xvEPA
United States
Environmental Protection
Agency
Air Pollution Training Institute
MD 20
Environmental Research Center
Research Triangle Park NC 27711
^PA 450/2-80-064
February 1980
Air
APTI
Course 427
Combustion Evaluation
Student Workbook
-------
United States
Environmental Protection
Agency
Air Pollution Training Institute
MD20
Environmental Research Center
Research Triangle Park NC 27711
EPA 450/2-80-064
February 1980
Air
APTI
Course 427
Combustion Evaluation
Student Workbook
Prepared By:
J. Taylor Beard
F. Antonio lachetta
Lembit U. Lilleleht
Associated Environmental Consultants
P.O. Box 3863
Charlottesville, VA 22903
Under Contract No.
68-02-2893
EPA Project Officer
James 0. Dealy
United States Environmental Protection Agency
Office of Air, Noise, and Radiation
Office of Air Quality Planning and Standards
Research Triangle Park, NC 27711
-------
Notice
This is not an official policy and standards document. The opinions, findings, and
conclusions are those of the authors and not necessarily those of the Environmental
Protection Agency. Every attempt has been made to represent the present state of
the art as well as subject areas still under evaluation. Any mention of products or
organizations does not constitute endorsement by the United States Environmental
Protection Agency.
Availability of Copies of This Docirtnent
This document is issued by the Manpower and Technical Information Branch, Con-
trol Programs Development Division, Office of Air Quality Planning and Standards,
USEPA. It was developed for use in training courses presented by the EPA Air Pollu-
tion Training Institute and others receiving contractual or grant support from the
Institute. Other organizations are welcome to use the document for training purposes.
Schools or governmental air pollution control agencies establishing training programs
may receive single copies of this document, free of charge, from the Air Pollution
Training Institute, USEPA, MD-20, Research Triangle Park, NC 27711. Others may
obtain copies, for a fee, from the National Technical Information Service, 5825 Port
Royal Road, Springfield, VA 22161.
-------
\ AIR POLLUTION TRAINING INSTITUTE
S 5SC?Z * MANPOWER AND TECHNICAL INFORMATION BRANCH
4 *"ffc * CONTROL PROGRAMS DE VELOPMENT DIVISION
OFFICE OF AIR QUALITY PLANNING AND STANDARDS
The Air Pollution Training Institute (1) conducts training for personnel working on the develop-
ment and improvement of state, and local governmental, and EPA air pollution control programs,
as well as for personnel in industry and academic institutions; (2) provides consultation and other
training assistance to governmental agencies, educational institutions, industrial organizations, and
others engaged in air pollution training activities; and (3) promotes the development and improve-
ment of air pollution training programs in educational institutions and state, regional, and local
governmental air pollution control agencies. Much of the program is now conducted by an on-site
contractor, Northrop Services, Inc.
One of the principal mechanisms utilized to meet the Institute's goals is the intensive short term
technical training course. A full-time professional staff is responsible for the design, development,
and presentation of these courses. In addition the services of scientists, engineers, and specialists
from other EPA programs governmental agencies, industries, and universities are used to augment
and reinforce the Institute staff in the development and presentation of technical material.
Individual course objectives and desired learning outcomes are delineated to meet specific program
needs through training. Subject matter areas covered include air pollution source studies, atmos-
pheric dispersion, and air quality management. These courses are presented in the Institute's resi-
dent classrooms and laboratories and at various field locations.
R. Alan Schueler /James A.Jahake
Program Manager Technical Director
Northrop Services, Inc. Northrop Sendees, Inc.
Jeanjf Schueneman
Chief, Manpower & Technical
Information Branch
in
-------
TABLE OF CONTENTS
CHAPTER TITLE PAGE
I. COMBUSTION CALCULATIONS 1-1
Problem I.I: Combustion of No. 6 Fuel Oil 1-1
Problem 1.2: Combustion of Gases 1-4
Problem 1.3: Available Heat 1-8
Problem 1.4: Liquid Waste Combustion in
Natural Gas-Fired Boiler 1-10
Problem 1.5: Combustion Calculations with
Heat Recovery 1-13
II. COMBUSTION SYSTEM DESIGN PROBLEMS II-l
Problem II. 1: Calculation of Furnace Volume II-l
Problem II.2: Furnace Volume Plan Review ...... II-2
Problem II.3: Calculation of Furnace
Gas Exit Temperature 11-3
III. EMISSION CALCULATIONS I III-l
Problem III.l: Pollutant Emissions from
Coal-Fired Power Plant . III-l
Problem III.2: Fuel Sulfur Limit Calculation .... III-6
TV. EMISSION CALCULATIONS II IV-1
Problem IV. 1: Excess Air Calculation Based
on Orsat Analysis IV-1
-------
CHAPTER TITLE PAGE
Problem IV.2: Use of F-factors to Compute
Emission Concentrations IV-6
Problem IV.3: Calculation of F-factor IV-7
Problem IV. 4: Calculation of Pollutant
Concentration IV-8
Problem IV.5: Correction of NOX Emission
Concentration to 3% 02 IV-10
V. AFTERBURNER DESIGN PROBLEMS V-l
Problem V.I: Afterburner Design for
Meat Smokehouse Effluent V-l
Problem V.2: Afterburner Design with
Combustion Oxygen from the Contaminated
Effluent V-6
VI. COMBUSTION SYSTEM CALCULATIONS VI-1
Problem VI.1: Fuel Requirements for
Combustion Installation VI-1
Problem VI.2: Combustion Improvement VI-5
in
-------
CHAPTER I
COMBUSTION CALCULATIONS
PROBLEM I.I: Combustion of No. 6 Fuel Oil
Assume perfect combustion of No. 6 fuel oil with stoichiometric air.
The gravimetric analysis of a sample of this fuel oil is:
88.52% carbon
10.87% hydrogen
0.40% sulfur
0.10% nitrogen
0.06% oxygen
0.05% ash
Compute;
1. The gravimetric analysis (weight percent) of the flue gases
2. Total volume of flue gases (at 500°F and 1 atmosphere) per
pound of oil burned
3. Volume percent of CO2 in dry flue gases
Solution to Problem I.I;
Select as a basis for calculation: 100 Ibs. of fuel oil burned. This is
chosen for convenience as the gravimetric analysis will give the amounts
of various elements directly. Answers can easily be scaled to the 1 Ib.
of oil as required in Part 2.
A tabular form of solution is presented on the next page, as this will
(i) help to organize thinking/ (ii) permit presentation of results in
a compact format, and (iii) avoid confusion
1-1
-------
TABLE I.I
FUEL
Element
(a)
C
H2
S
°2
N2
Ash
M.W.
(b)
12
2
32
32
28
—
Quantity
Ib. Ib-mole
(c) (d)
88.52
10.87
0.40
0.06
0.10
0.05
7.38
5.44
0.012
0.002
0.0036
—
COMB. AIR
02
Ib-mole
(e)
7.38
2.72
0.012
-0.002
-
REQ'D
N2
Ib-mole
(f)
27.8
10.2
0.045
-0.007
-
FLUE PRODUCTS
Qmpd.
(g)
co2
H2O
so2
°2
N2
M.W.
(h)
44
18
64
32
28
Ib-mole
(i)
7.38
5.44
0.012
-
38.0
Ib.
