operation of generation plants

8/13/2019 Operation of Generation Plants

1/12

EL 321EL 321

Electrical Power EngineeringElectrical Power Engineering

A. Professor. Professor Sajidajid Ansari ,nsari , P.Eng.Eng. PMP. PMP.

epartment of Electrical Engineeringepartment of Electrical EngineeringUsman Institute of Technologysman Institute of Technology

Barebones of AC Power SystemsBarebones of AC Power Systems

• Recall that AC impedance is a complex quantity made

up of real resistance and imaginary reactance.

AC Power AC Power

• AC Apparent Power is a complex quantity made up of

real active power and imaginary reactive power:

It is the real resistance (P) that leads to the

. It is the imaginary reactance (X) that leads to the

reactive power (gold colour).

The sum of the real power and the reactive power

yields the apparent power (green colour).


2/12

• The Active power is the power that is dissipated in the resistance of

the load. It uses the same formula used for DC:

Real (Active) Power (P)Real (Active) Power (P)Barebones of AC Power SystemsBarebones of AC Power Systems

Imag nary React ve Power Q : Imag nary React ve Power Q : • The reactive power is the power that is exchanged between reactive

components (inductors and capacitors). The formulas look similar to

those used by the active power, but use reactance instead of

resistances.

+Q for Capacitor+Q for Capacitor

--Q for InductorQ for Inductor

•The apparent power is the power that is “appears” to flow to theload. The magnitude of apparent power can be calculated using

similar formulas to those for active or reactive power:

Apparent Power (S) Apparent Power (S)

• The power triangle graphically shows

the relationship between real (P),

reactive (Q) and apparent power (S).

Power Triangle Power Triangle


• The power triangle also shows that we

can find real (P) and reactive (Q)

power, given S and the impedanceangle θ.


3/12

Total Power in AC Circuits Total Power in AC Circuits • The total power real (PT) and reactive

power (QT) is simply the sum of the real

and reactive powers for all of the

individual circuit elements.


• How elements are connected does not

matter for computation of total power.

• Sometimes it is useful to redraw the

circuit to symbolically express the real

and reactive power loads

Power Factor & Reactive Power (Q) ConceptsPower Factor & Reactive Power (Q) Concepts• The ratio between true power and apparent power is called the

power factor for the circuit.

• For the Purely Resistive Circuit, V& I are in Phase i.e. θ =0

between V & I and the power factor is 1 (perfect), because the

reactive power equals zero. Here, the power triangle would

S PCos f p

Power Apparent Power TrueFactor Power

/.

/

==

=

θ


,

power) side would have zero length.

• The Elements (L&C) have energy storage ½ Li2 & ½ CV2.

Since this energy is stored in the first half cycle goes back to

the source next half cycle that is why we called this power as

Reactive Power.

• For the Purely Inductive Circuit, the power factor is zero,

because true power equals zero. Here, the power triangle

would look like a vertical line, pointing up.

• The same could be said for a Purely Capacitive Circuit, the true

power must be equal to zero, making any power in the circuitpurely reactive. Here, the power triangle would look like a

vertical line, pointing down.

• As currentcurrent LAGSLAGS voltage in inductive circuitvoltage in inductive circuit and LEADSLEADS inin

capacitive circuit,capacitive circuit, the two reactive powers will have the

opposite sign. In other words:

Inductor -> Absorbs Reactive Power (-Q)

Capacitor -> Generates Reactive Power (+Q)


4/12

Power Factor Improvement / Correction Power Factor Improvement / Correction • The significance of power factor lies in the fact that utility companies supply

customers with volt-amperes, but bill them for watts. The relationship is (Watts

= volts x amperes x power factor). It is clear that power factors below 1.0

require a utility to generate more than the minimum volt-amperes necessary to

supply the power (watts). This increases generation and transmission costs.


oo power actor s cons ere to e greater t an . or .

• Utilities may impose penalties on customers who do not have good power factors

on their overall buildings.

• Watts, or real power, is what a customer pays for. VARS is the extra “power”

transmitted to compensate for a power factor less than 1.0. The combination of the

two is called "apparent" power (VA or volt-amperes).

• Poor power factor can be corrected, paradoxically, by adding another load to

the circuit drawing an equal and opposite amount of reactive power, to cancel

out the effects of the load's inductive reactance.

