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. 4
7.2 Heat and Mass Balance
qrad = εg σ(T
− Tw 4 )Aw
In the control volume, heat and mass balance equation are shown as below.
q + ṁ out hout − ṁ air hin ,air − ṁ fuel hfuel = 0 𝑚𝑚̇𝑜𝑜𝑜𝑜𝑜𝑜 ℎ𝑜𝑜𝑜𝑜𝑜𝑜 = 𝑚𝑚̇𝑎𝑎𝑎𝑎𝑎𝑎 ℎ𝑎𝑎𝑎𝑎𝑎𝑎 + 𝑚𝑚̇𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 ℎ𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓
From the LHV of a fuel: 𝑞𝑞 + 𝑚𝑚̇𝑜𝑜𝑜𝑜𝑜𝑜 ∆ℎ𝑜𝑜𝑜𝑜𝑜𝑜 − 𝑚𝑚̇𝑎𝑎𝑎𝑎𝑎𝑎 ∆ℎ𝑎𝑎𝑎𝑎𝑎𝑎 − 𝑚𝑚̇𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 ∆ℎ𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 − 𝑚𝑚̇𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝐻𝐻𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 = 0 Where,
𝑇𝑇 𝑏𝑏 ∆ℎ𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 = � 𝐶𝐶𝑝𝑝 𝑑𝑑𝑑𝑑 = 𝑎𝑎(𝑇𝑇 − 𝑇𝑇0 ) + (𝑇𝑇 2 − 𝑇𝑇0 2 ) 2 𝑇𝑇0
Reference temperature T0 is set to 25℃, or 298K, and coefficient “a” and “b” can be calculated by 𝑎𝑎 = � 𝑎𝑎𝑗𝑗 𝑀𝑀𝑗𝑗 , 𝑗𝑗
𝑏𝑏 = � 𝑏𝑏𝑗𝑗 𝑀𝑀𝑗𝑗 𝑗𝑗
(Mj is the mass fraction of each species of the combustion gas)
Gas
a
b
(J/kg K)
(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
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.
Fuel : Methane
T.?
Furnace
30kg/s, 25℃
q=?
Hfuel=4x107J/kg Convection is negligible
Air; 570kg/s
Tw=300℃;
EAR=0.11, 200℃
A=Aw=3,000m2 V=10,000m3
Where, The emissivity of gas is shown as below: Use εgT=k(pL-0.015)n H2O:CO2
1:2
Partial pressure ratio
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.
𝑇𝑇𝑖𝑖 + 1 = 𝑇𝑇𝑖𝑖 − 7.3.2
𝑓𝑓𝑖𝑖 𝑓𝑓́𝑖𝑖
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.
Flue gas Tin=1435K, 600kg/s
h=100W/m2K A=Aw=9000m2 V=250m3 Tw=300℃
T=? q=?
7.4 Solution 7.4.1 Solution For the furnace radiation:
Fuel : Methane 30kg/s, 25℃
F.O.T.?
Furnace
Hfuel=4x107J/kg Convection is negligible
Air; 570kg/s
Tw=300℃;
r=0.11, 200℃
A=Aw=3,000m2 V=10,000m3
(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 𝑓𝑓(𝑇𝑇) = εg × 5.67 × 10−8 × (T 4 − 5734 ) × 3000 − 30 × 4 × 104 0.07097 × (4732 − 2982 )� −570 × �1105.7 × (473 − 298) + 2 0.1066 +600 × [1225.2 × (T − 298) + × (T 2 − 2982 )] 2
(iv)
𝑓𝑓́(𝑇𝑇) = 6.8 × 10−4 𝜀𝜀𝑔𝑔 𝑇𝑇 3 + 64.07𝑇𝑇 + 7.35 × 105
Iteration
Begin with T1=1500K, Use εgT=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 εg(1500K)=540*(0.261*12-0.015)0.42/1500=0.580 & εg(1000K)=0.654 𝑓𝑓1 = 1.284 × 108 ; 𝑓𝑓́1 = 2.163 × 106 1.284 × 108 = 1441𝐾𝐾 → 𝑇𝑇2 = 1500 − 2.163 × 106 Using interpolation, εg(1441K)=0.589 𝑓𝑓2 = 1.197 × 107 ; 𝑓𝑓́2 = 2.026 × 106 → 𝑇𝑇2 = 1435𝐾𝐾 → 𝑇𝑇2 = 1435𝐾𝐾 → 𝑇𝑇4 = 1435𝐾𝐾(1162℃)
F.O.T. = 1435K (v)
q=? q = εg 𝜎𝜎�𝑇𝑇 4 − 𝑇𝑇𝑤𝑤 4 �𝐴𝐴𝑤𝑤 = 4.15 × 108 𝑊𝑊 q́́ = q/𝐴𝐴𝑤𝑤 = 138𝑘𝑘𝑘𝑘/𝑚𝑚2
7.4.2 For the superheaters
h=100W/m2K
Flue gas Tin=1435K, 600kg/s
A=Aw=9000m2 V=250m3 Tw=300℃Coals b
L=3.6V/A=0.1m pL=0.0261atm-m
(slightly out of range!)
