12.1 Boiler Horsepower (BHP), Heating Surface Area & Equivalent Evaporation Calculations

12.1 Boiler Horsepower (BHP), Heating Surface Area & Equivalent Evaporation Calculations

Quick Summary: In stationary power plant engineering, Boiler Horsepower (BHP) is an engineering and legal unit of thermal capacity rather than mechanical shaft output. One classical boiler horsepower represents the heat energy required to evaporate 34.5 pounds of water per hour from and at a temperature of 212°F into dry saturated steam, which equals 33,475 Btu/hr (or 9.8095 kW thermal). In the Commonwealth of Massachusetts, M.G.L. c. 146, § 48 — restated in 522 CMR 2.06(7) — is what actually fixes the horsepower used for licensing tiers and attendance thresholds, and it is a nameplate rule, not a heating-surface rule: the manufacturer's factory tag, or the tag's steam output ÷ 34.5, or the tag's Btu/hr input ÷ 41,840, or the tag's Btu/hr output ÷ 33,475. The familiar 10 sq ft (watertube) and 12 sq ft (HRT firetube) of heating surface per boiler horsepower are long-standing engineering conventions used to estimate capacity — not the statutory method. To determine true plant thermal output under varying operating temperatures and pressures, operators apply the Factor of Evaporation (FE) to convert actual steam production into Equivalent Evaporation (EE) and Developed Boiler Horsepower.

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1. The Classical Definition of Boiler Horsepower (BHP)

Unlike an internal combustion engine or steam turbine where horsepower measures mechanical work performed at a rotating shaft ($1\text{ mechanical hp} = 550\text{ ft-lb/sec} = 33,000\text{ ft-lb/min} = 745.7\text{ Watts} = 2,544.4\text{ Btu/hr}$), Boiler Horsepower (BHP) is exclusively a measure of thermal heat transfer capacity.

Historical Origin: The 1876 Centennial Exposition

The concept of boiler horsepower was formulated by a committee of judges at the 1876 Centennial Exhibition in Philadelphia. At that time, a typical industrial Corliss steam engine required approximately 30 pounds of steam per hour to produce one indicated mechanical horsepower when supplied with steam at 70 psig and receiving feedwater at 100°F. When converted to an equivalent heat transfer baseline evaporating water at atmospheric pressure (212°F) from feedwater at 212°F, this energy expenditure equated to 34.5 pounds of water evaporated per hour.

The Thermodynamic Benchmark: "From and At 212°F"

The modern engineering and legal definition adopted by the American Society of Mechanical Engineers (ASME) and incorporated into state examination standards defines one boiler horsepower as:

$\text{1 Boiler Horsepower (BHP)} = \text{The evaporation of } 34.5\text{ pounds of water per hour from and at } 212^\circ\text{F into dry saturated steam}$

To translate this definition into British Thermal Units per hour (Btu/hr), we reference the thermodynamic properties of saturated steam at standard atmospheric pressure (14.696 psia, saturation temperature 212°F):

• Enthalpy of saturated liquid ($h_f$) at 212°F: $180.17\text{ Btu/lb}$ (measured above the reference state of liquid water at 32°F).
• Enthalpy of saturated vapor ($h_g$) at 212°F: $1,150.5\text{ Btu/lb}$.
• Latent heat of vaporization ($h_{fg}$) at 212°F: $h_{fg} = h_g - h_f = 1,150.5 - 180.17 = 970.3\text{ Btu/lb}$

Multiplying the evaporation mass flow rate by the latent heat of vaporization gives the absolute thermal energy equivalent of one boiler horsepower:

$\text{Heat Output per BHP} = 34.5\text{ lb/hr} \times 970.3\text{ Btu/lb} = 33,475.35\text{ Btu/hr}$

In standard engineering practice, commercial trade literature, and Massachusetts licensing board examinations, this value is rounded to 33,475 Btu/hr.

Contrast Between Mechanical and Boiler Horsepower

Operating engineers must avoid confusing mechanical horsepower with boiler horsepower: $\frac{1\text{ Boiler Horsepower}}{1\text{ Mechanical Horsepower}} = \frac{33,475\text{ Btu/hr}}{2,544.4\text{ Btu/hr}} \approx 13.156$ One boiler horsepower represents more than 13 times the energy rate of one mechanical horsepower!

