II. B. Design Computation Relevant to Boiler A Capacity

II.B.1 Combustion Analysis:

Expected boiler performance:

MCR 220,000 lb/hr

Operating Steam Pressure - 250 psig with 92°F SH

Superheat Temperature - 506ºF

Feedwater Temperature - 221º

Ultimate Analysis of bagasse in % by weight

Australia practice,

Component % Weight

Carbon, C 23.50

Hydrogen, H2 3.25

Oxygen, O2 21.75

Moisture 50.00

Ash 1.50

Total 100.00

The combustible element in the bagasse are

Carbon & Hydrogen, burning it gives off

By weight:

C + O2 = CO2 and 2H2 + O2 = 2H2O

12 + 32 = 44 4 + 32 = 36

1 + 2.67 = 3.67 1 + 8 = 9

Consider combustion of one (1) pound bagasse without

excess air:

0.235 lb C requires 0.6274 lb O2 to produce 0.8624 lb CO2

0.0325lb H2 requires 0.26 lb O2 to produce 0.2925 lb H2O

1

However, the bagasse already contain 0.2175 lb O2. Hence,

the air must supply theoretical O2 of:

0.6274 lb + .26 lb – 0.2175 lb = 0.6699 lb O2

Assumed:

O2 (Oxygen) is 23% by weight of air

N2 (Nitrogen) is 76% by weight of air

Other Gases is 1% by weight of air

(Source: Encarta Premium 2007)

Therefore

Theoretical air required : 0.6699/0.23 = 2.9126 lbs

Nitrogen in air, N2 : 2.9126(0.77) = 2.2453 lbs

Allow 40% excess air to insure complete combustion. Flue

gas components for one (1) pound bagasse fuel

CO2 0.8624 lb Product of combustion

H2O 0.7925 lb Product of combustion &

moisture in bagasse

O2 (0.6699 X 0.40) 0.2679 lb Excess Oxygen

N2 (2.2453 X 1.40) 3.1434 lb Combustion of air

Total = 5.0662 lb Flue gas/lb bagasse

Calorific Value of Wet Bagasse:

with w moisture % bagasse 50, s sucrose % bagasse 2.5

Gross Calorific Value, GCV = 8,280(1-w) – 2,160s

GCV = 8,280(1-0.50)- 2,160(0.025)

= 4,140 - 54

2

GCV = 4,086 Btu/lb

Net Calorific Value, NCV = 7,650-8730(w)- 2,160(s)

= 7,650-8730(0.5)-2,160(0.025)

NCV = 3,231 Btu/lb

The projected flue gas temperature after economizer is

330°F. The mean specific heat at average flue gas

temperature T,

T = (330°F - 32°F)/2

= 149°F

N2: 0.246 + 0.000011T = 0.2476

O2: 0.214 + 0.000010T = 0.2155

H2O: 0.468 + 0.000087T = 0.4809

CO2: 0.199 + 0.000046T = 0.2058

Heat losses in one (1) pound of bagasse fuel:

a) Latent heat of H2O from combustion of H2 in the

bagasse.

b) Latent heat of H2O contained in bagasse.

Note: Heat losses in items a) and b) are already

accounted in the NCV formula.

c) Sensible heat loss of the flue gas leaving the air

heater , qa

Assume flue gas temperature before air pre-heater 680°F

(360°C), and after air pre-heater 520°F (271°C).

N2 : 0.2476(680-520)(3.1434) = 124.1 Btu/lb

O2 : 0.2155(680-520)(0.2679) = 9.2 Btu/lb

3

H2O : 0.4809(680-520)(0.7925) = 60.9 Btu/lb

CO2 : 0.2058(680-520)(0.8624) = 28.4 Btu/lb

Total sensible heat loss, qa = 222.6 Btu/lb

d) Losses in unburned solids

Factor α = coefficient to account unburned solids

= 0.975 for spreader stoker

e) Losses by radiation from furnace and esp. from boiler

Factor β = coefficient to account radiation &

convection

= 0.97 for normal rating

f) Losses due to incomplete combustion

Factor η = 0.96 average value and will be better

with lower moisture, lower excess

air, and higher furnace temp.

