
• IFSUMOF“P”ISINVESTEDNOW,
• ITSVALUEWILLGETDECREASE
• GEOMETRICALLYOVERTHEYEARS
• THERATEOFDECREASEIS“i”
• PRESENTWORTHIN“N”YEARS=P/(1+i)
N
• IFP=110:I=10%
• P.W.AFTER1YEAR=110/(1+.1)=110/1.1=100

Reliability of Power Plant
Total no. of hours a power plant is available for power generation is an
important factor in designing a power Plant.
This factor is called as availability factor or reliability factor.
No. of hours a plant is not in service due to stoppage of units due to
various causes while the unit in service are called as Forced outage
hours. Forced outage hours has to be below 2 % for efficient
operation. Forced outage is a measure of Reliabilty of Unit
In a year a power plant has to be stopped to undertake annual
maintenance works , which may vary to 30 days to 25 days.
In addition to these Plant may not be in service due to stoppage of
units on some causes while in operation , that is called forced outages.
Ratio total no. of hours units in service in a year to Total no. of hours
after deducting annual maintenance hours is availability factor.
Availability factor of a power plant is expected to maintain above 90
% to operate economical.

• As forced outages increase the operation and maintenance
costs, proven reliability of equipment and regular preventive
maintenance are essential not only to guarantee electric
supply but also for overall economical operation of plant
• Costs of power generation can be reduced by
 Selecting equipment of longer life and proper capacities
 Running the Plant at high load factor
 Increasing the efficiency of power plant
 Carrying out proper maintenance of equipments to avoid
breakdowns and forced outages
 Keeping proper supervision and trained manpower so as to
keep lesser break downs and efficient operation
 Using a plant of simple and proven design

SELECTION OF CAPACITY
TOTAL CAPACITY REQUIRED BASED ON
MAXIMUM DEMAND EXPECTED
GROWTH OF DEMAND ANTICIPATED
RESERVE CAPACITY REQUIRED

SIZE OF UNITS
• VARIATION OF LOAD DURING 24 HOURS
• TOTAL UNITS CONNECTED TO GRID
• MINI. STRAT-UP & SHUTDOWN PERIOD
• MANITEANCE PLANNED
• PLANT EFFICIENCY Vs SIZE
• COST/MW, SPACE /KW Vs SIZE

LOCATION OF THERMAL POWER PLANT
• AVAILABLITY OF FUEL
• AVAILABLITY OF
COOLING WATER
• LAND COST
• CHARACTER OF SOIL
• DISPOSAL OF ASH
• WIND DIRECTION &
WATER CURRENT
• RAIL & ROAD
CONNECTION
• ACCOMODATION FOR
STAFF
• ZONE -FREE FROM
EARTH QUAKE
• SECURITY
CONSIDERATIONS.

HYDRO ELEC. POWER PLANT
• HYDRO : BASED ON AVAILABLITY OF WATER
AND UTILISABLE WATER HEAD
• INITIAL COST WILL BE VERY HIGH
• RUNNING COST WILL BE LESS
• TAKES MORE TIME TO CONSTRUCT
• DIFFICULT TO CONVINCE PEOPLE
• POWER CAN BE PRODUCED BY TAKING LESS TIME
• MOSTLY MULTI PURPOSE LIKE … IRRIGATION,
FLOOD CONTROL , POWER GENERATION,
PUMPING STATION.

Problems
• 1. A Power Plant has the following annual factors :
Load factor = 0.75, Capacity factor= 0.6, Use Factor
=0.65. Max. demand is 60 MW.
• Estimate a) The Annual energy produced,
• b) the reserve capacity and C) hours of operation and
hours during which plant was not in service
Load factor = Average Load/Max. Load
0.75 = Average Load/60 MW
Average load = 60 x 0.75 = 45 MW
a) Annual Energy produced = Average load x 24 x 365
= 45x24x365 = 394200 MWH =394.2 x106
kWh

Capacity factor = Aver. Load/Installed capacity
0.6 = 45/installed capacity
Installed capacity = 45/0.6 = 75 MW
b) Reserve capacity = Installed – max. Load
= 75 – 60 = 15 MW
Use factor = Energy Produced/Max. energy possible
0.65 = 394.2 x 106
/75 x running hours
c)Running hours = 394.2x 106
/75x0.65 = 8086.154 hrs.
Unit not in service = 8760-8086.154 = 673.846hrs.

