[MEP Center Engineering]

Size the main exhaust for the fire or smoke-clearance case required by your code, run it at reduced speed under CO/NO₂ control in normal operation, and use jet fans only to move air towards the exhaust points. Normal-mode design is a dilution problem; fire mode is usually verified by CFD.

1. Airflow rates

• UK and many European practices: 6 air changes per hour for normal ventilation and 10 ach for smoke clearance (Approved Document B; BS 7346-7 for jet fan smoke control). Many national codes, including Turkey’s, follow similar figures; check yours.
• IMC 404 (US): 0.75 cfm/ft² (3.8 l/s·m²), with reduced operation permitted under CO and NO₂ detection control.
• ASHRAE Applications Handbook gives a contaminant-based method for CO dilution when the code allows it.

2. Worked example

Volume: 4000 m² × 3.0 m = 12,000 m³.

• Normal mode at 6 ach: 12,000 × 6 / 3600 = 20 m³/s (72,000 m³/h).
• Smoke clearance at 10 ach: 12,000 × 10 / 3600 ≈ 33.3 m³/s.
• IMC check: 3.8 l/s·m² × 4000 m² ≈ 15.2 m³/s.

Use two or more exhaust fans, for example 2 × 16.7 m³/s, so that one fan failure leaves half the capacity, with a separate standby where the authority requires it. In normal mode both fans run at about 60 % speed (flow is proportional to speed), which needs only about 0.6³ ≈ 22 % of full-speed power. With CO control they run lower or stop most of the day.

Make-up air of roughly 75 to 80 % of exhaust enters through ramps or dedicated shafts, located at the far end from the exhaust so the air sweeps the whole floor.

3. CO and NO₂ control

• Typical setpoints: low speed above about 25 to 30 ppm CO, high speed above about 50 ppm. As a reference, the EU occupational limit for CO is 20 ppm over 8 hours and 100 ppm for 15 minutes.
• Add NO₂ sensors where diesel vehicles park (typical stage around 1 to 3 ppm).
• Mount CO sensors at breathing height (about 1.5 to 1.8 m), at ramps, near exits and in dead corners, at the spacing given by the sensor’s coverage data.
• In fire mode the fire alarm overrides gas control.

4. Jet fan layout

• Place jet fans in series along the flow path towards the exhaust, clear of beams and services directly in front of the outlet, with their throw overlapping the next fan.
• Avoid blowing against the incoming make-up air or into dead-end bays without an exhaust point.
• Fire-mode fans and their cables must be rated for smoke temperature (EN 12101-3 classes such as F300 or F400, as the authority specifies) with secondary power supply.
• Verify smoke mode by CFD: visibility and temperature along escape routes and firefighting access, reversing the flow for different fire locations if the scheme requires it.

Common mistakes

• Sizing exhaust fans only for 6 ach while the fire strategy needs 10 ach.
• Make-up air entering right next to the exhaust.
• Too few CO sensors, or sensors near the ceiling or at the ramp mouth only.
• Ramps and bays with no air movement because jet fans ignore beam layout.

Related: Car Park Ventilation with Jet Fans: Quick Design. For deeper study: HVAC Systems Design Course.

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[MEP Center Engineering]

For a quick, defensible figure use the airflow per metre of hood for the duty of the appliances (IMC method) and check it against face velocity; for a detailed design use the appliance heat-load method of EN 16282 / VDI 2052. Supply make-up air for about 80 to 90 % of the exhaust at low velocity so the kitchen stays slightly negative.

1. Methods

• IMC (US): minimum exhaust per linear foot of hood depends on hood type and appliance duty (light, medium, heavy, extra-heavy). For a wall-mounted canopy over medium-duty appliances it is 300 cfm/ft (about 465 l/s per metre).
• Face velocity check: about 0.25 to 0.5 m/s across the hood face area for canopy hoods, higher for heavy duty.
• EN 16282 / VDI 2052: exhaust is built from the convective heat and moisture of each appliance, its simultaneity and the hood arrangement. It often gives lower values for light cooking and higher values for heavy charbroiling.
• ASHRAE 154 follows the hood-type and duty logic used in the IMC.

2. Worked example

• Hood: 3.0 m = 9.84 ft, wall canopy, medium duty (fryers, griddle, ranges).
• Exhaust: 300 cfm/ft × 9.84 ft ≈ 2950 cfm ≈ 1.39 m³/s (5020 m³/h).
• Face velocity check: 1.39 / (3.0 × 1.2) ≈ 0.39 m/s. Acceptable.
• Make-up air at 85 %: ≈ 1.18 m³/s (4270 m³/h).
• Exhaust duct at about 7.5 to 8 m/s: 1.39 / 8 ≈ 0.17 m² → for example 450 × 400 mm (0.18 m², 7.7 m/s).

3. Code checks

• NFPA 96 requires a minimum grease duct velocity of 500 fpm (2.5 m/s); design velocities of 7 to 10 m/s keep grease in suspension, but check noise and pressure drop.
• Canopy overhang: at least 6 in (150 mm) beyond the cooking surface on open sides, and the hood lip no more than 4 ft (1.2 m) above it (IMC).
• Grease ducts: continuous welded or listed, fire-rated enclosure where they pass through other compartments, access doors for cleaning, falling back to the hood.
• Fire suppression over cooking appliances, interlocked to shut off fuel and stop make-up air as the system requires.

4. Make-up air

• Introduce it through low-velocity diffusers or perforated plenums away from the hood edge; keep air velocity near the hood below about 0.25 m/s, or the plume is disturbed and spills.
• Temper it in winter. In large kitchens consider demand control (temperature and optical sensors) to reduce exhaust at low load.
• Interlock exhaust and make-up fans; the exhaust starts first and stops last.