The modern furnace easily exceed

0.90 and for well conducted

combustion, use 0.96 average.

g) Losses due to flue gas exiting the ID fan to the

chimney, qc

qc = [(1-w)(1.4 m – 0.13) + 0.5](t-32)

where: w = bagasse moisture,0.50

m = ratio of air supplied to

theoretically needed, 1.4

t = flue gas temp. at ID fan

exit, 330°F

qc = 407.5 Btu/lb

Net heat transferred to steam, M

M = (NCV – qc) α β η

4

Mi = Btu/hr

= (3,231 – 407.5) X

(0.975)(0.97)(0.96)

= 2,563.5 Btu/lb

Over-all boiler thermal efficiency, ρ

ρ = M / GCV X 100%

= 2,563.5 / 4,086 X 100%

= 62.7 %

II.B.2. Bagasse Consumption Calculation:

Steam outlet condition, h1

h1 = 1,265.08 Btu/lb 265 psia & 506°F

Feed water inlet condition, hf

hf = 189.23 Btu/lb 221°F

Heat absorbed by steam, Ms

Ms = Ws (h1 - hf) Btu/hr

= 220,000(1,265.08-189.23)

= 236,687,000 Btu/hr

Boiler heat input, Mi

Heat absorbed by the steam

Over-all boiler thermal eff.

= Ms / q

= 236,687,000 Btu/hr / 0.6273

= 377,310,696.6 Btu/hr

Bagasse consumption, Wb

Wb = Boiler heat input / GCV of bagasse

5

= 377,310,696/ 4,086

Wb = 92,342.3 lbs/hr

II.B.3. Flue Gas Condition At Air heater Outlet

At 520 air pre-heater outlet temperature and

approximately atmospheric pressure,

1. A mol is weight of the substance , lb/molecular

weight of the substance

2. A mol volume is 380 cu. Ft. at 60°F and 14.7 psia.

This is independent of the kind gas.

3. At T °R and p psia, the mole volume is

Mol volume = 10.72 T / p Where:

T = 520 + 460 absolute temp.

= 980°R

p = 14.7 psia

Mol volume = 10.72 (980) / 14.7

= 715 cu.ft/mol

Molar weight of flue gas per pound bagasse:

CO2 = 0.8624 / 44 = 0.0196

H2O = 0.7925 / 18 = 0.0440

C02 = 0.2679 / 32 = 0.0083

N2 = 3.1434 / 28 = 0.1123

Total = 0.1842 Mol/lb bagasse

Volume of flue gas per lb bagasse, Vf

Vf = (Mol volume)(Molar weight flue gas/lb bag.

= (715 cu.ft/mol)(0.1842 Mol/lb bagasse)

= 131.7 cu.ft/lb bagasse

Volume of flue gas, V

6

V = Volume of flue gas per lb bagasse X

Bagasse consumption lbs/hour

= 131.7 X 92,342.3

= 12,161,482.8 cu.ft/hr

= 202,691 cu.ft/min

Weight of flue gas, Wf

Wf = Bagasse consumption X Weight of flue gas

per lb bagasse, lb/hr

= 92,342.3 X 5.0662

Wf = 467,824.6 lbs/hr

= 212,647.5 kgs/hr

II.B.5. Evaporation Rate of the Boiler:

All water tube boilers are capable of normal rate of

4.4 lb/ft2/hr. Boiler with vertical bent tubes,

economizer, air heater and spreader stoker furnace a rate

of 8.7 can be achieved. If the economizer and air heater

is of generous dimension, a rate of 10 is possible.

Evaporation rate, τ

τ = Steam generated / boiler heating surface

= 220,000 lb/hr / 28,913 ft2

= 7.61 lb steam/ft2/hr

Note: This figure is within the 7-8 lb/ft2/hr typical of

spreader stoker furnace.

II.B.6. Furnace Design:

a) Combustion Rate:

7

Heat releases in the furnace is advisable to be about

25,000 Btu/cu.ft./hr and should not go above 40,000

reckoned on GCV. Furnace area heat release should be

within 600,000 to 900,000 Btu/ft2/hr of grate area.

