BOILER CALCULATION
Steam Rate SR = 3600/Wnet SR = 3600/(Wt – 3600/(Wt – Wp) Wp) SR = 3600/(889.62) SR = 4.05 kg/kWh
Heat Rate HR = 3600/ŋ HR = 3600/0.3452 HR = 10428.74 KJ/kWhr
Rated Capacity Typical Large Generators Efficiency Ranges form 75 – 75 – 85% 85% Use 85% efficiency:
Rated Capacity = 175 MW/0.85 Rated Capacity = 205.88 MW
Steam Flow, ms Assumptions: (Typical Efficiencies of Equipments) ηt = 80% ηg = 85% ηm = 95% Note: Turbine Rated Capacity = T.R.C T.R.C = Total Capacity / *ηt x ηg x ηm] T.R.C = 175 / [0.8 x 0.85 x 0.95] T.R.C = 270.90 MW T.R.C = Wt x ms ms = T.R.C / Wt
ms = 270.9 MW ( 1000 kJ/s / MW) / 889.62 kJ/kg ms = 304.51 kg/s x 3600 s/hr
ms = 1096243.34 kg/hr
Boiler Capacity BC = ms (h1-h18) = 304.51 kg/s (3381 – (3381 – 1070)KJ/kg 1070)KJ/kg BC = 703722.61 KJ/s
Coal flow rate Fuel to be Used B ituminous
Table 5-4 Analysis of Typical American Fuels by Morse (p.127) Proximate Analysis Moisture Ill. Bituminous
Volatile
Fixed
Matter
Carbon
33.62 3 3.62
41.34
13.31
Ultimate Analysis – Analysis – Moisture Moisture Included Ash
Sulfur
Hydrogen
Carbon
Nitrogen
Oxygen
11.73
3.75
5.19
59.07
0.95
19.31
High Heating Value of Illinois Bituminous Coal Re: Dulong’s Formula (S.I.) HHV = 33820 C + 144212 (H – (H – O/8) O/8) + 9304 S HHV = 33820 (0.5907) + 144212 (0 .0519 – .0519 – 0.1931/8) 0.1931/8) + 9304 (0.0375)
HHV = 24330.06 kJ/kg
Theoretical Air Fuel Ratio
A/F = 11.53 (0.5907) + 34.36 (0.0519 – (0.0519 – 0.1931/8) 0.1931/8) + 4.33 (0.0375)
A/F = 7. 927 kg air / kg fuel
Fuel Mass Flow Rate, mf Qa = mf Qf;
Qf = HHV; HHV of Illinois Bituminous Coal = 24330.06 KJ/kg
mf = msQa/Qf = ((304.51 kg/s)(2577.16 KJ/kg))/( 24330.06 KJ/kg)
ms = 270.9 MW ( 1000 kJ/s / MW) / 889.62 kJ/kg ms = 304.51 kg/s x 3600 s/hr
ms = 1096243.34 kg/hr
Boiler Capacity BC = ms (h1-h18) = 304.51 kg/s (3381 – (3381 – 1070)KJ/kg 1070)KJ/kg BC = 703722.61 KJ/s
Coal flow rate Fuel to be Used B ituminous
Table 5-4 Analysis of Typical American Fuels by Morse (p.127) Proximate Analysis Moisture Ill. Bituminous
Volatile
Fixed
Matter
Carbon
33.62 3 3.62
41.34
13.31
Ultimate Analysis – Analysis – Moisture Moisture Included Ash
Sulfur
Hydrogen
Carbon
Nitrogen
Oxygen
11.73
3.75
5.19
59.07
0.95
19.31
High Heating Value of Illinois Bituminous Coal Re: Dulong’s Formula (S.I.) HHV = 33820 C + 144212 (H – (H – O/8) O/8) + 9304 S HHV = 33820 (0.5907) + 144212 (0 .0519 – .0519 – 0.1931/8) 0.1931/8) + 9304 (0.0375)
HHV = 24330.06 kJ/kg
Theoretical Air Fuel Ratio
A/F = 11.53 (0.5907) + 34.36 (0.0519 – (0.0519 – 0.1931/8) 0.1931/8) + 4.33 (0.0375)
A/F = 7. 927 kg air / kg fuel
Fuel Mass Flow Rate, mf Qa = mf Qf;
Qf = HHV; HHV of Illinois Bituminous Coal = 24330.06 KJ/kg