2. A power station of 30 MW capacity has the
annual max. demand of 25 MW. It supplies
loads having max. demands of 10 MW, 8.5
MW,5MW and 4.5 MW. The annual load
factor is 45%. Estimate a) Average load,
b)Energy supplied in a year, c)the diversity
factor and
d) Demand factor.

• Peak Load = 25 MW
• Connected load = 10+ 8.5+5+ 4.5 = 28 MW
• Load factor = Aver.load/Peak load
• 0.45= Aver.Load/Peak load
• Aver.load=peak loadx0.45=25x 0.45=11.25MW
Energy produced=11.25x8760=98.550x 106
units
Demand factor = Peak load /Connected load
= 25/28 =0.8928
Diversity factor = Connected load/Peak load
= 28/25 = 1.12

• A central power station has annual load factor, capacity factor and use
factor as 60 %, 40 % and 45 % respectively. The maximum load is 15
MW. Estimate (a) the annual energy production and revenue earned if
the cost of energy is Rs. 1 /kWh, (b) the reserve capacity over and above
the peak load and (c) the hours per year the station is not in service.
Load factor = 0.6 ; Capacity factor = 0.4 ; use factor = 0.45 ; peak load =
15 MW
Load factor = Average load/Peak load
Average load = load factor x peak load = 0.6 x 15 = 9 MW
Total power generated in a year = Average load x 8760 hours
= 9 x 8760 = 78840 MWH
=78.84 x106
kWh
Cost of energy /unit = Rs.1
Total revenue earned = Total power generated x cost/unit
= 78.84 x106
x1 = Rs. 78.84 x106

Capacity factor = 0.4 = Average load/Installed capacity
= 9/installed capacity
Installed capacity = 9/0.4 = 22.5 MW
Reserve capacity = Installed capacity – Maximum load
= 22.5 – 15 = 7.5 MW
Use factor = Total power generated/ ( Installed capacity x Total
running hours)
0.45 = 78.84 x106
/22.5 x total running hours
Total running hours = 78.84 x106
/22.5 x0.45 = 7786 hours
Hours per year not in service = 24 x 365 – running hours
= 8760- 7786 = 974 hours.

A power station of 30 MW capacity has the maximum annual demand of 25
MW. It supplies loads having maximum demands of 10 MW, 8.5 MW, 5
MW and 4.5 MW. The annual load factor is 45 %. Estimate a) the average
load b) the energy supplied per year c) diversity factor and d) demand
factor.
Given Installed capacity = 30 MW, Maximum demand = 25 MW,
Demands connected loads in MW = 10, 8.5, 5, 4.5; Annual load factor = 0.45
Load factor = 0.45 = Average Load/Maximum load = 0.45
= Average Load/25
Average Load = 25 x 0.45 = 11.25 MW
Energy supplied in a year = Average load x 8760 = 11.25 x 8760 =98.55x106
kWH.
Load connected = 10 + 8.5+ 5 + 4.5 = 28 MW
Demand factor = Maximum load/ Connected load = 25/28 =0.89
Diversity factor = Connected load/Maximum load = 28/25 = 1.12

3. A Power Station has a maximum demand of
10,000 kW and the daily load on the Station
is as follows :
Time kW Time kW
6 a.m to 8 a.m. 3500 5 p.m. to 7 p.m. 8500
8 a.m. to 12 noon 8000 7 p.m. to 9 p.m. 10000
12 noon to 1 p.m. 3000 9 p.m. to 11 p.m. 4500
1 p.m. to 5 p.m. 7500 11 p.m. to 6 a.m. 2000
a) Draw the load curve and load duration curve
b) Choose the size and no. of generating Units
c) Draw up the operating schedule of the Units
d) What reserve plant would be necessary
e) Calculate Load factor, Plant capacity factor and Use factor of
the station.

