Water Supply · Calculation

How Do You Calculate Available Fire Flow at 20 psi?

Reviewed by a licensed Professional Engineer · NY, NJ, PA, MD, DE · Last reviewed · Code basis: NFPA 291 (2022) Ch 4; NFPA 13 (2022) Ch 28; AWWA Manual M17

General guidance, not a code determination for any project. Confirm the adopted code and local amendments with the authority having jurisdiction.

Available fire flow at 20 psi is calculated by taking the flow measured during a hydrant flow test and scaling it by the ratio of two pressure drops raised to the 0.54 power: the drop from static pressure to 20 psi, divided by the drop from static pressure to the residual pressure read during the test. The result is the flow, in gallons per minute (gpm), that the main could deliver at the gauge hydrant while holding 20 psi, and it is the single number that water supply reports, sprinkler calculations and fire pump selections are built on.

The method comes from National Fire Protection Association (NFPA) 291, Recommended Practice for Water Flow Testing and Marking of Hydrants (2022), Chapter 4, which sets out both the pitot discharge formula that turns a gauge reading into flow and the projection to 20 psi. The American Water Works Association (AWWA) Manual M17 is the parallel reference for hydrant practice on the purveyor’s side. This article covers the inputs, the two formulas, a worked example computed step by step, the errors that produce a wrong number and the cases where the simple projection is not enough. How the test itself is run in the street is a separate question, covered in DGA’s article on how a hydrant flow test is run and read under NFPA 291.

At a Glance: Available Fire Flow at 20 psi

  • What is calculated: the flow in gpm a water main can deliver at the test location while the residual pressure in the main stays at 20 psi, projected from one measured test point.
  • Method in one line: QR = QF × ((Pstatic − 20) ÷ (Pstatic − Presidual))0.54, after QF is computed from the pitot reading with Q = 29.84 × c × d2 × √p for each flowing outlet.
  • Inputs needed: static pressure, residual pressure, pitot pressure at each flowing outlet, outlet diameter, outlet coefficient, number of outlets flowing, and the elevation difference between the gauge hydrant and the system being served.
  • Code and section: NFPA 291 (2022) Chapter 4 for the method; NFPA 13, Standard for the Installation of Sprinkler Systems (2022), Chapter 28 for the hydraulic calculation that uses the result; AWWA Manual M17 for hydrant practice.
  • Typical result range (units): 1,000 to 3,000 gpm at 20 psi on an 8 in. to 12 in. looped municipal main; under 1,000 gpm on a 6 in. dead-end main; the worked example below lands at 2,232 gpm.
  • Most common input error: using the wrong outlet coefficient, which moves the measured flow by 12.5 to 29 percent and carries straight through to the projected flow.

What Inputs Does the Available Fire Flow Calculation Need?

Seven numbers go into the calculation, and every one of them comes from the field test except the outlet coefficient, which comes from looking at the shape of the hydrant outlet. The table lists each input, its units and where it comes from.

InputUnitsSource
Static pressure, PstaticpsiGauge on the residual (gauge) hydrant before any outlet is opened; read with the system at normal demand
Residual pressure, PresidualpsiSame gauge on the same hydrant while the flow hydrant is discharging and the stream has steadied
Pitot pressure, ppsiPitot tube held in the center of the stream at each flowing outlet, read at the same moment as the residual
Outlet diameter, din.Measured inside diameter of each flowing outlet; nominal 2.5 in. for a hose outlet, nominal 4 in. or 4.5 in. for a pumper outlet
Outlet coefficient, cnoneShape of the outlet where it meets the barrel: 0.9 smooth and rounded, 0.8 square and sharp, 0.7 projecting into the barrel (NFPA 291 (2022) Chapter 4)
Number of outlets flowingcountField record; each outlet gets its own pitot reading and its own flow, and the flows are summed
Elevation differenceft, converted at 0.433 psi per ftSurvey or site plan: the gauge hydrant elevation against the base of riser or the pump suction flange of the system being served

The static and residual readings have to come from the same gauge on the same hydrant, and that hydrant must not be the one flowing. A pitot reading taken at the gauge hydrant, or a residual read at the flow hydrant, produces a flow and a projection that do not describe the main. The gauge hydrant also has to sit between the supply and the flow hydrant, so the residual reading carries the friction loss the building’s service will see.

