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Lesson plan Grade 11 Math

Keldwick's Ivelbrook Avenue Main: A Log-Log Fit for Head Loss

A 55-minute grade 11 lesson where students test proportionality, take logarithms of flow and head-loss data, and recover a power model for an aging transmission main.

Objectives

  1. Students will test a data set for proportionality using ratios of the two measured quantities.
  2. Students will recover the exponent and coefficient of a power model from a log-log slope and intercept.
  3. Students will use the fitted model to predict head loss and to solve for a flow limit.

Vocabulary

Tuberculation
Rusty bumps that grow on the inside of old iron pipes. They make the pipe rougher and narrower, so less water can squeeze through.

Also: head loss, power model, log-log plot, transmission main

Teacher overview

Pushing water through a pipe costs pressure, and the amount lost rises much faster than the flow that causes it. Crews in the fictional Keldwick Municipal Water District measured head loss across a 2,000-foot section of the Ivelbrook Avenue transmission main at five steady flows, and this lesson uses that fictional data set to build a power model from a log-log transformation. The misconception this lesson targets is the assumption that doubling the flow doubles the head loss, which carries over from the proportional-reasoning work of earlier grades. Dividing head loss by flow at two rows of the table breaks that assumption immediately, because the quotient is five times larger at the high end than at the low end.

Procedure

Launch (6 min): Post the fictional Keldwick data table and ask students to predict the head loss at 400 gallons per minute given that 200 gallons per minute produces 2.0 feet. Record the predictions without comment. Many students predict 4.0 feet, which the table contradicts.

Testing proportionality (8 min): Have students compute head loss divided by flow for the top and bottom rows. The quotients are 0.010 and 0.050 feet per gallon per minute. Ask what a constant quotient would have meant and why these two values rule out a straight line through the origin.

Log transform and slope (14 min): Model taking the base-ten logarithm of both columns. Plot the transformed pairs on the board and show that they fall on a line. Find the slope between the first and last transformed points, which comes to 2.00, and interpret it as the exponent in a model of the form h equals k times Q raised to a power. Then solve for the intercept and convert it to the coefficient by raising ten to that power.

Guided practice (8 min): In pairs, students verify the model against the 600 gallons per minute row that was not used to build it, and report whether the predicted and measured values agree. Circulate and check that pairs are squaring the flow before multiplying rather than after.

Independent practice (14 min): Students work questions 1 through 13 on the handout, including the aged-pipe comparison in questions 10 and 11.

Closure and formative check (5 min): Students answer handout question 13 on a slip of paper, explaining what the two flow limits say about carrying capacity. Collect the slips and sort them for any answer claiming the aged main carries one-quarter as much water, which signals a student who divided the coefficients rather than taking a square root, and open the next session with that error.

Keldwick's Ivelbrook Avenue Main: A Log-Log Fit for Head Loss — Grade 11 Math
Name: ____________________________ Date: ______________

Student Handout

The fictional Keldwick Municipal Water District measured head loss across a 2,000-foot section of its Ivelbrook Avenue transmission main. The readings are below. Let Q stand for flow in gallons per minute and h for head loss in feet.

Flow Q (gpm) Head loss h (ft)
200 2.0
400 8.0
600 18.0
800 32.0
1,000 50.0
  1. Divide head loss by flow for the 200 gallons per minute row.
  2. Divide head loss by flow for the 1,000 gallons per minute row.
  3. Using your answers to questions 1 and 2, state whether head loss is proportional to flow, and give a one-sentence reason.
  4. Take the base-ten logarithm of Q and of h for the 200 and 1,000 gallons per minute rows. Find the slope of the line through those two transformed points, to two decimal places.
  5. Using your slope from question 4 and the transformed 200 gallons per minute point, find the intercept b in the equation log h = m log Q + b, to two decimal places.
  6. Find the coefficient k by computing 10 raised to the power b. Give the result to two significant figures in scientific notation.
  7. Write the power model for head loss h in feet as a function of flow Q in gallons per minute.
  8. Use your model to predict the head loss at 1,400 gallons per minute.
  9. A booster pump can overcome at most 12.5 feet of head loss across this section. Find the largest flow the main can carry within that limit.
  10. Forty years later the crew repeats the test and measures 128 feet of head loss at 800 gallons per minute. Assuming the exponent is unchanged, find the new coefficient k for the tuberculated main.
  11. Find the largest flow the tuberculated main can carry within the same 12.5-foot limit.
  12. Convert the original main's 50.0 feet of head loss at 1,000 gallons per minute into pounds per square inch, using 1 foot of head equals 0.433 psi.
  13. In one sentence, explain what your answers to questions 9 and 11 say about what tuberculation does to a main's carrying capacity.