Common mistakes

• Sizing by air changes of the kitchen volume instead of by hood and appliance duty.
• Make-up air through ceiling diffusers blowing straight at the hood.
• Grease duct velocity too low on long horizontal runs.
• Ignoring that the hood length must cover the whole cooking line plus overhang.

Related: Commercial Kitchen Hood Exhaust Airflow: Quick Sizing, Kitchen Hood Wet Chemical Systems: Design Basics.

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[MEP Center Engineering]

The force to open a door against a pressure difference is the door closer force plus the pressure force acting about the hinge. EN 12101-6 limits the opening force to 100 N at the handle; NFPA 101 / IBC allow 30 lbf (133 N). Check the maximum pressure with all doors closed and control it with a pressure-relief damper or a variable speed fan on a pressure sensor.

1. Door opening force

F = F_dc + (W · A · Δp) / (2 · (W − d))

• F_dc: force to overcome the door closer and friction (N)
• W: door width (m), A: door area (m²), Δp: pressure difference (Pa)
• d: distance from the handle to the latch edge (m)

2. Worked example

Door 0.9 × 2.1 m (A = 1.89 m²), handle 0.07 m from the edge, closer force 40 N, Δp = 50 Pa.

• Pressure term: (0.9 × 1.89 × 50) / (2 × 0.83) = 85.05 / 1.66 ≈ 51.2 N.
• F = 40 + 51.2 ≈ 91 N (about 20.5 lbf), below the 100 N limit.
• Maximum allowed pressure with this closer: Δp = (100 − 40) × 1.66 / 1.701 ≈ 58.6 Pa.

So the margin between design pressure and maximum pressure is small. A heavier closer of 60 N would allow only about 39 Pa. Specify the closer force on the door schedule and measure it at commissioning.

3. Design criteria (typical)

• EN 12101-6 (being replaced by EN 12101-13): pressure difference with doors closed typically about 45 to 50 Pa depending on system class; air velocity through the open fire-floor door 0.75 m/s or 2 m/s depending on class; opening force ≤ 100 N.
• NFPA 92 / IBC: minimum pressure difference 12.5 Pa (0.05 in wg) in sprinklered buildings, higher for unsprinklered; maximum governed by the 30 lbf opening force.

4. Leakage and fan sizing

Leakage through a closed door: Q = 0.83 · A_L · Δp^1/2. With an assumed leakage area of 0.01 m² per door at 50 Pa: 0.83 × 0.01 × 7.07 ≈ 0.059 m³/s per door, about 0.7 m³/s for 12 doors, before lift, wall and roof leakage. The open-door velocity case usually governs the fan size; add a margin for leakage uncertainty.

5. Keeping pressure under control

• Relief damper from the stair to outside, or a variable speed fan controlled by a differential pressure sensor; fast response matters when doors close.
• For tall stairs, multiple injection points, commonly every two to three floors, to limit pressure variation and stack effect.
• Separate duty and standby fans with automatic changeover, on secondary power.
• Air intake away from smoke exhaust outlets.

Common mistakes

• Fan sized for the open-door case with no relief, so closed-door pressure exceeds 80 Pa.
• Ignoring the closer force in the calculation.
• Commissioning on a calm day only; wind and temperature change the result.

Related: Stair Pressurisation: Quick Airflow Design. For deeper study: HVAC Systems Design Course.

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[MEP Center Engineering]

Size the set for the simultaneous (diversified) peak flow, not the sum of fixture flows, and build the head from static lift + required pressure at the most remote outlet + friction and component losses. Then check the lowest floors and add pressure reduction where the static pressure becomes excessive.

1. Design flow

• EN 806-3: add the loading units (LU) of all outlets and read the design flow from the standard’s LU-to-flow curve.
• IPC/UPC: add water supply fixture units (WSFU) and read the flow from the Hunter-based curve (flush-tank or flush-valve curve as applicable).
• Do not stack one diversity method on another: apply one method, once, to the whole building.

For the example below, assume the curve gives 4.0 l/s (14.4 m³/h, about 63 gpm). Use your own value; the procedure is what matters.

2. Pump head

1. Static lift from the pump to the highest outlet: 30 m.
2. Required flow pressure at the most remote outlet: 1.5 bar = 15 m (comfortable for thermostatic showers; the code minimum is lower, e.g. the 100 kPa assumed in EN 806-3, or the per-fixture minimum flow pressures in IPC 604.3).
3. Friction in pipes and fittings at peak flow plus meter, filter and check valves: 6 m (take it from your pipe sizing, not a guess).

Head = 30 + 15 + 6 = 51 m (≈ 5.0 bar, about 72 psi). With a flooded suction from the break tank you can ignore the small positive suction head and keep it as a margin.

3. Number of pumps and power

Use separate pumps with one standby, for example 3 × 50 %: two duty + one standby, each 2.0 l/s at 51 m. Do not use a twin-head pump as “duty + standby”; the standby must be a separate pump that can be isolated and removed while the others run.

Hydraulic power per pump = ρ·g·Q·H = 1000 × 9.81 × 0.002 × 51 ≈ 1.0 kW. With a pump efficiency of 0.65 the shaft power is ≈ 1.54 kW, so select the next motor size, 2.2 kW. Variable speed control on constant discharge pressure, a small membrane vessel to cover low draw-offs, dry-run protection on the tank and automatic duty rotation complete the set.