mf = msQa/Qf = ((304.51 kg/s)(2577.16 KJ/kg))/( 24330.06 KJ/kg)
mf = 32.26 kgfuel/s (3600/hr) mf = 116118.73 kgfuel/hr
Mass Flow Rate of Air mf = 32.26 kgfuel/s;
A/F = 7. 927 kg air/kg fuel
ma = mf (A/F) ma = (32.26 kgfuel/s)( 7. 927 kg air/kg fuel) ma = 255.73 kg/s
Factor of Evaporation FE = (hs – (hs – hf)/2257 hf)/2257 = (3381 – (3381 – 1070) 1070) / 2257 FE = 1.02
Developed Boiler Horsepower Developed Bo. HP = ms(hs – ms(hs – hf)/35322 hf)/35322 = 304.51 (3381 – (3381 – 1070) 1070) / 35322 Developed Bo. HP = 19.92 HP
Equivalent Evaporation EE = ms x FE = 304.51 (1.02) EE = 310.60 kg/s
Equivalent Specific Evaporation ESE = Bo. Economy x FE = (ms/mf) x FE = (304.51/32.26) x 1.02 ESE = 9.63
ECONOMIZER CALCULATION Values from CATT3 At 18MPa Vf = 0.00184 m3/kg Hf = 1732kj/kg Tsat = 357.1 C
At 3.53 MPa V1 = 0.001236m3/kg H1 = 1052 kj/kg T1 = 243.1 C S1 = 2.73 kJ/kg-K
REF: Power Plant Engineering by P. K. Nag, “It is assumed that the temperature of flue gas exhausting the economizer is relatively high to raise the air temperature on the air pre-heater. The mass flow rate of flue gas must also be relatively high so that the air would be dispersed at the top of the stack.” In these criteria, assumed temperature and flue gas flow rate are the ff: mfg = 500 kg/s Tg2 = 600 C
Solving for Tg1: QE = ms(ha – h18) = mfg(Cpfg)(ΔTg) QE = 304.51 kg/s (1732 kJ/kg – 1070 kJ/kg) QE = 201585.62 kJ/s
where: Cpfg = 1.214 kJ/kg-oC QE = mfg (Cpfg) (ΔTg) 201585.62 kJ/s = 500 kg/s (1.214 kJ/kg-C) (ΔTg) ΔTg = 332.10 C
Tg1 = 332.10 C + 600 C
Tg1 = 932.10 C
Solving for LMTD: T1 = 243.1 C Tsat = 357.1 C Tg1 = 932.10 C Tg2 = 600 C
ΔT1 = tg2 – t1 = 600 – 243.1 ΔT1 = 356.9 C
ΔT2 = tg1 – tsat = 932.10 – 357.1 ΔT2 = 575 C
LMTD = (ΔT2 - ΔT1) / ln (ΔT2 / ΔT1) LMTD = (575- 356.9) / ln(575/356.9)
LMTD = 457.31 C
Note: For Flue gas and water heat exchanging, U ranges from 30 to 100 W/m2C Therefore, Use Average U = 65 W/m2C.
Solving for the Total Area of Heat Transfer: A = Q / (LMTD x U) A = 201585.62 kJ/s / [(457.31 C)(65 W / m2-C)(1/1000)] A = 6781.65 m2
Note: Typical Outside Diameter of Economizer tubes ranges from 45 – 70 mm Let Inside Diameter be the minimum range, ID = 45 mm
Thickness of Tubes Material: Titanium Tensile Strength = 63000 psi Pressure = 18 MPa = 2611.40 psi ID = R = 45mm/2 = 22.5mm = 0.8858 in E = 1.0 (weld joints efficiency)
Re: t = PR / (SE – 0.6P);
REF: ASME code
t = 2611.40 (0.8858) / [63000 (1.0) – 0.6 (2611.40)] t = 0.0377 in = 0.956mm
t = 1 mm
Outside Diameter OD = ID + 2t OD = 45 mm + 2(1 mm)
OD = 47 mm
Number of Tubes REF: Power Plant Engineering by Nag, “Velocity of water, vw, must not exceed 1.2 m/s.”