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2. Thermodynamic, Electrical, and Radiation Equivalents

Operating engineers must navigate multiple engineering measurement systems when interfacing with building mechanical systems, electrical boilers, and central heating loops. The following thermodynamic equivalents are fundamental to power plant calculations:

Equivalent Direct Radiation (EDR) in Facility Heating

In commercial and institutional facility engineering, heating loads are historically sized in square feet of Equivalent Direct Radiation (EDR):

• Steam EDR: Defined as the emission of 240 Btu/hr per square foot of radiator surface when supplied with steam at 215°F in a room at 70°F. $\text{Steam EDR per BHP} = \frac{33,475\text{ Btu/hr}}{240\text{ Btu/hr/sq ft}} = 139.48\text{ sq ft EDR}$
• Hot Water EDR: Defined as the emission of 150 Btu/hr per square foot of radiator surface when supplied with mean water temperature of 180°F in a room at 70°F. $\text{Hot Water EDR per BHP} = \frac{33,475\text{ Btu/hr}}{150\text{ Btu/hr/sq ft}} = 223.17\text{ sq ft EDR}$

If a facility engineer is tasked with selecting a replacement boiler for a building containing 14,000 sq ft of steam EDR radiators, the required boiler output is: $\text{Required Output} = \frac{14,000\text{ sq ft EDR}}{139.48\text{ sq ft/BHP}} = 100.37\text{ BHP} \implies 100\text{ BHP minimum output}$

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3. How Massachusetts Actually Determines Boiler Horsepower (M.G.L. c. 146, § 48)

In Massachusetts, plant licensing tiers, attendance minimums, and watch engineer requirements (M.G.L. c. 146, §§ 46 and 49) all turn on aggregate boiler horsepower. Candidates routinely lose points here because they recite the textbook heating-surface conversions as if they were the statute. They are not. Section 48 is a nameplate rule.

3.1 Boilers — the Statutory Method (§ 48; restated at 522 CMR 2.06(7)(a))

"When liquid or gaseous fuel, electric or atomic energy or any other source of heat is used, the horsepower of a boiler shall be determined by either the manufacturer's factory tag affixed to the boiler or burner denoting horsepower, or calculated by one of the following formulae: the steam output capacity as listed on the manufacturer's tag divided by 34.5, the Btu/Hr Input listed on the manufacturer's tag divided by 41,840 or the Btu/Hr Output listed on the manufacturer's tag divided by 33,475."

  +-------------------------------------------------------------------------+
  |        MASSACHUSETTS STATUTORY BOILER HORSEPOWER (M.G.L. c. 146 § 48)    |
  |                                                                         |
  |  1. Manufacturer's factory tag horsepower, OR                           |
  |  2. Steam output capacity (lb/hr) on the tag      /  34.5    = BHP      |
  |  3. Btu/hr INPUT  listed on the tag               /  41,840  = BHP      |
  |  4. Btu/hr OUTPUT listed on the tag               /  33,475  = BHP      |
  +-------------------------------------------------------------------------+

Why three divisors, and why do they differ?

• 34.5 lb/hr is the Centennial evaporation rate for one boiler horsepower.
• 33,475 Btu/hr is that same evaporation expressed as heat delivered to the water — hence it is the output divisor.
• 41,840 Btu/hr is the corresponding input at an assumed 80% efficiency ($33,475 \div 0.80 = 41,844$). The statute therefore builds a standard 80% conversion efficiency into the input route, which is why the input and output divisors are not interchangeable.

Worked check: A gas-fired package boiler is tagged at 8,368,000 Btu/hr input. Statutory horsepower is $\text{BHP} = \frac{8,368,000}{41,840} = 200.0\text{ BHP}$ Aggregated with other units, that 200 BHP is the number that decides whether the plant sits in the § 46 periodic (9–250 HP), noncontinuous (251–500 HP), or continuous (501 HP and above) attendance tier.

Missing or Illegible Tag (§ 48; 522 CMR 2.06(7)(b)): If a tag is missing, damaged, or unclear, the licensed engineer in charge or on duty at the time shall notify the owner or user. The owner or user must then obtain a notarized letter signed by an officer of the manufacturer of the boiler or burner listing the maximum capacity in Btu/hr. That letter becomes the acceptable basis for calculating the horsepower of that particular boiler. Section 48 also states that the minimum safety valve relieving capacity shall be determined in accordance with the ASME Code — not from the horsepower figure.