4. Pressure on the lower floors

With about 5 bar at the pump discharge, the floors close to the plant room see almost the full pressure. EN 806-2 aims to keep the pressure at outlets at or below about 5 bar, and IPC 604.8 requires a pressure-reducing valve where the static pressure exceeds 80 psi (552 kPa). In practice many designers keep outlets below 4 to 5 bar for noise and tap comfort: either split the riser into pressure zones or fit PRVs on the lower floors.

Common mistakes

• Sizing on the sum of fixture flows: the set is oversized and cycles at low demand.
• Forgetting the losses of the water meter, filter and check valves.
• No standby pump, or a twin-head pump treated as standby.
• Ignoring the pressure on the lowest floors.
• Undersizing the break tank so the pumps run dry at peak.

Related: Booster Pump Set Sizing for Buildings, Domestic Water Pipe Sizing with Fixture Units, Pressure Reducing Valve Stations.

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[MEP Center Engineering]

Size the stack for its waste water flow so it stays within the capacity of a ventilated stack of that diameter, and size vents from the connected load and length. AAVs can replace many branch and stack vents, but they only admit air: the system still needs at least one open vent to atmosphere and a way to relieve positive pressure.

EN 12056-2 approach

1. Add the discharge units (DU) of the fixtures on the stack, using the values for your system type (System I is common in continental Europe and Turkey, System III in the UK).
2. Waste water flow: Q_ww = K·√ΣDU, with K = 0.5 for dwellings.
3. Choose a stack whose permitted flow for the chosen ventilation arrangement (primary or secondary ventilated, with or without swept entries) is at least Q_ww, and never smaller than the largest connected drain (DN100 where a WC is connected).

Worked example

A stack serves 10 flats, each with a 6 l WC (2.0 DU), basin (0.5), shower (0.6), kitchen sink (0.8) and washing machine (0.8), System I values.

• Per flat: 2.0 + 0.5 + 0.6 + 0.8 + 0.8 = 4.7 DU; 10 flats: ΣDU = 47.
• Q_ww = 0.5 × √47 = 0.5 × 6.86 ≈ 3.4 l/s.
• EN 12056-2 gives a primary ventilated DN100 stack with ordinary branch entries about 4.0 l/s, so DN100 with the stack vent continued to the roof at full size is adequate.

On the drawings, label each branch that serves two or more fixtures with its total DU and DN so the checker can follow the sizing.

IPC approach

• Vents are sized from the drainage fixture units (DFU) served and the developed length of the vent (IPC Chapter 9).
• A vent is not smaller than half the diameter of the drain it serves and not smaller than 1¼ in (32 mm).
• Terminals extend at least 6 in (150 mm) above the roof, more where the roof is occupied or snow is deep, and are not within 10 ft (3 m) horizontally of an openable window or air intake unless at least 2 ft (0.6 m) above it. Local amendments often increase these.

Air admittance valves

• Product standards: EN 12380 in Europe; ASSE 1050 (stack type) and ASSE 1051 (fixture and branch type) under the IPC.
• Install them accessible, in a ventilated space, above the drain they protect (IPC: branch type at least 4 in / 100 mm above the horizontal branch or fixture drain; stack type at least 6 in / 150 mm above the flood-level rim of the highest fixture).
• Keep at least one open vent to atmosphere on each building drain, normally the stack vent of the largest or furthest stack.
• AAVs do not relieve positive pressure. In tall stacks, where surges are expected, add relief venting or a positive pressure attenuator.
• Do not use them on sumps, ejector pits or grease interceptor vents unless the product and the local code explicitly allow it.

Common mistakes

• Reducing the stack vent diameter above the highest branch.
• Replacing every roof vent with AAVs, leaving no open vent.
• Placing AAVs in sealed boxes without access or ventilation.
• Terminating vents next to openable windows or fresh air intakes.

Related: Plumbing Vent Pipes: Why, Where and How Big, Drainage Pipe Sizing: EN 12056 DU vs IPC DFU.

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[MEP Center Engineering]

Store at 60 °C or above, keep the circulating loop so that no part falls below about 55 °C (50 °C is the absolute minimum in the UK guidance), and get at least 50 °C at every outlet within one minute. Keep cold water below 20 °C. Size the circulation flow from the heat loss of the loop, not by rule of thumb.

Typical target temperatures

• Storage and calorifier outlet: ≥ 60 °C. UK HSG274, German DVGW W 551 and ASHRAE Guideline 12 all use about 60 °C (140 °F) for storage.
• Circulation return: DVGW W 551 limits the temperature drop round the loop to 5 K, i.e. return ≥ 55 °C. HSG274 requires ≥ 50 °C, preferably 55 °C. ASHRAE Guideline 12 recommends about 51 °C (124 °F) or more.
• Outlets: ≥ 50 °C within one minute of opening (55 °C in healthcare under UK guidance).
• Cold water: below 20 °C within two minutes at outlets.
• Dead legs: DVGW limits the volume between the circulated pipe and the tap to 3 litres; keep unrecirculated branches as short as possible everywhere.

Lowering the storage temperature is only acceptable with a recognised alternative control method (for example continuous treatment) and the agreement of the authority; it is not a free energy saving.

Scald protection

Fit thermostatic mixing valves where scalding is a risk (showers, baths, basins for vulnerable users), as close to the outlet as possible so the pipe carrying mixed water stays short.

