6.4 Boiler Heat Balance

7/27/2019 6.4 Boiler Heat Balance

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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

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 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 combustion 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

f

w

W

hhW Q 121

In which W w = weight of fluid flowing through the boiler during the test, lb

h1 and h2 = fluid enthalpies entering and leaving the boiler respectively, 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 the boiler losses for a properly operated unit. Thus,

a g dg t t W Q 24.02

In which 0.24 = specific heat of the flue gas at constant pressure, Btu 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:

F t whent t MQ g f g f 57546010893 ,.

F t whent t MQ g f g f 5755010663 ,.

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,

ff hhHQ 24 9

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

F t whent t H Q g f g 575,46.010899 24



F t whent t H Q g f g 575,5.010669 24

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.

COCO

COC C Q abi

25 160,10160,10

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. abC C Q 600,146

If the unburned combustible is determined from the heating value of all of the refuse, then

r r HV W Q 6

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,

6543217 QQQQQQ HHV Q

Example No. 1

Calculate the boiler heat balance on the 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


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

f

w

W

hhW Q 121

hr per lbW w 030,202



hr per lbW f 260,22

lb per Btu F psig hh 1381743,476@2

lb per Btu F hh f 1.2957.324@1

fuel lb per BtuQ 9856

260,22

1.2951381030,2021

Q2 = energy loss due to dry flue gases


F t g 463

F t a 82

1602673

7811

2

222 S S C

COCO

N COOCOW abdg

The value of sulfur must be expressed as percentage of S in percentage.C ab in decimal.


C = 0.7181, A = 0.0827, and S = 0.0334

AW C C r ab

fuel of lb per lbhr per lb

hr per lbW r 1150.0

260,22

560,2

6858.00827.01150.07181.0 abC


N2 is 100% - 14.2% - 0.3% - 4.3% = 81.2%

1602673

7811

2

222 S S C

COCO

N COOCOW abdg

160

34.3

267

34.36858.0

3.02.143

2.813.073.482.1411

dg W

fuel lb per lbW dg 04.12


fuel lb per BtuQ 10978246304.1224.02

Q3 = energy loss due to moisture in fuel

F t whent t M Q g f g f 575,46.010893


M f = moisture in fuel = 0.0509 lb per lb of fuel

F t f 82 , F t g 463

f g f t t M Q 46.010893

fuel lb per BtuQ 628246346.010890509.03



Q4 = energy loss due to evaporating and superheating moisture formed by combustion of hydrogen

F t whent t H Q g f g 575,46.010899 24


H2 (ultimate) = 0.0523

Moisture (proximate) = 0.0509

0467.09

0509.00523.02 H

fuel lb per BtuQ 5138246346.010890467.094

Q5 = energy loss due to incomplete combustion of carbon to CO

COCO

COC C Q abi

25 160,10160,10

6858.0abC


fuel lb per BtuQ 1443.02.14

3.06858.0160,105

Q6 = energy loss due to combustible in refuse

abC C Q 600,146

fuel lb per BtuQ 4726858.07181.0600,146

Q7 = energy loss due to radiation and unaccounted for

6543217 QQQQQQ HHV Q

fuel lb per BtuQ 9384721445136210979856082,137

Boiler Heat Balance

Item Energy, Btu per lb fuel Percentage

Q 1 9856 75.34

Q 2 1097 8.38

Q 3 62 0.48

Q 4 513 3.92

Q 5 144 1.10

Q 6 472 3.61

Q 7 938 7.17

HHV 13,082 100.00

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6.4 Boiler Heat Balance