Fire pump selection calculator

Why fire pump selection is a curve problem

A fire pump is chosen against a curve, not a single point. The hydraulic calculation gives a demand point: a flow and the pressure needed at the pump discharge. A sound selection keeps that point under the pump curve, keeps the curve within the standard's shape limits and keeps shutoff pressure below the rating of the system components. This calculator checks all three and plots them.

The standards are never mixed: NFPA 20 mode applies only NFPA 20-2025 clauses, TS EN 12845 mode only TS EN 12845+A2:2026 clauses. All pressures are net: discharge-flange pressure minus suction-flange pressure (NFPA 20 clause 3.3.49.3).

Selection rules under NFPA 20

• 4.10.1: the greatest single demand of any system connected to the pump may not exceed 150 % of the pump's rated flow.
• A.4.10 (explanatory): use below 90 % of rated flow is not recommended, and above 140 % suction conditions can affect performance; hence the 90–150 % selection range on the plot.
• 6.2.1: a centrifugal pump must deliver at least 65 % of total rated head at 150 % of rated flow.
• 6.2.2: shutoff head may not exceed 140 % of rated head. The standard writes both ratios in terms of head; the calculator uses the ratio of net pressures, so it ignores the velocity-head difference between suction and discharge (A.3.3.49.3 notes that this is usually small). Near a limit, check against the maker's total head curve. A.6.2 explains that listed pumps range from 101 % to 140 % at shutoff and from 65 % to just below rated at 150 % flow.
• 4.10.2: the rated flow must be one of the values in Table 4.10.2 and the pump must be rated at a net pressure of 40 psi (2.7 bar) or more. Under NFPA 20 clause 1.6.4 metric values are converted and rounded from the primary value, so the calculator decides on 40 psi = 2.758 bar in both unit displays. Pumps above 5000 gpm (18,925 L/min) need individual review by the authority having jurisdiction or a listing laboratory (4.10.3).
• 4.7.7.1: net churn pressure plus the maximum static suction pressure, adjusted for elevation, may not exceed the pressure rating of the system components. A relief valve may not be used to meet this (4.7.7.2). For the elevation adjustment the calculator asks for the drop to the lowest component below the pump and adds it with the 0.433 psi/ft relation of A.3.3.30.
• 6.1.1.4 and 4.16.3: the centrifugal pumps of that chapter shall not be used where a static suction lift is required; with the supply below the pump and too little pressure, a vertical shaft turbine pump is needed (7.1.1). With positive suction the flange pressure must stay at 0 psi or more at 150 % flow (4.16.3.1); where the tank level is at or above the pump centre line, down to −3 psi (−0.2 bar) at 150 % flow is accepted (4.16.3.2).

Table 4.10.2 — centrifugal fire pump rated capacities

The L/min values are the standard's soft conversions (A.4.10.2); the calculator itself uses gpm × 3.7854.

Selection under TS EN 12845

TS EN 12845+A2:2026 refers pump-set performance to EN 17451 (clause 10.1), so EN mode applies no curve-shape limits like NFPA's 140 %/65 %. It checks the standard's own clauses:

• 10.7.3 (fully calculated systems): the pump set must meet the pressure and flow demand of the sprinkler system. As in Figure 7, the pump curve must stay above both the most unfavourable area demand point and the most favourable area Q_max point, so the tool asks for both. Per Table 14 the rating is the highest flow and pressure required for the most favourable area, and available NPSH must exceed required by at least 1 m at that flow (10.6.2.1).
• 10.7.1 and Table 16 (pre-calculated LH/OH fed from a tank): a nominal point and one or two characteristic points per hazard class and sprinkler height. Pressures are measured at the control valve set (Note 1), so the tool asks for the loss between pump outlet and valve set. Row labels are printed as merged cells, e.g. "OH1 dry or alternate" with "OH2 wet or pre-action". Interpretation: the tool reads from the table layout that both classes share the same three height rows; the standard does not state this separately.
• 8.2.1: water pressure may not exceed 12 bar at equipment connections such as sprinklers and alarm valves; pump outlets join that list where the height difference between the highest and lowest sprinklers does not exceed 45 m (8.2.1.2 a), including any driver speed rise at closed valve. In high-rise systems over 45 m, pump outlets may exceed 12 bar where the equipment is fit for it (8.2.2).
• 10.6 suction conditions: positive suction head is used wherever possible (at least two thirds of the effective tank capacity above the pump centre line, and the centre line no more than 2 m above the low water level; 10.6.1). Suction lift is used only where that is not practicable: no more than 3.2 m from the low water level to the pump centre line, a suction pipe of at least 80 mm with no more than 1.5 m/s at maximum demand flow, a foot valve and automatic priming (10.6.2.3, 10.6.2.4). With positive head the suction pipe is at least 65 mm with no more than 1.8 m/s (10.6.2.2).
• 10.7.5.2: the first pump starts no lower than 0.8 of closed-valve pressure, the second no lower than 0.6 of it; both are reported.