Worked example: circulation flow

The heat lost by the flow pipes of the loop must be replaced by the circulating water within the allowed temperature drop:

q = Φ / (c · ΔT)

• Insulated flow pipework: 120 m at an assumed 8 W/m → Φ = 960 W.
• Allowed drop: ΔT = 5 K (60 → 55 °C).
• q = 0.96 kW / (4.19 kJ/kg·K × 5 K) = 0.046 kg/s ≈ 165 l/h.

Size the circulation pipes for this flow at a low velocity (typically 0.2 to 0.5 m/s) and select the pump for the index loop resistance including the balancing valves. In a building with many risers, calculate the flow of each riser from its own losses and balance with thermostatic circulation valves set to the return target, with a test point on each riser return.

Common mistakes

• One large pump flow without balancing: the nearest risers take the flow and the far risers drop to 40 °C.
• Uninsulated or poorly insulated return pipes and valves.
• Long unrecirculated branches to showers or kitchens.
• Calorifier outlet set to 60 °C with a return that is never measured or monitored.
• Large storage that stratifies and leaves a cool volume at the bottom; provide destratification or a periodic heat-up of the full volume.

Related: Hot Water Recirculation and Legionella Temperatures, Hot Water Storage Cylinder Sizing.

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[MEP Center Engineering]

A compliant U-value describes the average or clear-field wall, but surface condensation and mould start at the coldest point: thermal bridges, corners and reveals, combined with high indoor humidity. Check the temperature factor f_Rsi of every critical junction against the value needed for your indoor humidity, then fix the bridges and control humidity with ventilation.

1. The criterion: temperature factor f_Rsi

f_Rsi = (θ_si − θ_e) / (θ_i − θ_e), where θ_si is the internal surface temperature (EN ISO 13788, with R_si = 0.25 m²K/W for this check). Mould is assessed at 80 % relative humidity at the surface, visible condensation at 100 %.

2. Worked example

Indoor 20 °C and 50 % RH, outdoor −5 °C.

• Vapour pressure indoors: 0.50 × 2337 Pa = 1169 Pa → dew point ≈ 9.3 °C.
• Mould limit: the surface may not be colder than the temperature at which 1169 Pa is 80 % of saturation (1461 Pa) → ≈ 12.6 °C.
• Required f_Rsi,min = (12.6 − (−5)) / (20 − (−5)) = 17.6 / 25 ≈ 0.70.

Clear wall with U = 0.40 W/m²K: f_Rsi = 1 − U·R_si = 1 − 0.40 × 0.25 = 0.90, θ_si = −5 + 0.90 × 25 = 17.5 °C. No problem.

Uninsulated column or beam face with f_Rsi = 0.60: θ_si = −5 + 0.60 × 25 = 10.0 °C. Above the 9.3 °C dew point, but below the 12.6 °C mould limit: mould, without visible water.

If the indoor humidity rises to 60 % RH, the dew point becomes ≈ 12.0 °C and the mould limit ≈ 15.4 °C: the same junction now runs wet. Humidity matters as much as insulation.

3. Remedies on the building side

• Insulate column, beam and slab-edge faces continuously with the wall; on existing buildings external insulation is the most effective cure.
• Insulate window reveals and set frames in the insulation plane.
• Treat balcony and canopy slabs with a thermal break or wrap them.
• Model critical junctions to EN ISO 10211 to get f_Rsi and the ψ-value; many national guides use f_Rsi ≥ 0.70 to 0.75 for dwellings.

4. Remedies on the heating and ventilation side

• Ventilate to remove moisture. Example: a household releasing about 10 kg/day (≈ 420 g/h), outdoor air at −5 °C and 90 % RH (≈ 2.9 g/m³). To hold 20 °C and 50 % RH (≈ 8.6 g/m³) requires 420 / (8.6 − 2.9) ≈ 73 m³/h of outdoor air; at 60 % RH about 56 m³/h. Continuous extract from kitchen and bathrooms with trickle vents in habitable rooms, or balanced ventilation with heat recovery, provides this reliably. Rates per EN 16798-1 or ASHRAE 62.2 as your code requires.
• Place emitters under windows, and keep furniture and curtains a few centimetres off cold external walls.
• Avoid deep night set-back or unheated bedrooms; a cold room with doors closed collects moisture from the rest of the flat.

Common mistakes

• Checking only interstitial condensation (Glaser) and never the surface factor.
• Accepting the code U-value for the wall and ignoring columns, beams and reveals.
• Painting over mould without changing ventilation or insulation.
• Sealing windows in a refurbishment without adding any controlled ventilation.

Related: ASHRAE 62.1 vs EN 16798: Ventilation Rates Compared, Radiator Sizing: Correction Factors. For deeper study: HVAC Systems Design Course.

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[MEP Center Engineering]

A heat pump works best at the lowest possible flow temperature, so underfloor heating (flow about 30 to 35 °C) is the natural partner. Radiators also work if they are sized for low temperature, typically 45 to 55 °C flow, which usually means two to three times the nominal output of a boiler-era radiator.

1. Why flow temperature matters

The ideal (Carnot) heating COP is T_flow / (T_flow − T_source) in kelvin. With outdoor air at 2 °C: at 35 °C flow, 308 / 33 ≈ 9.3; at 55 °C, 328 / 53 ≈ 6.2. Real machines reach roughly 40 to 50 % of this, so typically COP 3.5 to 4.5 at A2/W35 and 2.5 to 3.0 at A2/W55. A common rule: each 1 K lower flow temperature improves seasonal performance by about 2 to 3 %.

2. Worked example: one living room

Room: 20 m², design heat loss 1400 W (70 W/m²), 20 °C.