Boiler A furnace dimensions:

Grate area = 15.916 ft X 30.687 ft

= 488.41 ft2

Furnace volume = Grate Area X furnace height, cu.ft.

= 488.41 X 31.5

= 15,384.9 cu.ft

Total heat of bagasse based on GCV, Hb

Hb = GCV X Bagasse consumption, Btu/hr

= 4,086 Btu/lb X 92,342.3 lb/hr

= 377,310,637 Btu/hr

Furnace volume Heat release, FVHR

FVHR = Total heat of bagasse based on GCV /

Furnace volume, Btu/cu.ft/hr

= 377,310,637 / 15,384.9

= 24,524.7 Btu/cu.ft/hr

(This figure is within the value

acceptable for combustion rates)

Furnace area heat release, FAHR

FAHR = Total heat of bagasse based on GCV /

Grate Area, Btu/ft2/hr

= 377,310,637 / 488.41

= 772,528 Btu/ft2/hr

8

(This figure is also within range)

Weight of bagasse burnt per unit grate area, β

For spreader stoker furnaces operating at continuous

ratings, combustion rate per unit grate area for high

rate is between 160-200 lb/ft2/hr, preferably should not

exceed 175. For Boiler A as given by the following

formula,

β = Weight of bagasse burnt per hour /

area of the grate

= 92,342.3/ 488.41

= 189 lb/ft2/hr

(slightly above the recommended value,

Of 175, but still ok since suspension

firing is employed)

Ratio of boiler heating surface to grate area, σ

This ration is of the order 50-70. For spreader

stoker it is in the neighborhood of 50.

σ = heating surface of the boiler /

area of the grate

= 28,913 / 488.41

= 59 (slightly above the recommended)

II.B.7. Design of superheater

a) Steam velocity at superheater tubes:

Verify existing loop:

Data:

55 heating elements, 2” diameter 0.157” thick

250 psig outlet steam pressure, 506°F temp.

9

220,000 lb steam/hr (upgraded capacity)

Specific volume of steam, 2.0362 cu.ft./lb

Volumetric flow of steam, Vfs

Vfs = 220,000 X 2.0362 / 3600

= 124.43 cu.ft/sec

Flow area of 55 tubes with 1.4351” inside diameter, Af

Af = π X 1.4351 X 1.4351 X 55/4 (1/144)

= 0.6177 ft2

Velocity of steam passing the 55 tubes, Vs

Vs = Vfs / Af ft/sec

= 124.43 / 0.6177

= 201.44 ft/sec (within acceptable

range of 130-210 ft/sec for super

heated steam per European practice)

II.B.8. Temperature of steam at given heating surface

For 55 tubes 2” diameter with effective heating

length of 30 ft, the heating surface area is 728 ft2.

From Handbook by Hugot p.883, furnace temperature

obtained in practice for a bagasse fired boiler is:

Most inefficient furnace - 1500-1800°F

Highest recorded - 2,350°F

Continuously - 2.275°F

Most Common - 2,000°F

Super heater gas inlet - 1,500-1800°F

For Boiler A calculation assume 2,000°F furnace

temperature and 1,700°F superheater inlet gas

temperature.

10

a) Superheater outlet steam temperature, Ts

K S(2T1-t)/2 + p[ct-r(1-x)][1+S K/(2 ά P C)] Ts =

(K S/2) + p c [ 1 + K S /(2 ά P C) ]

Where: Ts = temperature of superheated steam, °F

S = heating surface area of superheater,

728 ft2, or 67.66 m2

K = coefficient of heat transfer, assume

11 Btu/ft2 hr °F or 60 kcal/m2 °C hr

t = temperature of saturated steam at drum

pressure of 280 psia, 411°F or 210.5°C

c = mean specific heat of superheated steam,

0.468 + 0.000087 (411°F)