ṁ fuel = 0; ṁ out = ṁ in =600kg/s
T=? q=?
𝑓𝑓(𝑇𝑇) = 𝜀𝜀𝑔𝑔 𝜎𝜎 �𝑇𝑇 (i)
(ii)
4
− 𝑇𝑇𝑤𝑤4 � 𝐴𝐴𝑤𝑤 + ℎ(𝑇𝑇 − 𝑇𝑇𝑤𝑤 )𝐴𝐴𝑤𝑤
̇ [𝑎𝑎𝑜𝑜𝑜𝑜𝑜𝑜 (𝑇𝑇 − 𝑇𝑇𝑖𝑖𝑖𝑖 ) + +𝑚𝑚𝑜𝑜𝑜𝑜𝑜𝑜
𝑏𝑏𝑜𝑜𝑜𝑜𝑜𝑜 2 (𝑇𝑇 − 𝑇𝑇𝑖𝑖𝑖𝑖 2 )] 2
Iteration 𝑇𝑇1 = 1000𝐾𝐾, 𝜀𝜀𝑔𝑔 = 0.096, 𝑓𝑓1 = 7.44 × 107 𝑓𝑓́1 = 1.96 × 106 → T2 = 961K → T3 = 961K → T4 = 961K(688℃)!
Heat transfer rate 𝑞𝑞 = 𝜀𝜀𝑔𝑔 𝜎𝜎�𝑇𝑇 4 − 𝑇𝑇𝑤𝑤 4 �𝐴𝐴𝑤𝑤 + ℎ(𝑇𝑇 − 𝑇𝑇𝑤𝑤 )𝐴𝐴𝑤𝑤 = 3.86 × 108 𝑊𝑊
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+A∙S](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 A∙S, where A is 2.16 for sulfur capture and is zero when no sulfur is captured as calcium sulfate.
-
Total dry air, Tda Tda = Mda∙EAR (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+1.375S∙R(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 𝑊𝑊𝑐𝑐 = 𝑀𝑀𝑤𝑤𝑤𝑤 − 0.2315𝑀𝑀𝑑𝑑𝑑𝑑 + 3.66𝐶𝐶 + 9𝐻𝐻 + 𝐿𝐿𝑞𝑞 𝑋𝑋𝑚𝑚𝑚𝑚 + 𝑁𝑁 + 𝑂𝑂 + 2.5𝑆𝑆(1 − 𝐸𝐸𝑠𝑠𝑠𝑠𝑠𝑠 ) +1.375𝑆𝑆𝑆𝑆(1 + 1.19�𝑋𝑋𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶3 ⁄𝑋𝑋𝐶𝐶𝐶𝐶 𝐶𝐶𝐶𝐶3 �) + 𝑎𝑎𝑐𝑐 𝐴𝐴𝐴𝐴𝐴𝐴(𝑘𝑘𝑘𝑘⁄𝑘𝑘𝑘𝑘 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏)
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: 𝐿𝐿𝑞𝑞 =
100𝑆𝑆 𝑋𝑋𝑋𝑋 32𝑋𝑋𝐶𝐶𝐶𝐶 𝐶𝐶𝐶𝐶3
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 : 32𝑋𝑋 𝐶𝐶𝐶𝐶𝐶𝐶 } 𝑅𝑅́ = {𝑅𝑅 − 56𝑆𝑆
Solid waste Produced
𝐶𝐶𝐶𝐶𝐶𝐶𝑂𝑂3 = 𝐶𝐶𝐶𝐶𝐶𝐶 + 𝐶𝐶𝑂𝑂2 − 1830(𝑘𝑘𝑘𝑘⁄𝑘𝑘𝑘𝑘 𝑜𝑜𝑜𝑜 𝐶𝐶𝐶𝐶𝐶𝐶𝑂𝑂3 )
𝑀𝑀𝑀𝑀𝑀𝑀𝑂𝑂3 = 𝑀𝑀𝑀𝑀𝑀𝑀 + 𝐶𝐶𝑂𝑂2 − 1183(𝑘𝑘𝑘𝑘⁄𝑘𝑘𝑘𝑘 𝑜𝑜𝑜𝑜 𝑀𝑀𝑀𝑀𝑀𝑀𝑂𝑂3 )
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
𝐿𝐿𝑤𝑤 = 136
𝐿𝐿𝑞𝑞 𝑋𝑋𝐶𝐶𝐶𝐶𝐶𝐶 𝑂𝑂3 𝑆𝑆𝑆𝑆𝑠𝑠𝑠𝑠𝑠𝑠 40𝐿𝐿𝑞𝑞 𝑋𝑋𝑀𝑀𝑀𝑀𝑀𝑀 𝑂𝑂3 𝑠𝑠 𝐸𝐸 + 56 � − �+ + 𝐿𝐿𝑞𝑞 𝑋𝑋𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 100 32 84 32 𝑠𝑠𝑠𝑠𝑠𝑠
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)
EFFICIENCY Output= Input – Losses
INPUT
Output ×100 Input Input − Losses = ×100 Input OUTPUT Efficiency = %
Boiler LOSSES
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
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).
