Boilers, Furnaces & Combustion

6% of exam

Fuel heating values, air-fuel ratio and excess air, combustion and thermal efficiency, boiler/furnace fuel rate, and flue-gas losses.

4 concepts

Exam areas

B. Boilers and Furnaces

Heating Values, Combustion Efficiency and Fuel Rate

Higher vs lower heating value, combustion vs thermal efficiency, and how to back the fuel-input rate (and boiler horsepower) out of a required output.

Almost every boiler or furnace problem on the PE Mechanical HVAC exam is a single energy balance read in one of two directions. Forward: a fuel burns, releases its heating value, and some fraction reaches the water or the air stream. Backward: a building needs a known output, and you must size the burner, the gas meter, or the fuel bill that delivers it. The trap that quietly sinks candidates is not the algebra — it is the heating-value basis and the efficiency basis. Mix higher heating value with a lower-heating-value efficiency, or charge a furnace's output against its input on the wrong basis, and the answer is off by ten percent before the arithmetic even starts. This concept builds the bookkeeping so that never happens. The reference is the NCEES PE Mechanical Reference Handbook — §10 Combustion and Fuels, with the fuel-input-rate relation in §10.6 and the boiler-horsepower conversion in §1 unit tables.

Higher versus lower heating value

When a hydrocarbon burns, the hydrogen in the fuel becomes water. Whether you count the heat released when that water vapor condenses is the entire difference between the two heating values. The higher heating value (HHV, also gross or total heating value) is measured with the product water condensed to liquid, so it includes the latent heat of vaporization; the lower heating value (LHV, or net heating value) leaves the water as vapor and omits that latent term. For natural gas the gap is about

10\%

of HHV — pure methane is

23{,}875\,\text{Btu/lb}

HHV versus

21{,}495\,\text{Btu/lb}

LHV (handbook §10.1 table). North American practice quotes fuel and rates fuel meters on HHV; much of the world and most combustion-textbook efficiencies use LHV, which is why a condensing boiler can be advertised at "

95\%

" (LHV) and "

86\%

" (HHV) and both be true.

HHV = LHV + m_{w}\,h_{fg}

Combustion efficiency versus fuel-to-load efficiency

Combustion efficiency is what is left after the heat that escapes up the stack:

\eta_{c} = 100\% - \text{(stack loss)}

. It measures how completely the burner extracts the fuel's energy into the flue path before the gases leave, and it is the number a combustion analyzer reports from a flue-gas sample. Watch the terminology trap in the handbook: the NCEES §10.3 "thermal efficiency" is printed two ways: first as

100\times

useful heat / heating value of fuel (the fuel-to-load definition), or as

(Q_{h}-Q_{fl})/Q_{h}\times 100\%

, where

Q_{fl}

is the flue-gas (stack) loss per unit volume of fuel gas only. That second form is stack-loss based and is numerically the same quantity this concept calls combustion efficiency. It does NOT debit radiation, convection, or blowdown. The broader figure that does subtract jacket radiation, convection, and — for boilers — blowdown is the fuel-to-load (overall boiler) efficiency,

\eta = \dot{Q}_{out}/(HHV\cdot \dot{m}_{f})

: useful delivered heat divided by fuel energy input. It sits below the stack-only number. Seasonal furnace performance is captured by AFUE, which also debits cyclic off-period and pilot losses. So as a general engineering hierarchy, combustion (stack-only) > fuel-to-load > AFUE — but if a problem uses the handbook's

(Q_h-Q_{fl})/Q_h

form, recognize that it is the stack-only value, not the fuel-to-load one. Do not use a flue-gas combustion efficiency where the problem asks for delivered heat.

\eta_{f\text{-}l} = \frac{\dot{Q}_{out}}{HHV\cdot \dot{m}_{f}}\times 100\%

Backing out the fuel-input rate

Sizing problems run the balance backward. Once you know the required output and the efficiency on a matching basis, the fuel input rate is just

\dot{Q}_{in}=\dot{Q}_{out}/\eta

, and the fuel quantity follows by dividing the input rate by the heating value — per unit mass for oil and coal, per unit volume for gas. The single most common error is dividing the output by the efficiency-times-heating-value in one step but then quoting the result against the wrong heating value. Keep input and output as separate Btu/h numbers, then convert input to fuel.