Load (kW) - duration
10000 - 2 hours
8500 - 2 hours
8000 - 4 hours
7500 - 4 hours
4500 - 2 hours
3500 - 2 hours
3000 - 1 hr.
2000 - 7 hrs.
Load requirements and its duration can be had from load
Duration curve
Max. load required = 10000 kW
Minimum load required = 2000 kW
Reserve capacity assumed = 2500 kW

4 units of 2500 kW for running and 2500 kW as stand
by , so 5 nos. of 2500 kW are to be selected.
Operating schedule :
1 unit of 2500 kW run for – 24 hours
2 units of 2500 kW run for – 17 hours
3 units of 2500 kW run for - 12 hours
4 units of 2500 kW run for - 8 hours
Instead of 2500 kW unit if 2000 kW is selected,
then no. of units required including reserve
of 2000 kW = 10000+2000/2000 = 6

Then operating schedule will be
1 unit of 2000 kW run for 24 hours
2 units of 2000 kW run for 17 hours
5 units of 2000 kW run for 4 hours.
Load factor = Average load/Max load
Average load = (10x2+8.5x2+8x4+7.5x4+4.5x2+
3.5x2+3x1+2x7)MWH/24 = 132/24 = 5.5 MWh
Load factor = 5.5/10 = 0.55

• Capacity factor = Average load /Installed Cap.
• Capacity factor = 5.5/12.5 (or) 5.5/12
=0.44 (or) 0.45833
Use factor = Energy produced/Max. energy possible
= 5.5 x24/12.5 x24 (or) 5.5x24/12 x24
=044 (or) 0.45833

4.Determine the Generating cost per unit of 80 MW
Power Station with the following data:
• Capital Cost = Rs.160 Cr.
• Annual fuel Cost = Rs.3.2 Cr.
• Annual wages = Rs. 3.6 cr.
• Interest and Depreciation = 10 %
• Annual load factor = 45 %
Assume capacity = Peak load = 80 MW
Average load = peak load x load factor = 80x0.45
= 36 MW
Energy produced in a year = 36x24x365 = 315.36x106
units

Generation cost = total cost/Total energy produced
Cost = Fixed cost + Running cost
Fixed Cost = (Int.+ Depreci.) x Constn. Cost
= 160x 107
x 0.1 =Rs. 160 x 106
Running Cost = Annual Fuel cost + Annual Mtce cost =
3.2 + 3.6 = 6.8 Cr.
= Rs. 68 x 106
Total cost = Rs. (160 + 68 ) x 106
Generation Cost = 228 x 106
/ 315.36 x 106
= Rs. 0.7229 = 72.29 paise

5. A Plant costing Rs. 65,000 has a useful life of
15 Years. Find the amount which should be
saved annually to replace the equipment at
the end of time
a) by the straight-line method
b) by the sinking fund method , if the annual
rate of compound interest is 5%. Assume
salvage value of the equipment is Rs.5000

Plant cost = A = Rs. 65000
Salvage Cost = G = Rs. 5000
No. of years = N = 15
Depreciation by straight line method :
Depreciation charge per year = (A-G)/N
= (65000 – 5000)/15
= 60000/15 = Rs.4000
Depreciation by sinking fund method :
•A’ = Capital cost – salvage value = 65000 – 5000 = Rs. 60000
•Rate of compound interest = I = 5 %
•No. of years = N = 15
•Amount to be kept side per year = P
•P = (A’ x i) /{(1+i)N
- 1} = (60000 x 0.05)/ {(1+.05)15
– 1 }
•P=3000/ (1.0515
– 1) = 3000/(2.07893 -1) = 3000/1.07893
= Rs. 2780 S.PALANIVEL ASSOCIATE PROF./MECH. ENGG

The incremental fuel costs for two generating units A and B of s power
plant are given by the following relations;
dFA /dPA = 0.06 PA + 11. 4
dFB /dPB = 0.07 PB + 10
where P is the power in MW and F is the fuel cost in rupees per hour.
Find the economic loading of the two units when the total load to be
supplied by the power station is 150 MW.
Find the increase in fuel cost per hour if the load is equally shared by the
two units.
Given dFA /dPA = 0.06 PA + 11. 4
dFB /dPB = 0.07 PB + 10
Total load = PA + PB = 150 MW
PA = 150 – PB