= 0.503 kcal/kg °C

r = latent heat of vaporization at the

boiler drum pressure of 280 psia,

454 kcal /kg

C = mean specific heat of flue gas

= 0.27 + 0.00003 Tm

Where Tm = (1,700°F-32°F)/2

= 834°F

C = 0.303 kcal/kg °C

p = weight of steam to be superheated,

220,000 lb/hr or 100,000 kg/hr

P = weight of flue gas passing the super-

heater, 467,824.6 lb/hr or

212,647.5 kg/hr

T1 = temperature of flue gas at entrance of

11

super heater, 1,700°F or 926.6 °C

T2 = temperature of flue gas leaving the super

heater

ά = coefficient, equal or less than 1,

generally 0.9

x = dryness fraction of the saturated steam

(0.8-0.98 in general), use 0.93

Substituting all the given value to the formula,

Ts = 265°C or 509°F

(the projected steam temperature at the

superheater outlet is 506°F)

b) Degree superheat, at the super heater outlet, ° SH

° SH = Ts – t

= 265.3°C – 210.5°C

= 54.8°C or 98°F

c) Temperature of flue gas leaving the superheater, T2

T2 = T1 – p [ (1-x)r + c(Ts - t) ] / ά P C

(eq. 42.63 p.907 Handbook by Hugot)

= 819.9°C or 1507.8°F

d) Pressure drop at superheater, Pd

Data: 55 elements, 2” OD 0.157 thick but inlet and

outlet spool connected to the superheater

headers has OD of 1.75” with thickness 0.157”

approximately 33 ft long, flow area per

element 0.01123 ft2, 220,000 lb/hr MCR at

12

250 psig operating pressure and 506°F super

heat temperature, density of steam at

operating condition 0.4911 cu.ft/lb.

Calculation:

Vfs = 124.43 cu.ft/sec (previously computed)

Af = 0.6177 ft2 (previously computed)

Ws = steam flow per element

= 220,000 lb/hr / 55 tubes

= 4,000 lb/hr

f = flow friction coefficient

= 0.0054 + 0.0375 z /( D V d )

(English units)

(eq.14-15, for the flow of steam and air

in pipes, p. 605 Power Plant by Morse)

where :

z = viscosity of steam

= 0.0083 + (2 t / 100,000)

(eq. 14-16, viscosity of

Steam, p. 606 Power Plant

by Morse), t = 506°F steam

superheat temperature.

= 0.01842 centipoises

D = tube inside diameter

= 1.4351 ”

V = Velocity of steam passing

the superheater tube, Vs

= 201.4 ft/sec

13

d = density of steam

= 0.4911 lb/cu.ft

Substituting all variables in the eq. for flow friction

coefficient, f

f = 0.005405

Friction loss in psi for the flow of steam or air, Pd

Pd = f L d V² / 6 g D

(eq. 14.11 p.605 Power Plant

by Morse)

where: (see above data)

f = 0.005405

L = tube length

= 32.5 ft

d = density of steam

= 0.4911 lb/cu.ft

V = flow velocity

= 201.4 ft/sec

g = 32.2 ft/sec²

D = tube ID

= 1.4351 ”

Substituting all the variables in the eq. for Pd,

Pd = 12.62 psi

II.B.9. Design of air pre-heater

This type of equipment recovers sensible heat from

combustion gases before it passes to the economizer

assembly as in the case of Boiler A. Combustion air

temperature of 482°F can be attained but not considered

14

wise to exceed 356°F, otherwise damage will be caused to

the grate bars and refractory in the furnace due to

resulting high temperature.

Verify heating surface area with the given air pre-

heater air inlet and outlet temperature.

Data:

Total surface area of 2.5” OD, 0.126” thick, 2,054

pieces, 12 ft long - 16,150 ft2

Gas flow - 467,824 lb/hr

Temperature of gas at outlet of air pre-heater,

520°F

Temperature of gas at inlet of air pre-heater,

680°F

Ambient temperature, 86°F

Tubes are arranged in a single bank 79 tubes

wide x 26 tubes deep, in two passes

Weight of air at 40% excess, Wa

Wa = 2.9126 X 1.4 X 92,342 lb/hr

= 376,537 lb/hr

S = α P C / k (r – 1) ln (T - t◦ ) / (To – t)

where S = total heating surface required

P = 467,824.6 lb/hr (flue gas)

p = Wa, 376,537 lb/hr

α = 0.92 – 0.95, say 0.935 for air

heaters of metal with effective

circulation.

