efficiency-boiler & turbine

8/13/2019 Efficiency-boiler & Turbine

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SP. ENTROPY,S

SP.ENTH

ALPY,H


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Need of efficiency & performance monitoring :

High cost of installation of new power plant

Rs. 3.5 to 4 Crore /MW installation

+Rs. 1.5 to 2 Crore /MW for T&D

Increased plant performance leads to increased plant

availability and vice versa

Maximising generation with minimum generation cost

For increasing station performance main areas :

Planned Maintenance Loss

Thermal Efficiency FactorsPlant Load factor

Forced outages

Plant Availability Factor

Optimising terminal conditions of the unit

MS parameters

RejectionParameters


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Areas of concentration for increasing Efficiency :

Heat rate of Turbine

Boiler Efficiency

DM water Make-upSpecific Oil Consumption

Excess air

Condenser Back Pressure


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Basic Concepts of Efficiency :

Overall Station Efficiency = Output

Input

Energy Sent Out (KW)Fuel Burnt (Kg) * Calorific Value of Fuel

=

Rankine Cycle

ENTROPY,S

TEMPER

ATURE,T

Boiler Efficiency((Steam Supplied in Kgs *

Total heat in superheated steam) -

Total heat of feed water))

Fuel Burnt (Kg) * Calorific Value

of Fuel (Kcal/Kg)

=

4

3

2

1

=BMs*h3-Mf*h1

Mc * C.V.

* 100


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Heat Balance Diagram Showing Losses :

0

100

PERCENTAGE

HEATINPUT

BOILER LOSSES

10 - 13 %

CONDENSER LOSSES

45 - 49 %

GENERATOR LOSS

2 - 4 %

USEFUL HEAT OUTPUT

34 - 39 %


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Weight of Air required for Combustion :

(i) Carbon C + O2= CO2

12 + 32 = 44

1 + 8/3 = 11/3 O2 = 8/3 C --------- (a)

Oxygen required = 8/3 times wt. Of Carbon

(ii) Hydrogen 2H2+ O2 = 2H2O

4 + 32 = 36

1 + 8 = 9O2 = 8H -----------(b)

Oxygen required = 8 times the wt. Of Hydrogen


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Weight of Air required for Combustion : contd...

(iii) Sulphur S + O2= SO2

32 + 32 = 64

1 + 1 = 2 O2 = 1 S --------- (c)

Oxygen required = Same as wt. of Sulphur

(iv) Combining formula (a), (b) & (c)Oxygen required / gm of fuel = 8/3 C + 8H + S ------- (d)

(v) Assuming all the Oxygen in the fuel will combine with

Hydrogen in the fuel the actual amount of Hydrogen

requiring air is (H - O/8)

(vi) Oxygen in gm/gm of fuel = 8/3C + 8(H - O/8) + S

Air in gm / gm of fuel = 4.31[ 8/3C + 8(H-O/8) + S ]


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Excess Air requirement :

Optimum Excess air = 20 % of Stoichiometric (perfect) air for

combustion

PERCENTAGE EXCESS AIR

PERCENTAGEHEATL

OSS

20

10

40 60 80 100

20

30

40

MINIMUM LOSS

EXCESS AIR FOR MINIMUM LOSS0

0


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Boiler Efficiency :

Direct method

Boiler Efficiency =

( Enthalpy of Steam - Enthalpy of Feed water)

* Steam flow

Fuel Burnt (Kg) * Calorific Value of Fuel

Indirect or losses method

Boiler Efficiency = 100 % - Total Loss in Percentage

Boiler Losses

Dry Flue Gas Loss

Wet Flue Gas Loss

Due to moisture in fuel

Due to Hydrogen in fuelUnburnt Carbon Loss

Rejection Loss in Ash

Radiation Loss

Unaccounted Loss


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Boiler Efficiency :

L2 = LOSS DUE TO MOISTURE IN FUEL = {100* M* Wg* ( hg - ha)}Wg * GCV

hg = SP. ENTHALPY OF VAPOUR AT AIRHEATER OUTLET IN Kcal/Kg(FOR AIha = SP. ENTHALPY OF WATER AT AIRHEATER INLET IN Kcal/Kg (FOR AIR