Therefore, Use vw = 1 m/s 2
ms = nπ ID /4 (vf /vw) 304.51 = nπ (0.045)2/[4(0.00184/1)] n = 352.29 tubes Use 353 tubes
Length of Tubes 2
A = nπ OD L = 7580.57 m 2
6781.65 m = π 353 (0.047m)L
L = 130.11 m
Number of Turns Assumption: W=5m Clearance = 20 mm nt = L / (W-2C) nt = 130.11 m / (5 m – 2 (0.02)) nt = 26.23
use 27 turns
Height of Economizer Use 60 mm pitch. H = nt x Pitch H = 27 x 0.06 m
H = 1.62 m
Length of Economizer L = n x Pitch L = 353 x 0.06 m
L = 21.18 m
Header Dimensions Inlet Header Length of header is the same as length of economizer L = 21.18 m
Note: Since water flows to the header constantly, it can be assumed that the mass of water inside the header is equal to the Steam flow rate.
ms = 304.51 kg o
at 7.97 MPa, 294.8 C 3
v = 0.001383 m /kg
V = v (mass) 3
V = 0.001383 m /kg (304.51 kg) 3
2
V = 0.421 m = π (D /4) (21.18 m) D = 0.159 m = 159 mm
Outlet Header At 18 MPa, Saturated Liquid 3
vf = 0.00184 m /kg 3
V = 0.00184 m /kg (304.51 kg) 3
2
V = 0.56 m = π (D /4) (21.18 m) D = 0.183 m = 183 mm
SUPERHEATER CALCULATIONS
o
Tg1= 932.10 C (temperature inlet of economizer) o
T3 = 357.1 C T4 = 538 oC H1 = 3381 kJ/kg Hb = 2509 kJ/kg o
3
v at 538 C = 0.01824 m /kg
Solving for T g1 Q SH = ms(h1 – hb) = mfgCpfgΔTg = 304.51 kg/s (3381 kJ/kg – 2509 kJ/kg)
Q SH = 265532.72 kJ/s where: o
Cpfg = 1.196 kJ/kg- C at 538C 265532.72 kJ/s = 500 kg/s (1.196 kJ/kg-C) ΔTg o
ΔTg = 444.03 C o
o
Tg1 = 444.03 C + 932.10 C
Tg1 = 1376.13 oC
Solving for LMTD o
o
ΔT1 = 1376.13 C – 538 C o
ΔT1 = 838.13 C
ΔT2 = 932.1 oC – 357.1 oC o
ΔT2 = 575 C
LMTD = (ΔT2 - ΔT1) / ln (ΔT2 / ΔT1) LMTD = (575- 838.13) / ln(575/838.13)
LMTD = 698.32 oC
Note:
2
For Flue gas and water heat exchanging, U ranges from 30 to 100 W/m C Therefore, Use Average U = 65 W/m2C.
Solving for the Total Area of Heat Transfer:
A = Q / (LMTD x U) o
2 o
A = 265532.72 kJ/s / (698.32 C x 65 W / m - C)(1/1000)
A = 5849.91 m2
Note: Typical Outside Diameter of Superheater tubes ranges from 50 – 75 mm.
Let Inside Diameter be the maximum range, ID=75mm
Solving for the thickness of tubes:
Material: Carbon Steel Tensile Strength = 78300 psi Pressure = 18 MPa = 2611.40 psi ID = R = 75mm/2 = 37.5mm = 1.4764 in E = 1.0 (weld joints efficiency)
t = PR / (SE – 0.6P)
from ASME code
t = 2611.40 (1.4764) / [78300 (1.0) – 0.6 (2611.40)] t = 0.05 in = 1.28 mm
t = 2 mm
Solving for the Outside Diameter:
OD = ID + 2t = 75mm + 2(2 mm)
OD = 79 mm
Solving for No. of Tubes:
2
ms = nπID /4(v/vs)
From Power Plant Engineering by P.K. Nag, “Velocity for very high pressure steam is approximately 10 m/s.” Therefore, Use vs = 10 m/s
2
3
304.51 kg/s = nπ(0.075 m) /4(0.01824 m /kg / 10 m/s) n = 125.72 tubes
n = 126 Tubes
Solving for the Length of Tubes:
A = nπODL 2 5849.91 m = 126π(0.079m)L
L = 187.07 m
Solving for No. of Turns: Assumption: W = 5m Clearance = 10 mm nt = L / (W-2C) nt = 187.07 m / (5m-2(0.01)) nt = 37.56 turns nt = 38 Turns
Solving for the Length of Superheater:
OD = 79 mm Use 60 mm pitch.