The condition for economic loading demands equal
incremental heat rate of the two units which leads to
dFA /dPA = dFB /dPB
0.06 PA + 11. 4 = 0.07 PB + 10
0.06 PA - 0.07 PB = 10 – 11.4 = - 1.4
0.06 (150- PB ) – 0.07 PB = - 1.4
9 -0.06 PB - 0.07 PB = - 1.4
-0.13 PB = - 10.4
0.13 PB = 10.4
PB = 10.4/0.13 = 80 MW : PA = 150 -80 = 70 MW
If load is equally shared by units ie. PA = PB = 150/2 = 75 MW

Then increase in fuel cost of unit A is given by
75
ƪ70 dFA x dPA = [ 0.06 PA
2
/2 + 11.4 PA ]75
70
dPA
= {( 0.06 x 752
)/2 + 11.4 x 75)} – { (0.06 x 702
)/2 + 11.4 x 70)}
=Rs. 78.75
Then increase in fuel cost of unit B is given by
75
ƪ80 dFB x dPB = [ 0.07 PB
2
/2 + 10 PB ]75
80
dPB
= {( 0.07 x 752
)/2 + 10 x 75)} – { (0.76 x 802
)/2 + 10 x 80)}
= Rs.- 77.125
Net increase in fuel cost = 78.75 – 77.125 = Rs. 1.63

The loads on a power plant with respect to time for 24 hours are given
below
Time (hours) :0-6 : 6-8: 8-12:12-14:14-18:18-22:22-24
Load (MW) : 40: 50 : 60 : 50: 70 :80 : 40
Draw the load curve and find out the load factor of the power station. If the
loads above 60 MW are taken by unit of 20 MW capacity, find out the
load factor and the use factor of the standby unit.
Total load in 24 hours = 6x 40 + 2x50+ 4x 60 +2x 50 + 4x 70 + 4x
80 + 2 x40 = 1360 MWH
Average load = 1360/24 =56.67 MW
Load factor of station = Average load/peak load = 56.67/80 =
0.70833=0.71
Loads taken by standby unit (load above 60 MW) are 70-60 =
10 MW between 14 and 18 hours ie for 4 hours, 80-60 = 20
MW between 18 and 22 hours ie for 4 hours.S.PALANIVEL ASSOCIATE PROF./MECH. ENGG

Total Loads taken by standby unit = 4 x 10 + 4 x 20 =
120 MWH
Average load = 120/8 = 15 MW
Load factor = average load/peak load = 15/20 = 0.75
Capacity use factor = Total power generated/
(Installed capacity x hours of operation)
=120/(20x8) = 120/160 =0.75

A central power station has annual load factor, capacity factor and use
factor as 60 %, 40 % and 45 % respectively. The maximum load is 15 MW.
Estimate (a) the annual energy production and revenue earned if the cost
of energy is Rs. 1 /kWh, (b) the reserve capacity over and above the peak
load and (c) the hours per year the station is not in service.
Load factor = 0.6 ; Capacity factor = 0.4 ; use factor = 0.45 ;
peak load = 15 MW
Load factor = Average load/Peak load
Average load = load factor x peak load = 0.6 x 15 = 9 MW
Total power generated in a year = Average load x 8760 hours
= 9 x 8760 = 78840 MWH =78.84 x106
kWh
Cost of energy /unit = Rs.1
Total revenue earned = Total power generated x cost/unit
= 78.84 x106
x1 = Rs. 78.84 x106
S.PALANIVEL ASSOCIATE PROF./MECH. ENGG KAMARAJ COLLEGE
OF ENGG. & TECH. VIRUDHUNAGAR(NEAR)

Capacity factor = 0.4 = Average load/Installed capacity =
9/installed capacity
Installed capacity = 9/0.4 = 22.5 MW
Reserve capacity = Installed capacity – Maximum load
= 22.5 – 15 = 7.5 MW
Use factor = Total power generated/ ( Installed capacity x Total
running hours)
0.45 = 78.84 x106
/22.5 x total running hours
Total running hours = 78.84 x106
/22.5 x0.45 = 7786 hours
Hours per year not in service = 24 x 365 – running hours
= 8760- 7786 = 974 hours.