M = MOISTURE CONTENT IN FED COAL IN % OF WEIGHT

L3 = LOSS DUE TO HYDROGEN IN FUEL = {9 * 100* H* Wg* ( hg - ha)}Wg * GCV

H = HYDROGEN CONTENT IN FED COAL IN % OF WEIGHT


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Boiler Efficiency :

L4 = LOSS DUE TO UNBURNT IN ASH = {100* Wg * U * A * KWg * GCV

U = WEIGHTED AVERAGE OF UNBURNT CONTENT IN %

A = ASH CONTENT IN FED COAL IN %

K = C. V. OF CARBON BURNT TO CO2 IN Kcal/Kg = 8139

L5 = DUE TO HEAT REJECTED IN ASH= [ 100* Cpg * A * {0.1*(Tba Ta) + 0.9*(Tg-Ta)}]

GCV

Tba = BOTTOM ASHING TEMP. IN K

L6 = RADIATION LOSS = 0.8 (ASSUMED)

L7 = UNACCOUNTD LOSS = 0.647 ( DESIGN FIGURE )

= 100[ L1 +L2 +L3 +L4 +L5 +L6 +L7]


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Other parameters affecting Boiler Efficiency :

CONTROL OF BLOW DOWN AND MAKE UP

AUXILIARY POWER CONSUMPTION

OPTIMIZATION OF OIL CONSUMPTION

AIR HEATER PERFORMANCE AND TRAMP AIR TO BOILER

PERCENTAGE EXCESS AIR

PERCENTAGEHEATLOSS

20

10

40 60 80 100

20

30

40

MINIMUM LOSS

HEAT LOSS DUE TO UNBURNT

0

0

HEAT LOSS DUE TO FLUE GAS

HEAT LOSS DUE TO UNBURNT GAS

TOTAL HEAT LOSS


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HEAT RATE :

TURBINE HEAT RATE = Qs * (Hs - Hf)

Eg

Qs = STEAM FLOW AT TURBINE INLET IN KG/HR

Hs = TOTAL HEAT OF STEAM AT TURBINE INLET IN KCAL/KG

Hf = TOTAL HEAT OF FEED WATER AT ECONOMISER INLET IN KCAL/KG

Eg = NET LOAD GENERATED IN KW

TURBINE EFFICIENCY = 860 * 100

HEAT RATE

PLANT HEAT RATE = 860 *100

TURBINE EFF. *BOILER EFF.


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CONDENSER PERFORMANCE :

Volume, m3/kg

Pressure,barabs.

1

4

2 3 4

8

12

16

0

0

20

p2

p3

p1

p4

EFFICIENCY = (H1-H2) / H1

= (T1-T2) / T1

DELTA T = CW O/L - CW I/L

TERMINAL TEMP. DIFF. (TTD) =

EXH. HOOD - CW O/L

CONDENSER VACUUM =

BAROMETRIC PR. - BACK PR.


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C : Frictional head= 4* f*L*V2 / 2*g*Dwhere , f= pipe surface roughness

L= length of pipe

V= flow velocityD= inner dia. of pipe

g= gravitational constant

E : Net positive suction head (NPSH)NPSH) available = (P - Pv)* 2.31/ sp. Gravity -losses +/- Z

Where , P= absolute pressure at liquid surfacePv= vapour pressure of liquid at pumping temp.

Losses = kinetic head + frictional head +

entrance loss

Z= static elevation from liquid level in suctiontank to the centre line of the first stage impeller of the pump

D : Gross total head= potential head + kinetichead + losses


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Cavitation starts

FLOWRATE Q

NPSH

NPSH) available

NPSH)required

As a general rule the NPSH ) available should be 30% higher than

the required NPSH at the operating point.


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NOTE: If NPSH)availableapproaches to zero than there will be

severe cavitation in the pump.

IMPROVEMENT IN NPSH AVAILABLE OF BFP1:Raise the deaerator height for more static head.

2: Incorporate slow speed booster pump to have lower

NPSH) required.3: Keep dp across the suction filter less than 0.5 kg/cm2.

IMPROVEMENT IN NPSH AVAILABLE OF CEP

1: Use larger size suction piping with larger radius bend insteadof elbows.

2: Use long radius suction bell mouths in case of vertical pumps.


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