Exercise 7: Thermal Design of A Simple Boiler

In this exercise, calculation procedure for boiler design is presented. This simplification is by

no means satisfactory for engineering design of the practical facility, but is a reasonable example

to discuss some of the fundamental thermal processes in the furnaces with considerations on fuel

and combustion.

7.1 Modes of Heat Transfer

The importance of heat transfer in boiler design is self-evident. In the boiler we are concerned

with the conversion of chemical energy in the fuel to thermal/pressure energy of a steam flow

and the effectiveness of this conversion process will be strongly dependant on the heat transfer

processes involved. We can identify three modes of heat transfer

Conduction Convection Radiation

all of which have a part to play in the boiler. However, conduction is not considered in simple

boiler calculation.

Heat transfer by convection relies on the bulk movement of a heated fluid in relation to a

surface. It features strongly in the design of plant items such as superheaters and reheaters, where

there are gas-to-metal and metal-to-steam convective heat transfers.

qconv = h(T Tw)Aw

Heat transfer by radiation does not rely on the presence of an intervening medium. Radiation is

the transfer of heat through electromagnetic radiation in the infra-red range. Its contribution may

amount to 95% of total heat transfer in a furnace part and 30% heat transfer in a tube bank.

qrad = g(T4 Tw4)Aw

7.2 Heat and Mass Balance

In the control volume, heat and mass balance equation are shown as below.

q + mout hout mair hin ,air mfuel hfuel = 0 = +

From the LHV of a fuel: + = 0 Where,

= = ( 0) + 20 (2 02)

Reference temperature T0 is set to 25, or 298K,

and coefficient a and b can be calculated by

=

, =

(Mj is the mass fraction of each species of

the combustion gas)

Total heat transfer is the sum of the radiation plus the convection;

q=qconv+qrad

7.3 Example of Simple Boiler Calculation

7.3.1 Boiler Furnace

All heat transfer is assumed to be by radiation in the furnace.

Methane feed rate is 30kg/s at 25. LHV of methane is 4*107J/kg. Air flow is 570kg/s. The excess

air ratio(EAR) is 0.11. The air temperature is 200. The wall temperature is 300. The furnace

area is 3,000m2. The furnace volume is 10,000m3.

Find the temperature of flue gas leaving the furnace and heat transfer in the furnace.

Gas a

(J/kg K)

b

(J/kg K2)

CH4 1182.8 3.4714

CO2 1170.8 0.09336

H2O 1945.2 0.36366

O2 1027.8 0.07360

N2 1129.6 0.07016

Where, The emissivity of gas is shown as below:

Use gT=k(pL-0.015)n

H2O:CO2

Partial pressure ratio 1:2 1:1 2:1

T k n k n k n

1000 444 0.34 416 0.34 444 0.34

1500 540 0.42 495 0.40 540 0.42

2000 572 0.51 509 0.48 572 0.51

L=3.6V/A where L is the depth of the gas layer and A is all surface area of gas layer.

p=(H2O partial pressure + CO2 partial pressure) atm

Furnace is considered as well-stirred reactor model.

T=Tout.

T can be draw by the Newton-Rhapson method.

+ 1 =

7.3.2 Tube bank The flue gas flow is 600kg/s at 1435K. Heat transfer coefficient is 100W/m2K. The wall

temperature is 300. The tube bank area is 3,000m2. The tube bank volume is 250m3.

Find the temperature of flue gas leaving the tube bank and heat transfer in the

furnace.

Furnace

Convection is negligible

Fuel : Methane

30kg/s, 25

Hfuel=4x107J/kg

T.?

q=?

Air; 570kg/s

EAR=0.11, 200

Tw=300;

A=Aw=3,000m2

V=10,000m3

h=100W/m2K

A=Aw=9000m2

V=250m3

Tw=300

Flue gas

Tin=1435K,

600kg/s

T=?

q=?

7.4 Solution

7.4.1 Solution For the furnace radiation:

(i) Check the air fuel ratio

afr = (2(1+EAR)(32+3.76*28))/16=19.05

afr = 570/30=19

(ii) Exhaust gas composition :

CO2+2H2O+EAR*2O2+(1+EAR)(2*3.76N2)

Total above mass; m total = 1*44 + 2*18 + EAR*2*32 + (1+EAR)*7.52*28 = 321

Mole fraction of the species CO2, H2O, O2, N2 ;

mf CO2 = 44/321 = 0.137

mf H2O= 36/321 = 0.112

mf O2 = 7.04/321 = 0.022

mf N2 = 233.72/321 = 0.728

aout= 0.137*1182.8 + 0.112*1170.8 + 0.022*127.8 + 0.728*1129.6 = 1138.1,

bout= 0.137*0.09336 + 0.112*0.36366 + 0.022*0.0736 + 0.728*0.07016 = 0.1062

Exhaust partial pressures of H2O and CO2 ::

P CO2 = 1 / (1 + 2 + 2*EAR + 2*(1+EAR) *3.76) = 1/11.5672 = 0.086 atm

P H2O = 2 / 11.5672 = 0.174 atm

(iii) The energy equation

Correct following relation

() = g 5.67 108 (T4 5734) 3000 30 4 104 570 1105.7 (473 298) + 0.070972 (4732 2982) +600 [1225.2 (T 298) + 0.10662 (T2 2982)]

Furnace

Convection is negligible

Fuel : Methane

30kg/s, 25

Hfuel=4x107J/kg

F.O.T.?