Underfloor heating. EN 1264 basic characteristic: q ≈ 8.92 × (θ_F,m − θ_i)^1.1. For 70 W/m²: θ_F,m − θ_i = (70 / 8.92)^1/1.1 ≈ 6.5 K, so a mean floor surface of about 26.5 °C, well under the 29 °C limit for occupied areas (that limit gives about 100 W/m²). With a tiled floor this is typically reached at about 35 °C flow. A thick carpet (R about 0.15 m²K/W) raises the required flow temperature noticeably.

Radiators. Radiator output is rated at ΔT 50 K (75/65/20 °C, EN 442). At 45/40/20 °C the mean water-to-air difference is 22.5 K (log mean 22.4 K), and with exponent n = 1.3:

factor = (22.5 / 50)^1.3 ≈ 0.35

Required nominal output = 1400 / 0.35 ≈ 4000 W at ΔT 50, against about 1400 to 1700 W for a boiler-era radiator. Keep each radiator at 180 cm or less and split larger outputs into two units, for example two radiators of about 2000 W under two windows, connected diagonally (flow top on one end, return bottom on the other) so the whole panel is warm at low ΔT.

3. How to decide

• New build, floors not yet laid: underfloor heating, designed for 30 to 35 °C flow.
• Existing house: calculate the room heat losses again (after any insulation upgrade), check each radiator at 45 to 55 °C flow and replace or add only those that fall short. Fan-assisted low-temperature radiators help in small rooms.
• Mixed systems: avoid a design where one undersized radiator forces the whole house onto a high flow temperature.
• Design ΔT for heat pumps is usually 5 K (sometimes up to 8 K), not the 20 K of a boiler: flow rates are about four times higher, so check pipe sizes and pump head.

Common mistakes

• Keeping 70/50 °C radiators and setting the heat pump to 60 °C.
• Using catalogue ΔT 50 outputs without the correction factor.
• Ignoring floor covering resistance in underfloor design.
• Sizing the heat pump without a bivalent point and backup strategy.

Related: Heat Pump Sizing and the Bivalent Point, Underfloor Heating: Quick Design Rules, Radiator Sizing: Correction Factors for Low Temperatures.

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[MEP Center Engineering]

The vessel must accept the expansion volume of the water plus a small water reserve between the pre-charge pressure (set by the static height) and the final pressure (set by the safety valve). Nominal volume = (expansion + reserve) × (p_e + 1) / (p_e − p₀), with pressures in bar gauge. The pressure window, not the water volume, is what usually drives the size.

Step by step (EN 12828 method)

1. System volume V_s: boiler, pipes, emitters, buffer tanks. 800 l in this example.
2. Expansion coefficient e from the cold fill temperature (about 10 °C) to the maximum temperature: water density 999.7 kg/m³ at 10 °C and 971.8 kg/m³ at 80 °C gives e ≈ 0.029.
3. Expansion volume: V_e = 800 × 0.029 = 23.2 l.
4. Water reserve: V_wr ≥ 0.5 % of V_s, minimum 3 l → 4 l.
5. Pre-charge p₀: static height 15 m = 1.5 bar, plus about 0.3 bar margin so the highest point stays under positive pressure → p₀ = 1.8 bar.
6. Final pressure p_e: for safety valves up to 5 bar, p_e = p_sv − 0.5 bar = 3.0 − 0.5 = 2.5 bar.
7. Nominal volume: V_n = (23.2 + 4) × (2.5 + 1) / (2.5 − 1.8) = 27.2 × 3.5 / 0.7 = 136 l → select the next standard size, for example 140 or 150 l.

Why sizes differ so much

The factor (p_e + 1) / (p_e − p₀) is very sensitive. If the safety valve can be 4 bar (check the rating of the boiler and every component), p_e = 3.5 bar and V_n = 27.2 × 4.5 / 1.7 ≈ 72 l, about half. In a tall building the static height eats the pressure window; a pressure-maintaining unit (pump or compressor type) is then often smaller and more stable than a very large membrane vessel.

ASHRAE form

For a diaphragm tank: V_t = V_s × [(v₂/v₁ − 1) − 3αΔt] / (1 − P₁/P₂), with absolute pressures. The pipe expansion term 3αΔt is small for steel, and 1 − P₁/P₂ equals (p_e − p₀)/(p_e + 1) in gauge bar, so both methods give the same answer for the same inputs.

Installation points

• Connect the vessel on the pump suction side, near the boiler or heat source, so the pump does not lower the pressure at the high points.
• Use a lockable isolating valve with a drain between vessel and system for servicing; the vessel must never be isolated in operation.
• Fill pressure when cold: slightly above p₀ (typically +0.2 to 0.3 bar) so the water reserve is in the vessel.
• Glycol systems, solar circuits and heat pumps with large buffers need their own expansion coefficients and temperature ranges.

Common mistakes

• Pre-charge left at the factory value (often 1.5 bar) regardless of static height.
• Forgetting buffer tanks in the system volume.
• Using the operating temperature instead of the maximum the system can reach.
• No water reserve, so the system loses pressure after the first venting.

Related: Expansion Vessel Sizing for Heating Systems. For deeper study: HVAC Systems Design Course.

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[MEP Center Engineering]

At 9 m, point smoke detectors are still within the usual height limits (BS 5839-1 general limit 10.5 m; NFPA 72 has no fixed limit but requires a stratification assessment). Beams 0.6 m deep are less than 10 % of the 9 m ceiling height, so both codes allow smooth-ceiling spacing. You need about 12 detectors under BS 5839-1 and about 15 under NFPA 72.