3.2 Steam Engines and Turbines — Also § 48

Engineer grades are limited by prime mover horsepower as well as boiler horsepower (a Third Class Engineer is capped at engines of 50 HP each; a Second Class Engineer at 150 HP each), so § 48 supplies rating rules for them too:

The turbine rule is a favorite oral-board question precisely because it is counter-intuitive: Massachusetts rates a small turbine by the external diameter of its steam supply pipe, not by shaft output.

3.3 Heating Surface Conversions — Useful Engineering Convention, Not Statute

The classical conversions below remain the standard way to estimate the capacity of an older, untagged boiler and to work textbook problems, and Massachusetts prep courses drill them. Treat them as engineering convention:

  +-------------------------------------------------------------------------+
  |   TRADITIONAL HEATING-SURFACE CONVERSIONS (ENGINEERING CONVENTION)      |
  |   -- NOT the method prescribed by M.G.L. c. 146, § 48 --                |
  |                                                                         |
  |  • Watertube Boilers:           10 sq ft of heating surface = 1 BHP     |
  |  • Traditional HRT Firetube:    12 sq ft of heating surface = 1 BHP     |
  |  • Modern Compact Firetube:     10 sq ft of heating surface = 1 BHP     |
  |  • Electric Boilers:            10 kW input (approx.)       = 1 BHP     |
  +-------------------------------------------------------------------------+

1. Watertube Boilers: conventionally rated at 10 square feet of water-heating surface per boiler horsepower. $\text{Nominal BHP}_{\text{watertube}} = \frac{\text{Total Heating Surface (sq ft)}}{10}$
2. Firetube Boilers: traditional Horizontal Return Tubular (HRT) and Scotch units are conventionally evaluated at 12 square feet per boiler horsepower; modern packaged firetubes with forced draft and extended passes are taken at 10 square feet per boiler horsepower (or the certified manufacturer's ASME rating).
3. Electric Boilers: the trade shorthand is 10 kW ≈ 1 BHP. The exact thermodynamic equivalent is $33,475 \div 3,412.14 = 9.81\text{ kW}$, and the ASME relieving-capacity constant of 3.5 lb/hr per kW gives essentially the same figure. But for Massachusetts jurisdictional purposes use § 48: an electric boiler's tag horsepower, or its Btu/hr input ÷ 41,840, or its Btu/hr output ÷ 33,475. Worked contrast: a 1,500 kW electric boiler has an input of $1{,}500 \times 3{,}412.14 = 5{,}118{,}210\text{ Btu/hr}$. Because an electric boiler converts essentially all input to output, § 48's output route gives $5{,}118{,}210 \div 33{,}475 = \mathbf{152.9\text{ BHP}}$, while the input route (which assumes 80% efficiency) gives $5{,}118{,}210 \div 41{,}840 = 122.3\text{ BHP}$. The rule-of-thumb 10 kW/BHP answer of 150 BHP is close to the output figure but is not the statutory calculation.

Statutory Licensing Tiers Under M.G.L. c. 146, § 49

The resulting horsepower dictates the license grade required to have charge of, or to operate, the plant. The § 49 limits are charge limits — note that neither fireman grade carries any horsepower cap at all:

• Second Class Fireman: may operate any boiler or boilers under the engineer or fireman in direct charge thereof. May not have charge of any boiler, at any horsepower.
• First Class Fireman: may have charge of and operate any boiler or boilers where the safety valve or valves are set to blow at not more than 25 psi, or operate high-pressure boilers under the engineer or fireman in direct charge. A first class fireman may operate a third class plant under the engineer in direct charge. There is no horsepower limit on the fireman grades — the limit is the 25 psi safety valve setting.
• Third Class Engineer: may have charge of and operate boilers not exceeding 150 HP in the aggregate on solid fuel, or 500 HP in the aggregate on liquid or gaseous fuel, electric or atomic energy or any other source of heat, and engines not exceeding 50 HP each; or operate a second class plant under the engineer in direct charge.
• Second Class Engineer: may have charge of and operate a boiler or boilers (no aggregate horsepower cap) and engines, no one of which shall exceed 150 HP; or operate a first class plant under the engineer in direct charge.
• First Class Engineer: may have charge of and operate any steam plant, without limit.

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4. Geometric Calculation of Boiler Heating Surface

Under ASME Section I and Massachusetts Department of Fire Services (DFS) inspection guidelines, heating surface is defined as:

$\text{Heating Surface} = \text{All boiler surface area that is exposed to hot combustion gases on one side and water/steam on the other side}$

Surface area in contact with hot gases on one side and steam on the other (such as superheater tubes) is designated as superheating surface, not water-heating surface.

