fired heater efficiency guide
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Tulsa Heaters Midstream 2
Fired Heater Efficiency
“Why is the efficiency of my fired heater important?”
Because inefficient heaters cost you money!
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Measuring Fired Heater EfficiencyThis guide will teach you the process of measuring and calculating your heater’s efficiency – so you can optimize your plant andsave money.
• Based on method outlined in API 560 Annex G
• Procedure intended for fired heaters burning liquid or gaseous fuels. Not recommended for solid fuels.
What do we mean by “efficiency”?
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Thermal VS Fuel Efficiency
THERMAL EFFICIENCY- total heat absorbed divided by total heat input
FUEL EFFICIENCY- total heat absorbed divided by heat input derived
from the combustion of the fuel only
NOTE: this definition differs from the traditional definition of fired heater efficiency, which generally refers to fuel efficiency
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Equipment Needed
• Temperature-measuring devices, such as thermocouples or thermometers, to measure the temperature of:• Fuel
• Ambient air
• Atomizing medium (if applicable)
• Flue gas
• Thermal fluid
• Flue-gas analytical devices to measure oxygen and combustible gases
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Before the Test
Establish and maintain operating conditions
Select and calibrate instrumentation
Perform any re-rating necessary to account for differences between design and test conditions
Ensure fuel is acceptable for the test
Ensure heater is operating properly with respect to the size and shape of flame, excess air & draught
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Testing
Test procedure:
Measurements:
• The heater shall be operated at a uniform rate throughout the test
• Data shall be taken at the start of the test, and every 2 hours thereafter
• The duration of the test shall extend until three consecutive sets of collected data fall within the prescribed limits
• Fuel gas quantity and heating value
• Flue-gas temperature and composition analysis
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ExampleHot oil heater for gas plant Ambient air temperature: 70°F
Relative humidity: 50%Fuel gas composition (vol%):• Nitrogen – 1.61• Carbon dioxide – 0.15• Methane – 98.17• Ethane – 0.008
Fuel gas LHV• 20,814 Btu/lb
Fuel gas HHV• 23,115 Btu/lb
Fuel gas temperature• 100°F
Fuel gas pressure• 50 psig
Excess Oxygen: 3% (dry)
Radiation heat loss: 1.5%
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Thermal Efficiency
Where:ℎ𝐿 lower heating value of the fuel (Btu/lb)
∆ℎ𝑎 heat correction due to air (Btu/lb)
∆ℎ𝑓 heat correction due to fuel (Btu/lb)
∆ℎ𝑚 heat correction due to atomizing medium (Btu/lb)
ℎ𝑟 assumed radiation heat loss (Btu/lb)
ℎ𝑠 calculated stack heat loss (Btu/lb)
𝒆 = 𝒉𝑳 + ∆𝒉𝒂 + ∆𝒉𝒇 + ∆𝒉𝒎 − (𝒉𝒓 + 𝒉𝒔
𝒉𝑳 + ∆𝒉𝒂 + ∆𝒉𝒇 + ∆𝒉𝒎
× 𝟏𝟎𝟎
Need to solve for all variables
Step 1:solve for heat loss through the stack (ℎ𝑠)
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Combustion Worksheet
Fuel Component
Column 1
Volume fraction
%
Nitrogen 1.61
Carbon dioxide
0.15
Methane 98.17
Ethane 0.08
TOTAL -
Total per pound of fuel
-
Insert fuel composition into combustion worksheet
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Combustion Worksheet
Fuel Component
Column 1
Volume fraction
%
Column 2
Relativemolecular
mass
Column 3 (1x2)
Total mass
(lb)
Column 4
Net heating value
(Btu/lb)
Column 5 (3x4)
Heatingvalue(Btu)
Nitrogen 1.61 28.0 0.4508 - -
Carbon dioxide
0.15 44.0 0.066 - -
Methane 98.17 16.0 15.70 21,500 337,704.80
Ethane 0.08 30.1 0.02 20,420 491.71
TOTAL - - 16.25 - 338,196.51
Total per pound of fuel
- - - - 20,814.55
(5)
Calculate total mass and heating value of fuel
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Combustion Worksheet
Fuel Component
Column 1
Volume fraction
%
Column 2
Relativemolecular
mass
Column 3 (1x2)
Total mass
(lb)
Column 4
Net heating value
(Btu/lb)
Column 5 (3x4)
Heatingvalue(Btu)
Column 6
Air required(lb air/lb)
Column 7(3x6)
Air required
(lbs)
Column 8
CO2
formed(lbs CO2/lb)
Column 9 (3x8)
CO2
formed (lbs)
