6.4Boiler Heat Balance
1. GeneralThe 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 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 the ASME short-form heat balance, all
expressed in units of Btu per pound of fuel:Q1 = energy absorbed by
boiler fluidQ2 = energy loss due to dry flue gasesQ3 = energy loss
due to moisture in fuelQ4 = energy loss due to evaporating and
superheating moisture formed by combustion of hydrogenQ5 = energy
loss due to incomplete combustion of carbon to COQ6 = energy loss
due to combustible in refuseQ7 = energy loss due to radiation and
unaccounted forAn 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
In which Ww = weight of fluid flowing through the boiler during
the test, lbh1 and h2 = fluid enthalpies entering and leaving the
boiler respectively, Btu per lbWf = 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 the boiler losses for a properly operated unit. Thus,
In which 0.24 = specific heat of the flue gas at constant
pressure, Btu per lb per deg Ftg = temperature of the gas leaving
the boiler, Fta = temperature of the air entering the boiler,
FObviously, 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. Wdg 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:
Where Mf = moisture in fuel, lb per lb of fueltf = 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,
In which H2 = weight of hydrogen in the fuel, lb per lb fuelh =
enthalpy of superheated vapor, Btu per lbhff = enthalpy of liquid
at the incoming fuel temperatureSince 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(tg-150) when the gas temperature
is less than 575 F. The enthalpy of the liquid (hff) is taken as
(tf 32). Combining these terms we arrive at the expressions
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.
7. Energy loss due to unconsumed carbon.All combustible in the
refuse may be assumed to be carbon, since the other 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.
If the unburned combustible is determined from the heating value
of all of the refuse, then
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, when they are
combined,
Example No. 1Calculate the boiler heat balance on the as fired
basis for the following data:Fuel: Cherokee County, Kansas, coalGas
analysis: 14.2% CO2, 0.3% CO, 4.3% O2Coal fired: 22,260 lb per
hrRefuse: 2,560 lb per hrWater: 202,030 lb per hrWater entering:
324.7 FSteam leaving: 476 psia, 743 FFuel and room temperature: 82
FGas temperature: 463 F
Given:Fuel: Cherokee County, Kansas, coalGas analysis: 14.2%
CO2, 0.3% CO, 4.3% O2Coal fired: 22,260 lb per hrRefuse: 2,560 lb
per hrWater: 202,030 lb per hrWater entering: 324.7 FSteam leaving:
476 psia, 743 FFuel and room temperature: 82 FGas temperature: 463
F
Required:Boiler heat balance
Solution:From Table 5.1, Fuel: Cherokee County, Kansas, coalHHV
= 13,082 Btu per lbQ1 = energy absorbed by boiler fluid
Q2 = energy loss due to dry flue gases
The value of sulfur must be expressed as percentage of S in
percentage. Cab in decimal.
From Table 5.1, Fuel: Cherokee County, Kansas, coalC = 0.7181, A
= 0.0827, and S = 0.0334
Gas analysis: 14.2% CO2, 0.3% CO, 4.3% O2N2 is 100% - 14.2% -
0.3% - 4.3% = 81.2%
Q3 = energy loss due to moisture in fuel
From Table 5.1, Fuel: Cherokee County, Kansas, coalMf = moisture
in fuel = 0.0509 lb per lb of fuel
,
Q4 = energy loss due to evaporating and superheating moisture
formed by combustion of hydrogen
From Table 5.1, Fuel: Cherokee County, Kansas, coalH2 (ultimate)
= 0.0523Moisture (proximate) = 0.0509
Q5 = energy loss due to incomplete combustion of carbon to
CO
Gas analysis: 14.2% CO2, 0.3% CO, 4.3% O2
Q6 = energy loss due to combustible in refuse
Q7 = energy loss due to radiation and unaccounted for
Boiler Heat Balance
ItemEnergy, Btu per lb fuelPercentage
Q1985675.34
Q210978.38
Q3620.48
Q45133.92
Q51441.10
Q64723.61
Q79387.17
HHV13,082100.00
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