Lesson plan Grade 12 Math

Bramleigh Hydrant Flow Test: Flow as a Square-Root Model

A 55-minute grade 12 math lesson in which students model hydrant discharge as a square-root function of pitot pressure and rate a loop against a needed fire flow.

Grade: Grade 12
Subject: Math
Material: Lesson plan
Time: 55 min
Prep: No prep
Materials: Scientific calculator, Printed Student Handout

Standards

HSN-Q.A.1 Common Core Math

Use units as a way to understand problems and to guide the solution of multi-step problems; choose and interpret units consistently in formulas; choose and interpret the scale and the origin in graphs and data displays.

How this material aligns: Handout questions 1, 3 and 8 carry inches, psi and gallons per minute through the hydrant discharge formula and the residual-pressure extrapolation without changing units midstream.

All materials for HSN-Q.A.1 →

HSA-CED.A.1 Common Core Math

Create equations and inequalities in one variable and use them to solve problems. Include equations arising from linear and quadratic functions, and simple rational and exponential functions.

How this material aligns: Questions 5 and 10 require students to write a one-variable equation from the square-root discharge model and solve it for an unknown pressure and an unknown test flow.

All materials for HSA-CED.A.1 →

HSN-Q.A.3 Common Core Math

Choose a level of accuracy appropriate to limitations on measurement when reporting quantities.

How this material aligns: Question 11 asks how much the reported discharge shifts when the pitot gauge can only be read to the nearest 2 psi, matching reported accuracy to instrument precision.

All materials for HSN-Q.A.3 →

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This material is in the public domain (CC0) and may be copied, edited, and shared freely, without attribution. Updated 2026-09-04.

Objectives

Students will model hydrant discharge as a square-root function of pitot pressure and find its constant.
Students will solve the discharge equation for the pressure or the test flow that meets a stated target.
Students will extrapolate a flow test to a 20 psi residual and compare the result with a needed fire flow.

Vocabulary

Key words: static pressure, residual pressure, pitot pressure, discharge coefficient, available fire flow

Teacher overview

A hydrant flow test tells a water utility how much water a section of main can actually deliver, and the arithmetic behind it rests on a square-root relationship rather than a proportional one. In this lesson students work the flow test that operators in the fictional town of Bramleigh ran on the Caldermoor Street loop, build the discharge model Q equals k times the square root of P, solve that model backward for an unknown pressure, and extrapolate the measured flow to a 20 psi residual. The misconception worth naming before students start is the assumption that doubling the pressure at an open outlet doubles the flow through it; because discharge varies with the square root of the pitot reading, doubling the pressure raises the flow by a factor of about 1.41, and students who miss that will consistently overstate what a main can deliver.

Procedure

Hook (5 min): Tell the class that a fire marshal wants to know whether a loop of water main can supply a proposed warehouse, and that the only way to find out is to open a hydrant and measure. Ask students to predict what happens to the flow out of a hydrant outlet if the pressure pushing on it doubles. Record the predictions without comment; most classes predict that the flow doubles.

Direct instruction (12 min): Define static pressure, residual pressure, and pitot pressure using a sketch of two hydrants on one main. Present the discharge formula, in which Q in gallons per minute equals 29.83 times the discharge coefficient c, times the square of the outlet diameter d in inches, times the square root of the pitot pressure P in psi. Model it once with numbers the handout does not use: for a fictional test with c equal to 0.80, d equal to 2.5 inches, and a pitot reading of 25 psi, the constant is 149.15 and the discharge is 149.15 times 5, or about 746 gpm. Then present the extrapolation to a 20 psi residual, in which the measured discharge is multiplied by the ratio of the two pressure drops raised to the 0.54 power, and model it with a static reading of 74 psi and a residual of 62 psi, which gives a drop ratio of 4.5 and about 1,680 gpm available.

Guided practice (10 min): Read the Bramleigh test record together and have students label which number is static, which is residual, and which is the pitot reading. Before any computation, ask for a thumbs-up or thumbs-down on the claim that the flow available at a 20 psi residual will be larger than the discharge measured during the test. Move on once about four students in five agree and can give the reason, which is that the loop can be drawn down further than the test drew it; if fewer are correct, re-draw the two pressure drops on the board and re-poll.

Independent practice (21 min): Students complete the Student Handout individually or in pairs. Circulate and watch for two errors: squaring the pressure instead of taking its square root, and entering the 0.54 exponent as a multiplier. Ask any student who finishes early to check question 10 by substituting the answer back into the available-flow formula.

Closure and exit ticket (7 min): Each student writes one sentence explaining why raising the pitot reading from 16 psi to 36 psi does not raise the discharge by the same factor as the pressure. Collect the cards and scan for a reference to the square root rather than to proportionality before dismissal.

Bramleigh Hydrant Flow Test: Flow as a Square-Root Model — Grade 12 Math

Name: ____________________________ Date: ______________

Student Handout

The town of Bramleigh is fictional, and so is every reading below. Operators from the Bramleigh Water Department ran a flow test on the Caldermoor Street loop, which would serve a proposed warehouse.

Test record: static pressure at the test hydrant was 68 psi before any hydrant was opened; the flow hydrant had one 2.5-inch outlet with a discharge coefficient c of 0.90; the pitot pressure at the open outlet read 16 psi; the residual pressure at the test hydrant while water flowed read 52 psi; and the engine company needs 1,500 gpm at the site.

