Fahrurodin Syah Alfa Roby
Soal :
1. Sebuah boiler berbahan batu bara dengan data-data seperti table 2.
Tentukanlah efisiensi boiler dengan indirect method
Fuel firing rate
Steam generation rate
Steam pressure
Steam temperature
=
=
=
=
Feed water temperature
%CO in Flue gas
2
%CO in flue gas
Average
flue
temperature
Ambient temperature
=
gas
Humidity in ambient air
=
=
0.55
o
31 C
0.0204 kg / kg dry
air
o
70 C
Surface temperature of =
boiler
Wind velocity around the =
boiler
Total surface area of boiler =
GCV of Bottom ash
GCV of fly ash
Ratio of bottom ash to fly
ash
Fuel Analysis (in %)
Ash content in fuel
Moisture in coal
Carbon content
Hydrogen content
Nitrogen content
Oxygen content
GCV of Coal
60000 kg/hr
20000 kg/hr
5 MPa
o
400 C
o
98 C
14
=
=
=
=
=
=
=
=
=
=
o
190 C
3.5 m/s
2
90 m
800 kCal/kg
452.5 kCal/kg
90:10
8.63
31.6
43
2.04
1.5
13.23
3800
kCal/kg
Jawaban :
Step 1 Find theoretical air requirement
Theoretical air required for
complete combustion
= [(11,6 x C) + {34,8x ( H2 - O2 / 8 )} + (4,35 x S ) ] /100 Kg/Kg of coal
=[(11,6 x 43) + {34,8 x (2,04 - 13,23 /8 )} + (4,35 x 0 ) ] /100 Kg/Kg of coal
= 5,12 Kg/Kg of coal
Step 2 Find theoretical CO2 %
Moles of N 2+
Moles of C
Moles of C
% CO2 at theoretical condition( CO2)t =
Wt of N 2
Mol . wt of N 2
theortical air
Where, Moles of N2
Moles of N2
Where moles of C
5,12 x 77 /100
28
0,43
12
0,015
28
= 0,0358
0,358
0,1405+ 0,0358
(CO2)t
(CO2)t
= 0,2031 x 100% = 20,31 %
Step 3 To find Excess airsupplied
Actual CO2 measured in flue gas = 14.0%
= 0,2031
Wt of N 2 fuel air
Mol . wt of N 2
= 0,1405
7900 x [ ( C O 2 ) t( C O2 ) a]
% Excess air supplied (EA)
( C O2 ) a x [100( C O2 ) t ]
7900 x [20,3114 ]
14 x [ 10020,31]
x 100 %
= 44,68 %
Step 4 To find actual mass of air supplied
Actual mass of air supplied = {1 + EA/100} x theoretical air
= {1 + 44,68 /100} x 5.12
= 7,41Kg/Kg of coal
Step 5 To find actual mass of dry flue gas
Mass of dry flue gas = Mass of CO2 + Mass of N2 content in the fuel+ Mass of N2 in the
combustion air supplied + Mass of oxygen in flue gas
Mass of dry flue gas =
0,43 x 44
12
+ 0,015 +
7,41 x 77
100
( 7,415,12 ) x 23
100
= 7,83 Kg / Kg of coal
Step 6 To find all losses
f T a
T
1. % Heat loss in dry flue gas (L1) =
x 100 %
m x Cp x
7,83 x 0,23 x ( 19031 )
=
3800
L1 = 7,535 %
x 100 %
2. % Heat loss due to formation of water from H2 in fuel (L2)
f T a
T
(L2) =
x 100 %
9 x H 2 x {584 +C p
9 x 0,0204 x {584+0,,45( 19031)
3800
= 3,17 %
3. % Heat loss due to moisture infuel (L3)
(L3) =
f T a
T
}
M x {584+C p
31,6 x {584+ 0,45 ( 19031 ) }
x 100
3800
= 5,45 %
4. % Heat loss due to moisture in air (L4)
L6
f T a
T
}
ASS x Humidity x C p
7,41 x 0,0204 x 0,45 ( 19031 ) }
x 100
3800
= 0,28 %
5. % Heat loss due to partial conversion of C to CO (L5)
x 100 %
L5
CO x C
5744
x
CO + C O2 GCV of fuel
0,05 x 0,1465
0,55+14
x 100 %
0,05 x 0,1465
x 100
3800
= 2,418
6. Heat loss due to radiation and convection (L6)
= 0.548 x [ (343/55.55)4 (303/55.55)4 ] + 1.957 x (343- 303)1,25 x sq.rt of
[(196.85 x 3.5 + 68.9) /68.9]
= 645,45 w/m2
= 645 x 0,86 = 555,09
Total radiation and convection
loss per hour
% radiation and convection loss
L6
=555,09 x 90 = 49.958,16
49.958,16 x 100
3800 x 60000
= 0,02 %
7. % Heat loss due to unburnt in fly ash
% Ash in coal
= 8.63
Ratio of bottom ash to fly ash
= 90:10
GCV of fly ash
= 452.5 kCal/kg
Amount of fly ash in 1 kg of coal
= 0.1 x 0.0863 = 0.00863 kg
Heat loss in fly ash
= 0.00863 x 452.5 = 3.905 kCal / kg of coal
% heat loss in fly ash
= 3.905 x 100 / 3800
L7
= 0.10 %
8. % Heat loss due to unburnt in bottom ash
GCV of bottom ash
= 800 kCal/kg
Amount of bottom ash in 1 kg of coal
= 0.9 x 0.0863
= 0.077 kg
Heat loss in bottom ash
= 0.077 x 800
% Heat loss in bottom ash
= 62.136 kCal/kg of coal
= 62.136 x 100 / 3800
L8 = 1.63 %
Boiler efficiency by indirect method = 100 (L1+ L2+ L3+ L4+ L5+ L6+ L7+ L8)
= 100 - ( 7,535 + 3,17 + 5,45 + 0,28 + 2,418 + 0,02 +
0,10 +1,635 )
= 79,392 %