• Inductive reactance can only be canceled by capacitive reactance, so we have to

add a capacitor in parallel to our example circuit as the additional load.• The effect of these two opposing reactances in parallel is to bring the circuit's

total impedance equal to its total resistance (to make the impedance phase angle

equal, or at least closer, to zero).

Example 1Example 1

For the circuit shownFor the circuit showna.a. Determine the unknown real (P2) and reactive powers (Q3) in the circuit.Determine the unknown real (P2) and reactive powers (Q3) in the circuit.b.b. Determine total apparent power.Determine total apparent power.


.. ..d.d. Is the unknown element in Load #3 an inductor or capacitor?Is the unknown element in Load #3 an inductor or capacitor?


5/12

Example 2Example 2

A 120V, 60Hz, 1Ø consumer in Toronto hasA 120V, 60Hz, 1Ø consumer in Toronto has (uncorrected) reactive(uncorrected) reactivepower of 119.998 VAR (inductive). Will you please calculate for him thepower of 119.998 VAR (inductive). Will you please calculate for him thecorrect capacitor size to produce the same quantity of (capacitive)correct capacitor size to produce the same quantity of (capacitive)


reac ve power.reac ve power.

Example 3Example 3

A 220V, 50Hz, 1Ø consumer in Karachi has 5kW load @0.8p.f lagging.A 220V, 50Hz, 1Ø consumer in Karachi has 5kW load @0.8p.f lagging.Due to addition of some inductive load,Due to addition of some inductive load, p.fp.f is reduced to 0.7.is reduced to 0.7.

(a)(a) How much extra current the KESC will have to su l due toHow much extra current the KESC will have to su l due to


this inductive load.this inductive load.

(b)(b) To avoid penalty from the KESC the consumer connected aTo avoid penalty from the KESC the consumer connected acapacitor generating 2.5capacitor generating 2.5 kVARkVAR to his load. Calculate the newto his load. Calculate the new p.fp.f ofofthe consumer.the consumer.

Comment on the result and explain each step of calculation.Comment on the result and explain each step of calculation.


6/12

Example 4Example 4 (Assignment # 4)(Assignment # 4)

A 220V, 50Hz, 1Ø consumer has 15kW load @0.7A 220V, 50Hz, 1Ø consumer has 15kW load @0.7 p.fp.f lagging.lagging.(a)(a) Calculate the reactive power of the load.Calculate the reactive power of the load.


.. . ,. ,Comment on the result and explain each step of calculation.Comment on the result and explain each step of calculation.

In power generation or the operation of power plants environment,In power generation or the operation of power plants environment,

specific terms are commonly used. Let us discuss these terms.specific terms are commonly used. Let us discuss these terms.

1 1- - Connected Load Connected Load --

Operation of Generation PlantsOperation of Generation Plants

connected to the power system.connected to the power system.

2 2- - Firm PowerFirm PowerIt is the power intended to be always available even under emergencyIt is the power intended to be always available even under emergency

conditionsconditions (The load that is served, on a guaranteed basis, 100 percent of the time).(The load that is served, on a guaranteed basis, 100 percent of the time).

3 3- - Cold Reserve Cold Reserve

It is that reserve generating capacity available for service but not inIt is that reserve generating capacity available for service but not inoperation.operation.

4 4- - Spinning Reserve Spinning Reserve It is that generating capacity connected to the bus and ready to takeIt is that generating capacity connected to the bus and ready to take

load.load.


7/12

1 1- - Diversity Factor / Simultaneity Factor (Ks) Diversity Factor / Simultaneity Factor (Ks) Diversity Factor = Sum of Individual Max. Demand/Max. Demand on Power Station.Diversity Factor = Sum of Individual Max. Demand/Max. Demand on Power Station.

OR OR Diversity Factor = Installed load /Diversity Factor = Installed load / Running loadRunning load

Factors Affecting the Cost of GenerationFactors Affecting the Cost of Generation

Thus, in this example,Thus, in this example,D.F = 130 kW / 90 kWD.F = 130 kW / 90 kW

TimeTime LoadLoad TotalTotal8 am – 4 pm 60 kW (motor load) 60 kW

. = .. = .