BS 5839-1 (and similar EN 54-14 national practice)

1. Height: 9 m < 10.5 m, so point smoke detectors are acceptable. For heat detectors the limits are lower and depend on the grade.
2. Beams: a downstand deeper than 10 % of the ceiling height is treated as a wall. Here 0.6 m < 0.9 m, so the ceiling counts as flat.
3. Coverage: no point may be more than 7.5 m horizontally from a smoke detector. On a rectangular grid, the half-diagonal of each cell must be ≤ 7.5 m.
4. Layout: 4 columns × 3 rows gives cells of 10 m × 8 m. Half-diagonal: √(5² + 4²) = 6.4 m ≤ 7.5 m. Corners are 5 m and 4 m from the nearest detector, again 6.4 m. Total: 12 detectors.

NFPA 72

1. Beams: depth less than 10 % of the ceiling height (0.1 H = 0.9 m), so smooth-ceiling spacing applies. Detectors may be on the ceiling or on the bottom of the beams.
2. Nominal spacing S = 9.1 m (30 ft), and every point of the ceiling must be within 0.7 S = 6.4 m (21 ft) of a detector.
3. Layout: 40 / 9.1 = 4.4, so 5 columns at 8 m; 24 / 9.1 = 2.6, so 3 rows at 8 m. Half-diagonal √(4² + 4²) = 5.7 m ≤ 6.4 m. Total: 15 detectors.

If the beams were deeper

With 1.0 m beams (more than 10 % of H): BS 5839-1 treats each 6 m bay as a separate space, each with its own detectors. NFPA 72, with a beam spacing of 6 m ≥ 0.4 H = 3.6 m, requires a detector in each beam pocket. Both routes give a detector line in every bay, often 7 bays × 3 = 21 detectors. At that point a beam detector or aspirating system is usually cheaper and easier to maintain.

High-ceiling considerations

• Stratification: a warm layer under the roof can stop smoke from reaching it. NFPA 72 asks you to consider it. In practice, add detection at a lower level (beams at 60 to 80 % of height) or use aspirating sampling. See Optical Beam Smoke Detectors: Where and How to Place Them and Aspirating Smoke Detection: When and How.
• Maintenance access at 9 m: plan for lifts or remote-test detectors.
• Racking up to the ceiling changes the airflow; recheck with the final storage layout.

Common mistakes

• Using the BS 7.5 m radius and the NFPA 9.1 m spacing interchangeably.
• Ignoring the beam depth until the ceiling is built.
• Placing detectors closer than 0.5 m to walls or beams (BS 5839-1), or less than 100 mm (4 in) from a wall (NFPA 72).

A comparison of the two codes: EN 54-14 vs NFPA 72.

For deeper study: Fire Detection and Alarm System Design course.

Check the edition of the code that applies to your project.

[MEP Center Engineering]

The address limit is rarely the governing one. In practice the loop is limited by the fault-tolerance rules (isolators), by the loop current and voltage drop of powered devices such as sounders, and by spare capacity. 160 devices on one loop can work, but only with enough isolators and a voltage-drop check in the alarm condition.

Step 1: Spare capacity

Design to about 80 % of the panel limit, so later additions do not need a new loop. Here 160 of 250 addresses (64 %) is fine. Many designers also split by building wing or fire compartment, so a loop failure is easy to understand on site.

Step 2: Isolators and fault tolerance

• EN 54-2 requires that a single fault in a transmission path does not remove more than 32 detectors and/or call points.
• BS 5839-1 adds that a single open or short circuit must not disable protection in an area larger than one zone (and a zone is limited to 2 000 m²).
• NFPA 72 uses pathway classes (Class A, X). Short-circuit isolation is commonly specified so that one fault takes out no more than one floor or about 50 devices; check the requirement in your edition and with the authority.
• For 160 devices: 160 / 32 = 5 segments at least, so at least 4 field isolators in addition to the panel’s loop-end isolators. In practice, put one at every zone or floor boundary, or use devices with built-in isolators.

Step 3: Voltage drop in alarm

Example: 1.5 mm² copper loop cable, R₂₀ = 12.1 Ω/km per conductor, loop length 1 500 m, alarm current from loop sounders 150 mA. Assume the panel delivers 21 V at end of battery discharge and the devices need 17 V, a 4 V allowance. (Use your manufacturer’s actual figures.)

• Loop resistance with one end feeding (after a break, which the design must survive): 2 × 1.5 × 12.1 = 36.3 Ω.
• Whole load lumped at the far end: 0.150 × 36.3 = 5.4 V, which fails.
• Evenly distributed load: about half, 2.7 V, which passes. Real layouts are in between, so check with the manufacturer’s loop calculator.
• With 2.5 mm² (7.41 Ω/km): 2 × 1.5 × 7.41 = 22.2 Ω, lumped 3.3 V, which passes even in the worst case.

Step 4: Other limits

• Cable capacitance and inductance limits from the panel manufacturer, which matter for long loops or screened cable.
• Fire-resistant cable, for example to BS 8434 or the EN 50200 class required, is needed where the cable must keep working during a fire.
• Isolator count per loop, because each isolator adds resistance.

Common mistakes

• Filling a loop to 100 % of the addresses.
• Putting all sounders at the far end of one loop.
• Checking voltage drop only in quiescent condition, or only with the loop fed from both ends.

Related: Fire Alarm Standby Battery Calculation and Manual Call Point Placement.

For deeper study: Fire Detection and Alarm System Design course.

Check the edition of the code that applies to your project.