(j)
325
97.9
0.77
-
1064
wgt %
Oc>
21.8
6.6
0.05
-
71.5
H
I
to
Total
100.00
10.11
38.94
50.8
1,488
48.15
Notes for Column Headings;
(c) From gravimetric analysis of fuel
(d) = (c) * (b)
(e) From basic chemistry, i.e.:
C + O2 -»• C02
H2 + *iO2 •*• H2O
S + O2 •* SO2
Oxygen in fuel reduces air requirements.
Excess air, if any, is usually specified
as % of theoretical and added to the total.
99.95
(f) = (0.79/0.21) x (e)
(g) Products corresponding to complete combustion
of various oxidizable elements in the fuel
(i) Pound-moles of products from the amount of
combustibles in (d). Note that oxygen present
only if excess air added, and nitrogen is the
total of (f), including any from excess air.
(j) = (h) x (i)
(k) = (j) x 100/E(j)
-------
Part 1.
Gravimetric analysis of flue gases given by Column (k) of the table.
Part 2.
Ideal gas law used to calculate the volume of flue gases (Equation 2.6,
p. 2-8 of the Student Manual) .
V =» n RT / p
where n = 50.8 Ib-moles flue gases m 0i5Q8 ^.^eg/^
100 Ib oil
from Table I.I
R - 0.7302 atm-f t3/ (Ib-mole °R)
from Attachment 2-2, p. 2-24 of the Student Manual
T - 500°P + 460 - 960°R
p » 1.0 atm.
V = (0.508) (0.7302) (906) / (1.0) - 356 ft3
Part 3.
Dry flue gases (from Table I.I)
Compound Ib-moles
CO2 7.38
SO2 0.012
°2
38.0
Total 45 . 4 Ib-moles
Vol. % C02 = 7.38 x 100 / 45.4 = 16.3 %
1-3
-------
PROBLEM 1.2: Combustion of Gases
Consider a gaseous fuel composed of 60% H2 and 40% CH* by volume.
Determine:
1. The volume of air required for complete combustion of 1,000 scfm
of the above gases with 100% theoretical air
2. The pounds of air required for burning 1.00 pounds of fuel
3. The volumetric analysis of flue gases (products), including
water vapor (assume no water is condensed)
4. The gravimetric analysis of the reactants (fuel gas and air
mixture)
5. The partial pressure of the water vapor in the flue for a total
pressure of 13.7 psia
Solution to Problem 1.2;
Complete and balance the combustion equation using 1 Ib-mole of gas
as the basis.
0.60 H2 + 0.40 CH4 + a 03 + b N2 •*• c C02 + d H20 + b N2 (A)
To balance the equation c «
d o 0.60 + 2 (0.40)
a • c + d/2 =»
b - (0.79/0.21) a
Thus:
0.60 H2 + 0.40 CH4 + 02 + 4'14 N2 •* C02 + H2O + N2 (B)
**?" 0.60 + 0.40 + + 4.14 + + + 4.14 (C)
Volumes:
1.00 5.25
1-4
-------
Rel.
Mass:
1.20 + 6.40 +
7.60
115.9
i
115.9
(D)
Part -1.
From Equation (C) note that 5.24 volumes of air required for complete
combustion of 1.00 volumes of this fuel gas.
Therefore:
Vol. of air ** (5.24 scfm air/scfm gas) (
scfrn gas)
scfm air
Another approach makes use of Equation 2.4, p. 2-7 of the Student Manual,
which for gases containing only H2 and CH4 reduces to:
2.38
-------
Part 3.
From Equation (C) above, total volume of flue products is:
+ + 4.14 -
(F)
C02 H20 N2
Volume % of flue products:
% CO2 « x 100 / (Result of Equation F above)
% H20 » x 100 / ( ) - %
% H2 - 4.14 x 100 / j )_ - _%
Part 4.
Tabulate the left-hand aide of Equation (B) above:
Reactant Relative Mass
H2 1.20
CB4 6.40
02 __
H2 115.9
Itotal 100-°
Hote; Wgt. % of Reactant i - (Mass of i) x 100 / Total Mass
Part 5.
Partial pressure of a gaseous component is given by Equation 2.9, p. 2-10
of the Student Manual.
(nH20) X (P)/ntotal
(G)
1-6
-------
where n = Ib-moles or volume from Equation (C)
p = total pressure of flue products
Thus:
PH 0 = ( Ib-moles H20) (14.7 psia) / (__ Ib-moles of flue gases)
PH20 *
1-7
-------
PROBLEM 1.3:
Available Heat
Consider a boiler which burns 10,000 standard cubic feet per hour of a
waste gas with higher heating value of 258 Btu/scf.
Determine;
1. The gross heating value per hour for complete combustion
2. The available heat if the flue gases leave the boiler heat
exchanger at 500°F and complete combustion is achieved with
theoretical combustion air
3. The available heat from the same boiler if 20% excess air had
been used and flue gas exit temperature was still 500°F
Solution to Problem 1.3;
Part 1.
Gross heating value per hour
, Btu/scf) (Fuel rate, scf/hr)
(258 Btu/scf) (10,000 scf/hr)
2,580,000 Btu/hr
(A)
Part 2.
Use Attachment 2-9, p. 2-31 of the Student Manual to estimate the avail-
able heat, QA , from the above fuel with flue gases at 500°F.
Interpolate between curves in Attachment 2-9 at identical flue gas tem-
peratures using the following ratio:
OR
Desired fuel
OH
(B)
Reference fuel
Choosing Producer Gas as the reference fuel:
-------
With waste gas: QH = 258 Btu/scf, from Equation (B) above:
QA = (258 Btu/scf) (0.80) = 206 Btu/scf
Total heat available from waste gases = (10,000 scfh) (206 Btu/scf)
= 2,060,000 Btu/hr
Part 3.
Attachment 2-10, p. 2-32 of the Student Manual, gives available heat as
the percent gross heating value with various amounts of excess air.
With flue gases at 500°P and 20% excess air, read
(QA/QH). x 100 - 79% (C)
Thus, heat available per hour with 20% excess air is:
QA = (79/100) (2,580,000 Btu/hr) - 2,038,000 Btu/hr
1-9
-------
PROBLEM 1.4:
Liquid Waste Combustion in Natural Gas-Fired Boiler
Combustible liquid waste from a manufacturing process is to be burned
in a boiler which is fired with 1,059 Btu/scf higher heating value
natural gas at a rate of 5,000 scfh. The liquid waste is equivalent to
10 Ib/h of benzene.
Determine;
1. The total gross heating value to the boiler per hour.
2. The amount of combustion air required to burn the waste liquid.
Assume a 20% excess of theoretical air and express your results
in scfm.
3. The amount of available heat from the boiler if the flue gases
leave the heat exchanger at 600°F and complete combustion is
achieved with 20% excess air.
Flow Diagram
NATURAL GAS
IMiM^HMMHMBBM^H^^^HBaiHMHB^MMB
5000 ft'/hr
1 STACK GASES
S 600°F
WASTE
BENZENE
10 Ib/hr
HEAT AVAILABLE
1-10
-------
Solution to Problem 1.4:
Choose as a basis:
1 hour of operation.
Part 1.
Gross heating value of natural gas t QH * 1,059 Btu/scf.