Example 5:Example 5:A subA sub--station has three outgoing feeders:station has three outgoing feeders:feeder 1 has maximum demand 10 MW at 10:00 am,feeder 1 has maximum demand 10 MW at 10:00 am,feeder 2 has maximum demand 12 MW at 7:00 pm andfeeder 2 has maximum demand 12 MW at 7:00 pm andfeeder 3 has maximum demand 15 MW at 9:00 pm,feeder 3 has maximum demand 15 MW at 9:00 pm,While the maximum demand of all three feeders is 33 MW at 8:00 pm. FindWhile the maximum demand of all three feeders is 33 MW at 8:00 pm. FindDiversity Factor?Diversity Factor?

-

6 pm - 10 pm 30 kW (pumping) + 40 kW lighting) 70 kW

10 pm – 8 am 30 kW (pumping) 30 kW

Solution:Solution:Here,Here, thethe sumsum ofof thethe maximummaximum demanddemand ofof thethe individualindividual subsub--systemssystems(feeders)(feeders) isis 1010 ++ 1212 ++ 1515 == 3737 MW,MW, whilewhile thethe systemsystem maximummaximum demanddemand isis3333 MWMW.. TheThe diversitydiversity factorfactor isis 3737//3333 == 11..1212..

Diversity factor / Simultaneity Factor (Ks): Diversity factor / Simultaneity Factor (Ks):

which indicates the maximum demand of the individual subwhich indicates the maximum demand of the individual sub--system occurssystem occurssimultaneously.simultaneously.

Example 6:Example 6: A distribution feeder serves 5 houses, each of which has aA distribution feeder serves 5 houses, each of which has apeak demand of 5 KW. The feeder peak turns out to be 20 kW. Findpeak demand of 5 KW. The feeder peak turns out to be 20 kW. FindDiversity Factor?Diversity Factor?

Solution:Solution:The diversity is 20/25 or 0.8, which is less than 1.The diversity is 20/25 or 0.8, which is less than 1.

This happens due to the timing differences between the individualThis happens due to the timing differences between the individualheating/cooling appliance usages in the individual customers.heating/cooling appliance usages in the individual customers.

As supply availability decreases, the diversity factor will tend to increaseAs supply availability decreases, the diversity factor will tend to increasetoward 1.00. This can be demonstrated when restoring service aftertoward 1.00. This can be demonstrated when restoring service afteroutages (called “cold starts”) as the system initial surge can be muchoutages (called “cold starts”) as the system initial surge can be muchgreater than the historical peak loads.greater than the historical peak loads.


8/12


8585%%

180 MW’s180 MW’s

300300MW’sMW’s

15%15%

INTERMEDIATEINTERMEDIATEINTERMEDIATEINTERMEDIATE

PEAKINGPEAKINGPEAKINGPEAKING

HrHr -- 74467446HrHr -- 13141314

BASEBASEBASEBASE220220MW’sMW’s

2 2- - Load Curves Load Curves

TheThe dailydaily variationsvariations inin loadload fromfrom timetime toto timetime – – hourlyhourly oror half half--hourlyhourly areare givengiven onon whatwhat isis calledcalled ‘LOAD ‘LOAD CURVE’ CURVE’ whichwhich showsshows


graphicallygraphically thethe variationsvariations of of loadload withwith respectrespect of of timetime.. MuchMuchusefuluseful informationinformation cancan bebe obtainedobtained fromfrom loadload curves,curves, ee..gg, ,•• thethe areaarea underunder thethe curvecurve givesgives thethe numbernumber of of unitsunits

generatedgenerated perper dayday..•• TheThe ratioratio of of thethe areaarea underunder thethe curvecurve toto thethe totaltotal areaarea of of

rectanglerectangle inin whichwhich itit isis containedcontained givengiven thethe loadload factorfactor forforthethe dayday..

demanddemand onon thethe stationstation onon thatthat dayday..•• TheThe areaarea underunder thethe curvecurve divideddivided byby thethe numbernumber of of hourshours

givesgives thethe averageaverage demanddemand onon thethe generatinggenerating stationstation..ItIt maymay bebe notednoted thatthat thethe dailydaily loadload curvecurve forfor aa stationstation isis notnotthethe samesame forfor allall thethe daysdays..