L = nt x Pitch L = 38 x 0.06 m
L = 2.28 m
Solving for the Width of Superheater:
W = n x Pitch W = 126 x 0.06 m
W = 7.56 m
Solving for Header Dimensions:
For Inlet Header:
Length of header is the same as the width of superheater L = 2.28 m
Note: Since water flows to the header constantly, it can be assumed that the mass of water inside the header is equal to the steam flow rate.
ms = 304.51 kg at 18 MPa, saturated gas 3
v = 0.00749 m /kg
V = v x mass 3
V = 0.00749 m /kg (304.51 kg) 3
2
V = 2.28 m = πD /4 (2.28 m) D = 1.128 m
D = 1128 mm
For Outlet Header:
L = 2.28 m Mass = 304.51 kg
o
At 18 MPa, 538 C 3
v = 0.01824m /kg 3
V = 0.01824 m /kg (304.51 kg) 3
2
V = 5.55 m = πD /4 (2.28 m) D = 1.761 m
D = 1761 mm
AIR PREHEATER CALCULATIONS o
Tg1 = 600 C (inlet temperature of flue gas) o
Tair-inlet = 26 – 30 C (Mean temperature of air of in Bataan)
Use average temperature of 28 oC.
From Power Plant Engineering by P.K. Nag, o
“The temperature of heat air ranges from 280 to 400 C.”
o
Therefore, Use average temperature of 340 C.
o
Tleaving-air = 340 C (Average temperature of leaving air)
Solving for T g2: Q AP = maCp(Tleaving-air – Tair-inlet) o
o
Q AP = 255.73 kg/s (1.005 kJ/kg-C)(340 C – 28 C)
Q AP = 80186.70 kW
Q AP= mfgCpfg(Tg1 - Tg2) 80186.7 kW = 600 kg/s (1.151 kJ/kg-C)ΔTg o
ΔTg = C
o
o
Tg2 = 600 C - 161.111 C o
Tg2 = 483.89 C
Solving for LMTD: o
o
ΔT1 = 600 C - 340 C o
ΔT1 = 260 C
o
o
ΔT2 = 483.89 C – 28 C o
ΔT2 = 455.89 C
LMTD = (ΔT2 - ΔT1) / ln (ΔT2 / ΔT1) LMTD = (455.89 - 260) / ln(455.89 /260)
LMTD = 348.83 oC
Solving for the Area of Heat Transfer: Note: 2
For gas to air heat transfer, U ranges from 30 to 60 W/m C.
Therefore, 2
Use Average U = 45 W/m C.
Solving for the Total Area of Heat Transfer: A = Q / (LMTD x U) o
2 o
A = 80186.7 kJ/s / (348.83 C x 45 W / m - C) (1/1000)
A = 5108.36 m2
Solving for the No. of tubes: Assumptions: Length of tubes = 15 m OD = 0.05 m (Typical OD of air pre heater tubes is about 50 mm.)