* y»s
Gross heating value of benzene is obtained from Attachment 2-1, p. 2-23
of the Student Manual:
OH, benzene " Btu/lb
Total gross heat input to the boiler is:
Natural gas (5,000 scfh) (1,059 Btu/scf) » Btu/hr
Benzene (10 Ib/hr) ( Btu/lb) a Btu/hr
Total » Btu/hr
Part 2.
Attachment 2-1, p. 2-23 of the Student Manual, gives the combustion air
requirement for benzene (Substance No. 21) as Ib air/lb benzene
or ___________ scf air/scf benzene.
Density of benzene J p. is Ib/scf.
benzene —————————
Theoretical air required to burn benzene type waste completely
VA, t = ^enzene/Pbenzene1 <^oretical scf air/scf benzene)
10 Ib/hr
V, = x scf air/scf benzene)
A. L
' ( Ib/scf benzene)
a scf air/hr
1-11
-------
Air requirements with 20% excess air:
V
air
scf air/hr)
scf air/hr
Part 3.
Refer to Attachment 2-10, p. 2-32 of the Student Manual. Bead avail-
able heat as percent of gross heating value with flue gases at 600°F
and 20% excess air as %
100
UJ
UJ
600 °F
Heat available from the boiler
Btu/hr) (
Btu/hr
[from Part l]
[from Attachment 2-10]
1-12
-------
PROBLEM 1.5: Combustion Calculations with Heat Recovery
Part A
A chemical plant has installed an industrial boiler to produce process
steam. 'Hie boiler is fired with natural gas of the following composi-
tion by volume: 90% methane, 5% ethane, and 5% nitrogen. The boiler
is designed to burn 80,000 cubic feet per hour (delivered at 60°F) of
natural gas at 10% excess combustion air.
Determine;
1. The gross heat input to the boiler, Btu per hour.
Assumptions; (a) natural gas and combustion air enter the
boiler at 60°F; (b) heat losses from the boiler due to
radiation and convection are negligible.
2. The combustion air requirement, cubic feet per hour (at 60°F,
30 inches mercury pressure). Assume average atmospheric con-
ditions are 60°F and 30 inches mercury pressure.
3. The available heat for process steam if the flue gases leave
the boiler heat exchanger at 400°F.
Part j>
The personnel of the chemical plant are now considering the addition of
an air preheater to the boiler to preheat combustion air. Calculations
show that the flue gases leaving the heat exchanger section would enter
the air preheater at the following conditions: 1,500,000 cubic feet per
hour at 400°P. The air preheater will be designed to reduce the flue
gas temperature to 350°F. At conditions of complete combustion, calcu-
lations show the flue gases entering the air preheater to be of the
following composition by volume: 8.8% CO2, 1.7% 02, 72.3% N2» and
17.2% H2O. (Note: Calculations show the water vapor flow rate in the
flue gases equals 7,400 Ibs. per hour.)
Determine;
4. The heat recovered (H.R.) from the flue gases by_ the air pre-
heater based on the flue gas analysis and flue gas flow rate.
(Note; Use Equation 1 shown below.)
5. The increase in combustion air temperature resulting from the
use of an air preheater. (Notet Use Equation 2 shown below.)
6. The combustion air temperature after passing through the air
preheater.
1-13
-------
H.R. » 2. (flow rate of each component) x (difference
in
heat content of each)
(1)
Btu
H-R'Air * °'24 it> PF x temperature increase x air flow rate (2)
LOW DIAGRAM FOR BOILER:
COMBUSTION AIR
STEAM
COMBUSTION
ZONE
HEAT
EXCHANGER
HATER
1,500,0-50
ft»/hr
AMBIENT AIR
(60 °F)
350°F
FLUE GASES
AIR
PREKEATER
8.8% CO.
NATURAL GAS
72.3% N2
17.2% H20 ^ 7400 Ib/hr
1-14
-------
Solution to Problem 1.5:
Basis: 1 hour of operation
Part 1.
Volumetric
Substance Flow Rate
scfh
(a)
Gross Htg.
Value, Btu/scf
(b)
Heat Input
Btu/hr
Methane— CH4 (0.90)(80,000) =
Ethane— C2Ef> (0-05) (80,000) a
Nitrogen— N2 (0.05)(80,000) =
Totals:
scfh
Btu/hr
Note; (a) From Attachment 2-1, p. 2-23 of the Student Manual.
(b) Obtained by multiplying volumetric flow rate by the
corresponding gross heating value.
Part 2.
Combustion air requirements:
Combustible
Substance
CH4
C2H6
Volume,
scfh
72,000
4,000
Theor. Air*
scf air/scf gas
Actual Air (10% •xcess)
scf air/scf scf air/hr
Total Air
*From Attachment 2-1, p. 2-23 of the Student Manual.
scfh
1-15
-------
Part 3.
For available heat as percent of gross heating value, use Attachment 2-10,
p. 2-32 of the Student Manual.
Read for 400°F flue gases and 10% excess air %.
Available heat from the boiler
gross Btu/hr)( % gross/100)
Btu/hr.
Part 4.
Need to calculate flow rate of combustion products in Ib/hr. First correct
flue gas flow rate from 400°F to standard temperature of 60°F, using
Charles' law (Equation 2.7, p. 2-9 of the Student Manual).
'flue, 60°F * (1.500,000 ft3/hr) (460 + °F) / (460 +
£t3/hr
Mass flow rate of component « (volume fraction) (total volume flow) (density)
Component
C02
°2
N2
H20
•total
Fraction
1.00
Density
Ib/ft3
Mass Flow
Ib/hr
7,400
Enthalpy, Btu/lb
at 400°F
Btu/lb
1,212
Btu/hr
8.97 x 106
H400 »
at 350°F
Btu/lb
H350 -
Btu/hr
Note: Densities available from Attachment 2-1, p. 2-23 of the Student Manual.
.Enthalpies from Attachment 2-7, p. 2-29 of the Student Manual.
1-16
-------
Heat recovered from cooling flue gases = ^H4oo ~ H35cP Btu/nr-
H.R.Air
Part 5.
Refer to Equation (2) of the Problem Statement, which on rearrangement
gives:
AT . « (H.R. ' ) / (0.24 Btu lb°F x Air Flow Rate, Ib/hr)
Air Air
Obtain density of air from Attachment 2-1, p. 2-23 of the Student Manual,
_£p compute:
Air Flow Rate, Ib/hr - (Volumetric Air Flow, scfh)(density, Ib/scf)
- ( . scfh) ( Ib/scf)
Ib/hr.
Substituting into expression for AT^.^ :
( Btu/hr)
AT = - °F
Air (0.24 Btu/lb-°F) ( . Ib/hr)
Part 6.
Air temperature after preheater - 60°F + ATAir
» 60 + = °F
1-17
-------
CHAPTER II
COMBUSTION SYSTEM DESIGN PROBLEMS
PROBLEM II.1: Calculation of Furnace Volume
Consider the design of a pulverized coal-fired furnace which operates with
an average energy release rate of 25,000 Btu/hr per cubic foot of furnace
volume. The furnace produces steam with an energy output of 55 x 10^ Btu/
hr and a thermal efficiency of 80%.