What Is the Method for Projecting the Test to 20 psi?

The method has two formulas and five steps. The first formula converts each pitot reading into a flow; the second scales the total measured flow to the flow that would exist at 20 psi residual. Both are standard hydraulics, and NFPA 291 (2022) Chapter 4 is the source of the method as applied to hydrant tests.

Figure 1: The Two Formulas Behind Available Fire Flow

Flow from a pitot reading: Q = 29.84 × c × d2 × √p, where Q is the flow from one outlet in gpm, c is the outlet coefficient (0.9, 0.8 or 0.7), d is the outlet diameter in inches and p is the pitot pressure in psi. The total test flow QF is the sum of Q for every flowing outlet.

Projection to 20 psi: QR = QF × (hr0.54 ÷ hf0.54), where hr = Pstatic − 20 psi (the drop the main is allowed to take) and hf = Pstatic − Presidual (the drop the test produced). QR is the available flow at 20 psi residual.

Elevation correction: where the system served sits above or below the gauge hydrant, project to 20 psi from the hydrant readings first, then shift the whole supply curve down or up by 0.433 psi for every foot of difference, so the curve is expressed at the base of riser while the 20 psi floor stays a pressure in the main.

With a 2.5 in. outlet, 29.84 × d2 is a constant 186.5, so Q = 186.5 × c × √p: a 0.9 outlet at 38 psi pitot gives 1,034 gpm and the same reading on a 0.8 outlet gives 920 gpm. The 0.54 exponent is the inverse of the 1.85 exponent in the Hazen-Williams relationship between flow and friction loss.

Figure 2: From Pitot Reading to Available Flow in Five Steps

1. Convert each pitot reading to flowApply Q = 29.84 × c × d2 × √p to each flowing outlet with its own coefficient and diameter.
2. Sum the outletsAdd the flows from every outlet that was open when the residual was read. That sum is the test flow QF.
3. Compute the 0.54 power ratioTake hr = static − 20 and hf = static − residual, divide, and raise the quotient to the 0.54 power.
4. Project to 20 psiMultiply QF by the ratio from step 3. The product is the available flow at 20 psi, QR.
5. Correct for elevation and plotShift all three points, static, test point and 20 psi point, by 0.433 psi per ft to the base of riser, then plot them on N1.85 graph paper or tabulate them.
Steps 1 and 2 produce the measured flow, steps 3 and 4 the projected flow, and step 5 the supply curve that NFPA 13 (2022) Chapter 28 hydraulic calculations and fire pump selections read from. On N1.85 paper the three points fall on one straight line.

The 20 psi residual is the reporting point because it is the lowest pressure most water purveyors will allow a main to be drawn down to while fire flow is taken. A positive pressure in the main keeps groundwater and cross-connections from being pulled into the drinking water supply and keeps the pipe from collapsing under external load. Many purveyors write 20 psi into their rules as the minimum a fire pump may draw the main down to, and some set a higher floor, which is why the projection is reported at 20 psi and then checked against the purveyor’s number on each project.

The fifth step matters more than it looks. A sprinkler system calculated at the base of riser 12 ft above the gauge hydrant sees 5.2 psi less static and residual than the hydrant gauge read, and the whole supply curve shifts down by 5.2 psi, while the flow available at 20 psi in the main does not change. The same correction runs the other way when the building sits below the street.

Worked Example: 74 psi Static, 61 psi Residual, One 2.5 in. Outlet

A flow test on a 10-inch municipal main in front of a 2-story commercial building reads 74 psi static at the gauge hydrant. With one 2.5-inch outlet open on the flow hydrant downstream, the pitot reads 38 psi and the residual at the gauge hydrant settles at 61 psi. The flow hydrant outlet is smooth and rounded where it meets the barrel, so the coefficient is 0.9. The gauge hydrant and the base of riser are at the same elevation, so no correction is applied.