The "Gases Touch" Rule: Internal vs. External Tube Diameter

A critical distinction tested on the Massachusetts licensing examinations is the geometric surface measured for boiler tubes:

• Firetube Boilers: Hot combustion gases flow through the inside of the tubes. The heat transfer boundary where gases transfer energy to the metal is the tube's inner surface. Therefore, firetube heating surface calculations must use the inside diameter ($d_i$) of the tube: $\text{Heating Surface per Firetube} = \pi \times d_i \times L$
• Watertube Boilers: Hot combustion gases sweep across the outside of the tubes. The heat transfer boundary in contact with combustion gas is the tube's outer surface. Therefore, watertube heating surface calculations must use the outside diameter ($d_o$) of the tube: $\text{Heating Surface per Watertube} = \pi \times d_o \times L$

(Where $d$ is diameter in feet, and $L$ is length in feet exposed to hot gases). If tube diameter is given in inches, divide by 12 to convert to feet: $\text{Surface Area (sq ft)} = \pi \times \left(\frac{d\text{ (inches)}}{12}\right) \times L\text{ (feet)} = \frac{\pi \times d \times L}{12} \approx 0.2618 \times d\text{ (inches)} \times L\text{ (feet)}$

Total Heating Surface of a Firetube (HRT) Boiler

For a complete Horizontal Return Tubular (HRT) boiler, the total heating surface comprises three distinct geometric components:

1. Tubes: Inner surface of all firetubes: $N \times (\pi \times d_i \times L)$.
2. Exposed Shell: The lower portion of the cylindrical shell exposed to direct furnace gases (typically the lower half, or $\frac{1}{2}$ of the cylindrical circumference: $0.5 \times \pi \times D_{\text{shell}} \times L$).
3. Tubesheets: The net exposed flat area of the front and rear tubesheets: Total tubesheet area exposed to gas minus the cross-sectional area of all tube holes.

Worked Example 1: Firetube Boiler Heating Surface & Statutory BHP

Problem: A Horizontal Return Tubular (HRT) firetube boiler contains 72 firetubes. Each tube has an internal diameter of 3.0 inches ($0.25\text{ ft}$) and an effective gas-contact length of 18 feet. The lower half of the outer shell is exposed to combustion gases, having a shell diameter of 6 feet ($72\text{ inches}$) and length of 18 feet. The combined net exposed tubesheet area (front and rear heads, after subtracting all tube holes) is 35 square feet. Determine:

1. The heating surface of the tubes.
2. The heating surface of the shell.
3. The total water-heating surface.
4. The conventional nominal horsepower using traditional HRT guidelines (12 sq ft/BHP) and modern firetube rules (10 sq ft/BHP). (Remember: for a jurisdictional figure Massachusetts uses the § 48 nameplate method; heating surface is an estimating convention.)

Step 1: Calculate tube heating surface $\text{Circumference per tube} = \pi \times d_i = \pi \times \left(\frac{3}{12}\right) = \pi \times 0.25 = 0.7854\text{ ft}$ $\text{Surface area per tube} = 0.7854\text{ ft} \times 18\text{ ft} = 14.137\text{ sq ft}$ $\text{Total tube heating surface} = 72 \times 14.137\text{ sq ft} = 1,017.88\text{ sq ft}$

Step 2: Calculate exposed shell heating surface $\text{Shell heating surface} = \frac{1}{2} \times (\pi \times D_{\text{shell}} \times L) = 0.5 \times (\pi \times 6\text{ ft} \times 18\text{ ft}) = 0.5 \times 339.29 = 169.65\text{ sq ft}$

Step 3: Calculate total water-heating surface $\text{Total Heating Surface} = A_{\text{tubes}} + A_{\text{shell}} + A_{\text{tubesheets}}$ $\text{Total Heating Surface} = 1,017.88 + 169.65 + 35.00 = 1,222.53\text{ sq ft}$

Step 4: Calculate conventional nominal boiler horsepower

• Traditional HRT convention (12 sq ft / BHP): $\text{BHP} = \frac{1,222.53\text{ sq ft}}{12\text{ sq ft/BHP}} = 101.88\text{ BHP} \approx 101.9\text{ BHP}$
• Modern packaged firetube convention (10 sq ft / BHP): $\text{BHP} = \frac{1,222.53\text{ sq ft}}{10\text{ sq ft/BHP}} = 122.25\text{ BHP} \approx 122.3\text{ BHP}$

Plant Licensing Consequence: Under M.G.L. c. 146, § 49, a Third Class Engineer may have charge of boilers up to 150 HP in the aggregate burning solid fuel, or 500 HP in the aggregate on oil, gas, electric or atomic energy. A plant containing this single HRT boiler (about 102 BHP by the 12 sq ft convention) operating on coal falls within the operating scope of a Third Class Engineer. For the jurisdictional filing, confirm the figure against the manufacturer's tag under § 48 — and, at 102 BHP aggregate, the plant sits in the § 46 periodic attendance band (9 to 250 HP).

