Performance Management Of Boiler, Turbine , Cycle#l3

Performance management of Boiler, Turbine , Cycle Efficiency and other performance Parameters.

Concept of Efficiency • Efficiency = ( Output / Input) X 100 • It is same as Productivity

B

A 100

80%

80

70%

C 56

Over All Efficiency = 28/100 = 28%

Much Lower Than Lowest Block i.e. “C” Overall Efficiency= Eff of A * Eff of B * Eff of C

50%

28

Power Plant Output of Station Efficiency= 100

X

Input of Station Output= Energy Sent Out Input = Fuel Burnt in Kg X CV of Fuel (Kcal/Kg) 1 KW = 860 Kcal = 3600

TERMS & DEFINITIONS

• Connected Load: It is the sum of Loads of all Installed equipment in Consumers Premises • Maximum Demand: It is the maximum Load used by consumer at any time. • Demand Factor: It is defined as the ratio of maximum demand to connected Load. • Load Factor: Ratio of average load to maximum demand.

PLANT CAPACITY FACTOR Defined as the ratio of actual energy produced in Kilowatt Hour to the maximum possible energy that could have been produced during a defined period i.e. total period of operation including shut downs and outages. E PLF = -----------C*T E: Energy Produced in actual period C: Capacity of the Plant in KW T: Total numbers of Hours in Given Period

PLANT USE FACTOR Defined as the ratio of energy produced in Kilowatt Hour to the maximum possible energy that could have been produced during actual period of operation. E PLF = -----------C*t E: Energy Produced in actual period C: Capacity of the Plant in KW t: Total numbers of Hours in Given Period

Diversity Factor • It is defined as the ratio of sum of individual maximum demand to the simultaneous maximum demand. • It is always more than unity • Power station should be capable of supplying simultaneous maximum demand

EXAMPLE 1

A Power plant has a maximum Demand of 80 MW and daily Load curve is defined as follows:

Time 0-6 6-8 (Hours )Load 40 50 (MW)

8-12 1214 60 50

1418 70

1822 80

2224 40

(A)Determine the Load Factor of Station? (B) What is the Load Factor of stand by equipment rated at 25 MW that takes up all load in excess of 60 MW? Also calculate its use factor?

(A) Total Load =40*6+50*2+60*4+50*2+70*4+80*4+40* 2 = 1360 MW Average Load = 1360/ 24 = 56.67 MW Maximum Demand = 80 MW Load Factor = 56.67/80 = 0.71

(B) Energy Generated by Stand by Equipment= = 10*4+ 20*4 = 40 + 80 = 120 MW Time of Operation = 8 Hours Average Load = 120/8 = 15 MW Load Factor = 15/ 20 = 0.75 Use Factor = 15/25 = 0.60

EXAMPLE 2

Find the diversity Factor of a power station which supplies the following Loads Load A: Motor Load of 100KW between 10 AM and 6PM Load B: Lighting load of 60 KW between 6 PM and 10PM Load C: Pumping load of 40 KW between 4 PM and 10AM

Total Load = 100 + 60 + 40 = 200 KW 10AM to 4 PM = 100 (M) = 100 KW 4PM to 6PM = 100(M)+ 40(P)= 140KW 6PM to 10 PM = 60(L)+40(P) = 100 KW 10PM to 10AM= 40 (P) = 40 Simultaneous max demand = 140

POWER PLANT BASICS

Power Plant Cycle • Works on Rankine Cycle • Over all efficiency= Multiplication of Boiler, Turbine, Cycle and Generator Efficiency.

Main Parts of Power Plant • Boiler • Turbine • Generator

Boiler • • • • • •

Furnace Milling System Fans for Air Supply Electrostatic Precipitator Drum Water and Steam Circuit in Side Boiler

Turbine • • • • •

HPT IPT LPT Condenser Pump

Heat Rate • Requirement of Heat per unit energy generation Heat Rate=

Heat added to Boiler (Kcal) Electrical Energy Sent Out (KWhr)

Efficiency Loss Areas • Improper Fuel Combustion • Problems in Milling System • Heat Losses due to improper Insulation • Leakages in Water and Steam Circuit • Condenser • Pumps • Heaters