Calculate:
1. The furnace volume for the steam generator.
Solution for Problem II.I:
1. Determine the fuel energy input required in order to realize the
given energy output
?1 5S x 106 Btu/hr
H ~ 0.80
Btu/hr
2. Refer to Table 9.6 in the Student Manual, p. 9-10, to obtain
the average design value for the heat release rate of 25,000
Btu/hr ft^ for pulverized coal firing.
Furnace Volume
25,000 Bt% 25,000 Btu ,
hr ftT hr ft3
ft3
II-l
-------
PROBLEM II.2: Furnace Volume Plan Review
An industrial organization proposes to build a 100,000 pounds-per-hour
steam generator. The furnace is to be fired by a chain-grate stoker
with continuous ash removal, similar to that shown in Attachment 9-4 of
the Student Manual, p. 9-19. The furnace is 12 ft. wide (across the
front), 14 ft. deep, and 28 ft. high. The volume corresponding to
these dimensions includes the superheater volume, which is small enough
to be neglected in the calculation. The fuel for the proposed unit is
to be the high-volatile bituminous coal described in Attachment 3-11
of the Student Manual, p. 3-20. The steam generator will require
6 tons per hour of this coal to achieve its full steam capacity.
Determinet
1. If the furnace volume is adequate.
Solution for Problem II.2;
1. Calculate the furnace volume using the dimensions given:
Furnace volume - (length) x (width) x (height) - (superheater volume)
-( ) x ( )x( )-( 0.0)
ft3
2. Calculate the energy release rate per cubic foot for the speci-
fied fuel and design capacity.
(coal firing rate) x (higher heating value)
Energy release rate »
(Furnace volume)
Bttt/hr ft3
3. Compare the value obtained above to that given in Table 9.6 on
p. 9-10 of the Student Manual.
II-2
-------
PROBLEM II.3: Calculation of Furnace Gas Exit Temperature
A reheat steam generator design has energy utilization based on the
total energy input (higher heat value) as follows:
1. Energy absorbed in radiant boiler 49.5%
2. Energy absorbed in convection superheater 20.8%
3. Energy absorbed in economizer 6.6%
4. Energy absorbed in steam reheater 8.0%
5. Energy absorbed in air preheater 5.0%
6. Furnace heat losses 3.0%
7. Flue gas and other losses 7.1%
100.0%
The unit is fired with pulverized coal, using the coal described as the
"as received" coal listed in Attachment 3-12 on p. 3-21 of the Student
Manual. The unit operates with 15% excess air; and the combustion air
is preheated to 300°F.
Calculate;
1. The temperature of the gas leaving the furnace.
Solution for Problem II. 3;
1. Determine the theoretical air required to burn the coal speci-
fied, using Equation 4.1 on p. 4.4 of the Student Manual.
The coal is 75% carbon, 5% hydrogen, 2.3% sulfur, 1.5% nitrogen,
6.7% oxygen, 2.5% moisture, and 7.0% ash.
°2
Theoretical Air - Afc » 11.53 (C) + 34.34 (H2 - -g-) + 4.29 (S)
= 11.53 ( ) + 34.34 ( - ) + 4.29 (
8
* Ibs per Ib of coal
II-3
-------
2. Calculate total air.
% EA
Total Air - Aa = (1.0 + ~ioo") At
(1.0 + ) x (
100
Ibs per Ib of coal.
3. Estimate the amount of flue gas produced using Equation 4.2 on
p. 4-5 of the Student Manual:
Theoretical flue gas » G
- noncombustibles ) 4- m A
Choose a basis of one pound of fuel* so that mf « 1:
% Ash
G - (1.0 - — — — ) + (1.0) x At
Ib gas per Ib of coal
Actual flue gas - Gf • • (G + Ag) - G + ^^ x A,.
(ior)x<
Ibs gas per Ib of coal
4. Calculate the useful energy, Qu, absorbed in the furnace region
(radiant boiler in this case).
« (fraction of energy absorbed in radiant boiler) x (HHV)
- 0.495 x ( )
a> Btu/lb of coal
II-4
-------
5. Note that Q is also related to the energy input as follows:
Qu * (lower heating value) - (losses) - (energy in the gases
leaving furnace)
which is given by Equation 4.8 on p. 4-7 of the Student Manual.
Qu = H - QL - Gf Cp (tf - ta)
a. The energy, H, is obtained from
H » HHV - energy of the water in flue gas
- HHV - Qv ,
where:
water in flue gas « 9.0 x (H2 in fuel) + (as-fired
moisture)
- 9.0 x ( ) + ( )
» Ibs H2O/lb coal
and the energy in this water is
Qv • (Ibs of water per Ib fuel) x (latent heat of
vaporization)
(1,000
now
H - HHV - Qv
Ibs fuel
Btu per Ib coal
Btu per Ib of coal
II-5
-------
b. Bie losses, Q- , are:
Qj^ - fraction of energy lost from furnace x (HHV)
- ( ) x ( )
» Btu per Ib of coal
c. The furnace gas tenperature is calculated by substituting
values obtained from Q^ , H , Qj,, Gf together with a
value for Cp - 0.26 Btu/lb °P and ta - 300°P:
C,, - H - Qj, - Gf Cp (tf -
- ( ) (0.26 g=5^) (tf - 300)
therefore:
II-6
-------
CHAPTER III
EMISSION CALCULATIONS I
PROBLEM III.l: Pollutant Emissions from Coal-Fired Power Plant
Plans call for a 500 MW power plant to have a dry-bottom design and to
burn pulverized coal. The overall thermal efficiency is designed to
be 34%. The coal specified in the plans contains 1.3% sulfur, 22% ash,
and has a 12,500 Btu/lb HHV.
Compute;
1. The input energy required when the unit is operated at the rated
capacity.
2. The coal firing rate at the rated capacity.
3. The pollutant mass rate for emissions of:
a. SO2
b. Participates
C.
d. HC
e. CO
4. The process emissions per million Btu of energy input for
a. S02
b. Participates
5. The degree of control required to meet a 1.2 Ib SO2/10 Btu per-
formance standard for S02-
6. The degree of control required to meet a 0.1 Ib particulate/10^
Btu performance standard for particulates.
III-l
-------
Solution to Problem III.l;
1. Plant electric output rating and thermal efficiency can be used
to find energy input from Equation 4.9, on p. 4-8 of the Student
Manual.
0, - energy in » energy out , QS
thermal ef f r\
( ) MWe 3 Btu
3413 x 10-
MWe hr
Btu
hr
2. With the value of Qg and the coal HHV , the coal-firing rate
is given by:
/ l Btu
. _. _ < < }
mf « coal fxred,
hr HHV per ton / » Btu
)
Jon
hr
3a. The pollutant mass rate for S02 can be obtained using the coal-firing
rate and the emission factor for S02 (refer to Student Manual, p. 5-30
for emission factors)
lb S02 ton coal
38 X S
ton coal * hr
. 38 . < , i
ton coal hr
Ib SO2/hr
III-2
-------
b. (PMR) . = 17 x A "> Part" x ~v ton coal
part. ton coal f
= 17 x ( ) lb Part. x ( ton coal
ton coal hr
lb part./hr
c. Similarly, the PMR for NO would be
18 * X * mf toTl coal
ton coal Kr
18 "* N0x x
ton coal
lb
Similarly the PMR's for CO and HC are:
d. (PMR)HC - ( )
ton coal
HC ton coal
) x ( ) lb HC/nr
lb HC/hr
e. (PMR)CO - ( ) * x m- ton °°al
co ton coal ^:
) x ( ) lb CO/hr
lb CO/hr
III-3
-------
4a. The S02 process emissions per million Btu energy input will be com-
puted from the SO- pollutant mass rate and the input energy rate:
E
SO2
(PMR)S02
OH
lb SO2
) hr
) Btu
lb S02
10* Btu
Ine particulate emissions per million Btu energy input will be
computed similarly:
Epart
(PMR)part
OH
( ) lb particulates/hr
( ) Btu/hr
106 Btu
The computations presented above can be used to compute the degree
of control required to meet a given emission standard. For this
problem the performance standards are listed on p. 5-20 in the Student
Manual. For a solid-fuel-fired power plant which is 250 x 106 Btu/hr
or larger, the S02 standard is 1.2 lb SO2/106 Btu.