Column 10
H2O formed
(lbs H2O/lb)
Column 11 (3x10)
H2Oformed
(lbs)
Column 12
N2
formed (lbs N2/lb)
Column 13 (3x12)
N2
formed (lbs)
Nitrogen 1.61 28.0 0.4508 - - - - - - - - - -
Carbon dioxide
0.15 44.0 0.066 - - - - - - - - - -
Methane 98.17 16.0 15.70 21,500 337,704.80 17.24 270.79 2.74 43.04 2.25 35.34 13.25 208.12
Ethane 0.08 30.1 0.02 20,420 491.71 16.09 0.39 2.93 0.07 1.80 0.04 12.37 0.30
TOTAL - - 16.25 - 338,196.51 - 271.18 - 43.11 - 35.38 - 208.42
Total per pound of fuel
- - - - 20,814.55 - 16.69 - 2.65 - 2.18 - 12.83
(5) (7) (9) (11) (13)
Calculate products of combustion for fuel
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Completed Combustion Worksheet
Fuel Component
Column 1
Volume fraction
%
Column 2
Relativemolecular
mass
Column 3 (1x2)
Total mass
Column 4
Net heating value
(Btu/lb)
Column 5 (3x4)
Heatingvalue(Btu)
Column 6
Air required(lb air/lb)
Column 7(3x6)
Air required
(lbs)
Column 8
CO2
formed(lbs CO2/lb)
Column 9 (3x8)
CO2
formed (lbs)
Column 10
H2O formed
(lbs H2O/lb)
Column 11 (3x10)
H2Oformed
(lbs)
Column 12
N2
formed (lbs N2/lb)
Column 13 (3x12)
N2
formed (lbs)
Nitrogen 1.61 28.0 0.4508 - - - - - - - - - -
Carbon dioxide
0.15 44.0 0.066 - - - - - - - - - -
Methane 98.17 16.0 15.70 21,500 337,704.80 17.24 270.79 2.74 43.04 2.25 35.34 13.25 208.12
Ethane 0.08 30.1 0.02 20,420 491.71 16.09 0.39 2.93 0.07 1.80 0.04 12.37 0.30
TOTAL - - 16.25 - 338,196.51 - 271.18 - 43.11 - 35.38 - 208.42
Total per pound of fuel
- - - - 20,814.55 - 16.69 - 2.65 - 2.18 - 12.83
(5) (7) (9) (11) (13)
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Relative Humidity
Correction for relative humidity:
where:
𝑃vapor vapor pressure of water at ambient temperature
(from steam tables)
𝑃air 14.696 psi
=𝑃vapor
𝑃air×
𝑅𝐻
100×
18
28.85moisture in air
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Relative Humidity
Correction for relative humidity:
where:
𝑃vapor vapor pressure of water at ambient temperature
(from steam tables)
𝑃air 14.696 psi
=𝑃vapor
𝑃air×
𝑅𝐻
100×
18
28.85moisture in air
=0.364
14.696×
50
100×
18
28.85
= 0.0077 lbs of moisture per lb of air (a)
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Relative Humidity
Correction for relative humidity:
where:
𝑃vapor vapor pressure of water at ambient temperature
(from steam tables)
𝑃air 14.696 psi
=𝑃vapor
𝑃air×
𝑅𝐻
100×
18
28.85moisture in air
=0.364
14.696×
50
100×
18
28.85
= 0.0077 lbs of moisture per lb of air (a)
=air required
1 − moisture in air (a)
(7)
= 16.82 lbs of wet air per lb of fuel
=16.69
1 − 0.0077
(b)
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Relative Humidity
Correction for relative humidity:
where:
𝑃vapor vapor pressure of water at ambient temperature
(from steam tables)
𝑃air 14.696 psi
=𝑃vapor
𝑃air×
𝑅𝐻
100×
18
28.85moisture in air
=0.364
14.696×
50
100×
18
28.85
= 0.0077 lbs of moisture per lb of air (a)
=air required
1 − moisture in air (a)
(7)
= 16.82 lbs of wet air per lb of fuel
=16.69
1 − 0.0077
(b)
= lbs wet air per lb of fuel(b) – air required(7)
= 0.1295 lbs of moisture per lb of fuel (c)
= 16.82 – 16.69
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Relative Humidity
Correction for relative humidity:
where:
𝑃vapor vapor pressure of water at ambient temperature
(from steam tables)
𝑃air 14.696 psi
=𝑃vapor
𝑃air×
𝑅𝐻
100×
18
28.85moisture in air
=0.364
14.696×
50
100×
18
28.85
= 0.0077 lbs of moisture per lb of air (a)
=air required
1 − moisture in air (a)
(7)
= 16.82 lbs of wet air per lb of fuel
=16.69
1 − 0.0077
(b)
= lbs wet air per lb of fuel(b) – air required(7)
= 0.1295 lbs of moisture per lb of fuel (c)
= H2O formed(11) + lbs of moisture per lb of fuel(c) + atomizing steam
= 2.31 lbs of H2O per lb of fuel (d)
= 2.17 + 0.1295 + 0
= 16.82 – 16.69
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Excess Air
Correction for excess air:
=(28.85 ×%O2)(
N2 formed28
+CO2 formed
44+
H2O formed18
)
20.95 −%O2[ 1.6028 ×lbs H2O
lbs air required+ 1]
lb excess air per lb of fuel
NOTE: If oxygen samples are extracted on a dry basis, a value of zero shall be inserted for line (e) where a value is required from lines (c) and (d). If oxygen samples are extracted on a wet basis, the appropriate calculated value shall be inserted.