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5. Factor of Evaporation (FE) and Equivalent Evaporation (EE)

In operating power plants, boilers rarely operate at atmospheric pressure (0 psig / 212°F) or receive feedwater at 212°F. Instead, a power boiler might generate steam at 150 psig (saturation temperature 366°F) from feedwater delivered by a deaerator at 220°F, or an economizer at 280°F.

Because more heat energy is required to produce a pound of steam at high pressure from cold water than from and at 212°F, comparing raw mass flow rates ($W_a$, pounds of steam produced per hour) between different plants is thermodynamically meaningless. To compare boiler performances on an equal basis, steam generation must be normalized to standard Centennial conditions using the Factor of Evaporation (FE).

Mathematical Formulation of Factor of Evaporation

The Factor of Evaporation (FE) is the dimensionless ratio of the heat required to produce one pound of steam under actual operating conditions to the heat required to evaporate one pound of water from and at 212°F (which is $970.3\text{ Btu/lb}$):

$FE = \frac{h_s - h_f}{970.3}$

Where:

• $h_s = \text{Total enthalpy of steam leaving the boiler (Btu/lb)}$.
  • For dry saturated steam: $h_s = h_g$ at boiler operating pressure.
  • For wet steam with quality $x$: $h_s = h_f + x \cdot h_{fg}$.
  • For superheated steam: $h_s = h_{\text{superheat}}$ at operating pressure and final steam temperature.
• $h_f = \text{Enthalpy of liquid feedwater entering the boiler (Btu/lb)}$.
  • For liquid water at temperatures below 250°F, $h_f$ is closely approximated by: $h_f \approx T_{\text{feedwater}} - 32^\circ\text{F}$.
• $970.3 = \text{Latent heat of vaporization of water at atmospheric pressure (212°F) in Btu/lb}$.

Equivalent Evaporation (EE)

Multiplying the actual boiler steaming rate by the Factor of Evaporation yields the Equivalent Evaporation (EE), representing the mass of water the boiler would have evaporated if it were operating "from and at 212°F":

$W_e = EE = W_a \times FE = W_a \times \left[\frac{h_s - h_f}{970.3}\right]$

Where $W_a$ is the actual evaporation rate in pounds of steam per hour (lb/hr).

Developed Boiler Horsepower

Once equivalent evaporation is known, dividing by the Centennial constant ($34.5\text{ lb/hr per BHP}$) determines the actual Developed Boiler Horsepower:

$\text{Developed BHP} = \frac{W_e}{34.5} = \frac{W_a \times (h_s - h_f)}{34.5 \times 970.3} = \frac{W_a \times (h_s - h_f)}{33,475}$

This fundamental equation links mass flow, thermodynamic steam enthalpy, feedwater enthalpy, and boiler horsepower in a single direct calculation.

Percentage of Rated Boiler Capacity

Modern industrial boilers frequently operate well above their nominal nameplate heating surface rating (e.g., $150%$ to $300%$ of rated capacity) due to forced draft, waterwalls, and high combustion turbulence. The Percentage of Rating is calculated as:

$\%\text{ of Rated Capacity} = \left(\frac{\text{Developed BHP}}{\text{Rated Nameplate BHP}}\right) \times 100\%$

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6. Comprehensive Worked Calculation: Actual Steam Flow to Developed BHP

Problem Statement

An industrial watertube boiler operates at a continuous drum pressure of 150 psig (atmospheric pressure is 14.7 psia; absolute pressure = $164.7\text{ psia}$). The boiler generates 32,000 lb/hr of dry saturated steam. Feedwater enters the boiler from a pressurized deaerator at a temperature of 215°F.