A power station is said to have a use factor of 47 % and capacity factor of
40 %. For how many hours in a year was the power station not in service.
Given use factor = 0.47 ; Capacity factor = 0.4
Use factor = kWH generated in a year / (Installed capacity x
operating hours)
Capacity factor = kWH generated in a year/(Installed capacity x
8760 hours)
0.4/0.47 = (kWH generated in a year/Installed capacityx8760) /
(kWH generated in a year/Installed capacity x running hours )
0.4/0.47 = Operating Hours /8760
Operating Hours = 0.4 x 8760 /0.47 = 7455 hours
Hours not in service = 8760 – operating hours = 8760- 7455
= 1305 hours.

The following data pertain to a power plant ;
Installed capacity = 200 MW; Capital cost = Rs. 4000 x 106
; Annual cost of
fuel, taxes and salaries =Rs. 600 x106
; Rate of interest = 5 % of the
capital; Rate of depreciation = 6 % of the capital ; Annual load factor =
65 %; Capacity factor = 55 % energy used in running the plant auxiliaries
= 4 % of total units generated. Determine a) the reserve capacity and b)
the cost of power generation
Load factor = Average load/peak load = 0.65
Capacity factor = Peak load / Installed capacity = 0.55
Peak load = 0.55 x Installed capacity = 0.55 x 200 = 110 MW.
Reserve capacity = Installed capacity – Peak load
= 200 -110 = 90 MW
Load factor = 0.65 = Average load / peak load
Average load = 0.65 x 110 MW = 71.5 MW

Energy generated = Average load x 8760 hrs. = 626.34x 106
kWH
Power generated used in auxiliaries = 4 %
Total units exported = 0.96 x 626.34 x106
units
= 601.286 x 106
units.
Total generation cost = Fixed Cost + Running Cost
Fixed Cost = % of (Rate of Interest + depreciation) x Capital cost
= % (5 + 6) x 4000x106
= 0.11 x 4000 x 106
= Rs. 440 x 106
Running cost = 600 x 106
Total generation cost = 440 x 106
+ 600 x 106
= Rs. 1040 x 106
Considering total energy produced
Cost of generation = Total generation cost / Energy generated
= 1040 x 106
/626.34 x 106
= Rs. 1.66
Considering auxiliaries consumption total energy exported
Cost of generation = 1040 x 106
/601.286 x 106
= Rs. 1.73S.PALANIVEL ASSOCIATE PROF./MECH. ENGG

• A power plant has the following annual factors;
• Load factor = 70%; capacity factor = 50 % ; use factor = 60 %; max.
demand = 20 MW Estimate a) annual energy production; b) reserve
capacity over and above the peak load and c) the hours per year during
which plant is not in service.
Load factor = Average load/Max. demand
Average load = Load factor x Max. demand = 0.7 x 20 MW= 14 MW.
Annual energy produced = Average load x 8760 hours = 122.640 x 106
units.
Capacity factor = Max. demand / Installed capacity = 0.5
Installed capacity = Max. demand/ Capacity factor = 20/0.5 = 40 MW
Reserve capacity = Installed capacity – Max. demand = 40 -20 = 20 MW.
Use factor = Total energy produced/ Installed capacity x Running hours =
0.6
Running hours = 14 x 8760/(40 x 0.6) =5110 hours
Hours per year during which plant is not in service = 8760 – 5110 = 3650
hours.

• The connected loads of a consumer consist of 10 electric bulbs of 60 W
each and 2 electric heaters of 1000 W each. His maximum demand is
1500 W. On an average, he uses 8 bulbs for 5 hours a day and one heater
for 3 hours a day. Estimate his average load, monthly energy
consumption and load factor.
Total load in a day = 8 bulbs x 5 hours x 60 W + 1 heater x 3
hours x 1000 W
= 8x 5 x 60 + 1 x 3 x 1000 = 2400 + 3000 = 5400 kWh
Average load = 5400 kWh/24 = 225 kWh
Monthly consumption = 30 days x average load = 30 x 225 kWh
= 6750 kWh
Load factor = Average load/max. demand = 225/1500 = 0. 15

Unit5ppts

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Similar to Unit5ppts

Similar to Unit5ppts (20)

More from PALANIVEL SUBBIAH

More from PALANIVEL SUBBIAH (13)

Recently uploaded

Recently uploaded (20)

Unit5ppts

Editor's Notes