# Boiler Calculation

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BOILER CALCULATION

Steam Rate

SR = 3600/Wnet

SR = 3600/(Wt – 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 – 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

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

ms = nπ ID2/4 (vf/vw)

304.51 = nπ (0.045)2/[4(0.00184/1)]

n = 352.29 tubes

Use 353 tubes

Length of Tubes

A = nπ OD L = 7580.57 m2

6781.65 m2 = π 353 (0.047m)L

L = 130.11 m

Number of Turns

Assumption:

W = 5 m

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

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

at 7.97 MPa, 294.8 oC

v = 0.001383 m3/kg

V = v (mass)

V = 0.001383 m3/kg (304.51 kg)

V = 0.421 m3 = π (D2/4) (21.18 m)

D = 0.159 m = 159 mm

At 18 MPa, Saturated Liquid

vf = 0.00184 m3/kg

V = 0.00184 m3/kg (304.51 kg)

V = 0.56 m3 = π (D2/4) (21.18 m)

D = 0.183 m = 183 mm

SUPERHEATER CALCULATIONS

Tg1= 932.10 oC (temperature inlet of economizer)

T3 = 357.1 oC

T4 = 538 oC

H1 = 3381 kJ/kg

Hb = 2509 kJ/kg

v at 538 oC = 0.01824 m3/kg

Solving for Tg1

QSH = ms(h1 – hb) = mfgCpfgΔTg

= 304.51 kg/s (3381 kJ/kg – 2509 kJ/kg)

QSH = 265532.72 kJ/s

where:

Cpfg = 1.196 kJ/kg-oC at 538C

265532.72 kJ/s = 500 kg/s (1.196 kJ/kg-C) ΔTg

ΔTg = 444.03  oC

Tg1 = 444.03 oC + 932.10 oC

Tg1 = 1376.13 oC

Solving for LMTD

ΔT1 = 1376.13 oC – 538oC

ΔT1 = 838.13 oC

ΔT2 = 932.1 oC – 357.1 oC

ΔT2 = 575 oC

LMTD = (ΔT2 – ΔT1) / ln (ΔT2 / ΔT1)

LMTD = (575- 838.13) / ln(575/838.13)

LMTD = 698.32 oC

Note:

For Flue gas and water heat exchanging, U ranges from 30 to 100 W/m2 C

Therefore,

Use Average U = 65 W/m2C.

Solving for the Total Area of Heat Transfer:

A = Q / (LMTD x U)

A = 265532.72 kJ/s / (698.32 oC x 65 W / m2- oC)(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:

ms = nπID2/4(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

304.51 kg/s = nπ(0.075 m)2/4(0.01824 m3/kg / 10 m/s)

n = 125.72 tubes

n = 126 Tubes

Solving for the Length of Tubes:

A = nπODL

5849.91 m2 = 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

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

v = 0.00749 m3/kg

V = v x mass

V = 0.00749 m3/kg  (304.51 kg)

V = 2.28 m3 = πD2/4 (2.28 m)

D = 1.128 m

D = 1128 mm

L = 2.28 m

Mass = 304.51 kg

At 18 MPa, 538 oC

v = 0.01824m3/kg

V = 0.01824 m3/kg (304.51 kg)

V = 5.55 m3 = πD2/4 (2.28 m)

D = 1.761 m

D = 1761 mm

AIR PREHEATER CALCULATIONS

Tg1 = 600 oC (inlet temperature of flue gas)

Tair-inlet = 26 – 30 oC (Mean temperature of air of in Bataan)

Use average temperature of 28 oC.

From Power Plant Engineering by P.K. Nag,

“The temperature of heat air ranges from 280 to 400 oC.”

Therefore, Use average temperature of 340 oC.

Tleaving-air = 340 oC (Average temperature of leaving air)

Solving for Tg2:

QAP = maCp(Tleaving-air – Tair-inlet)

QAP = 255.73 kg/s (1.005 kJ/kg-C)(340 oC – 28 oC)

QAP = 80186.70 kW

QAP= mfgCpfg(Tg1 ­- Tg2)

80186.7 kW = 600 kg/s (1.151 kJ/kg-C)ΔTg

ΔTg = oC

Tg2 = 600 oC – 161.111 oC

Tg2 = 483.89 oC

Solving for LMTD:

ΔT1 = 600 oC – 340 oC

ΔT1 = 260 oC

ΔT2 = 483.89 oC – 28 oC

ΔT2 = 455.89 oC

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:

For gas to air heat transfer, U ranges from 30 to 60 W/m2 C.

Therefore,

Use Average U = 45 W/m2C.

Solving for the Total Area of Heat Transfer:

A = Q / (LMTD x U)

A = 80186.7 kJ/s / (348.83 oC x 45 W / m2- oC) (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 preheater tubes is about 50 mm.)

A = nπODL

5108.36 m2 = 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

H1 = f (L/D)(v2/2g)

H1 = 0.005(40m/2.5)(12 m/s)2/2(9.81 m/s2)

H1 = 0.587 m

H2 = 10 m

H3 = v2/2g = (12 m/s)2/2(9.81 m/s2)

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

V = 218.03 m3/s

Note:

At 28 oC, Specific weight (ϒ) of air is 11.502 N/m3.

P = VϒHT

P = 218.03 m3/s(11.502 N/m3)(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:

Permissible Range: 3000 – 4500 kCal/m2-K-hr

Multiplier: 3-5, So use 4 as multiplier.