From above the calculated £„ - lb SO2/106 Btu
Therefore,
III-4
-------
ESO2 ~ Standard
Degree of control needed = x 100%
x 100%
% reduction of the
uncontrolled value
6. Similarly the particulates standard is 0.1 lfc/10 Btu and the
estimated uncontrolled particulates was
. v Ib particulates
•
- Standard
Degree of control needed - — - = - x 100%
100%
% reduction of the
uncontrolled value
III-5
-------
PROBLEM III.2: Fuel Sulfur Limit Calculation
A 22-degree API fuel oil is to be burned subject to a maximum S02
emission standard of 0.8 Ib S02/106 Btu input.
Determine:
1. The maximum sulfur composition of the 22-degree API fuel oil
which meets the standard without flue gas desulfurization.
Solution to Problem III. 2;
From Student Manual, Attachment 3-5, p. 3-15, find:
total heat of combustion at constant volume = _ Btu/lbm
One should note that SO2 is 1/2 oxygen and 1/2 sulfur by weight.
Therefore,
.5 Ib S (0.8 Ibp S02) j _ ) Btu
Ib S02 106.Btu U>m oil
oil
% S in the oil
III-6
-------
CHAPTER IV
EMISSION CALCULATIONS II
PROBLEM IV.1: Excess Air Calculation Based on Orsat Analysis
The effluent of a combustion unit is characterized by:
Orsat Analysis: 11% CO2, 9% O2, 2% CO
SO2: 120 ppn
Gas Flow: 200,000 acfm @ 27.5 in.Hg pressure,
340°F temperature, 8% moisture by volume
Particulates: 400 grain per acf
Calculate:
1. The percentage of excess air in the flue gas using Equation 1
from Attachment 5-4, p. 5-23 of the student manual.
2. The SO2 emissions in micrograms per cubic meter (yg/m3).
3. The volume of the dry flue gas at the observed conditions.
4. The volume of the dry flue gas at standard conditions which
are a pressure of 30.00 inches of mercury and a temperature
of 60°F.
5. The particulate concentration based on standard pressure and
temperature.
6. The particulate emission concentrations corrected for
a. 50% excess air
b. 12% CO2
c. 6% 0.
IV-1
-------
Solution for Problem IV.1;
1. Refer to Student Manual, p. 5-23, to obtain the equation for
computing the excess air from an Orsat analysis.
(05_ - 0.5 C0_)
EA - fE P x 100%
0.264 N2p - (02p - 0.5 C0p)
The nitrogen in the product gas, N2 , may be calculated as
follows:
N2 - 100 - %C02 - %O2 - %CO
- 100 - ( ) - ( ) - (
- % N2
Substitute the above value into equation for EA:
( ) - 0.5 ( )
EA
0.264 ( ) - ( - 0.5 ( ))
% excess air
2. Convert 120 SO2 ppm to Ug/m using Equation 5.8 on p. 5-5 of
the Student Manual:
1 ppm - 40.8 x (MW) HJ.
m3
1 ppm S02 = 40.8 x ( )
*
120 ppm SO, = 120 x (
IV-2
-------
Reduce the —=• units to 52- , by noting that
1,000 x U£ - 1
mo
mg.
Therefore
120 ppm SO2 * (
3-
SSL
m3
3. Calculate the volume of the gas as follows:
vo dry * 7o wet
) (1.0 -
cfm
4. Refer to the Student Manual, Equation (1) on p. 5-22 and
reduce Vo ^. to Vs ^ry , using P8 - 30.00 Hg and Ts - 520 R
Vs - V0
J L
( )
scfm
Likewise reduce the particulate loading concentration to that
at the standard conditions
IV-3
-------
) x
J I
( )
J I
( )
grain/scf
6a. Refer to the Student Manual/ p. 5-23 and use Equations (2) and (3)
to calculate the particulate concentrations on a 50% excess-air
basis
F
50V
1 -
1.50 0 - 0.133 N - 0.75 CO
0.21
1 -
1.50 ( ) - 0.133 (
-0.75 (
0.21
C50V
( )
'50V
grain/scf
b. Correct to 12% CO2, using Equations (6) and (7) on p. 5-23 of the
Student Manual
:12V
*^2p ( )
0.12 ( )
( )
'12V
F12V (
grain/scf
Correct to 6% C^ using Equations (10) and (11); however, note that
Equation (10) snould be modified for the net O2 (after the CO is
oxidized):
6V
0.21 - (02p - 0.5 (Cpp))
0.15
TV-4
-------
0.21 - ( - 0.5 ( ))
0.15
C
6V = - = - - - = _ grain/scf
F6V ( >
IV-5
-------
PROBLEM IV.2: Use of F-factors to Compute Emission Concentrations
The effluent from a bituminous coal-fired source is found to have a
particulate concentration, Cvs , equal to 2.0 grains/scf (dry basis)
The flue gas oxygen is 9% and the carbon monoxide is 2%, as measured
on a dry basis.
Calculate;
1. The particulate emissions in the units of (grains/million Btu)
using the F-factor technique
Solution to Problem IV.2;
From Attachment 5-4, p. 5-25, of the Student Manual, find:
dscf
106 Btu
with the given C^s value and the computed F^ , use Equation (5.32),
p. 5-16, to calculate E , the particulate emissions, grains/106 Btu
F 20.9
Fd [20.9 -
-------
PROBLEM IV.3: Calculation of F-factor
F-factors are useful in the calculation of emissions from combustion
sources. Consider a bituminous coal having the "as-fired" ultimate
analysis of 75% carbon, 5% hydrogen, 6.7% oxygen, 1.5% nitrogen, 2.3%
sulfur, 7.0% ash, and 2.5% free moisture. .The heating value of this
coal is 13,000 Btu/lb.
Calculate;
1. The F-f actor, F, , using the Equation 5.28 on p. 5-15 of the
Student Manual and compare this value with that given in
Attachment 5-5, p. 5-25, of the Student Manual.
Solution to Problem IV. 3:
The equation for the F-f actor, Fd , is
(3.64 H9 + 1.53 C + 0.57 S + 0.14 N - 0.46 02) fi a8cf
° v*owj.
p «
d HHV 106 Btu
[3.64 ( ) + 1.53 ( ) + 0.57 ( ) + 0.14 ( ) - 0.46 ( )]
d»cf
106 BtU
IV-7
-------
PROBLEM IV. 4: Calculation of Pollutant Concentration
Bituminous coal is burned completely at a rate of 5 ton/hr with excess
air. An evaluation of the effluent yields the following data:
Orsat Analysis: C02 9.1%
02 10.6%
CO 0.0%
Volume Flow: 26,000 scfm
Pollutant Mass Rate: 130,000 grains/min.