StepCalculationResult
1. Square the diameterd2 = 2.5 × 2.56.25
2. Square root of the pitot pressure√386.164
3. Test flow, QF29.84 × 0.9 × 6.25 × 6.164 = 167.85 × 6.1641,034.7 gpm (1,034 gpm rounded)
4. Allowed drop, hr74 − 2054 psi
5. Test drop, hf74 − 6113 psi (17.6 percent of static)
6. Raise each drop to 0.54540.54 = 8.620; 130.54 = 3.995Ratio 2.158
7. Available flow at 20 psi, QR1,034.7 × 2.1582,232 gpm

The main can deliver about 2,230 gpm at the gauge hydrant while holding 20 psi. The three points that define the supply curve are 74 psi at 0 gpm, 61 psi at 1,034 gpm and 20 psi at 2,232 gpm. Those are the numbers a sprinkler designer plots the demand point against and a fire pump designer reads suction pressure from; DGA’s article on how to plot sprinkler demand against the supply curve takes the curve from here.

The arithmetic can be checked in a few seconds with the hydrant flow test calculator, which accepts the same seven inputs, sums multiple outlets and returns the projected flow and the plotted curve. On a test with two outlets open, each outlet gets its own pitot reading and its own line in step 3, and the two flows are added before step 6.

Why Does a Small Residual Drop Make the Projection Unreliable?

Suppose the same test had been run on a day when the flow hydrant was partly obstructed and the residual only fell to 71 psi, a drop of 3 psi, or 4.1 percent of static. The test drop hf is now 3 psi instead of 13 psi, and the ratio (54 ÷ 3)0.54 = 180.54 = 4.763. The projected flow becomes 1,034.7 × 4.763 = 4,928 gpm, more than double the first result from a test on the same main.

The second number is not evidence that the main got stronger; it is evidence that the projection has become hypersensitive to gauge error. A 1 psi error in the residual reading moves the first projection from 2,232 gpm to between 2,145 gpm and 2,331 gpm, a swing of about 190 gpm. The same 1 psi error on the 3 psi drop moves the second projection to between 4,219 gpm and 6,134 gpm, a swing of about 1,900 gpm, which is larger than the entire first projection.

That is why NFPA 291 (2022) Chapter 4 expects the residual to drop by a meaningful percentage of the static, at least 10 percent in the 2022 edition (the 2019 edition said 25 percent), and why AWWA Manual M17 practice looks for a drop of at least 10 psi. A test that does not reach the drop is not wrong, but its projection is not usable for design: the fix is to open a second outlet or a second hydrant and run the test again. The table and the figure below show both cases side by side.

QuantityCase A: residual 61 psiCase B: residual 71 psi
Test drop, hf13 psi (17.6 percent)3 psi (4.1 percent)
Ratio (54 ÷ hf)0.542.1584.763
Projected flow at 20 psi2,232 gpm4,928 gpm
Projected flow with residual read 1 psi high or low2,145 to 2,331 gpm4,219 to 6,134 gpm
Projected flow as a multiple of the test flow2.2 times4.8 times
Usable for designYes, with the projection past 2,000 gpm treated with cautionNo; retest with more outlets flowing

Figure 3: Supply Curve from the Worked Example on N1.85 Paper

020406080 05001,0001,5002,0002,400 Flow, gpm (1.85 power scale) Pressure, psi Static 74 psi at 0 gpm Test point: 1,034 gpm at 61 psi Case B: 1,034 gpm at 71 psi (3 psi drop) 2,232 gpm available at 20 psi Case B still near 60 psi at 2,400 gpm 20 psi residual floor Case A supply line: 74 psi static, 61 psi at 1,034 gpm Case B, dashed: same test flow, residual 71 psi (projection not usable)
On a 1.85 power flow axis the Case A supply line runs straight from 74 psi at 0 gpm through 61 psi at 1,034 gpm to 20 psi at 2,232 gpm. The Case B line, built from a 3 psi drop, is so flat that it is still near 60 psi at 2,400 gpm and would not reach 20 psi until about 4,900 gpm, a projection that a 1 psi gauge error moves by about 1,900 gpm.