Therefore, Use 3750 kCal/m2-K-hr as the average overall thermal coefficient.

Convert kCal/m2-K-hr to kW/m2-K:

U = 3750 kCal/m2-K-hr (1 BTU / 0.252 kCal) (1 kW-hr / 3413 BTU)

U = 4.36 kW/ m2-K x (4)

U = 17.44 kW/ m2-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

The average temperature of sea in South China Sea that is beside Bataan is 27 oC.

Assumption:

Outlet Temperature of cooling water = 55 oC

Solving for LMTD:

ΔTmax = 85.94 oC – 27 oC = 58.94 oC

ΔTmin = 85.94 oC – 55 oC = 30.94 oC

LMTD = (ΔTmax – ΔTmin) / ln (ΔTmax / ΔTmin)

LMTD = (58.94 – 30.94) / ln (58.94 / 30.94)

LMTD = 43.45 oC

Solving for total Area of heat transfer:

Qc = 329352.12 kJ/s

A = Qc / (LMTD x U)

A = 329352.12 kJ/s / (43.45 oC x 17.44 kW / m2-K)

A = 434.67 m2

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

Vf = 1.01×10-3 m3/kg

Vw = mw * Vf

Vw = 2809.44 kg/s (1.01×10-3 m3/kg)

Vw = 2.84 m3/s

Determining the Cooling Water Volume Flow Rate, VW

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

Vf = 1.01×10-3 m3/kg

Vw = mw * Vf

Vw = 2809.44 kg/s (1.01×10-3 m3/kg)

Vw = 2.84 m3/s

ρ = 990.1 kg/m3

ϒ = 990.1 kg/m3 x 9.81 N/kg(1KN/1000N)

ϒ = 9.71 kN/m3

Assumptions:

Length of Pipe = 250 m approx.

Note: Cooling Water to be used if coming from South China Sea

P = Vf x ϒ x H

P = 2.84 m3/s x 9.71 kN/m3 x 22.5 m

P = 620.65 kJ/s

P = 621 KW

PUMPS

Solving for Pump Capacity:

Pump 1

Q­p1 = ms (1-m1-m2-m3-m4-m5-m6)(v11)(P12-P11)

Qp1 = 304.51(1.033)(0.52)(3.53 – 0.06)

Qp1 = 567.59 kJ/s

Assumption:

TDH = 10 m

Note: Specific Weight = 9.81 kN/m3

V = 567.59 (kN-m)/s / (9.81 kN/m3 * 10 m)

V = 5.79 m3/s

V = 20829.99 m3/hr = 91710

Use Torishima Pump. (See A-6)

Pump 2:

Q­p2 = ms (1 – m6)(v17)(P18 – P17)

Qp2 = 304.51(0.66)(1.236)(18 – 3.53)

Qp2 = 3594.45 kJ/s

Assumption:

TDH = 15m

Note: Specific Weight = 9.81 kN/m3

V = 3594.45 (kN-m)/s / (9.81 kN/m3 * 15m)

V = 24.43 m3/s

V = 87937.62 m3/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 emergency 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,

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, QFG

AIR AND FUEL NEEDED FOR COMPLETE COMBUSTION

Table 5-4 Analysis of Typical American Fuels by Morse

Proximate Analysis

Ultimate Analysis – Moisture Included

Moisture

Volatile Matter

Fixed Carbon

Ash

Sulfur

Hydrogen

Carbon

Nitrogen

Oxygen

Semi-Anthracite

1.28

12.84

73.69

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 air-fuel 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 require 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 can 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

Fly ash density ranges from 1,000 – 1,400 kg/m3

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

Bottom ash density ranges from 1,200 – 1,500 kg/m3

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 = (m3 x 864.5 kg/m3  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) b2

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

PUMP CALCULATION

Wp1 = (hb9-h9)(1-m2-m3-m4-m5) = (1 – 0.0089 – 0.0803 – 0.0619 – 0.0192)(361.2 – 359.8)(674.32)

= 783.29 KJ/s

Wp2 = (hb12-h12)(1-m2) = (1 – 0.0089)(836.5 – 825.5)(674.32)

= 6805.30 KJ/s

Assumption

TDH1 = 10 m

TDH2 = 20 m

Note: Specific Weight = 9.81 kN/m3

V1= 783.29 (kN-m)/s / (9.81 kN/m3 * 10m)

V1= 7.98 m3/s

V1=28744.98 m3/hr

V2 = 6805.30 (kN-m)/s / (9.81 kN/m3 * 20m)

V2 = 34.69 m3/s

V2 = 124867.89 m3/hr

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Boiler Capacity

BC = ms (h1-h18) = 304.51 kg/s (3381 – 1070)KJ/kg

BC = 703722.61 KJ/s

Coal flow rate

Fuel to be Used Bituminous

Table 5-4 Analysis of Typical American Fuels by Morse (p.127)

Proximate Analysis

Ultimate Analysis – Moisture Included

Moisture

Volatile Matter

Fixed Carbon

Ash

Sulfur

Hydrogen

Carbon

Nitrogen

Oxygen

Ill. Bituminous

13.31

33.62

41.34

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 – O/8) + 9304 S

HHV = 33820 (0.5907) + 144212 (0.0519 – 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 – 0.1931/8) + 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 – hf)/2257 = (3381 – 1070) / 2257

FE = 1.02

Developed Boiler Horsepower

Developed Bo. HP = ms(hs – hf)/35322 = 304.51 (3381 – 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

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