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

Answer Key

  1. Answer: 0.010 ft per gpm.
    Dividing gives 2.0 divided by 200 = 0.010 feet of head loss for each gallon per minute.

  2. Answer: 0.050 ft per gpm.
    Dividing gives 50.0 divided by 1,000 = 0.050 feet of head loss for each gallon per minute.

  3. Answer: No.
    A proportional relationship has a constant quotient, and this quotient is five times larger at 1,000 gallons per minute than at 200 gallons per minute, so no straight line through the origin fits the data.

  4. Answer: 2.00.
    The transformed points are (2.301, 0.301) and (3.000, 1.699). The slope is (1.699 - 0.301) divided by (3.000 - 2.301), which is 1.398 divided by 0.699 = 2.00.

  5. Answer: -4.30.
    Substituting gives 0.301 = 2.00(2.301) + b, so b = 0.301 - 4.602 = -4.301, which rounds to -4.30.

  6. Answer: 5.0 x 10^-5.
    Raising ten to the power -4.301 gives 0.00005, written 5.0 x 10^-5 to two significant figures.

  7. Answer: h = (5.0 x 10^-5)Q squared.
    The slope from question 4 is the exponent and the value from question 6 is the coefficient, so head loss in feet equals 0.00005 times the square of the flow in gallons per minute.

  8. Answer: 98 ft.
    Squaring gives 1,400 squared = 1,960,000, and multiplying gives 0.00005 x 1,960,000 = 98 feet.

  9. Answer: 500 gpm.
    Solving 12.5 = 0.00005 times Q squared gives Q squared = 250,000, so Q = 500 gallons per minute.

  10. Answer: 2.0 x 10^-4.
    Dividing gives 128 divided by 800 squared = 128 divided by 640,000 = 0.0002, written 2.0 x 10^-4.

  11. Answer: 250 gpm.
    Solving 12.5 = 0.0002 times Q squared gives Q squared = 62,500, so Q = 250 gallons per minute.

  12. Answer: 21.65 psi.
    Multiplying gives 50.0 x 0.433 = 21.65 pounds per square inch.

  13. Answer: The tuberculated main carries half the flow.
    Quadrupling the coefficient cuts the flow at a fixed head-loss limit in half, because flow depends on the square root of the limit divided by the coefficient, and the square root of one-fourth is one-half.

Differentiation

Support

Give these students the logarithm columns already computed for all five rows of the fictional Keldwick table, so their first task is finding a slope rather than managing a calculator sequence. Provide the model with the coefficient in place and only the exponent blank, and let them check questions 8 and 9 against the printed table before moving on.

Extension

Ask these students to compute the residual for every row of the fictional Keldwick table after fitting, then repeat the whole analysis on a second data set in which one head-loss reading has been altered by a stuck gauge. Have them explain which step of the log-log method reveals the bad reading most clearly.

ELL supports

Pre-teach "head loss," "tuberculation," and "carrying capacity" with a labeled cross-section drawing of a clean pipe beside a pipe narrowed by mineral buildup. Provide sentence frames for questions 3 and 13, such as "Head loss is not proportional to flow because the quotient ___," and "When the coefficient grows, the flow at a fixed limit ___."

Extensions

Students can compare the fictional Keldwick exponent of 2.00 with the exponent of 1.85 that the Hazen-Williams equation uses for water in pipe, and explain why a slope read from five noisy field readings would rarely land on either value exactly. A second task is to rewrite the model with flow in cubic feet per second rather than gallons per minute, and to describe what changes in the coefficient and what does not.