Thermodynamic data from ASME Steam Tables:

• Enthalpy of saturated vapor ($h_g$) at $164.7\text{ psia}$: $h_s = 1,195.6\text{ Btu/lb}$.
• Enthalpy of liquid water ($h_f$) at $215^\circ\text{F}$: $183.2\text{ Btu/lb}$ (approximated as $215 - 32 = 183.0\text{ Btu/lb}$; use steam table value $183.2\text{ Btu/lb}$).
• Latent heat of vaporization at $212^\circ\text{F}$: $970.3\text{ Btu/lb}$.

The boiler has a certified water-heating surface of 4,500 square feet.

Determine:

1. The Factor of Evaporation (FE).
2. The Equivalent Evaporation (EE) in lb/hr.
3. The Developed Boiler Horsepower.
4. The conventional nominal rated horsepower based on heating surface.
5. The operating percentage of rated boiler capacity.

Step-by-Step Solution

Step 1: Calculate the net heat absorbed per pound of steam ($q$) $q = h_s - h_f = 1,195.6\text{ Btu/lb} - 183.2\text{ Btu/lb} = 1,012.4\text{ Btu/lb}$

Step 2: Calculate the Factor of Evaporation (FE) $FE = \frac{h_s - h_f}{970.3} = \frac{1,012.4\text{ Btu/lb}}{970.3\text{ Btu/lb}} = 1.043387 \approx 1.0434$ (Interpretation: Each pound of steam generated under these actual plant conditions requires 4.34% more energy than evaporating a pound of water from and at 212°F).

Step 3: Calculate Equivalent Evaporation (EE) $EE = W_a \times FE = 32,000\text{ lb/hr} \times 1.043387 = 33,388.38\text{ lb/hr} \approx 33,388\text{ lb/hr}$

Step 4: Calculate Developed Boiler Horsepower Method A (using equivalent evaporation): $\text{Developed BHP} = \frac{EE}{34.5} = \frac{33,388.38\text{ lb/hr}}{34.5\text{ lb/hr/BHP}} = 967.78\text{ BHP}$

Method B (using total heat transfer directly): $\text{Developed BHP} = \frac{W_a \times (h_s - h_f)}{33,475} = \frac{32,000\text{ lb/hr} \times 1,012.4\text{ Btu/lb}}{33,475\text{ Btu/hr/BHP}} = \frac{32,396,800\text{ Btu/hr}}{33,475\text{ Btu/hr/BHP}} = 967.79\text{ BHP}$ Both methods confirm: Developed BHP = 967.8 BHP.

Step 5: Calculate conventional nominal rated horsepower By the traditional watertube convention of 10 sq ft of heating surface per BHP: $\text{Nominal Rated BHP} = \frac{\text{Heating Surface}}{10} = \frac{4,500\text{ sq ft}}{10\text{ sq ft/BHP}} = 450.0\text{ BHP}$

Step 6: Calculate operating percentage of rated capacity $\%\text{ Rating} = \left(\frac{\text{Developed BHP}}{\text{Rated BHP}}\right) \times 100\% = \left(\frac{967.79\text{ BHP}}{450.0\text{ BHP}}\right) \times 100\% = 215.06\%$ The boiler is developing 215.1% of its conventional nominal rating, a standard operating performance for modern watertube steam generators equipped with membrane waterwalls and forced draft.

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7. Field Rules of Thumb & Engineering Approximations

On Massachusetts Department of Fire Services oral examinations, inspectors frequently expect candidates to demonstrate mental arithmetic and rapid field approximations:

1. Steam to BHP Rule of Thumb: In the field, where steam tables are not immediately accessible, operators approximate: 30 pounds of steam per hour equals approximately 1 boiler horsepower. Validation: Because modern boilers operate at factors of evaporation between 1.08 and 1.18, $34.5 / 1.12 \approx 30.8\text{ lb/hr actual steam per BHP}$.
2. Btu to Steam Conversion: One pound of steam at typical operating pressures absorbs approximately 1,000 Btu from feedwater ($h_s - h_f \approx 1,000\text{ Btu/lb}$). Application: To generate 10,000 lb/hr of steam, the boiler must absorb roughly $10,000,000\text{ Btu/hr}$ into the water.
3. Electric Boiler Output: An electric steam boiler generates approximately 3.45 to 3.5 pounds of steam per hour for each kilowatt (kW) of electrical input, so roughly $10\text{ kW} \approx 1\text{ BHP} \approx 34.5\text{ lb/hr EE}$. ASME Section I fixes the relieving-capacity constant for electric boilers at 3.5 lb/hr per kW. For a Massachusetts jurisdictional horsepower figure, fall back on § 48.