C = specific heat of the gas

= 0.27 + 0.00003 T, T = 600°F

15

= 0.288 Btu/lb °F

c = specific heat of air

= 0.24 Btu/lb °F

r = α P C / p c

= 1.397

k = coefficient of heat transfer in the

air heater, 3 – 4 Btu/ft2/h/°F, say

3.7 Btu/ft2/hr/°F

T = 520°F

t = 356°F

to = 86°F

To = 680°F

Substituting the values,

S = heating surface required, ft2

= 25,135 ft2

But the actual heating surface is only 16,150 ft2.

Examining the variables, it is the air and flue gas

temperature at the air pre heater outlet that would

change with the change in total heating surfaces.

Substituting again in the equation for total heating

surface required, with S = 16,150 ft2, and finding t,

heated air temperature after air pre-heater,

t = 320°F

II.B.10 Design of steam drum and water drum

Given:

Steam drum inside diameter - 5’-0”

Steam drum thickness - 1-1/4”

16

Approximate straight length - 33’-10-1/4”

Water drum inside diameter - 3’-9”

Water drum thickness - 1”

Approximate straight length - 32’-10-1/4”

a) Ligament efficiency of longitudinal joints, E

i) Screen, roof and down comer tubes

E = (Pt – dh) / Pt (100%)

where:

Pt = pitch of tube hole

= 6-1/4”

dh = diameter of hole

= 3”

E = (6.25 – 3) / 6.25 (100%)

= 52%

ii) Main bank tubes

E = (Pt – dh) / Pt (100%)

where:

Pt = pitch of tube hole

= 5-1/2”

dh = 2-1/2”

E = (5.5 – 2.5) / 5.5 (100%)

= 54.54%

The minimum ligament efficiency is at longitudinal

joint at screen, roof and down comer tubes, 52%.

b) From PSME Code of 1993, p.119, Article 7.3, section

7.3.1, the Maximum Allowable Working Pressure, MAWP for

steam drum

17

MAWP = TS x t x E / R x FS

where:

TS = Ultimate tensile strength of

shell plate @ 60,000 psi or

413.7 N/mm²

t = thickness of the drum

= 1-1/4” or 31.75 mm

E = minimum efficiency of drum

longitudinal joint

= 52%

R = ½ of the inside diameter of

the weakest course of the drum

= 5 ft x 12 x 25.4 / 2

= 762 mm

FS = Allowable safety factor

= 5

Substituting all the variables in the equation, MAWP

413.7 N/mm² x 31.75 mm x 0.52 MAWP =

762 mm x 5

= 1.79 Mpa or 260 psi

c) Verification water drum design

i) Efficiency of longitudinal joint, Ewd

Ewd = (Pt – dh) / Pt (100%)

= 54.54% (computed previously)

ii) Maximum Allowable Working Pressure for water

drum, MAWP

TS x t x Ewd MAWP =

18

R x FS

where:

TS = 413.7 N/mm²

t = 25.4 mm

R = 3’-9” /2

= 22.5” or 571.5 mm

FS = 5

MAWP = 2 Mpa or 290 psi

II.B.11. Verification of design of draught fans

a) Induced draft fan

Given:

Flue gas volume MCR - 169,000 cu.ft/min

Temperature at air pre-heater outlet - 520°F

Allow 20% margin on fan capacity-202,800 cu.ft/min

Previous fan water gage design – 9.36”

Previous fan Hp MCR - 370

Previous RPM - 920

Present fan Hp after upgrading - 640

Present RPM - 920

a.1) Gas pressure required:

(Total draft of a gas loop from p.477 Power

Plant Engineering MKS Units by Morse)

19

i) Furnace (D2) - - (5 mm) W.G.

ii) Boiler exit (D3) - - (30 mm) W.G.

iii)Air pre-heater outlet- - (80 mm) W.G.

iv) Dust collector (D3) - - (80 mm) W.G.

v) Economizer (D3) - - (80 mm) W.G.

vi) Scrubber exit (D3) - - (100 mm) W.G.

vii) Breeching/chimney/ducts- - (5 mm) W.G.

Total draft required – 380 mm

Add 40% to fan pressure:

380 mm X 1.4 = 532 mm

Note: item vii) draft represented by D4 was computed

separately for four different portions from outlet of air

pre-heater to outlet of induced draft fan using the

formula given below:

Air pre-heater outlet to connecting duct to

economizer - 0.0036 cm W.G.