How is it calculated?

Three points come from the maker's curve: shutoff, rated and 150 % flow. A quadratic through them is used only to read intermediate values; it does not replace the certified curve. All pressures read from it are therefore shown as approximate (≈); where the margin is under 5 % of the required pressure the verdict becomes CONDITIONAL and asks you to confirm the pressure at that flow on the maker's certified curve. The 5 % is a tool safety threshold, not a value from the standard.

The power figure is an estimate from the peak hydraulic power between 0 and 150 % flow. NFPA 20 clauses 4.7.6 and 4.11.3 require the driver to cover the maximum load at any flow, including beyond 150 %, so size the driver from the maker's data.

Worked example (NFPA 20)

The hydraulic calculation needs 8.0 bar at the pump discharge at 2,000 L/min; lowest suction pressure is 0.3 bar and static suction with the tank full is 0.6 bar. Net required is 8.0 − 0.3 = 7.70 bar. Ratings in Table 4.10.2 that fit the 90–150 % range are 400, 450 and 500 gpm. Trying a 500 gpm (1,893 L/min) pump rated at 8.0 bar, with 125 % shutoff (10.0 bar) and 72 % at 150 % flow (5.76 bar):

• Demand/rated = 2,000/1,893 = 106 % → 4.10.1 met.
• Curve coefficients c = −1.653, b = −0.347; H(1.057) ≈ 7.79 bar ≥ 7.70 bar → the demand sits under the approximate curve with only 0.09 bar to spare (1.2 % of the requirement). Being under 5 %, the verdict is CONDITIONAL.
• Shutoff 125 % ≤ 140 % and 72 % ≥ 65 % at 150 % flow → 6.2.1 and 6.2.2 met.
• Maximum pressure 10.0 + 0.6 = 10.6 bar (no components below the pump); 4.7.7.1 is met for components rated at 12.1 bar.
• Hydraulic power is 25.2 kW at rated and 27.8 kW at peak; at 70 % efficiency the estimated shaft power is about 39.7 kW.

Common mistakes

• Treating gross pressure as net: forgetting to subtract suction pressure (at the lowest water level) undersizes the pump.
• Forgetting shutoff pressure: a steep curve meets demand easily, but churn plus maximum suction may exceed component ratings.
• Choosing a rating far above the demand: a pump whose demand is below 90 % of rated runs where A.4.10 advises against.
• Mixing standards: NFPA's 65 % rule on an EN 12845 project, or Table 16 on an NFPA project, gives the wrong answer. In Türkiye the fire regulation (BYKHY) ties sprinkler design to TS EN 12845.
• Using the curve approximation instead of the maker's curve: the three-point curve is for preliminary selection; the final check is made on the certified test curve.

Frequently Asked Questions

How far can the demand exceed rated flow under NFPA 20?

Under NFPA 20-2025 clause 4.10.1 the greatest single demand of any connected system may not exceed 150 % of rated flow. A.4.10 also advises against applying the pump below 90 % of rated flow, so the practical selection range is 90–150 %.

What is the limit on shutoff (churn) pressure?

NFPA 20 clause 6.2.2 limits shutoff pressure to 140 % of rated pressure. Clause 4.7.7.1 also requires net churn pressure plus maximum static suction pressure to stay within the component pressure rating, and TS EN 12845+A2 clauses 8.2.1 and 8.2.1.2 a) set a 12 bar limit at pump outlets, including the closed-valve condition and any driver speed rise, where the height difference between the highest and lowest sprinklers does not exceed 45 m; 8.2.2 covers high-rise systems above 45 m.

How much pressure must the pump give at 150 % flow?

At least 65 % of total rated head, per NFPA 20 clause 6.2.1; the calculator checks it as a ratio of net pressures. TS EN 12845+A2 gives no such curve-shape rule in its own text; it refers pump-set performance to EN 17451 (clause 10.1).

Can any rated flow be chosen under NFPA 20?

No. Clause 4.10.2 requires one of the ratings in Table 4.10.2 (21 values from 25 gpm to 5000 gpm) and a rating at a net pressure of 40 psi (2.7 bar) or more. The L/min figures in the table are soft conversions; under clause 1.6.4 the primary value is 40 psi (2.758 bar), which the calculator uses for the decision.

Is the power figure good enough to size the motor?

No, it is an estimate. Hydraulic power is Q×H/600 converted to shaft power with the efficiency you enter. NFPA 20 clause 4.11.3 requires the nameplate to show peak power demand including flows beyond 150 %; size the motor or engine from the maker's certified data.

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