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Excess Air
Correction for excess air:
=(28.85 ×%O2)(
N2 formed28
+CO2 formed
44+
H2O formed18
)
20.95 −%O2[ 1.6028 ×lbs H2O
lbs air required+ 1]
lb excess air per lb of fuel
NOTE: If oxygen samples are extracted on a dry basis, a value of zero shall be inserted for line (e) where a value is required from lines (c) and (d). If oxygen samples are extracted on a wet basis, the appropriate calculated value shall be inserted.
=(28.85 ×%O2)(
N2 formed28
+CO2 formed
44+
H2O formed18
)
20.95 −%O2[ 1.6028 ×lbs H2O
lbs air required+ 1]
(13) (9) (d)
(c)
(7)
=(28.85 × 3)(
12.8328
+2.6544
+018
)
20.95 − 3[ 1.6028 ×0
16.69+ 1]
= 2.50 lbs of excess air per lb of fuel (e)
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=lb of excess air per lb of fuel
air required× 100
Excess Air
Correction for excess air:
=(28.85 ×%O2)(
N2 formed28
+CO2 formed
44+
H2O formed18
)
20.95 −%O2[ 1.6028 ×lbs H2O
lbs air required+ 1]
lb excess air per lb of fuel
NOTE: If oxygen samples are extracted on a dry basis, a value of zero shall be inserted for line (e) where a value is required from lines (c) and (d). If oxygen samples are extracted on a wet basis, the appropriate calculated value shall be inserted.
=(28.85 ×%O2)(
N2 formed28
+CO2 formed
44+
H2O formed18
)
20.95 −%O2[ 1.6028 ×lbs H2O
lbs air required+ 1]
(13) (9) (d)
(c)
(7)
=(28.85 × 3)(
12.8328
+2.6544
+018
)
20.95 − 3[ 1.6028 ×0
16.69+ 1]
= 2.50 lbs of excess air per lb of fuel (e)
(7)
(e)
=2.50
16.69× 100
= 14.98 lbs excess air (f)
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=lb of excess air per lb of fuel
air required× 100
Excess Air
Correction for excess air:
=(28.85 ×%O2)(
N2 formed28
+CO2 formed
44+
H2O formed18
)
20.95 −%O2[ 1.6028 ×lbs H2O
lbs air required+ 1]
lb excess air per lb of fuel
NOTE: If oxygen samples are extracted on a dry basis, a value of zero shall be inserted for line (e) where a value is required from lines (c) and (d). If oxygen samples are extracted on a wet basis, the appropriate calculated value shall be inserted.