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What Are the Common Errors in the Available Fire Flow Calculation?

Most wrong projections trace to one of eight errors, and the first three account for the majority DGA’s engineers find in review. The table lists each error, how it shows up in the numbers and the fix.

ErrorHow it shows upFix
Wrong outlet coefficientA 0.9 coefficient applied to a square-edged outlet overstates the flow by 12.5 percent; 0.9 applied to a projecting outlet overstates it by 29 percent, and the error carries through to the projection unchangedLook at the inside of the outlet where it meets the barrel and record its shape on the field sheet; use 0.9, 0.8 or 0.7 per NFPA 291 (2022) Chapter 4
Forgetting to square the diameterQ = 29.84 × 0.9 × 2.5 × √38 gives 414 gpm instead of 1,034 gpm; the supply looks 60 percent weaker than it isWrite d2 as its own line (6.25 for a 2.5 in. outlet, 16 for a 4 in. outlet) before multiplying
Pitot read at the gauge hydrantThe field sheet shows a pitot reading and a residual from the same hydrant; the flow and the pressure describe different conditions and the projection is meaninglessStatic and residual come from the gauge hydrant; the pitot comes from the flow hydrant; never from the same one
Mixing psi and feet of headA 12 ft elevation difference entered as 12 psi instead of 5.2 psi shifts the whole curve by 6.8 psiConvert feet to psi at 0.433 psi per ft (or psi to feet at 2.31 ft per psi) before any addition or subtraction
Projecting past the test rangeA 600 gpm test projected to 2,400 gpm at 20 psi, 4 times the test flow, with no second point to confirm the slopeTreat a projection beyond about 1.5 to 2 times the test flow as provisional and retest with more outlets if the design depends on it
Static read with the flow hydrant openThe “static” is really a residual at some unknown flow; the drop looks small and the projection runs highRead and record the static before any outlet is cracked, then open the flow hydrant and wait for the stream to steady before the residual
Ignoring elevationA base of riser 30 ft above the gauge hydrant is credited with 13 psi it does not have; the sprinkler calculation passes on paper and fails in the fieldRecord the elevation of the gauge hydrant and the base of riser and apply the 0.433 psi per ft correction before plotting
Summing pitot pressures instead of flowsTwo outlets at 20 psi pitot each are entered as one outlet at 40 psi; the flow is understated by 29 percentCompute the flow for each outlet separately and add the flows, never the pressures

The coefficient error is the one that survives longest because it is invisible in the arithmetic. A spreadsheet that takes 0.9 as the default will produce a plausible number for every test, and only a photograph of the outlet or a field note about its shape reveals that 0.8 was the right value. Whether the test itself was set up correctly, from hydrant selection to gauge calibration, is covered in the hydrant flow test article linked above.

Where Does the Simple Projection Stop and a Professional Engineer’s Calculation Begin?

The 0.54 power projection assumes that the main behaves like a single pipe with a fixed friction characteristic between the source and the gauge hydrant, and that the flow test sampled that behavior. Several situations break that assumption, and in each the result needs a licensed Professional Engineer (PE) to build the supply curve from more than one test point or from a hydraulic model of the distribution system.

SituationWhy the simple projection is not enough
Looped or gridded main with multiple feedsAs flow rises, more of the grid contributes and the effective friction characteristic changes; a single test point does not fix the slope, and the projection can run either high or low
Boosted or pumped distributionA purveyor’s booster station or elevated tank may switch state during a fire; the supply curve has to be built for the condition the purveyor will guarantee, not the one found on test day
Test hydrant on a different pressure zoneA pressure-reducing valve or zone boundary between the test hydrants and the building means the test describes the wrong main; the test is moved or the zone hydraulics are modeled
Purveyor floor above 20 psiWhere the purveyor limits drawdown to 25 psi or 30 psi, the available flow is recalculated at that residual with the same formula and hr = static minus the purveyor’s floor; the 20 psi number is not the design number
Projected flow more than 1.5 to 2 times the test flowThe worked example projects to 2.2 times its test flow; a design that depends on flows near 2,000 gpm from that main would be confirmed with a two-outlet retest before the pump is sized
Fire pump suction from the mainThe pump pulls the main down at 150 percent of rated flow; the supply curve has to show positive pressure there and satisfy the purveyor’s floor, which is a check in DGA’s article on how to size a fire pump

