6.4
Boiler Heat Balance
1. General The proper operating ratio of air to fuel for a boiler can be established only by computing heat losses from test data. Items computed in a boiler heat balance are in terms of Btu per pound of fuel for solid, liquid, and gaseous fuels. To make a heat balance from one boiler comparable to one from from another boiler the energy terms are converted into percentages where 100% is the heating value of the fuel. The items in the boiler heat balance may be computed per pound of coal on the as fired or dry basis, but the energy values will be different for each. However, when transferred to percentages, there will be no difference for the two bases. Energy supplied to the boiler by 1 lb of fuel is distributed among the following items in t he ASME short-form heat balance, all expressed in units of Btu per pound of fuel: Q1 = energy absorbed by boiler fluid Q2 = energy loss due to dry flue gases Q3 = energy loss due to moisture in fuel Q4 = energy loss due to evaporating and superheating moisture formed by combustion of hydrogen Q5 = energy loss due to incomplete c ombustion of carbon to CO Q6 = energy loss due to combustible in refuse Q7 = energy loss due to radiation and unaccounted for An explanation of each of these items follows: 2.
Energy absorbed by boiler fluid. The useful output of the steam generator is the heat transferred to the fluid. Sometimes it is advantageous to divide this item into the heat transferred to the fluid by the boiler proper, the air preheater, the economizer, and the superheater. For our purpose, we shall combine all these subdivisions into Q1
W w h2 h1 W f
In which W w w = weight of fluid flowing through the boiler during the test, lb h1 and h2 = fluid enthalpies entering and leaving the boiler re spectively, Btu per lb W f = weight of fuel burned during test It is preferable to operate the boiler without blowdown during the test. However, if the test is of such long duration that blowdown is necessary, the equation should be altered appropriately. Q1 expressed as a percentage of the higher heating value of the fuel is the boiler efficiency. 3.
Energy loss due to dry flue gas. This loss is the greatest of any of t he boiler losses for a properly operated unit. Thus, Q2 0.24W dg t g t a
In which 0.24 = specific heat of the t he flue gas at constant pressure, Bt u per lb per deg F
t g = temperature of the gas leaving the boiler, F t a = temperature of the air entering the boiler, F Obviously, this loss is a function of the flue-gas temperature, but it is sometimes uneconomical to reduce the flue-gas temperature to too low a value. A low gas-outlet temperature can be obtained only by a large heat transfer surface and a low-temperature fluid to which the energy may be transferred. Air preheaters and economizers furnish the low-temperature fluid and additional surface for reduction of the flue-gas temperature. Since the incoming air temperature is beyond human control, the only variable is the weight of dry gases per pound of fuel. W dg should be kept as small as possible, consistent with complete combustion, by control of the excess air. 4.
Energy loss due to evaporating and superheating moisture in fuel. Moisture entering the boiler with the fuel leaves as a superheated vapor in the same way as does the moisture from the combustion of hydrogen. Therefore the formula for calculating this loss may be derived in the same way as that for Q4: Q3 M f 1089 0 46t g t f .
Q3 M f 1066 0 5t g t f .
,
,
when t g 575 F when t g 575 F
Where M f = moisture in fuel, lb per lb of fuel t f = temperature of fuel, F 5.
Energy loss due to evaporating and superheating moisture formed by combustion of hydrogen. This loss is higher for gaseous fuels containing relatively large percentage of hydrogen than for the average low-hydrogen coal. Water formed by burning hydrogen leaves the boiler in the form of superheated vapor, and its energy cannot be released to the boiler fluid until the vapor can be condensed. With flue-gas temperatures of 300 F or more and the vapor at a partial pressure less than atmospheric, condensation is impossible within the boiler. Q4 represents the loss of energy due to the inability of the boiler to condense this superheated vapor to a liquid at a temperature corresponding to the temperature of the incoming air. Thus, Q4 9H2 h h ff
In which H2 = weight of hydrogen in the fuel, lb per lb fuel h = enthalpy of superheated vapor, Btu per lb h ff = enthalpy of liquid at the incoming fuel temperature Since the partial pressure of the superheated vapor would be difficult to determine, and since this loss of energy is usually small, the equation above may be simplified by assuming that the vapor pressure corresponds to a saturation temperature of 150 F. Then the enthalpy of the superheated vapor is equal to the enthalpy of the saturated vapor (1126.1 Btu per lb) plus the energy needed to superheat the vapor. The latter term is taken as 0.46(t g-150) when the gas temperature is less than 575 F. The enthalpy of the liquid (h ff ) is taken as (t f – 32). Combining these terms we arrive at the expressions Q4 9 H 2 1089 0.46t g t f ,
when t g 575 F
Q4 9 H 2 1066 0.5t g t f ,
when t g 575 F
The proper value of H2 to be used in the equation for Q4 is the amount of hydrogen in the fuel that is available for combustion. Ultimate analyses given in Table 1 list all the hydrogen in the fuel, including the hydrogen present in the fuel in the form of moisture. To obtain the value of H2 for the above equation, deduct one-ninth of the weight of moisture from the hydrogen listed in Table 1. The weight of moisture may be found from the proximate analysis.
6.