=(28.85 ×%O2)(
N2 formed28
+CO2 formed
44+
H2O formed18
)
20.95 −%O2[ 1.6028 ×lbs H2O
lbs air required+ 1]
(13) (9) (d)
(c)
(7)
=(28.85 × 3)(
12.8328
+2.6544
+018
)
20.95 − 3[ 1.6028 ×0
16.69+ 1]
= 2.50 lbs of excess air per lb of fuel (e)
=percent excess air
100× lbs moisture per lb fuel + lb H2O per lb fuel
(7)
(e)
=2.50
16.69× 100
= 14.98 lbs excess air (f)
(f)(c) (d)
=14.98
100× 0.1295 + 2.31
= 2.33 total lbs H2O per lb of fuel (corrected for excess air) (g)
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Stack Loss
ComponentColumn 1
Component formed(lb per lb of fuel)
Carbon dioxide 2.65
Water vapor 2.33
Nitrogen 12.83
Air 2.50
Total 20.31
(9) from combustion worksheet
(13) from combustion worksheet
(g) from excess air worksheet
(e) from excess air worksheet
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Stack Loss
ComponentColumn 1
Component formed(lb per lb of fuel)
Column 2
Enthalpy at T(Btu/lb formed)
Carbon dioxide 2.65 100
Water vapor 2.33 192
Nitrogen 12.83 120
Air 2.50 110
Total 20.31 -
Exit flue-gas temperature, 𝑇𝑒: 500°F
Values taken from enthalpy tables in API 560, Figures G.6 and G.7 for each flue-gas component
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Stack Loss
ComponentColumn 1
Component formed(lb per lb of fuel)
Column 2
Enthalpy at T(Btu/lb formed)
Column 3
Heat content(Btu/lb of fuel)
Carbon dioxide 2.65 100 265.31
Water vapor 2.33 192 446.72
Nitrogen 12.83 120 1,539.27
Air 2.50 110 274.98
Total 20.31 - 2,526.28
Exit flue-gas temperature, 𝑇𝑒: 500°F
ℎ𝑠 = heat content at 𝑇𝑒 = 2,526.28 Btu/lb of fuel
Step 2:solve for additional heat losses
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Thermal Efficiency
Where:ℎ𝐿 lower heating value of the fuel (Btu/lb)
∆ℎ𝑎 heat correction due to air (Btu/lb)
∆ℎ𝑓 heat correction due to fuel (Btu/lb)
∆ℎ𝑚 heat correction due to atomizing medium (Btu/lb)
ℎ𝑟 assumed radiation heat loss (Btu/lb)
ℎ𝑠 calculated stack heat loss (Btu/lb)
𝒆 = 𝒉𝑳 + ∆𝒉𝒂 + ∆𝒉𝒇 + ∆𝒉𝒎 − (𝒉𝒓 + 𝒉𝒔
𝒉𝑳 + ∆𝒉𝒂 + ∆𝒉𝒇 + ∆𝒉𝒎
× 𝟏𝟎𝟎
Tulsa Heaters Midstream 32
Heat Losses
Heat loss due to air:
∆ℎ𝑎 = 𝑐𝑝a × (𝑇𝑎 − 𝑇𝑑) × ( 𝑚𝑎 𝑚𝑓)
where:
𝑐𝑝a specific heat of air
𝑇𝑎 temperature of air (°F)
𝑇𝑑 temperature of design air (°F)
𝑚𝑎 𝑚𝑓 the sum of 𝑚𝑎 and 𝑚𝑓, expressed as
pounds of air per pound of fuel (from lines (b) and (e) on the excess air and relative humidity work sheet)
= 𝑐𝑝a × (𝑇𝑎 − 𝑇𝑑) × ( 𝑚𝑎 𝑚𝑓)
= 0.24 × (70 − 60) × (16.81 + 2.50)
∆ℎ𝑎 = 46.37 Btu/lb
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Heat Losses
Heat loss due to fuel gas:
∆ℎ𝑓 = 𝑐𝑝fuel × (𝑇𝑓 − 𝑇𝑑)
where:
𝑐𝑝fuel specific heat of fuel gas
𝑇𝑓 temperature of fuel gas (°F)
𝑇𝑑 temperature of design fuel gas (°F)
= 𝑐𝑝fuel × (𝑇𝑓 − 𝑇𝑑)
= 0.587 × (100 − 60)
∆ℎ𝑓 = 23.47 Btu/lb
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Heat Losses
Heat loss due to atomization medium:
∆ℎ𝑚 = ∆𝐸 × ( 𝑚𝑠𝑡 𝑚𝑓)
where:
∆𝐸 enthalpy difference
𝑚𝑠𝑡 mass of steam (lb)
No atomization steam in this case.