Compute;
1. The particulate concentration corrected to 50% excess air.
2. The particulate concentration corrected to 12% O^.
3. The particulate concentration corrected to 6% 02.
Solution to Problem IV.4:
1. Find the particulate concentration, C^g , using the flow and the
pollutant mass rate from Equation 5.21, p. 5-14 of the Student
Manual;
m PMR
—
( ) grains/min.
) scfm
grains/scf
Correct the concentration to 50% excess air using Equations (2) and
(3) on p. 5-23 of the Student Manual
50v
( 1.5 O2p - 0.133 N2p - 0.75 C0p
0.21
0
1 - 1.5 ( ) - 0.133 ( ) - 0.75 (
0.21
IV-8
-------
c
C50v
P50vg ( )
grains/scf at 50% EA
2. Correct the concentration to 12% C02 using Equations (6) and (7)
on p. 5-23 of the Student Manual
co2p
L2v
r!2
grains/scf at 12% CO2
3. Correct the concentration to 6% O2 using Equations (10) and (11)
on p. 5-24 of the Student Manual
F6V
- °2P . 0.21-
0.15 0.15
grains/scf at 6% O2
IV- 9
-------
PROBLEM IV.5: Correction of NOX Emission Concentration to 3% 02
Limiting the excess air during combustion is an important technique for
controlling the NOX emissions. In order to provide a more meaningful
basis for comparison, the resulting emissions will be corrected to a
standard basis of 3% O2 (or 3% excess 02). Consider the NOx emissions
of 200 and 300 ppra from an oil-fired power plant under the stack gas con-
ditions A and B, respectively (which have different conditions of excess
air).
Dry Volume Basis
V*WJJUJL UJ-Wil
A
B
co2, %
13.3
9.7
o2, %
2.2
7.3
N2, %
84.5
83.0
NOjj , ppm
200
300
Determine:
1. The excess air corresponding to conditions A and B.
2. the correction factor to be used in correcting NOX emissions
from their actual condition to the basis of 3% 02.
3. The corresponding values of NOX at the standard basis of 3%
oxygen.
Solution to Problem IV. 5;
1. Find the excess air for conditions A and B using Equation (1)
on p. 5-23 of the Student Manual.
%EA
02p - 0.5 C0p
0.264 N2 - (O2p - 0.5 C0p)
x 100%
For condition A:
%EA
) - 0.5 (
)
0.264 ( ) - (
- 0.5 (
for condition A.
x 100%
IV-10
-------
For condition B:
%EA
) - 0.5 (
0.264 (
- 0.5
x 100%
for condition B.
2. The volume correction factor for flue gas 02 is derived front
0.21 - O
•02V
2p
std
0.21 - 0
:3v
2p
0.21 - 0.03
0.21 -
°2P
3. Use the correction factor developed above, to correct the measured
emissions at conditions A and B to the 3% 02 standard basis:
% excess air
For 200 ppm NO- at
-3v
200 ppm
21 -
.18
ppm corrected to 3%
For 300 ppm NOX at
% excess air
'3v
F
3v
.21 - ( )
.18
ppm corrected to 3% O,
IV-11
-------
CHAPTER V
AFTERBURNER DESIGN PROBLEMS
PROBLEM V.I: Afterburner Design for Meat Smokehouse Effluent
Consider a meat smokehouse discharging 1,000 scfm effluent at 150°F,
which needs to be treated to control a very low concentration of pollu-
tants at the parts-per-million level. This could be accomplished by
thermal incineration at 1,200°F for at least 0.3 seconds. The follow-
ing are reasonable assumptions:
1. The amount of combustibles in effluent gases is very low;
there is no contribution to the heating value due to their
oxidation.
2. Effluent gases have the same thermal properties as air.
3. Intake combustion air is available at 60°F.
Determine;
1. The natural gas required for preheating the contaminated
effluent to 1,200°F using all fresh combustion air intake.
2. The afterburner throat diameter to give 20 ft/sec throat
velocity for good mixing.
3. The diameter and the length of the afterburner for a minimum
L/D ratio of 2 and afterburner chamber velocity of 12 ft/sec.
£££ST a— «- «-«•«i.*»0'
Waste Effluent
1,000 scfm
at 150°F
Natural Gas, G
gas
Combustion Air,
at 60°F
V-l
-------
Solution to Problem V.I;
Choose as a basis for calculation:
1 hour operation
Part 1.
•
a. Calculate waste effluent flow rate, m (Ib/hr)
•
m = (volume flow rate) (density)
Since assumed effluent to have properties of air, density from Attach-
ment 2-1, p. 2-23 of the Student Manual.
m = (1,000 scfm) (0.0766 Ib/scf) (60 min/hr) • 4,600 Ib/hr
b. Calculate the heat required to increase the effluent waste stream
temperature from 150°F to 1,200°P, allowing for 10% loss (i.e., multiply
by 1.10):
Q » 1.10 m AH
Enthalpy difference, AH, obtained by using Attachment 2-7, p. 2-29
of the Student Manual:
Enthalpy of air at 1,200°F is: 288.5 Btu/lb
Enthalpy of air at 150°F is: 21.6 Btu/lb
Therefore, AH » 266.9 Btu/lb
Therefore,
(1.10) (4,600 Ib/hr) (266.9 Btu/lb) = 1.35 x 106 Btu/hr,
V-2
-------
c. Available heat from natural gas, QA (Btu/scf)
Assume: Gross heating value of natural gas = 1,059 Btu/scf
Theoretical combustion air « 10.0 scf air/scf gas
Combustion products = 11.0 scf/scf gas.
From Attachment 2-9, p. 2-31 of the Student Manual, obtain for 1,200°F
flue gas temperature:
QA * 690 Btu/scf
(This is the amount of heat remaining after the combustion products
from 1 scf of gas are raised to the afterburner temperature. This
heat is then available for heating the waste effluent to the same
afterburner temperature.)
d. Natural gas needed, G (scfh):
- (1.35 x 106 Btu/hr)/(690 Btu/scf gas) - 1,960 scf gas/hr.
Part 2.
a. Volume of combustion products at 1,200°F, Gp (ft3/sec):
Gp = (1,960 scf gas/hr)(11.0 scf prod/scf gas)(460 + 1,200,°R)/(460 + 60,°R)
- 68,800 ft3/hr - 19.1 ft3/sec.
b. Volume of waste effluent at 1,200°F, GE (ft3/sec)
Gg - (1,000 sc£m)(460 + 1,200,°R)/(460 + 60,°R)
3,190 ft3/min " 53.2 ft3/sec.
c. Total volumetric flow of gases to the afterburner chamber through the
throat:
G^ „ - G,, + G_ » 19.1 + 53.2 - 72.3 ft3/sec.
tOt i* E
V-3
-------
d. Afterburner throat area Athroat = ^d ' 4
Throat diameter d = (4Athroat/ir)^ (A)
Now the velocity through the throat is:
vthroat = Gtot/Athroat W
Combining Equations (A) and (B) above to eliminate the throat area
and solving for throat diameter, d :
(C)
T vtnroat
For required throat velocity of 20 ft/sec:
|(4/TT) (72.3 ft3/S€c) / {20 ft/sec)! * - 2.15 ft
Part 3.