A = nπODL 2
5108.36 m = nπ(0.05 m)(15 m) n = 2168.05
n = 2169 tubes
FORCED DRAFT FAN CALCULATIONS
Fan Capacity ma = 255.73 kg/s f = 0.005 (air against steel) air duct diameter = 2.5 m Length = 40 m (length from fan to furnace) Elevation = 10 m v = 12 m/s
Solving for Heads 2
H1 = f (L/D)(v /2g) 2
2
H1 = 0.005(40m/2.5)(12 m/s) /2(9.81 m/s ) H1 = 0.587 m
H2 = 10 m
2
2
2
H3 = v /2g = (12 m/s) /2(9.81 m/s ) H3 = 7.34 m
HT = H1 + H2 + H3 HT = 0.587 m + 10 m + 7.34 m
HT = 17.93 m
Solving for Forced Draft Fan Capacity: Re: PV=mRT V = mRT / P V = [255.73 kg/s (0.287 kJ/kg-K)(28 + 273)K ] / 101.325 kPa 3
V = 218.03 m /s
Note: o
3
At 28 C, Specific weight (ϒ) of air is 11.502 N/m . P = VϒHT 3
3
P = 218.03 m /s(11.502 N/m )(17.93 m) P = 44964.22 W
P = 44.96 kW
CONDENSER CALCULATIONS
Condenser Capacity Qc = (h10 – h11)(1-m1-m2-m3-m4-m5-m6)(ms) Qc = (2431 – 359.8)(0.5222)(304.51) Qc = 329352.12 KW
Specification of Titanium Tube to be used:
Determine the Overall thermal coefficient: 2
Permissible Range: 3000 – 4500 kCal/m -K-hr Multiplier: 3-5, So use 4 as multiplier. 2
Therefore, Use 3750 kCal/m -K-hr as the average overall thermal coefficient. 2
2
Convert kCal/m -K-hr to kW/m -K: 2
U = 3750 kCal/m -K-hr (1 BTU / 0.252 kCal) (1 kW-hr / 3413 BTU) 2
U = 4.36 kW/ m -K x (4) 2
U = 17.44 kW/ m -K
From Power Plant Theory and Design by Potter, p. 351, Use Standard Size of D = 1” BWG 15
Note: BWG stands for Birmingham Wire Gauge.
Permissible Water Velocity = 7 – 10 ft/s Use Water Velocity = 8 ft/s = 2.4384 m/s Determine Steam Inlet Temperature: At 0.06 MPa, Tsat = 85.94 oC
o
The average temperature of sea in South China Sea that is beside Bataan is 27 C.
Assumption: o
Outlet Temperature of cooling water = 55 C
Solving for LMTD:
ΔTmax = 85.94 oC – 27 oC = 58.94 oC o
o
o
ΔTmin = 85.94 C – 55 C = 30.94 C
LMTD = (ΔTmax - ΔTmin) / ln (ΔTmax / ΔTmin) LMTD = (58.94 - 30.94) / ln (58.94 / 30.94) o
LMTD = 43.45 C
Solving for total Area of heat transfer:
Q c = 329352.12 kJ/s A = Q c / (LMTD x U) o
2
A = 329352.12 kJ/s / (43.45 C x 17.44 kW / m -K)
A = 434.67 m
2
Solving for No. of Tubes: Assumption: No. of Pass = 2 L = 3.05 m (Maximum Length)
No. of Tubes = A / (πDL x No. of Pass) D = 1” = 0.0254 m L = 3.05 m No. of tubes = 434.67 / (π x 0.0254 x 3.05)(2) No. of tubes = 892.99 tubes
No. of tubes = 893 Tubes
Condenser Water Velocity From fig 8-9 of Power Plant Theory and Design page 351 by Philip Potter For inlet water of 26 °C = 78.8 F @ 1 in diam. C = 263
Get the coefficient of heat transfer,
Water Velocity
√
7.92 ft/s = 8 ft/s
Corrected Coefficient of Heat Transfer From Power Plant Theory and Design by Philip Potter, page 356. “It is good to practice to use 7 – 7.5 fps velocity." Use 7 fps to have more heat transfer.