[MEP Center Engineering]

Under BS 5839-1: C = 1.25 × (T₁ × I₁ + D × I₂ × T₂), with T₂ = 0.5 h and D = 1.75. For this panel that gives 16.6 Ah, so use two 12 V 18 Ah batteries. Under NFPA 72 (24 h plus 5 min alarm plus a 20 % margin) the result is 13.2 Ah, and the same 18 Ah battery is a sensible choice.

BS 5839-1 method

• T₁: standby time, 24 h (longer, often 72 h, where a fault may go unnoticed, for example in buildings not continuously occupied and without remote monitoring).
• I₁: quiescent current, 0.45 A.
• I₂: alarm current, 2.8 A. T₂: 0.5 h.
• D = 1.75: derating for the high discharge rate in alarm. 1.25: ageing factor.

C = 1.25 × (24 × 0.45 + 1.75 × 2.8 × 0.5) = 1.25 × (10.8 + 2.45) = 16.6 Ah. Next standard size: 17 or 18 Ah. Choose 18 Ah for a small reserve.

NFPA 72 method

• 24 h standby, then 5 min in alarm (15 min for an emergency voice/alarm communication system).
• Capacity: 24 × 0.45 + 2.8 × 5/60 = 10.8 + 0.23 = 11.0 Ah.
• Add the 20 % safety margin: 11.0 × 1.2 = 13.2 Ah. Next standard size above 12 Ah: 17 or 18 Ah.

Charger and recharge

EN 54-4 and BS 5839-1 require recharge to 80 % within 24 h. NFPA 72 requires recharge within 48 h. Rough charger current: 0.8 × 18 / 24 = 0.6 A for recharge, plus the 0.45 A quiescent load, about 1.05 A. Check that the panel’s power supply is rated for this and for the largest battery it allows.

Where the current figures come from

• Quiescent: panel, each loop device (sum from datasheets), remote displays, interface modules and relays held energised.
• Alarm: the above plus every sounder, beacon and voice-alarm load, and door-holder relays if they draw current in alarm. Many beacons draw more than the sounders.
• Loads powered by separate power supplies need their own battery calculations. Do not forget them.

Common mistakes

• Leaving out the 1.25 and 1.75 factors, which halves the battery size.
• Using typical device currents instead of the maximum alarm currents.
• Choosing a battery larger than the panel’s charger is rated for.
• Battery enclosures in hot locations: capacity and life fall quickly above 25 °C.

Step-by-step article: Fire Alarm Standby Battery Calculation. Voice-alarm loads: Voice Alarm Speaker Spacing and Intelligibility.

For deeper study: Fire Detection and Alarm System Design course.

Check the edition of the code that applies to your project.

[MEP Center Engineering]

Start from the commodity classification and the storage height. Under both EN 12845 and NFPA 13, ceiling-only protection is allowed up to limits that depend on commodity, storage height and ceiling height. Beyond those limits you need in-rack sprinklers, or ceiling-level ESFR/CMSA sprinklers listed for that commodity and geometry. Plastic goods at 7.5 m under 9.5 m are usually near or beyond the ceiling-only limits, so check ESFR early.

Step 1: Classify

• EN 12845: high-hazard storage (HHS) with the commodity in Category I to IV, and the storage configuration (ST1 to ST6). Plastics often push goods to Category III or IV.
• NFPA 13: Class I to IV or Group A plastics (expanded or unexpanded, cartoned or exposed). Cartoned unexpanded Group A plastic is a common worst case.
• Get the classification from the owner in writing. It drives everything else.

Step 2: Choose the scheme

1. Ceiling-only density/area: look up the design density and area of operation for the class, height and configuration. If the tables allow it, this is the simplest.
2. Ceiling plus in-rack: required above the ceiling-only limits. It is less sensitive to roof height but fixed to the rack layout.
3. ESFR (or NFPA CMSA): ceiling-only for high storage, but strict about roof slope, obstructions, clearance and ceiling height. See ESFR Sprinklers: Quick Rules. Given the architect’s wish, this is often the best fit.

Worked hydraulic illustration

Suppose the applicable table gives 12.5 mm/min over 260 m², with 9 m² per sprinkler. (These values are for illustration only.)

• Flow per sprinkler: 12.5 × 9 = 112.5 L/min.
• Pressure at K80: (112.5 / 80)² = 1.98 bar. At K115: (112.5 / 115)² = 0.96 bar. A larger K-factor lowers the pump head.
• Sprinklers in the area: 260 / 9 = 28.9, so 29. Minimum ceiling flow: 29 × 112.5 = 3 263 L/min, before hydraulic imbalance, in-rack demand and hose allowance.

The real demand is always higher than this minimum, because remote sprinklers run at higher pressure. Use full hydraulic calculation and add hose streams to size the tank and pump: Fire Water Tank Sizing.

Rack layout rules to put on drawings

• Clearance between the sprinkler deflector and the top of storage: at least 1 m under EN 12845 for storage. NFPA 13 requires 457 mm (18 in) for standard sprinklers and 914 mm (36 in) for ESFR and CMSA.
• Flue spaces (NFPA 13, double-row racks): transverse flues about 75 mm (3 in), longitudinal flue about 150 mm (6 in). Solid shelves or flue blockage change the protection scheme.
• A maximum storage height line on the elevation, and “no storage above this line” signage.

Common mistakes

• Designing for a commodity class lower than what will actually be stored.
• Choosing ESFR and then adding ducts and lights that obstruct it.
• Ignoring the roof height. Ceiling-only criteria depend on it, not only on storage height.

Code differences: EN 12845 vs NFPA 13: Key Differences.

For deeper study: Fire Protection Design course.

Check the edition of the code that applies to your project.

