Presentation on Heat & Mass Balance Diagram

2. Index
PLANT BLOCK DIAGRAM
PLANT RANKINE CYCLE
HIP HEAT & MASS BALANCE
LP HEAT & MASS BALANCE
HEAT RATE
Important Formulae

3. Plant BLOCK DIAGRAM
HRH P=
LPT Inlet
41, T =
566, G= P= 9.45,
1660.39, G=
H= 1363.63,
Main 3594.6 H=
Steam P = 3161.5
HPT
LPT
LPT
IPT
242, T = Gr.
566, G =
1984.36 LPT Outlet,
H = 3396 CRH P= G= 1125.25,
Condense
45, T = H = 2411.4 r
Los BOILE Vc. Pr =
R 318.7, G
s =
.1 bar
Hotwel
1660.39
H= DRTR, P = 0.916,
Feed T=176.1 CEP
Water 2999.1
T= 280°C
, G = 2044
TDBF CP
U
P
MDBFP
G
S
C
HP- HP- HP-
LP-5 LP-6 LP-7 LP-8
1 2 3

4. RH Outlet , T =566 ° C, P = 40.4,
G = 1660.39, H = 3594.6
HPT Inlet , T =566 ° C, P
Plant Rankine Cycle = 242, G= 1984.36, H = 3396
Expansion in HP
Turbine, Stages = 9,
Total Power Output =
186 MW
T= Expansion in IP
Temperature Turbine, Stage = 6,
HP HTR-1 Outlet,
Total Power Output =
T=280,H=1229.6 = Feed Water
206 MW
Temp.
HP HTR-2 Outlet,
T=255.5,H=1114 IP Outlet, P = 9.45, G=1363.63,
HP HTR-3 Outlet,
3161.5
T=213.6,H=925.8 RH Inlet , T =318.7 ° C, P =
DRTR Outlet, T=175.6,H=
45, G = 1660.39, H = 2999.1
746.2
LP HTR 8 Outlet,
T=138.9,H= 585.2
LP HTR 7 Outlet,
T=121.9,H= 512.9
LP HTR 6 Outlet, T=104, H= Expansion in LP
437.1 Turbine, Double Flow,
LP HTR 5 Outlet, T=79.2, Each Side Stages = 6,
H= 332.8 Total Power Output =
268 MW
LP HTR 5 Inlet,
T=46.5, H= 194.6 LP Outlet, G=1125.25, H = 2411.4
CEP Inlet, P =
Condenser Vaccum Pr. = 0.1 Bar, make
0.1, T = 46.1, H =
up water = 60 G, Emergrncy Drain = .75
192.9
G, Extractin Drains = 239.7 G
T in ° C,
P = Bar, S = Entropy
Flow = G,
Enthalpy = H

5. Heat Rate
 The heat rate is usually expressed as the amount of heat energy needed to produce
a given amount of electrical energy.
 It can also be expressed as rate of Heat input required to produce unit shaft output.
 Guarantees Heat Rate = 1898.8 Kcal / KWHr or 8068.786 Kj / KW.h ( Conditions :-
660 MW, 76mm Hg TMCR, 0% Make up)
 Heat Rate for Actual Condition ( 660 MW, 76 mm Hg TMCR, 3 % Make up) = 8011.7
KJ/kW.h
 Heat Rate = H/E
 H = heat supplied = Heat in HP Turbine + Heat in IP Turbine
 H = M*∆h ( Kj / Hr)
 M = Mass Flow ( Tons/ Hr) or 1000Kg/Hr
 ∆h = Enthalpy Change ( Kj / Kg)
 E = Energy output ( KW)
 So HR = M*∆h/E Kj / KwHr

6. Vidarbha Heat Balance Diagram FOR HIP
TURBINE
P=242, H = 3396, T = 566, G = 1984.36
.63G
A GV
.15G
B P=40.44, H = 3594.6, T = 566, G =
1660.39
To LP
P=9.45, Turbin
H=3161.5, e HP
G=
1363.63 IP Turbine
TurbineG= 13.39 G= 29.88,
P= 183.48,
LOS
P=9. P=21 H=3337.3 P=44. Boiler S
46 .6 93
.10 P=64
T=351.1, .6 .11
G
H=3161.5, G
16.49G
G= 81.79 L M N
T=468, T=318.7,H=2999.1,G=1660.39
R P H=3395.2,
126.81 G= 90.03
T=280,
G
X A H H=1229.6,
To BFPT 15.37G G= 2044
L T=318.7, T=366.7,
H=2999.1, H=3081.7,
G= 155.78 G= 121.65
T=182, T=213.6, T=255.5,
T=138.9,
H=790.7 H=925.8 H=1114
H= 585.2
DRTR HTR-3 HTR-2 HTR-1
From
LPH
T=175.6, T=261,
T=187.5,
H=746.2 T=219.2, H=1140
H=796.5
G=2044 H=939.8 G=121.65
G=383.94
BFP G=293.91

7. HEAT BALANCE DIAGRAM FOR LP TURBINE
From IP Turbine
H P=9.45,
Fl H=3161.5,
M G=
as 1363.63
R
h
LPT
LPT
ta S
n
k
.26 .26
H= 2411.4, G =
0.75G G G
1125.25
T=189.5,
H=2846.9
T S T=83.2,
, G=
S T
H=2596.8
T=132.5, 48.15
Condenser , G=
H=2846.9 T=244,
79.93
, G= H=2952.1
Make up Water = 48.15 , G=
60G 46.75
.89 T=138.9,
T=46.1, G H=585.2,
H=192.9
T=46.8,
G
H=196
To
S LPH-8 LPH-7 LPH-6 LPH-5 DRTR
T=46.5,
H=194.6
C
CEP G=1562.7
4 T=52.4,
T=52.4,
T=52.4, T=52.4, H=219.2
H=219.2
H=219.2 H=219.2 G=239.79
G=239.79
G=239.79 G=239.79

8. Heat Rate
 The heat rate is usually expressed as the amount of heat energy needed to produce
a given amount of electrical energy.
 It can also be expressed as rate of Heat input required to produce unit shaft output.
 Guarantees Heat Rate = 1898.8 Kcal / KWHr or 8068.786 Kj / KW.h ( Conditions :-
660 MW, 76mm Hg TMCR, 0% Make up)
 Heat Rate for Actual Condition ( 660 MW, 76 mm Hg TMCR, 3 % Make up) = 8011.7
KJ/kW.h
 Heat Rate = H/E
 H = heat supplied = Heat in HP Turbine + Heat in IP Turbine
 H = M*∆h ( Kj / Hr)
 M = Mass Flow ( Tons/ Hr) or 1000Kg/Hr
 ∆h = Enthalpy Change ( Kj / Kg)
 E = Energy output ( KW)
 So HR = M*∆h/E Kj / KwHr

9.  Boiler Efficiency
◦ η = Output / Input
◦ η = Steam Output * Enthalpy / Coal Input
◦ Η = M ₁* H / GCV * M₂
◦ Were M ₁ = Flow
◦ Where M₂ = Coal Flow rate
GCV for the design coal is 3300 Kcal/Kg
 Specific Heat Consumption
◦ It is defined as rate of steam flow ( kg/ s) required to produce unit shaft output.
◦ S.R. = Total Steam Flow / Total Unit Generated
 Cost of Power
◦ Cost of Power = HR*Fuel Price / GCV

10. Thank You