Volumetric gas input and meter corrections

Gas is metered by volume, but a cubic foot is only a fixed amount of energy at a stated temperature and pressure. The handbook's §10.6 gas-input relation corrects the metered volumetric flow rate from meter conditions back to the standard basis on which the heating value is quoted (

T_s=520\,^\circ\text{R}

P_s=14.735\,\text{psia}

). Warm, low-pressure gas is less dense, so a raw volumetric reading overstates the energy unless you scale it by

(T_s/T)(P/P_s)

. On a sea-level problem at

60\,^\circ\text{F}

the correction is essentially unity, but the exam likes to hand you an elevated meter temperature or a few psig of supply pressure to see whether you apply it.

\dot{Q} = HHV\cdot VFR\cdot \frac{T_s\,P}{T\,P_s}

Boiler horsepower and output ratings

Boiler horsepower is a historical capacity unit, not a shaft power: one boiler horsepower is the heat to evaporate

34.5\,\text{lb/h}

of water "from and at"

212\,^\circ\text{F}

, which the handbook unit table lists as

33{,}470\,\text{Btu/h}

(commonly rounded to

33{,}475

). It is an output rating — divide a boiler's gross output in Btu/h by

33{,}470

to get boiler horsepower, or divide the steam rate in lb/h (from and at

212\,^\circ\text{F}

) by

34.5

. For real feedwater that enters below

212\,^\circ\text{F}

, the actual output per pound of steam is larger than the latent heat alone, so the rated steam capacity at real conditions is lower than the nameplate "from and at" figure.

1\ \text{BHP} = 33{,}470\ \tfrac{\text{Btu}}{\text{h}} = 34.5\ \tfrac{\text{lb}}{\text{h}}\ \text{(from \& at }212\,^\circ\text{F)}

Exam strategy

First thing on any boiler problem: write down which heating value and which efficiency basis you are on, and force them to match before dividing. If the problem gives an LHV efficiency and an HHV fuel value (or vice versa), convert one with

\eta_{HHV}=\eta_{LHV}\,(LHV/HHV)

. Keep

\dot{Q}_{in}

and

\dot{Q}_{out}

as separate Btu/h numbers — output is what the load needs, input is what the meter sees. Mind the efficiency labels: the handbook's §10.3

(Q_h-Q_{fl})/Q_h

"thermal efficiency" form is stack-loss based (the combustion-efficiency value), whereas a fuel-to-load efficiency additionally debits radiation and blowdown — so read which one a problem hands you. For boiler horsepower, remember it is an output unit (

33{,}470\,\text{Btu/h}

); never multiply it by the efficiency to get output. When a gas meter temperature or pressure is stated, apply the

(T_s/T)(P/P_s)

correction; when it is not, assume standard conditions and move on.

Key equations

Higher vs lower heating value

HHV = LHV + m_{w}\,h_{fg}

$m_w$ = mass of water formed per unit fuel (lb water / lb fuel), $h_{fg}\approx 1{,}060\,\text{Btu/lb}$ at the heating-value reference temperature (~60 °F, where the product water is condensed to liquid). HHV counts the latent heat of the product water; LHV does not.

Fuel-to-load (overall) efficiency

\eta_{f\text{-}l} = \dfrac{\dot{Q}_{out}}{HHV\cdot \dot{m}_{f}}

Useful delivered heat divided by fuel energy input; debits stack, radiation, and blowdown losses. $\dot{Q}_{out}$ in Btu/h, $\dot m_f$ in lb/h (or use $HHV_{vol}$ with volumetric flow). Use HHV with HHV-basis fuel data. This is the handbook's first §10.3 "thermal efficiency" form (useful heat / heating value of fuel); its alternative $(Q_h-Q_{fl})/Q_h$ form, below, is stack-only.

Combustion efficiency (= handbook §10.3 stack-loss form)

\eta_{c} = 100\% - L_{stack} = \dfrac{Q_h - Q_{fl}}{Q_h}\times 100\%

What a flue-gas analyzer reports; equals 100% minus the dry-gas and moisture stack losses. The NCEES handbook §10.3 prints this stack-loss form as its alternative ("or") definition of "thermal efficiency," after useful heat / heating value of fuel. Always higher than the fuel-to-load efficiency, which also subtracts radiation and blowdown.