Energy loss due to incomplete combustion. Products formed by incomplete combustion may be mixed with oxygen and burned again with a further release of energy. Such products of incomplete combustion that are present in flue gas are CO, H2, and various hydrocarbons. Carbon monoxide is the only one of these gases that can be determined conveniently in the power-plant test. Therefore, the loss due to incomplete combustion refers specifically to the incomplete combustion of carbon to carbon monoxide. A formula for the weight ratio of carbon burned to CO per pound of fuel was developed. The difference in the energy release due to burning carbon to carbon monoxide rather than to carbon dioxide is given as 10,160 Btu per lb of carbon. Q5 10,160C i 10,160C ab
7.
CO CO2 CO
Energy loss due to unconsumed carbon. All combustible in the refuse may be assumed to be carbon, since the othe r combustible parts of coal would probably be distilled out of the fuel before live embers would drop into the ashpit. Any unburned carbon in the flue gas (fly ash) or in the ashpit refuse is included. Q6 14,600C C ab
If the unburned combustible is determined from the heating value of all of the refuse, then Q6 W r HV r
8.
Unaccounted-for and radiation loss.
This loss is due to radiation, incomplete combustion resulting in hydrogen and hydrocarbons in the flue gas, and unaccounted-for losses. Under the ASME code, the radiation loss may be estimated separately and not combined with the unaccounted-for loss. However, w hen they are combined, Q7 HHV Q1 Q2 Q3 Q4 Q5 Q6
Example No. 1 Calculate the boiler heat balance on t he as fired basis for the following data: Fuel: Cherokee County, Kansas, coal Gas analysis: 14.2% CO2, 0.3% CO, 4.3% O2 Coal fired: 22,260 lb per hr Refuse: 2,560 lb per hr Water: 202,030 lb per hr Water entering: 324.7 F Steam leaving: 476 psia, 743 F Fuel and room temperature: 82 F Gas temperature: 463 F Given: Fuel: Cherokee County, Kansas, coal Gas analysis: 14.2% CO2, 0.3% CO, 4.3% O2 Coal fired: 22,260 lb per hr Refuse: 2,560 lb per hr Water: 202,030 lb per hr Water entering: 324.7 F Steam leaving: 476 psia, 743 F Fuel and room temperature: 82 F Gas temperature: 463 F
Required: Boiler heat balance Solution: From Table 5.1, Fuel: Cherokee County, Kansas, coal HHV = 13,082 Btu per lb Q1 = energy absorbed by boiler fluid Q1
W w h2 h1 W f
W w 202,030 lb per hr
W f 22,260 lb per hr
h2 h @ 476 psig ,743 F 1381 Btu per lb h1 h f @ 324.7 F 295.1 Btu per lb
Q1
202,0301381 295.1 22,260
9856 Btu per lb fuel
Q2 = energy loss due to dry flue gases Q2 0.24W dg t g t a
t g 463 F
t a 82 F W dg
11CO2 8O2 7CO N 2 S S C ab 3CO2 CO 267 160
The value of sulfur must be expressed as percentage of S in percentage. C ab in decimal. From Table 5.1, Fuel: Cherokee County, Kansas, coal C = 0.7181, A = 0.0827, and S = 0.0334 C ab C W r A W r
2,560 lb per hr 22,260 lb per hr
0.1150lb per lb of fuel
C ab 0.7181 0.1150 0.0827 0.6858
Gas analysis: 14.2% CO2, 0.3% CO, 4.3% O2 N2 is 100% - 14.2% - 0.3% - 4.3% = 81.2% W dg W dg
11CO2 8O2 7CO N 2 S S C ab 3CO2 CO 267 160 1114.2 84.3 70.3 81.2
3.34 3.34 0.6858 267 160
314.2 0.3
W dg 12.04 lb per lb fuel Q2 0.24W dg t g t a
Q2 0.2412.04463 82 1097 Btu per lb fuel
Q3 = energy loss due to moisture in fuel
Q3 M f 1089 0.46t g t f ,
when t g 575 F
From Table 5.1, Fuel: Cherokee County, Kansas, coal M f = moisture in fuel = 0.0509 lb per lb of fuel t f 82 F , t g 463 F Q3 M f 1089 0.46t g t f
Q3 0.05091089 0.46463 82 62 Btu per lb fuel
Q4 = energy loss due to evaporating and superheating moisture formed by combustion of hydrogen
Q4 9 H 2 1089 0.46t g t f ,
when t g 575 F
From Table 5.1, Fuel: Cherokee County, Kansas, coal H2 (ultimate) = 0.0523 Moisture (proximate) = 0.0509 H 2 0.0523
0.0509 9
0.0467
Q4 90.04671089 0.46463 82 513 Btu per lb fuel
Q5 = energy loss due to incomplete combustion of carbon to CO Q5 10,160C i 10,160C ab
CO CO2 CO
C ab 0.6858
Gas analysis: 14.2% CO2, 0.3% CO, 4.3% O2
0.3 144 Btu per lb fuel 14.2 0.3
Q5 10,1600.6858
Q6 = energy loss due to combustible in refuse Q6 14,600C C ab Q6 14,6000.7181 0.6858 472 Btu per lb fuel
Q7 = energy loss due to radiation and unaccounted for Q7 HHV Q1 Q2 Q3 Q4 Q5 Q6 Q7 13,082 9856 1097 62 513 144 472 938 Btu per lb fuel
Item Q 1 Q 2 Q 3 Q 4 Q 5 Q 6 Q 7 HHV
Boiler Heat Balance Energy, Btu per lb fuel 9856 1097 62 513 144 472 938 13,082 -
End -
Percentage 75.34 8.38 0.48 3.92 1.10 3.61 7.17 100.00