Connecting duct to economizer - 0.09 cm W.G.

Economizer to induced draft fan - 0.0238 cm W.G.

Induced draft fan to chimney - 0.34 cm W.G.

Total D4 0.45 cm W.G.

or say 5 mm W.G.

D4 = (d f V² H) / 10 (2gR), cm

(eq. 12-3 p. 477 Power plant Eng’g by

Morse MKS units)

where:

D4 = draft loss due to friction in

in ducts, breechings, chimney

20

d = flue gas density at IDF inlet

temperature 300°F,

= 1.25 cu.m/kg

f = 0.014 for flue against

steel and concrete.

H = length of conduit, m

g = acceleration due to gravity,

9.7 m/sec

R = hydraulic radius of cross

Section, m (area/perimeter)

V = velocity of flue gas at

different cross sections

a.2) Fan power requirement, kWf

kWif = Vf x Ps / 102 (ή)

where:

Vf = flue gas volume flow, cu.m/sec

= 202,800 cu.ft/min

= 95.8 cu.m/sec

Ps = static pressure required, mm W.G.

= 532 mm W.G

ή = drive efficiency, 85%

kWif = 95.8 x 532 / 102 (0.85)

= 587 kW

a.3) The actual fan design is double inlet double width

backward curve blades centrifugal fan.

b) Forced draft fan:

Given:

21

Weight of air, Wa @ 40% excess air - 405,007 lb/hr

Density of air, δa @ 30°C (86°F) - 0.0727 lb/cu.ft

i) Volume of air @ MCR = Wa / 60 ( δ )

= 405,007 / 60 (0.0727)

= 92,721 cu.ft/min

ii) Required capacity = 92,721 cu.ft/min x 1.2

= 111,265 cu.ft/min

= 52.56 cu.m/sec

iii)Air pressure required:

Furnace - + (37 mm)

Air duct loss - + (25 mm)

Changes in direction- + (25 mm)

Across air heater - + (37 mm)

Total air pressure required = 127mm

Add 40% on air pressure required = 177.8

iv) Required power of the fan, kWfd

kWfd = Vf x Ps / 102 (ή); ή = motor eff.

= 52.56 x 177.8 / 102 (0.85)

= 107 kw, say 150 Hp

v) The actual fan design is double inlet double width

backward curve blades centrifugal fan 150 Hp 860 RPM

c) Secondary air fan:

22

This is the over-fire air fan that at least will

deliver about 20% of combustion air required inside the

furnace.

Given:

Weight of combustion air - 92,721 cu.ft/min

Temperature of air - 86°F

Density of air at 86°F - 0.0727 lb/cu.ft

c.1) Required capacity = 92,721 (0.15) (1.2)

= 16,689.8 cu.ft/min

7.88 cu.m/sec

c.2) Required draft, Ps

i) Furnace - + (400 mm)

ii) Air duct loss - + (30 mm)

iii)Changes in direction - + (30 mm)

Total required - + (460 mm)

Add 25% on req. draft - + (575 mm)

c.3) Power requirement, kWsf

kWsf = Vf x Ps / 102 (0.85)

where:

Vf = 7.88 cu.m/sec

Ps = required draft, 723 mm

ή = 0.85 ( drive eff.)

kWsf = 10.83 x 575 / 102 (0.85)

= 52.3 kW or 75 hp

II.B.12 Specification of feed water pump:

Given:

Steam generation (MCR) - 220,000 lb/hr

23

(This is upgraded capacity)

Steam drum pressure - 280 psia

Pressure loss due to economizer - 15 psi

Pump suction line size - 6”

Pump discharge line size - 5”

a) Volume required = total capacity required/density

where:

density = 1/specific volume @ 250°F

= 1/0.017001

= 58.82 lb/cu.ft

MCR - 220,000 lb/hr

+ 10% for emergency make - 22,000 lb/hr

+ 10% for exceptional load 1 hr - 22,000 lb/hr

+ 5% for blow downs - 11,000 lb/hr

Total capacity required - 275,000 lb/hr

Volume required = 275,000 lb/hr / 58.82 lb/cu.ft

Volume required = 4,675 cu.ft/hr

= 2.208 cu.m/min

b) Head required:

i) Suction Head

Velocity head = v² / 2g

where:

v = volume flow /area

= 6.6 ft/sec

Velocity head = (6.6)(6.6) / 2 (32.2)

24

= 0.676 ft

Static head = 25 ft

Friction loss= loss due to 2 valves, 3 pc. 90°

elbows, 50 ft of 6” pipes.