9/12

3 3- - Load Factor Load Factor Load Factor = Average load. /Maximum load during a given periodLoad Factor = Average load. /Maximum load during a given period

It can be calculated for a single day, for a month or for a year. Its value is alwaysIt can be calculated for a single day, for a month or for a year. Its value is always

less than one. Because maximum demand is always more than average demand.less than one. Because maximum demand is always more than average demand.Load Factor = Load that a piece of equipment actually draws / Load it could draw (full load)Load Factor = Load that a piece of equipment actually draws / Load it could draw (full load)

Example 7: Example 7: Motor of 20 hp drives a constant 15 hp load whenever it is on.Motor of 20 hp drives a constant 15 hp load whenever it is on.The motor load factor is thenThe motor load factor is then


15/20 = 75%.15/20 = 75%. It is used for determining the overall cost per unit generated. Higher the loadIt is used for determining the overall cost per unit generated. Higher the load

factor, lesser will be the cost per unit.factor, lesser will be the cost per unit. Load factor is term that does not appear on your utility bill, but does affectLoad factor is term that does not appear on your utility bill, but does affect

electricity costs. Load factor indicates how efficiently the customer is using peakelectricity costs. Load factor indicates how efficiently the customer is using peakdemand.demand.

Load Factor = ( energy (kWh per month) ) / (Load Factor = ( energy (kWh per month) ) / ( peak demand (kW) x hours/month )peak demand (kW) x hours/month )

A high load factor means power usage is relatively constant. Low load factor shows that A high load factor means power usage is relatively constant. Low load factor shows thatoccasionally a high demand is set. To service that peak, capacity is sitting idle for long periods,occasionally a high demand is set. To service that peak, capacity is sitting idle for long periods,thereby imposing higher costs on the system. Electrical rates are designed so that customersthereby imposing higher costs on the system. Electrical rates are designed so that customers

. .For ExampleFor ExampleCustomer ACustomer A – – High Load FactorHigh Load Factor

82% load factor = (3000 kWh per month x 100%) / 5 kW x 730 hours/month.82% load factor = (3000 kWh per month x 100%) / 5 kW x 730 hours/month.Customer BCustomer B – – Low Load FactorLow Load Factor41% load factor = (3000 kWh per month x 100%) / 10kW x 730 hours/month.41% load factor = (3000 kWh per month x 100%) / 10kW x 730 hours/month. To encourage efficient use of installed capacity, electricity rates are structuredTo encourage efficient use of installed capacity, electricity rates are structured

so the price per kWh above a certain load factor is lower. The actual structure ofso the price per kWh above a certain load factor is lower. The actual structure ofthe price blocks varies by rate.the price blocks varies by rate.

Factors Affecting the Cost of GenerationFactors Affecting the Cost of GenerationTWO OTHER FACTORS mentioned frequently are as follows :4 4- - PLANT PLANT CAPACITY CAPACITY FACTOR FACTOR = Actual Energy Produced /Maximum Possible Energy That Could Have Been Provided (Based OnInstalled Plant Ca acit

5 5- - PLANT PLANT USE USE FACTOR FACTOR = Actual Energy Produced / PlantCapacity X No. Of Hours The Plant Has Been In Operation

The The cost cost of of energy energy generated generated in in MINIMUM MINIMUM UNDER UNDER THE THE FOLLOWING FOLLOWING condition condition • When the size (and hence the corresponding capital costs ) of

generating plant are kept low. This can be achieve by a good

,• The daily output of each generating unit is large (which means a

good load factor for the station). The reduction in cost of generation with good load factors is due to the fact that overallworking costs become low, the fixed changes having beendistributed over more units of energy generated.


10/1211

Example 8 Example 8 CalculateCalculate the size of a main feeder from substationthe size of a main feeder from substationswitchgear that is supplying five feeders with connected loadsswitchgear that is supplying five feeders with connected loadsof 400, 350, 300, 250 and 200 kilovoltof 400, 350, 300, 250 and 200 kilovolt--amperes (amperes (kVAkVA) with) with

demand factors of 95, 90, 85, 80 and 75 percent respectively.demand factors of 95, 90, 85, 80 and 75 percent respectively.Use a diversity factor of 1.5.Use a diversity factor of 1.5.

o u on o u on CalculateCalculate demand for each feeder:demand for each feeder:

•• 400400 kVAkVA ×× 95% = 38095% = 380 kVAkVA•• 350350 kVAkVA ×× 90% = 31590% = 315 kVAkVA•• 300300 kVAkVA ×× 85% = 25585% = 255 kVAkVA•• 250250 kVAkVA ×× 80% = 20080% = 200 kVAkVA•• 200200 kVAkVA ×× 75% = 15075% = 150 kVAkVA

TheThe sum of the individual demands is e ual to 1 300sum of the individual demands is e ual to 1 300 kVAkVA If the feeder were sized at unity diversity, then 1,300If the feeder were sized at unity diversity, then 1,300 kVAkVA

÷÷ 1.00 = 1,3001.00 = 1,300 kVAkVA

However, using the diversity factor of 1.5, theHowever, using the diversity factor of 1.5, the kVAkVA = 1,300= 1,300kVAkVA ÷÷ 1.5 = 8661.5 = 866 kVAkVA for the feeder.for the feeder.

TransformerTransformer supplying the main feeder plus wiringsupplying the main feeder plus wiringmethods and equipment can be sized from this kilovoltmethods and equipment can be sized from this kilovolt--ampere rating.ampere rating.

ExampleExample 9 9 A A hydro electric station has to operate with a mean headhydro electric station has to operate with a mean headof 30 m and is supplied from a reservoir lake which drains aof 30 m and is supplied from a reservoir lake which drains a

catchments area of 250 sq. kmcatchments area of 250 sq. km – – over which the average rainfall isover which the average rainfall is125 cm per annum. If 70% of the rainfall can be utilized and the125 cm per annum. If 70% of the rainfall can be utilized and theexpected load factor for the station is 80%.expected load factor for the station is 80%. Calculate the power inCalculate the power inkW for which the station should be designed.kW for which the station should be designed. Assume that no head is Assume that no head islost in pipes penstocks etc. The mechanical efficiency of the turbineslost in pipes penstocks etc. The mechanical efficiency of the turbinesis 90% and the efficiency of the generators is 95%.is 90% and the efficiency of the generators is 95%.

Assuming no spares Assuming no spares what should be the type andwhat should be the type and h.ph.p. ratings of two. ratings of twoturbines for the stationturbines for the station..

Solution: Solution: We know that P = (0.736 / 75)P = (0.736 / 75) Qw Qw h h η η kWkWWhere Q = Discharge m3/secw= density of water = 1000 kg/m2 (2)h = Head, m

Water available in m3 = 0.70 x 250 x 106 x 125 = 218 x 108

This quantity is available in the whole year. So thatquantity available per second = 218 x 108 / 8760 x 3600Q= 6.93 m3 (1) Available head = h = 30 m (3)


11/1211

And η= Overall efficiency= ηa x ηt x ηpSuffix a, t, and p denote Alternator, Turbine and Penstock

respectively. Here we have,η = 0.95 x 0.90 x 1 = 0.855 (4)

= = . . .This is average output of the generating units.

Recall that 75 kg-m/sec = 1 metric hp = 736W = 0.736kWWith a load factor 80% Rating of generators (total)= 1750 / 0.80 = 2200 kW and for each turbine = (2,200 / 2)= 1100 kW/(0.736 kW x 0.95) = 1570 metric hp

As the head is low propeller type turbines are recommended.

Answer : So the choice of units is Adopt 2 generators each of 1,100 kW maximum rating and anequal number of propeller turbines, their maximum rating being1,570 metric hp each.

ExampleExample 1010 A A power system having a M.D of 100 MW has a loadpower system having a M.D of 100 MW has a loadfactor of 30%. It is to be supplied by either of the following schemes:factor of 30%. It is to be supplied by either of the following schemes:

a)a) A steam power station in conjunction with a hydro A steam power station in conjunction with a hydro--station, thestation, thelater supplying 100x10later supplying 100x1066 units per annum with a maximum outputunits per annum with a maximum outputto 40 MWto 40 MW

b)b) A steam station capable of supplying the whole load A steam station capable of supplying the whole loadc)c) A hydro station capable of supplying the whole load A hydro station capable of supplying the whole load

nuc ear stat on capa e o supp y ng t e w o e oanuc ear stat on capa e o supp y ng t e w o e oaThe following data may be assumed:The following data may be assumed:

Steam Hydro Nuclear

1. Capital cost kW installed capacity Rs. 600 Rs. 1,500 Rs. 2,000

2. Interest and Depreciation on capital cost 12% 10% 10%

3. Operating costs per unit 5 P 1 P 2 P

4. Transmission cost per unit Negligible 0.25 P Negligible

Neglecting spares, calculate the overall cost per unit generated inNeglecting spares, calculate the overall cost per unit generated ineach case.each case.

e)e) How will be overall cost be affected for cases (b) , (c) and (d) ifHow will be overall cost be affected for cases (b) , (c) and (d) ifthe load factor were 90%.the load factor were 90%.

f)f) What shall conclusion do you draw from the results?What shall conclusion do you draw from the results?