Afterburner chamber velocity specified at 12 ft/sec. Thus chamber dia-
meter, O , obtained from Equation (C) above with v throat rePlace(* by
Vchamber • 12 ft/sec
D = (4/TT) (72.3 ft3/sec) / (12 ft/sec)] - 2.77 ft
Length of afterburner chamber (L/D 21 2}
Minimum L - 2D - (2) (2.77) - 5.54 ft (D)
Check residence time, t
t * L/V " (5'54 ft) / (12 ft/sec) - 0.46 sec (E)
Since t = 0.46 sec is greater than the minimum required residence
time of 0.30 see, the above design is satisfactory.
V-4
-------
Note; Natural gas requirements can be reduced by:
(i) heat recovery from clean gases to preheat incoming
waste effluent, and
(ii) using oxygen from the waste effluent stream for
combustion, thereby reducing primary air require-
ments for the auxiliary fuel.
This latter option is illustrated in Problem V.2.
V-5
-------
PROBLEM V.2: Afterburner Design with Combustion Oxygen from the
Contaminated Effluent
Assume that the meat smokehouse effluent in Problem V.I has also the
same composition as air (21% by volume oxygen) except for the minute
concentration of contaminants. Repeat the calculations of Problem V.I,
but use the oxygen from the smokehouse effluent for combustion of the
auxiliary fuel as much as possible.
Reasonable assumptions are: a mixing-plate type burner (see Attach-
ment 7-6) will be used in this application. A ring baffle, which was
used in Problem V.I, will therefore not be necessary to obtain good
mixing between the auxiliary fuel combustion products and the effluent
to be incinerated.
Determine;
1. The hypothetical available heat for this afterburner application.
2. The natural gas requirements and the fraction of combustion oxy-
gen available from the effluent.
3. The afterburner dimensions as in Problem V.l-3.
Solution to Problem V.2;
Preliminary Notes on Hypothetical Available Heat Calculations:
Let X » fraction of theoretical air for burning auxiliary fuel entering
through the burner (primary or fresh intake air)
1 - X » fraction of theoretical air from waste effluent
Bg * heat content (enthalpy) of effluent at final temperature
EE m Cp AT = (0.24 Btu/lb-°P) (T-60, °P) (A)
W » weight of combustion air from effluent
W = "W (1 * X) PAir
V-6
-------
Heat content, Q , of that combustion air at final afterburner temperature
W HE
(1 - X) p
(C)
Since this amount of heat, Q , is no longer needed to heat up fresh intake
(primary) air, it will be available to heat the rest of the contaminated
effluent. Thus we have a "hypothetical" available heat, Ql :
Q; - Q. + A (1 - X) p
A A Tn
where QA obtained from sources such as Attachment 2-9, p. 2-31 of
the Student Manual
- 0.0766 Ib air/scf
=» 10.0 scf air/scf natural gas burned (typically)
» calculated from Equation (A)
(D)
(E)
For a natural gas with 1,059 Btu/scf gross heating value and the above
burning characteristics, the hypothetical available heat as a function
of the afterburner temperature is:
Afterburner Temperature
op,
Hypothetical Available Heat
Qa i Btu/scf gas
600
800
1,000
1,200
1,400
1,600
1,800
830 + 100 (1 - X)
785 + 136 (1 - X)
740 + 173 (1 - X)
690 + 210 (1 - X)
645 + 246 (1 - X)
600 + 283 (1 - X)
550 + 320 (1 - X)
V-7
-------
Part 1.
Assume first that no primary air is needed, i.e. X = 0, and all combustion
air comes from the waste effluent. This needs to be checked; if assump-
tion is not justified/ adjust value of X and go through the calculations
again.
Hypothetical available heat for T = 1,200 F:
Q = + (1-0) - Btu/scf gas
A
Part 2.
Auxiliary natural gas fuel needed
Ggag =» (Heat to raise effluent to 1,200°P)/QA
= (1.35 x 106 Btu/hr)/( _ Btu/scf gas)
» _ scf gas/hr
Theoretical air needed to burn auxiliary gas :
scf gas/hr) (10.0 scf air/scf gas)/ (60 min/hr)
scfm air.
Compare the above Gair with volumetric flow rate of waste effluent (which
is equivalent to air) .
If G=,. _ < G __, . , then assumed value of X justified and proceed
air effluent
to next part.
If Gaj.r > Gef fluent ' then adjust X accordingly and repeat above
calculations .
V-8
-------
Part 3.
Auxiliary fuel combustion products at 1,200°F , G*
G' = (Ggas scf 9as/hr> (11-0 scf prod./scf gas) (460 + 1200)/ (460 + 60)
- ft3/hr
» ft3/sec
Waste effluent volume at 1,200°?, 63 , after removing portion already
accounted for in auxiliary fuel burning:
(1,000 - G.) ( * ° * ) -— — - ftVsec
—~—^^—~
— ** —3m m r « » * ^**
3X1 460 + 60 sec
Total volumetric flow to afterburner:
ft3/sec.
Afterburner Chamber Diameter (Equation C from Problem V.I):
D - ["(4/10 (G^t) / (12 ft/sec) I * - - ft.
Afterburner length:
L - 2D - (2) ( _ ) - _ _ ft-
Residence time (Equation (E) from Problem V.I):
t - L/Vchanber - ( _ _ ft) / (12 ft/sec) - - sec
_> 0.3; hence O.K.
V-9
-------
CHAPTER VI
COMBUSTION SYSTEM CALCULATIONS
PROBLEM VI.1: Fuel Requirements for Combustion Installation
A steam generator is rated at 400,000 Ibs of steam per hour. Steam (99%
dry) leaves the boiler at 1,500 psia pressure and enters a superheater.
Steam leaves the superheater at 1,400 psia pressure and a temperature
of 1,000°F. The feedwater for this unit enters the economizer at 300°F
and leaves at 400°F. The overall thermal efficiency of the steam genera-
tor is 74%. The energy and water losses associated with blowdown may be
neglected.