Air; 570kg/s

r=0.11, 200

Tw=300;

A=Aw=3,000m2

V=10,000m3

() = 6.8 1043 + 64.07 + 7.35 105 (iv) Iteration

Begin with T1=1500K, Use gT=k(pL-0.015)n

Then from the foregoing table

T k n

1000 444 0.34

1500 540 0.42

2000 572 0.51

L=3.6V/A=12m; p=(0.087+0.174)atm

g(1500K)=540*(0.261*12-0.015)0.42/1500=0.580

& g(1000K)=0.654 1 = 1.284 108; 1 = 2.163 106 2 = 1500 1.284 1082.163 106 = 1441 Using interpolation, g(1441K)=0.589 2 = 1.197 107; 2 = 2.026 106 2 = 1435 2 = 1435 4 = 1435(1162) F.O.T. = 1435K

(v) q=? q = g4 4 = 4.15 108 q = q/ = 138/2

7.4.2 For the superheaters

L=3.6V/A=0.1m

pL=0.0261atm-m (slightly out of range!) mfuel = 0; mout = min =600kg/s

h=100W/m2K

A=Aw=9000m2

V=250m3

Tw=300Coals b

Flue gas

Tin=1435K,

600kg/s

T=?

q=?

() = 4 4 + () + [ ( ) + 2 (2 2)] (i) Iteration

1 = 1000, = 0.096, 1 = 7.44 107 1 = 1.96 106 T2 = 961K T3 = 961K T4 = 961K(688)! (ii) Heat transfer rate

= 4 4 + ( ) = 3.86 108 qrad : qconv=1:9

APPENDIX 7: Boiler Design Calculation

7.1 Combustion (stoichiometric) calculation

7.1.1 Outline

Combustion (stoichiometirc) calculations provide much of the basic information necessary of the

design of a boiler plant. They help find the amount of fuel to be fed for the required thermal

output of the plant. The specifications of fans and blowers are based on the air required for

burning of gasifying that quantity of fuel. Combustion calculations also give the amount of

limestone required to achieve a certain amount sulfur capture. Finally the solid and gaseous

pollutants produced from the combustion are computed from this. Most of the calculations are

based on overall chemical reactions.

7.1.2 Calculation

Heating value

- HHV=33,823C+144,249(H-O/8)+9,418S(kJ/kg)

- LHV=HHV-22,604H-2581Mf(kJ/kg)

Air flow rate calculation

- Theoretical dry air,Mda

Mda=[11.53C+34.34(H-O/8)+4.34S+AS](kg/kg coal)

For each unit mass of sulfur converted to calcium sulfate, an additional

amount of dry air A, is required. So the extra air for a unit weight of coal is AS,

where A is 2.16 for sulfur capture and is zero when no sulfur is captured as

calcium sulfate.

- Total dry air, Tda

Tda = MdaEAR (kg/kg burned),

- Total wet air, Mwa

Mwa = Tda(1+Xm),

In standard air this weight fraction of moisture Xm, is about 0.013 kg/kg air,

and Xm is the weight fraction of moisture in the air.

- Total flue gas weight

N2 Equation: Nitrogen from fuel and air = N+0.768Mda EAR

H2O Equation: 9H+ EAR MdaXm+Mf+LqXml

CO2 Equation: 3.66C+1.375SR(1+1.19(XMgCO3/XCaCO3))

SO2 Equation: 2S(1-Esor)

O2 Equation: O+0.2314Mda(EAR -1)+(1-Esor)S/2

Fly ash Equation: Ash*ac , ac is the fraction of the ash in coal as it appears as fly ash (0.1-0.5).

Total flue gas mass, Wc

= 0.2315 + 3.66 + 9 + + + + 2.5(1 ) +1.375(1 + 1.193 3 ) + ( )

Sorbent requirement

If the coal ash contains a negligible amount of calcium oxide, the sorbent required Lq, to retain

the sulfur in a unit weight of fuel is found from the following equation: = 100323

Where S is the weight fraction of sulfur in coal, and XCaCO3, is the weight fraction of CaCO3 in

the sorbent. R is defined as the calcium to sulfur molar ratio in the feed of sorbent and coal

respectively.