- 2 valves fully open - 1m X 2

- 3 90° elbow - 2m x 3

- 60 ft of 6” pipe - 18m

Total equivalent length- 28m

= 2 f L v² / g D

(eq. 14-10 p.605 Power Plant

Eng’g by Morse MKS Units)

where:

f = coefficient of friction

0.424 = 0.0035+0.0007562(z/DVS)

(eq 14-13 p.605 Power Plant

Eng’g by Morse MKS Units)

z = 0.1850 centipoise for

water at 121°C as

extrapolated from Table

14-8 Viscosities p. 606

Power Plant by Morse)

D = pipe ID, 6” or 0.152 m

V = velocity at suction

= 6.6 ft/sec, 2 m/sec

S = specific gravity @

121°C

= 58.82/62.4

= 0.943

Substitute variables in equation for f,

25

f = 0.00412

Then, substitute variables in equation for Friction loss,

Friction loss = 2 f L v² / g D

= 0.625 m or 2 ft

Suction head = 25 ft - 0.676 ft – 2.05 ft

= 22.274 ft

ii) Discharge head

o Velocity head= v² / 2g

where:

v = volume flow/area

4,675 cu.ft/hr(4)144(1)hr =

π (5)² ft2 (3600) sec

= 9.5 ft/sec

Then, velocity head = 1.4 ft

o Pressure head= drum pressure x 2.31/sp. grav.

where:

drum pressure = 280 psi

sp. grav. = δ2 / δ1

= 58.82/62.4

= 0.943

Pressure head= 280 x 2.31 / 0.943

= 685.89 ft

o Static head = 35 ft (estimated)

o Friction loss= losses due to

* 3 valves (full open)- 1m x 3

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* 1 valve (3/4 open) – 7m x 1

* 8 90° elbow - 3m x 8

* 130 ft of 5” pipe - 40m

Total equivalent length- 34m

= 2 f L v² / g D

(eq. 14-10 p.605 Power Plant

Eng’g by Morse MKS Units)

where:

f = coefficient of friction

0.424 = 0.0035+0.0007562(z/DVS)

(eq 14-13 p.605 Power Plant

Eng’g by Morse MKS Units)

z = 0.1850 centipoise for

water at 121°C as

extrapolated from Table

14-8 Viscosities p. 606

Power Plant by Morse)

D = pipe ID, 5” or 0.127 m

V = velocity at suction

= 9.5 ft/sec, 2.89 m/sec

S = specific gravity @ 121°C

= 58.82/62.4

= 0.943

Substitute variables in equation for f,

f = 0.004079

Then, substitute variables in equation for Friction loss,

Friction loss = 2 f L v² / g D

= 1.88 m or 6 ft

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Discharge head = 685.9 ft + 1.4 ft + 6 ft

= 693.3 ft Then, Head required = Discharge head – Suction Head

= 693.3 ft – 22.274 ft

= 671 ft or 204.5 m

c) Power requirement = Q dw H / 4,500,000 (η)

where:

Q = 2,208 l/min (liters/min)

dw = 943 kg/cu.m

H = 204.5 m

η = 0.90 (motor efficiency)

2,208 x 943 x 204.5 Power requirement = Hp 4,500,000 x 0.9 x 0.70

= 150 Hp, say 200 Hp

d) But the actual pump rating is:

GPM (Boiler rating) - 400

GPM (Installed capacity) - 472

Rated head - 878 ft

RPM - 3,600

Hp (Turbine driven) - 190

HP (Electric motor) - 250

Single stage Worthington, single suction with

auto leak off non return valve at the discharge.

Therefore the existing pump is ok for the upgraded boiler.

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