Compute;
1. The rate of energy delivered to the:
(a) economizer,
(b) boiler,
(c) superheater, and
(d) the total delivered
2. The fuel energy required, million Btu/hr
3. The fraction of the fuel energy which is absorbed in the
{a) economizer,
(b) boiler, and
(d) superheater.
vi-l
-------
SCHEMATIC DIAGRAM FOR PROBLEM VI.1
H
m = 400,000
s Ibs/hr
tj = 300 F
h - 269 7 B/lt
1
Economizer
E
3,,
t2= 400 F
h,= 375.1
* B/lb
Boiler
B
p.= 1,500
psia
t3=596.4F
v — Q Q
A3~ '"
(
Super -
Heater
S
In
p = 1,400
psia
t4= 1,000 F
h.= 1,493.5
4 B/lb
7
h3= 611.5 + 0.99(557.2)
= 1,163.1 B/lb
-------
Solution to Problem VI.1:
From the steam tables one may determine the enthalpy values of the feed-
water and steam:
Economizer inlet:
Economizer exit:
Boiler exit:
Superheater exit:
h =
t2 =
P3 -
t3 =
X =
300°F,
400°F,
1,500
596.39
.99
1^ = 269.7 Btu/lb
h2 = 375.1 Btu/lb
1,500 psia h3 = 611.5 + X (557.2)
- 611.5 + .99 (557.2)
» 1163.1 Btu/lb
1,400 psia
1,000°F
1493.5 Btu/lb
1. Compute the energy delivered to each section using Equation 4.13
on p. 4-10 of the Student Manual.
a. Economizer:
SE
400,000 lb geam (
Btu/hr
Btu
b. Boiler:
SB
- h2)
400,000 "> f!eam (
hr
Btu/hr
Btu
Ib
c. Superheater:
ms (h4 *
400,000 .^b gteam
nr
Btu
Tir~
Btu/hr
-------
d. Total:
Qs = Qs + Qs + Qs
ST SE SB SS
) -I- ( ) + ( ) Btu/hr
Btu/hr
2. The fuel energy input required may be determined using Equation
4.9 on p. 4-8 of the Student Manual.
QS () Btu/hr
Btu/hr
3. The fraction of the fuel energy which is absorbed:
a. Economizer:
QH
b. Boiler:
2s
SB
OH (
c. Superheater:
OH
VI-4
-------
PROBLEM VI.2: Combustion Improvement
Combustion modification of a boiler resulted in changing the excess air
as may be determined from the following Orsat analyses of the flue gas:
Gas Before Modification After Modification
C02 10.1% 15.0%
02 8.3 3.1
QO 0.1 0.0
The fuel fired was lignite coal which has the following analysis:
0.22% S, 6.39% H2, 37.37% C, 0.61% N2, and 44.99% 02. The heating value
is 6,010 Btu/lb and the proximate analysis is: 36.93% moisture, 24.92%
volatile matter, 27.72% fixed carbon, and 10.43% ash.
The unit operates 7,700 hr per year with an average load of 5.3 tons of
coal per hour with a fuel cost of 75$ per 106 Btu. Assume that before
and after the modification, flue gas temperature was 355°P» the refuse
was 0.1062 Ib per Ib of coal; and the average combustion air was at
75°F.
Compute;
1. The excess air
(a) before the modification,
(b) after the modification.
2. The theoretical air required to burn a pound of the specified
coal.
3. The theoretical flue gas produced from firing a pound of coal.
4. The actual flue gas produced per pound of coal
(a) before the modification,
(b) after the modification.
5. The change in flue gas energy loss per pound of coal.
6. The value of the annual savings from reduced flue gas losses,
which occur because of the modification
VI-5
-------
Solution to Problem VI.2;
1. Compute excess air knowing that:
N2 = 100% - C02% - 02% - C0%
a. Before modification:
N, = 100% - ( ) - ( ) - (
Determine %EA from Equation (1), p. 5-23 of the Student Manual:
( ) - 0.5 ( )
EA » x 100%
0.264 N2 - (02p - 0.5 CO )
- 0.5
0.264 ( 81.5 ) - ( 8.3 - 0.5 ( 0.1 ))
b. After modification:
N2 = 100% - ( ) - ( ) - (
EA - ! > - °'5 < > x 100%
0.264 ( ) - ( - 0.5 ( ))
2. The theoretical air required is found from Equation 4.1 on p. 4-4
of the Student Manual
02
A. = 11.53 C + 34.34 (H, - —— ) + 4.29 S
fc 8
11.53 ( ) + 34.34 ( - J \
Ib air/lb coal
VI-6
) + 4.29 (
8
-------
3. The theoretical flue gas per pound of coal fired may be obtained
from Equation 4.2 on p. 4-5 of the Student Manual, with m^ » 1:
- nonconbustible) + nif A£
) + 1 ( )
lb gas/ Ib cnnl
4a. Before the modification the actual flue gas per pound of coal
was
- EA (Afc) + G
- ( ) X { ) + ( )
*. _ ' lb gas/ lb coal fired
4b. After the modification the actual flue gas was:
Gf -
lb gas/ lb coal fired
5. As it was stated that the average ambient and fine gas
tures did not change after the modification, the difference
in flue gas energy loss may be determined using Equation 4.12
on p. 4-8 of the Student Manual
(Gf before ' ^ after>
. ,_ __ ' Bttt . .
( - ) x (0.25 — > x (
Btu/lb coal fired
VI-7
-------
6. The value of the annual savings resulting from reduced flue
gas losses will be:
Annual = cost x AQ Btu x Ib coal x time _hr_
savings Btu y Ib coal E hr year
10° Btu * v ' Ib coal x ( x 200 ) hr
year
per year
VI-8
-------
TECHNICAL REPORT DATA
(Please read Instructions on the reverse before completing)
1. REPORT NO.
EPA-450/2-80-064
2.
3. RECIPIENT'S ACCESSION NO.
4. TITLE AND SUBTITLE
APTI COURSE 427
COMBUSTION EVALUATION
Student Workbook
5. REPORT DATE
February 1980
6. PERFORMING ORGANIZATION CODE
7. AUTHOR(S)
J. Taylor Beard, F. Antonio lachetta, Lembit U. Lillele
8. PERFORMING ORGANIZATION REPORT NO
It
I. PERFORMING ORGANIZATION NAME AND ADDRESS
Associated Environmental Consultants
P. 0. Box 3863
Charlottesville, Virginia 22903
10. PROGRAM ELEMENT NO.
B18A2C
11. CONTRACT/GRANT NO.
68-02-2893
12. SPONSORING AGENCY NAME AND ADDRESS
U. S. Environmental Protection Agency
Manpower and Technical Information Branch
Research Triangle Park, NC 27711
13. TYPE OF REPORT AND PERIOD COVERED
Student Workbook
14. SPONSORING AGENCY CODE
15. SUPPLEMENTARY NOTES
EPA Project Officer for this workbook is James 0. Dealy
EPA RTP NC 27711
'
16. ABSTRACT
This Student Workbook is used in conjunction with Course #427, "Combustion Evaluation1
as applied to air pollution control situations. The workbook was prepared by the
EPA Air Pollution Training Institute (APTI) to provide problem solving exercises of
typical combustion calculations as a supplement to the course lecture materials.
Sample problems are presented for:
Combustion fundamentals
Combustion design
Emission calculations
Pollutant concentrations
Afterburner design
Fuel requirements
Note: There is also an Instructors Guide to be used in conducting the training
course (EPA-450/2-80-065) and a Student Manual for reference and additional
subject material - (EPA-450/2-80-063)
7.
KEY WORDS AND DOCUMENT ANALYSIS
DESCRIPTORS
b.lDENTIFIERS/OPEN ENDED TERMS C. COSATI Field/Group
Combustion
Air Pollution Control Equipment
Personnel Development-Training
Incinerators
Nitrogen Oxides
Exhaust gases
Emissions
Training Programs
Fuels
13B
51
68A
3. DISTRIBUTION STATEMENT Unlimited. Available
rromi . .
National Technical Information Service
5285 Port Royal Road
gr.T-fnofield. Virginia 22161
19. SECURITY CLASS (ThisReport}
Unclassified
21. NO. OF PAGES
63
20.!
22. PRICE
EPA Form 2220-1 (9-73)
VI-9
------- |