√ √
COOLING TOWER CALCULATIONS Solving for Cooling Water Flow Rate: Q = mcwCp∆T Where: Q = 329352.12 kJ/s Cp = 4.1868 kJ/kg-K
mcw = Q / Cp∆T mcw = 329352.12 kJ/s / (4.1868 kJ/kg-K) (55-27) K
mcw = 2809.44 kg/s
Volume Needed For The Cooling Tower P = 0.06 MPa -3
3
Vf = 1.01x10 m /kg Vw = mw * Vf -3
3
Vw = 2809.44 kg/s (1.01x10 m /kg) 3
Vw = 2.84 m /s
Determining the Cooling Water Volume Flow Rate, V W
Cooling Tower Make-Up Water
From Power Plant Engineering by Frederick Morse, Page 179. “The makeup water is 2-5% of the water flow.” Use 3% make-up water Vwc = 2.84 * 0.03
Cooling Water Pump Capacity P = 0.06 MPa -3
3
Vf = 1.01x10 m /kg Vw = mw * Vf Vw = 2809.44 kg/s (1.01x10-3 m3/kg) 3
Vw = 2.84 m /s 3
ρ = 990.1 kg/m
3
ϒ = 990.1 kg/m x 9.81 N/kg(1KN/1000N) 3
ϒ = 9.71 kN/m
Assumptions: Friction Head = 15 m Static Head = 7.5 m Length of Pipe = 250 m approx.
Note: Cooling Water to be used if coming from South China Sea
P = Vf x ϒ x H 3
3
P = 2.84 m /s x 9.71 kN/m x 22.5 m P = 620.65 kJ/s P = 621 KW
PUMPS
Solving for Pump Capacity:
Pump 1 Q p1 = ms (1-m1-m2-m3-m4-m5-m6)(v11)(P12-P11) Q p1 = 304.51(1.033)(0.52)(3.53 - 0.06)
Q p1 = 567.59 kJ/s
Assumption: TDH = 10 m 3
Note: Specific Weight = 9.81 kN/m
3
V = 567.59 (kN-m)/s / (9.81 kN/m * 10 m) 3
V = 5.79 m /s 3
V = 20829.99 m /hr = 91710
Use Torishima Pump. (See A-6)
Pump 2: Q p2 = ms (1 - m6)(v17)(P18 - P17) Q p2 = 304.51(0.66)(1.236)(18 - 3.53)
Q p2 = 3594.45 kJ/s Assumption: TDH = 15m 3
Note: Specific Weight = 9.81 kN/m
3
V = 3594.45 (kN-m)/s / (9.81 kN/m * 15m) 3
V = 24.43 m /s 3
V = 87937.62 m /hr = 387200
Use Torishima Pump. (See A-6)
COAL STORAGE
Mass Of The Active Pile Coal Storage To Accumulate The Coal For a 15 Days Supply
mf = 51.26 kgfuel/s(3600s/hr) =
Volume Of The Active Pile Coal Storage To Accumulate The Coal Approximate coal weight is 800 - 929 kg/m3 The average is 864.5 kg/m3
Active Pile Coal Storage Area Use storage depth of 4 m
Bunker/Silo Capacity Storage silos are typically 10 to 90 ft (4 to 30 m) in diameter and 30 to 275 ft (10 to 84 m) in height
Bunker/Silo Capacity for 1 Day Operation
Height Of The Bunker For A Diameter Of 2m
BACK-UP FUEL
Back-Up Coal Mass From Power Plant Engineering by Frederick Morse page 441 “A storage of 10% of the annual consumption might suffice for most cases or e mergency reserved.” For thermal power plants, there are 45 days allotted for maintenance purposes per year.
of coal that will be used for back –up
Volume Of The Back – Up Coal Storage To Accumulate The Coal Approximate coal weight is 800 - 929 kg/m3 The average is 864.5 kg/m3
Back-Up Coal Storage Area: Use storage depth of 10 m
STACK SPECIFICATION from Sual power plant H= 226 m
Dry average Flue gas temperature of 300 °C at the Stack and 32 °C outside air temperature
Density Of Flue Gas,
() Density Of Air,
Draft Head, hw
( )
Average Air Outlet Velocity From Power Plant Theory and Design by Philip Potter, page 306 “Outlet velocities must be in the range of 15 – 30 fps. For safety purposes use the average velocity:
Specific Volume Of Flue Gas
Volume Flow Rate at 5m Diameter For a stack diameter of 5m, determining the volume flow rate, Q FG
AIR AND FUEL NEEDED FOR COMPLETE COMBUSTION
Table 5-4 Analysis of Typical American Fuels by Morse Proximate Analysis Moisture SemiAnthracite
1.28
Ultimate Analysis – Moisture Included
Volatile
Fixed
Matter
Carbon
12.84
73.69
Ash
Sulfur
Hydrogen
Carbon
Nitrogen
Oxygen
12.21
2.01
3.74
77.29
1.37
3.36
Theoretical Air Fuel Ratio
Since the Ultimate Analysis of coal is not available, an approximate formula to obtain the theoretical airfuel ratio where the heating value of the fuel is known:
Percentage Excess Air Needed for combustion
From Power Plant Theory and Design by Philip Potter page 191. “The amount of excess air that is required by a boiler depends on many factors, including the type of burner, fuel, and combustion chamber. Well-designed pulverized coal boilers may operate with complete combustion as slightly less than 15% excess air; i.e., 15% more than the theoretical air calculated from the chemical equations. Other boilers may re quire 50% or even 100% excess air”. Use 50% excess air for safety purposes.