Fuel-input rate from output

\dot{Q}_{in} = \dfrac{\dot{Q}_{out}}{\eta}

Required fuel energy input (Btu/h). Use the efficiency on the same heating-value basis as the fuel data.

Fuel quantity (mass or volume)

\dot{m}_{f} = \dfrac{\dot{Q}_{in}}{HHV}\qquad \dot{V}_{f} = \dfrac{\dot{Q}_{in}}{HHV_{vol}}

Mass rate (lb/h) for oil/coal using mass HHV (Btu/lb); volumetric rate (ft³/h or gal/h) for gas/oil using volumetric HHV (Btu/ft³ or Btu/gal).

Gas input rate with meter correction

\dot{Q} = HHV\cdot VFR\cdot \dfrac{T_s\,P}{T\,P_s}

Handbook §10.6. $T_s=520\,^\circ\text{R}$ , $P_s=14.735\,\text{psia}$ ; $T$ , $P$ are absolute meter conditions; $VFR$ is metered volumetric flow (ft³/h).

Efficiency basis conversion

\eta_{HHV} = \eta_{LHV}\,\dfrac{LHV}{HHV}

Convert a lower-heating-value efficiency to the higher-heating-value basis (and inverse). For methane $LHV/HHV = 0.900$ .

Boiler horsepower

\text{BHP} = \dfrac{\dot{Q}_{out}}{33{,}470} = \dfrac{\dot{m}_{s}}{34.5}

Output rating. $\dot Q_{out}$ in Btu/h; $\dot m_s$ in lb/h of steam from and at $212\,^\circ\text{F}$ . 1 BHP also $\approx 9.81\,\text{kW}$ .

Boiler steam output (real feedwater)

\dot{Q}_{out} = \dot{m}_{s}\,(h_{s} - h_{fw})

$h_s$ = enthalpy of generated steam, $h_{fw}$ = feedwater enthalpy (Btu/lb). Reduces to the latent heat only when feedwater is saturated liquid at 212 °F.

Hot-water output (standard coefficient)

\dot{Q} = 500\cdot \text{gpm}\cdot \Delta T

Sensible heat of water; $500 = 8.33\,\tfrac{\text{lb}}{\text{gal}}\times 60\,\tfrac{\text{min}}{\text{h}}\times 1\,\tfrac{\text{Btu}}{\text{lb}\,^\circ\text{F}}$ . $\Delta T$ in °F; valid for water near room temperature.

Worked examples

Natural-gas furnace fuel rate

Problem. A gas furnace must deliver $80{,}000\,\text{Btu/h}$ of heat to a house. The furnace AFUE (seasonal fuel-to-load efficiency, HHV basis) is $92\%$ and the supplied natural gas has a higher heating value of $1{,}020\,\text{Btu/ft}^3$ at standard conditions. Find the fuel input rate and the gas consumption in ft³/h.

Solution. Work the balance backward. Fuel energy input:

\dot{Q}_{in}=\frac{\dot{Q}_{out}}{\eta}=\frac{80{,}000}{0.92}=86{,}960\,\text{Btu/h}.

Gas volume (HHV basis matches the fuel data):

\dot{V}=\frac{\dot{Q}_{in}}{HHV_{vol}}=\frac{86{,}960}{1{,}020}=85.3\,\text{ft}^3/\text{h}.

Final answer: input

8.70\times 10^{4}\,\text{Btu/h}

, gas rate

85.3\,\text{ft}^3/\text{h}

. Sanity check: at

92\%

the input is just above the output, and

85.3\,\text{ft}^3/\text{h}\times 1{,}020\approx 87{,}000\,\text{Btu/h}

recovers the input — consistent.

\dot{V}=\dfrac{80{,}000/0.92}{1{,}020}\approx 85.3\ \text{ft}^3/\text{h}

Oil-fired boiler: fuel rate and boiler horsepower

Problem. A steam boiler delivers a gross output of $5.0\times 10^{6}\,\text{Btu/h}$ at a fuel-to-load efficiency of $82\%$ burning No. 2 fuel oil with a higher heating value of $138{,}000\,\text{Btu/gal}$ . Find the oil consumption (gal/h) and the boiler's output rating in boiler horsepower.