[MEP Center Engineering]

Put the FDC where the fire service can see it and connect quickly: on the street or fire-access side, near a hydrant and the appliance parking position, and unobstructed. A rear location 70 m from a hydrant will normally be rejected. The number of 65 mm (2½ in) inlets follows the system demand, and the piping needs a check valve and an automatic drain, with no shut-off valve in the FDC line.

Location

• NFPA 14 and many local codes require the FDC within 30 m (100 ft) of a hydrant connected to an approved supply, unless the authority agrees otherwise. Many fire services set their own distance and side of the building.
• UK and similar practice (BS 9990, Approved Document B): the fire appliance must be able to park within 18 m of the riser inlet, and the inlet must be visible from the appliance.
• Mounting height (NFPA 13/14): 457 to 1 219 mm (18 to 48 in) above finished ground.
• Keep it clear of loading bays, parked vehicles, falling glass and landscaping. Give it a sign stating the system and the zones served.

Inlets

Use the local rule. A common basis is one 65 mm (2½ in) inlet per 946 L/min (250 gpm) of demand. For 1 900 L/min: 1 900 / 946 = 2.0, so a two-inlet (Siamese) connection. High-rise standpipes often need two remotely located FDCs. Confirm the thread or coupling type with the fire service.

Worked check: pressure the fire service must supply

1. Demand at the base of the riser: 1 900 L/min at 5.0 bar.
2. FDC line: DN100 steel (internal diameter about 105 mm), 30 m equivalent length including the check valve and fittings, C = 120. Hazen-Williams: p = 6.05 × 10⁵ × Q^1.85 / (C^1.85 × d^4.87) = 0.0142 bar/m, giving 0.43 bar.
3. Elevation: the riser base is 4 m below the FDC, a gain of 4 × 0.098 = 0.39 bar.
4. Required at the FDC: 5.0 + 0.43 − 0.39 = 5.04 bar at 1 900 L/min. Put this figure on the FDC sign or in the fire strategy if the fire service asks for it.

Valves and piping

• A check valve in the FDC line close to the connection to the system. No shut-off valve between the FDC and the system (NFPA 13).
• An automatic drain (ball drip) between the FDC and the check valve, so the line does not freeze.
• Connect the FDC line on the system side of the main control valve, so the fire service can still supply the system if that valve is closed. Follow the arrangement the code shows for your system type.
• Where a backflow preventer is on the incoming supply, the FDC connects downstream of it. See Backflow Preventers on Fire Lines.

Common mistakes

• Locating the FDC for architectural reasons, out of sight of the access road.
• A gate valve installed in the FDC line “for maintenance”.
• No drain, so the line fills and freezes.

More detail: Fire Department Connection (FDC): Location and Arrangement.

For deeper study: Fire Protection Design course.

Check the edition of the code that applies to your project.

[MEP Center Engineering]

Under NFPA 20 the pump must deliver at least 65 % of rated net pressure at 150 % of rated flow, and churn (zero flow) pressure must not exceed 140 % of rated. Corrected to rated speed, this pump delivers about 1 503 gpm at 66.5 psi, which is above the 65 psi minimum, so it passes at the 150 % point.

Pass criteria (NFPA 20)

Point	Flow	Net pressure requirement
Churn	0	≤ 140 % of rated: ≤ 9.65 bar (140 psi)
Rated	3 785 L/min (1 000 gpm)	≥ 6.9 bar (100 psi)
Overload	5 678 L/min (1 500 gpm)	≥ 65 % of rated: ≥ 4.48 bar (65 psi)

Net pressure is discharge pressure minus suction pressure, both corrected for gauge elevation. Compare the field curve with the certified shop curve, not only with these three minimums. NFPA 25 asks for investigation when annual results fall more than 5 % below the acceptance or nameplate values.

Speed correction (affinity laws)

Diesel engines rarely run at exactly rated speed, so correct the readings: Q_r = Q × (N_r/N), P_r = P × (N_r/N)².

• Flow: 1 520 × 1 760 / 1 780 = 1 503 gpm (5 689 L/min).
• Pressure: 68 × (1 760 / 1 780)² = 68 × 0.9776 = 66.5 psi (4.58 bar).
• Result: 66.5 psi ≥ 65 psi at ≥ 1 500 gpm, so it passes. Do the same correction at churn and at 100 %.

Test arrangement

• A flowmeter loop or a test header with 65 mm (2½ in) hose valves. The header must pass at least 150 % of rated flow. One hose valve per 946 L/min (250 gpm) is the usual basis, so about six for this pump.
• Calibrated gauges on suction and discharge, and a tachometer.
• Suction: at 150 % flow the suction pressure must not fall below the NFPA 20 minimum (0 bar gauge at the flange in general, with a small negative value allowed from a suction tank). Water utilities often set a higher minimum on the main.
• Record motor current and voltage for electric pumps (no overload at 150 %), and engine temperatures for diesel pumps.

EN 12845 note

EN 12845 also requires a stable, non-overloading pump curve, with a drive rated for the maximum power along the curve, and its own flow test arrangements. Two pumps are required for superior or duplicate supplies, and they are separate pumps: a duty pump and an independent standby pump, each with its own driver. Never use a twin-pump unit for this.

Common mistakes

• Judging the pump on discharge pressure without subtracting suction pressure.
• Skipping speed correction on diesel pumps.
• A test header too small to reach 150 %, so the overload point is never tested.

Room layout and test lines: Fire Pump Room Design Checklist.

For deeper study: Fire Protection Design course.

Check the edition of the code that applies to your project.

[Author]

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