Two formulas, with Q in gallons per minute, d in inches, and every pressure in psi:

Discharge from an open outlet: Q equals 29.83 times c times d squared times the square root of P.
Available flow at a 20 psi residual: that discharge multiplied by the quantity (static pressure minus 20) divided by (static pressure minus the residual pressure), all raised to the 0.54 power.

Using the discharge formula with c equal to 0.90, d equal to 2.5 inches, and a pitot pressure of 16 psi, find the discharge from the flow hydrant. Round to the nearest gpm.
For this outlet the discharge formula collapses to Q equals k times the square root of P. Find k, using c equal to 0.90 and d equal to 2.5 inches. Round to two decimal places.
Using your value of k, predict the discharge from the same outlet if the pitot pressure were 36 psi. Round to the nearest gpm.
Going from a 16 psi pitot reading to a 36 psi reading multiplies the pressure by 2.25. By what factor does the discharge change?
Find the pitot pressure that would produce a discharge of 900 gpm from this outlet, using your k from question 2. Round to the nearest tenth of a psi.
Find the pressure drop the test produced, using the static pressure of 68 psi and the residual pressure of 52 psi.
Find the drop from the static pressure of 68 psi down to a 20 psi residual.
Using the available-flow formula with your question 1 discharge, the drop from question 7, and the drop from question 6, find the flow this loop can deliver at a 20 psi residual. Round to the nearest gpm.
The engine company needs 1,500 gpm. Using your answer to question 8, how many gpm short is the loop?
Suppose the loop had to deliver exactly 1,500 gpm at a 20 psi residual, with the same 16 psi drop from a static 68 psi. What discharge would the flow test have had to measure? Round to the nearest gpm.
The pitot gauge is marked in 2 psi steps, so the operator's 16 psi reading could have been 18 psi. Using the same k, how many more gallons per minute would an 18 psi reading give than the 16 psi reading? Round each discharge to the nearest gpm before subtracting.

From Water for Schools (waterforschools.org). This material is in the public domain (CC0) and may be copied, edited, and shared freely.

Answer Key

Answer: 671 gpm.
For the fictional Bramleigh test, 29.83 times 0.90 times 6.25 is 167.79, and the square root of 16 is 4, so the discharge is 167.79 times 4, which is 671.2, or 671 gpm.
Answer: 167.79.
The constant collects everything that does not change: 29.83 times 0.90 times 2.5 squared, which is 29.83 times 0.90 times 6.25, or 167.79 gallons per minute for each unit of the square root of the pressure.
Answer: 1,007 gpm.
The square root of 36 is 6, so the discharge is 167.79 times 6, which is 1,006.8, or 1,007 gpm.
Answer: 1.5.
The discharge changes by the square root of the pressure ratio, and the square root of 2.25 is 1.5. Checking against questions 1 and 3, 1,007 divided by 671 is also 1.5.
Answer: 28.8 psi.
Setting 900 equal to 167.79 times the square root of P gives a square root of 5.364, and squaring that gives 28.77, or about 28.8 psi.
Answer: 16 psi.
The drop during the test is the static 68 psi minus the residual 52 psi, which is 16 psi.
Answer: 48 psi.
The drop down to a 20 psi residual is 68 psi minus 20 psi, which is 48 psi.
Answer: 1,214 gpm.
The pressure-drop ratio is 48 divided by 16, or 3, and 3 raised to the 0.54 power is about 1.810. The available flow is 671 times 1.810, which is about 1,214 gpm.
Answer: 286 gpm.
The loop delivers about 1,214 gpm at a 20 psi residual, and 1,500 minus 1,214 is 286 gpm short of what the engine company needs.
Answer: 829 gpm.
The same drop ratio still multiplies the test discharge by about 1.810, so the required test discharge is 1,500 divided by 1.810, which is about 829 gpm. Substituting back, 829 times 1.810 returns roughly 1,500 gpm.
Answer: 41 gpm.
An 18 psi reading gives 167.79 times the square root of 18, which is 167.79 times 4.243, or 712 gpm. That is 712 minus 671, or 41 gpm more than the 16 psi reading produced.

Differentiation

Support

Give these students the constant k of 167.79 already computed, so question 1 becomes a single square root and a single product, and provide a filled-in line showing the two pressure drops for the fictional Bramleigh test, 48 psi and 16 psi, labeled on a sketch of the main. Allow them to stop after question 9.

Extension

Ask students to write the available flow for the fictional Caldermoor Street loop as a function of the static pressure alone, holding the 16 psi test drop and the 671 gpm test discharge fixed, then to find what static pressure would have made the loop meet the 1,500 gpm requirement.

ELL supports

Pre-teach "static," "residual," and "pitot" with a labeled two-hydrant diagram of the fictional loop, marking where each gauge sits and what it reads during the test. Provide the sentence frame "The flow at a 20 psi residual is larger than the test flow because the pressure drop is ___" for the guided-practice check.

Extensions

Have students look up the fire flow their own school building would need from a local fire code table, then work backward through the available-flow formula, as they did in question 10 for the fictional Bramleigh loop, to state the test discharge and static pressure a nearby hydrant would have to show to meet it.