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

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)

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

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

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

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

Outlet Header

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

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

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

For Outlet Header:

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

Solving for Heads

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