Actual Air Fuel Ratio The weight of the air supplied for combustion is necessarily in excess of what is theoretically required. The volumetric analysis of the dry flue gas c an be used to calculate the actual weight of air.
Percentage Excess Air Corrected
Mass Flow Rate of Air Needed For Complete Combustion
] [
Average Temperature Needed To Dry Of The Coal From Power Plant Theory and Design by Philip Potter page 247 ”Air supplied to a stoker-fired furnace should not exceed 300 F or 350 F at the most; otherwise, stoker war page and maintenance will be too high. Pulverized units can employ 500 F to nearly 700 F air if the coal is wet.
Since pulverized coal is to be used,
Air Volume Flow Rate
ASH Ash To Be Collected Per Day Ash content of 18% for the fuel to be used.
FLYASH
Fly Ash To Be Collected Located Under The Precipitator From Power Plant Engineering by Frederick Morse page 341. For pulverized coal there is “entrainment of 60-70% of the ash as Fly ash”
Use 70% for safety purposes
Average Density Of Fly Ash 3
Fly ash density ranges from 1,000 – 1,400 kg/m
Volume Of The Fly Ash Silo
Height Of The Fly Ash Silo For A Diameter Of 3m Two bottom ash silos should be installed
Since the height of a single bottom ash silo is too high. It is recommended to use 3 silos to be collected twice a day.
BOTTOM ASH Bottom Ash To Be Collected Located Under The Boiler Since 70% of the ash is the fly ash therefore the other 30% that will complete the total number of ash is the bottom ash.
Average Density Of Bottom Ash 3
Bottom ash density ranges from 1,200 – 1,500 kg/m
Volume Of The Bottom Ash Silo
Height Of Bottom Ash Silo For A Diameter Of 3m Two bottom ash silos should be installed
CONVEYOR
Speed = 150 m/min No. of hours to fill the silo = 5 hours
Conveyor Capacity = (
3
3
m x 864.5 kg/m x 1ton/1000kg) / 5 hours
Conveyor Capacity = 132.87 tons / hour
Length of Conveyor: L1 = L2 = 40 m
From PPE by Morse, Table 12-2:
Where S = speed of Conveyor, m/min b = width of belt, cm Conveyor Capacity = capacity of conveyor, ton/hr
Solving for width of belt:
132.87 tons/hr= 0.000404(150m/min) b
2
b = 46.82 cm
Solving for the Driveshaft Horsepower: Re: P = [{(L + 45.72) /9000} {0.06KS + T}]
Where: K = horsepower constant T = capacity of conveyor, ton/hr S = speed of conveyor From PPE by Morse, Table 12-3:
K = 31.3 P = [{(40 + 45.72)/9000} {0.06(31.3)(150) + 132.87}]
P = 3.95 hp
Solving for the Tripper Horsepower:
b= 46.82 cm THP = 1.03+0.0045T THP = 1.03+0.0045(132.87)
THP = 1.63 hp
Belt type = 5 ply, 32 duck
Note: No. of plies provided by the table are for 4, 6 and 8 plies. Therefore interpolate the value of Minimum Pulley diameter for a 5 ply belt.
Minimum Pulley Diameter = 76.2 cm