∆ℎ𝑚 = 0
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Radiation Losses
Heat loss due to radiation:
ℎ𝑟 = ℎ𝐿 ×%radiation loss ℎ𝑟 = 20,814 × 0.015
ℎ𝑟 = 312.24 Btu/lb
Step 3:solve for thermal and fuel efficiencies
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Thermal Efficiency
Where:ℎ𝐿 20,814 Btu/lb
∆ℎ𝑎 46.37 Btu/lb
∆ℎ𝑓 23.47 Btu/lb
∆ℎ𝑚 0 Btu/lb
ℎ𝑟 312.24 Btu/lb
ℎ𝑠 2,526.28 Btu/lb
𝒆 = 𝒉𝑳 + ∆𝒉𝒂 + ∆𝒉𝒇 + ∆𝒉𝒎 − (𝒉𝒓 + 𝒉𝒔
𝒉𝑳 + ∆𝒉𝒂 + ∆𝒉𝒇 + ∆𝒉𝒎
× 𝟏𝟎𝟎
Tulsa Heaters Midstream 38
Thermal Efficiency
Where:ℎ𝐿 20,814 Btu/lb
∆ℎ𝑎 46.37 Btu/lb
∆ℎ𝑓 23.47 Btu/lb
∆ℎ𝑚 0 Btu/lb
ℎ𝑟 312.24 Btu/lb
ℎ𝑠 2,526.28 Btu/lb
𝒆 = 𝒉𝑳 + ∆𝒉𝒂 + ∆𝒉𝒇 + ∆𝒉𝒎 − (𝒉𝒓 + 𝒉𝒔
𝒉𝑳 + ∆𝒉𝒂 + ∆𝒉𝒇 + ∆𝒉𝒎
× 𝟏𝟎𝟎
𝒆 = 𝒉𝑳 + ∆𝒉𝒂 + ∆𝒉𝒇 + ∆𝒉𝒎 − (𝒉𝒓 + 𝒉𝒔
𝒉𝑳 + ∆𝒉𝒂 + ∆𝒉𝒇 + ∆𝒉𝒎
× 𝟏𝟎𝟎
𝒆 =)𝟐𝟎, 𝟖𝟏𝟒 + 𝟒𝟔. 𝟑𝟕 + 𝟐𝟑. 𝟒𝟕 + 𝟎 − (𝟑𝟏𝟐. 𝟐𝟒 + 𝟐, 𝟓𝟐𝟔. 𝟐𝟖
𝟐𝟎, 𝟖𝟏𝟒 + 𝟒𝟔. 𝟑𝟕 + 𝟐𝟑. 𝟒𝟕 + 𝟎× 𝟏𝟎𝟎
𝒆 = 86.4%
Tulsa Heaters Midstream 39
Fuel Efficiency
Where:ℎ𝐿 20,814 Btu/lb
∆ℎ𝑎 46.37 Btu/lb
∆ℎ𝑓 23.47 Btu/lb
∆ℎ𝑚 0 Btu/lb
ℎ𝑟 312.24 Btu/lb
ℎ𝑠 2,526.28 Btu/lb
𝒆𝒇 = 𝒉𝑳 + ∆𝒉𝒂 + ∆𝒉𝒇 + ∆𝒉𝒎 − (𝒉𝒓 + 𝒉𝒔
𝒉𝑳× 𝟏𝟎𝟎
Tulsa Heaters Midstream 40
Fuel Efficiency
Where:ℎ𝐿 20,814 Btu/lb
∆ℎ𝑎 46.37 Btu/lb
∆ℎ𝑓 23.47 Btu/lb
∆ℎ𝑚 0 Btu/lb
ℎ𝑟 312.24 Btu/lb
ℎ𝑠 2,526.28 Btu/lb
𝒆𝒇 = 𝒉𝑳 + ∆𝒉𝒂 + ∆𝒉𝒇 + ∆𝒉𝒎 − (𝒉𝒓 + 𝒉𝒔
𝒉𝑳× 𝟏𝟎𝟎
𝒆𝒇 = 𝒉𝑳 + ∆𝒉𝒂 + ∆𝒉𝒇 + ∆𝒉𝒎 − (𝒉𝒓 + 𝒉𝒔
𝒉𝑳× 𝟏𝟎𝟎
𝒆𝒇 =)𝟐𝟎, 𝟖𝟏𝟒 + 𝟒𝟔. 𝟑𝟕 + 𝟐𝟑. 𝟒𝟕 + 𝟎 − (𝟑𝟏𝟐. 𝟐𝟒 + 𝟐, 𝟓𝟐𝟔. 𝟐𝟖
𝟐𝟎, 𝟖𝟏𝟒× 𝟏𝟎𝟎
𝒆𝒇 = 86.7%
* Based on $2.75/MMBtu gas price
Tulsa Heaters Midstream 42
Conclusion
Knowing how to check your heater’s efficiency gives you the
knowledge and power to improve your facility and optimize
your heater. As we have seen, improving efficiency can help
save your facility a lot of money.
What are you waiting for? Go check and start saving now!
Tulsa Heaters Midstream 43
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