In each case the formula still does the arithmetic; what changes is which pressures and flows go into it and how many test points are needed to trust the slope. Whether the projected supply is enough on its own or the building needs a pump is the question DGA’s article on how engineers decide whether a fire pump is needed answers from the same curve.

Does the Calculation Change by Jurisdiction?

No. The pitot formula, the 0.54 power projection and the 20 psi reporting point are the same in New York State, New York City, New Jersey, Pennsylvania, Maryland and Delaware, because every jurisdiction adopts an edition of NFPA 13 whose hydraulic calculation chapter (Chapter 27 in 2019, Chapter 28 in 2022 and 2025) reads the supply from a flow test run and projected the same way. What varies is the water purveyor’s rule on how far a main may be drawn down and whether a fire pump may take suction from it: the New York City Department of Environmental Protection (DEP), the investor-owned and municipal purveyors in New Jersey, and the Washington Suburban Sanitary Commission (WSSC) in Maryland each publish their own conditions. DGA checks the purveyor’s minimum residual on each project and reports the available flow at 20 psi and, where the purveyor sets a higher floor, at that floor as well.

Key Takeaway: Available fire flow at 20 psi is the test flow multiplied by ((static − 20) ÷ (static − residual))0.54, with the test flow computed from each pitot reading as 29.84 × c × d2 × √p. On a test reading 74 psi static, 61 psi residual and 1,034 gpm flowing, the main delivers 2,232 gpm at 20 psi. The number is only as good as the drop that produced it: a 3 psi drop on the same main projects to 4,928 gpm and is not usable for design.

Frequently Asked Questions

Why is available fire flow reported at 20 psi?

Because 20 psi is the residual pressure most water purveyors treat as the lowest a public main may be drawn down to while fire flow is taken. A positive pressure in the main keeps contaminated water from being drawn in through cross-connections and keeps the pipe from collapsing, and many purveyors write 20 psi into their rules as the minimum a fire pump may pull the main down to.

Can the projected flow at 20 psi be larger than the flow measured during the test?

Yes. The projection extends the test along the main’s friction characteristic to a larger drop, so the projected flow is always larger than the test flow when the residual was above 20 psi. In the worked example a 1,034 gpm test projects to 2,232 gpm, and the projection is treated as provisional once it exceeds about 1.5 to 2 times the test flow.

Does the outlet coefficient really change the result that much?

Yes. The coefficient multiplies the flow directly, so a 0.9 coefficient applied to an outlet that should have been 0.8 overstates the test flow by 12.5 percent, and the same 12.5 percent carries through to the projected flow at 20 psi. The coefficient comes from the shape of the outlet where it meets the hydrant barrel, per NFPA 291 (2022) Chapter 4.

What if the purveyor does not allow the main to be drawn down to 20 psi?

Conditional: the same formula is used with the allowed drop equal to the static pressure minus the purveyor’s floor instead of static minus 20 psi. A purveyor that limits drawdown to 30 psi on the worked example main would reduce the available flow from 2,232 gpm at 20 psi to about 2,000 gpm at 30 psi, and that lower number governs the design.

This article is general information, not a water supply analysis for any project. Purveyor rules, code editions and local amendments change; confirm the adopted edition of NFPA 13 and NFPA 291 and the purveyor’s minimum residual pressure with the authority having jurisdiction before a system or pump is designed.

Need the Available Fire Flow Confirmed for a Design?

Send the hydrant flow test sheet, the site plan with elevations and the purveyor’s name. DGA computes the available flow at 20 psi and at the purveyor’s floor, builds the supply curve and tells the design team whether the main carries the demand or a pump is needed. Project briefs get confirmation and preliminary code feedback within one business day.

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