12/1211

Solution:Solution:Total no of units generated per annum = 100,000 x (.30) x 8760 = 262.8 x 10Total no of units generated per annum = 100,000 x (.30) x 8760 = 262.8 x 1066 kWhkWh

a)a) Steam station in conjunction with hydroSteam station in conjunction with hydro--stationstationUnits supplied by hydro-station (Given) = 100x 106

Units supplied by steam station (262.8 – Hydro) = 162.8 x 106

Installed capacity of hydro-stat ion (Given) = 40,000 kWInstalled capacity of steam station (100 – Hydro) = 60,000 kW

Fixed cost for hydro-station= 40000x1500x10/100 = Rs. 600x104

Fixed cost for steam station= 60000x 600x12/100 =Rs. 432x104

b)b) Steam station aloneSteam station aloneFixed costs = 100,000 x 600 x 12 / 100 = Rs. 720 x 10 4

Running costs = 262.8 x 106 x 5 / 100 = Rs. 1314x 104

Overall cost per unit= Rs. (720+ 1314) x 104 / 262.8x 106

= 2034 x 104 x 100 / 262.8 x 106 = 7.75 P7.75 P

.

Running cost for hydro station =100x106(1+0.25)/100=Rs.125x104

Running cost for steam station= 162.8x106x5/100 = Rs.814 x104

Total running cost = Rs. 939 x 104

Overall cost per unit = Rs. (1032+939)x10 4 /262.8x106

= 1971 / 262.8 x 10 4 x 100 / 106 = 7.50 P7.50 P

c)c) Hydro station aloneHydro station aloneFixed costs = 100,000 x 1500 x 10 / 100 = Rs. 1500 x 10 4

Running costs = 262.8 x106 x (1+0.25)/100 = Rs.328x 104

Overall cost per unit= Rs. (1500+ 328) x 104 / 262.8x 106

= 1828 x 104 x 100 / 262.8 x 106 = 6.95 P6.95 P

d)d) Nuclear station aloneNuclear station aloneFixed costs = 100,000 x 2000 x 10 / 100 = Rs. 2000 x 10 4

Running costs = 262.8 x106 x 2 / 100 = Rs.525x 104

Overall cost er unit

e)When Load Factor is 90% unit generatede)When Load Factor is 90% unit generatedTotal no of units generated per annum =Total no of units generated per annum =100,000 x (.90) x 8760 = 788.4 x 10100,000 x (.90) x 8760 = 788.4 x 1066 kWhkWh

= Rs. (2000+ 525) x 104 / 262.8x 106

= 2525 x 104 x 100 / 262.8 x 106 = 9.61 P9.61 P

Fixed cost remain the same but the running cost willFixed cost remain the same but the running cost willincrease 3 times.increase 3 times.

Overall cost per unit from Steam station= Rs. (720+ 3 x 1314) x 104 / 788.4 x 106 = 5.93 P5.93 P

Overall cost per unit from Hydro station= Rs. (1500+ 3 x 328) x 104 / 788.4 x 106 = 3.15 P3.15 P

Overall cost per unit from Nuclear station= Rs. (2000+ 3 x 525) x 104 / 788.4 x 106 = 4.54 P4.54 P

f)f) ConclusionConclusionThe results show the nuclear generation isThe results show the nuclear generation iscompetitive only when a very high loadcompetitive only when a very high loadfactor exists (this being to high capitalfactor exists (this being to high capitalcosts). At low load factors it is verycosts). At low load factors it is veryuneconomical and hydro generation is good.uneconomical and hydro generation is good.

SajidajidAnsarinsari , P.Eng.Eng. PMP. PMP.A. Professor UIT).. Professor UIT).

operation of generation plants

Documents