Solution. Fuel input:

\dot{Q}_{in}=\frac{5.0\times 10^{6}}{0.82}=6.10\times 10^{6}\,\text{Btu/h}.

Oil rate:

\dot{V}=\frac{6.10\times 10^{6}}{138{,}000}=44.2\,\text{gal/h}.

Boiler horsepower is an OUTPUT rating, so use the output, not the input:

\text{BHP}=\frac{5.0\times 10^{6}}{33{,}470}=149\,\text{BHP}.

Final answer:

44.2\,\text{gal/h}

and

149\,\text{BHP}

. Sanity check:

44.2\,\text{gal/h}\times 138{,}000\times 0.82\approx 5.0\times 10^{6}\,\text{Btu/h}

output — closes the loop. Using the input in the BHP formula would wrongly inflate the rating to 182 BHP.

\text{BHP}=\dfrac{5.0\times10^{6}}{33{,}470}\approx 149

Metered gas with temperature and pressure correction

Problem. A gas meter reads $120\,\text{ft}^3/\text{h}$ of natural gas. The gas in the meter is at $90\,^\circ\text{F}$ and $15.5\,\text{psia}$ , and its higher heating value is $1{,}030\,\text{Btu/ft}^3$ at the standard basis ( $520\,^\circ\text{R}$ , $14.735\,\text{psia}$ ). Find the true energy input rate.

Solution. Apply the §10.6 correction. Absolute meter temperature

T=90+460=550\,^\circ\text{R}

\dot{Q}=HHV\cdot VFR\cdot\frac{T_s\,P}{T\,P_s}=1{,}030\times120\times\frac{520\times15.5}{550\times14.735}.

The correction factor is

\dfrac{520\times15.5}{550\times14.735}=0.9945

, so

\dot{Q}=1{,}030\times120\times0.9945=1.23\times10^{5}\,\text{Btu/h}.

Final answer:

1.23\times10^{5}\,\text{Btu/h}

. Sanity check: warmer-than-standard gas (raises

T

, lowers density) and slightly elevated pressure nearly cancel, giving a factor just under 1 — so the corrected energy is a hair below the raw

120\times1{,}030=123{,}600\,\text{Btu/h}

, as expected.

Common pitfalls

•Mixing heating-value bases: dividing an HHV-basis fuel value by an LHV efficiency (or vice versa) introduces a ~10% error for natural gas. Convert with $\eta_{HHV}=\eta_{LHV}(LHV/HHV)$ first.
•Misreading the efficiency label: the NCEES handbook §10.3 prints "thermal efficiency" both as useful heat / heating value and as $(Q_h-Q_{fl})/Q_h$ ; the second form is stack-loss based — numerically the combustion efficiency, NOT a radiation/blowdown-inclusive figure. The broader fuel-to-load (overall boiler) efficiency does debit radiation and blowdown and is the lowest; use it for delivered-heat problems, and use the stack-only value only for flue-gas questions.
•Multiplying boiler horsepower by efficiency to get output. Boiler horsepower is already an OUTPUT rating ( $33{,}470\,\text{Btu/h}$ ); efficiency relates output to fuel input, separately.
•Treating the 34.5 lb/h steam rate as actual capacity. That figure is 'from and at 212 °F'; with cold feedwater the real per-pound output is larger, so actual steam capacity is lower than the nameplate.
•Skipping the gas-meter $(T_s/T)(P/P_s)$ correction when the meter temperature or pressure is given. A warm, pressurized meter can shift the energy several percent.
•Using gauge pressure in the meter correction. The relation needs absolute pressure (psia) and absolute temperature (°R).
•Quoting fuel mass for gas or fuel volume for coal. Match the heating-value units: Btu/lb with mass rate, Btu/ft³ or Btu/gal with volumetric rate.

References

NCEES PE Mechanical Reference Handbook — §10 Combustion and Fuels — §10.1 heating-value tables; §10.3 thermal efficiency (useful-heat and stack-loss forms); §10.6 gas-input rate
NCEES PE Mechanical Reference Handbook — §6 Steam (boiler plant basics)
ASHRAE Handbook—Fundamentals, Ch. 28 Combustion and Fuels — Source of the handbook heating-value tables
ASHRAE Handbook—HVAC Systems and Equipment, Boilers chapter — Boiler ratings and efficiency definitions

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