Sometimes the coal ash contains an appreciable amount of calcium oxide, which removes a part

fed, the inherent Ca/S ratio is 32XcaO/56S. Therefore the limestone required for removal of same

amount of sulfur (EsorS) will be reduced by the above amount. Thus , R is to be replaced by R as

below : = { 3256 }

Solid waste Produced

3 = + 2 1830( 3 )

3 = + 2 1183( 3 )

From above reaction we find that the sorbent decomposes into MgO and CaO. Out of this a

part of the CaO is converted into CaSO4. The spent sorbent would thus contain CaSO4,

unconverted CaO, unconverted MgO, and inert components of the sorbent. The weight of spent

sorbent produced per unit weight of coal burned Lw, is the sum of CaSO4, CaO, MgO, and inerts.

Spent sorbents = calcium sulfate + calcium oxide + magnesium oxide + inert

= 136 32 + 56 3100 32 + 40384 + Where Lq is sorbent fed per unit weight of coal burned and is given by Lq equation.

The total solid waste contains, in addition to the spent sorbent Lw, coal and ASH, and unburned

carbon (1-Ec), less the CaO content XCaO, of coal converted to CaSO4 and included in Lw. The

solid waste produced per unit weight of fuel burned is thus:

Wa=[Lw+ASH+(1-Ec)-XCaO]

Where, Ec is the combustion efficiency expressed as a fraction.

7.2 Heat balance on the basis of higher heating value of coal, HHV (Boiler efficiency calculation by heat loss method)

7.2.1 Outline

The combustion heat generated in the furnace is equal to the sum of all heat losses and the

enthalpy gain of water/steam in the boiler. Heat losses are usually expressed as a fraction of

percentage of the potential combustion heat. They are expressed in terms of the heat losses

associated with the burning of 1 kg coal and its higher heating value (HHV).

EFFICIENCY

Output= Input Losses

Boiler

1. Dry flue gas loss through stack 2. Moisture losses( in sorbent, fuels, air, and hydrogen burning)

3. Calcination losses

4. Sulfating credit it is a negative loss

5. Unburnt carbon loss

6. Convection radiation loss

7. Sensible heat loss in fly ash and bottom ash

8. Fan credit it is also a negative loss

9. Unaccounted loss

INPUT

OUTPUT

LOSSES

% 100OutputEfficiencyInput

=

100Input LossesInput

=

The following components of the loss are discussed below:

1. Dry flue gass loss through stack

2. Moisture losses( in sorbent, fuels, air, and hydrogen burning)

3. Calcination losses

4. Sulfating credit it is a negative loss

5. Unburnt carbon loss

6. Convection radiation loss

7. Sensible heat loss in fly ash and bottom ash

8. Fan credit it is also a negative loss

9. Unaccounted loss

7.2.2 Calculations

Dry flue gas heat loss

Tf is the temperature of the flue gas leaving the stack

Ta is the ambient temperature

Mdf is dry flue gas mass per kg fuel burned .

Moisture losses

A. Loss due to moisture in Air , = [] { (100 ) + + }

We assume that the saturation temperature of water is 100 at stack pressure.

Qlatent is the latent heat of vaporization.

QSH is the enthalpy of superheating equal to Cg(Tf-100)

Mda is the stoichiometric amount of dry air required for 1 kg of fuel fired.

Xm is the moisture fraction in the air.

EAC is the execss air coefficient(= 1 + excess air fraction).

Cm is the specific heat of water.

B. Moisture loss from Coal , = [2]{ (100 ) + + }

C. Moisture loss from burning hydrogen in coal , = 9[]{ (100 ) + + }

Calcination Loss

When the boiler uses limestone to capture sulfur, two additional terms, calcinations loss,

and sulfation credit are considered in the heat balance.

3 = 3 1830

3 = 3 1183

Sulfation Credit

The calcined limestone (CaO) reacts with sulfur dioxide producing calcium sulfate (CaSO4)

according to the following exothermaic reaction:

2 + + 122 = 4 + 15,141 kg

The resulting heat gain is

= 15,141 = 15,141/

Esor is the fraction of the sulfur content.

S convered into CaSO4

Unburnt Carbon loss

Ash contains carbon * Solid waste produced * Carbon reaction heat / HHV

Other heat loss

Other heat losses have to be estimated according to previous experience as they cannot

be calculated directly. These losses include convective-radiation loss, ash sensitive heat

loss, fan credit, and other unaccounted losses.

Lothers = convection radiation loss + ash sensitive loss + + unaccounted loss

7.3 Mass balance on the basis of higher heating value of coal, HHV The mass balance specifies important items such as the division of fuel ash and spent limestone

between the particulate collectors and bed drain. This also requires special attention in a CFB

boiler, especially with sulfur-capture capability.

- Fuel heat input Qi,: Qi=Thermal capacity/Boiler efficiency(MW) - Coal feed required mc: mc=Qi/HHV

- Total airflow: mc*Mwa

- Coal ash: Ma=ASH*mc

- Total air: Mwa(Total wet air)*mc

- Flue gas wt: mc*Mwa