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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

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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

Tulsa Heaters Midstream 37

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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