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 ① 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
여기에 수식을 입력하 십시오 .
② Moisture losses A.
Loss due to moisture in Air
𝐿𝐿𝑚𝑚 ,𝑎𝑎𝑎𝑎𝑎𝑎 =
𝑀𝑀𝑤𝑤𝑤𝑤 [𝐸𝐸𝐸𝐸𝐸𝐸]𝑋𝑋𝑚𝑚 {𝐶𝐶𝑚𝑚 (100 − 𝑇𝑇𝑎𝑎 ) + 𝑄𝑄𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 + 𝑄𝑄𝑆𝑆𝑆𝑆 } 𝐻𝐻𝐻𝐻𝐻𝐻
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
𝐿𝐿𝑚𝑚 ,𝑓𝑓𝑢𝑢𝑢𝑢𝑢𝑢 = C.
[𝐻𝐻2 𝑂𝑂]{𝐶𝐶𝑚𝑚 (100 − 𝑇𝑇𝑎𝑎 ) + 𝑄𝑄𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 + 𝑄𝑄𝑆𝑆𝑆𝑆 } 𝐻𝐻𝐻𝐻𝐻𝐻
Moisture loss from burning hydrogen in coal
𝐿𝐿𝑚𝑚 ,ℎ = ③ Calcination Loss
9[𝐻𝐻]{𝐶𝐶𝑚𝑚 (100 − 𝑇𝑇𝑎𝑎 ) + 𝑄𝑄𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 + 𝑄𝑄𝑆𝑆𝑆𝑆 } 𝐻𝐻𝐻𝐻𝐻𝐻
When the boiler uses limestone to capture sulfur, two additional terms, calcinations loss,
and sulfation credit are considered in the heat balance.
𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝐶𝐶𝐶𝐶𝐶𝐶𝑂𝑂3 = 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑀𝑀𝑀𝑀𝑀𝑀𝑂𝑂3 = ④ Sulfation Credit
𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑜𝑜𝑜𝑜 𝐶𝐶𝐶𝐶𝐶𝐶𝑂𝑂3 × 1830 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 × 𝐻𝐻𝐻𝐻𝐻𝐻
𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑜𝑜𝑜𝑜 𝑀𝑀𝑀𝑀𝑀𝑀𝑂𝑂3 × 1183 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 × 𝐻𝐻𝐻𝐻𝐻𝐻
The calcined limestone (CaO) reacts with sulfur dioxide producing calcium sulfate (CaSO4) according to the following exothermaic reaction: 1 𝑆𝑆𝑂𝑂2 + 𝐶𝐶𝐶𝐶𝐶𝐶 + 𝑂𝑂2 = 𝐶𝐶𝐶𝐶𝐶𝐶𝑂𝑂4 + 15,141 𝑘𝑘𝑘𝑘⁄kg 2 The resulting heat gain is
𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 =
𝑘𝑘𝑘𝑘 𝑜𝑜𝑜𝑜 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 × 15,141 = 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 × 15,141/𝐻𝐻𝐻𝐻𝐻𝐻 𝑘𝑘𝑘𝑘 𝑜𝑜𝑜𝑜 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑓𝑓𝑓𝑓𝑓𝑓 × 𝐻𝐻𝐻𝐻𝐻𝐻
Esor is the fraction of the sulfur content. S convered into CaSO4 ⑤ Unburnt Carbon loss
Ash contains carbon * Solid waste produced * Carbon reaction heat / HHV ⑥ 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)
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Coal feed required mc:
mc=Qi/HHV
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Total airflow:
mc*Mwa
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Coal ash:
Ma=ASH*mc
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Total air:
Mwa(Total wet air)*mc
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Flue gas wt:
mc*Mwa
