Lesson plan Grade 9 Math
Ridgeport Pump Station: Total Dynamic Head and Pressure
A grade 9 lesson where students add static lift and friction to find total dynamic head, then convert between feet of head and psi for a fictional pump station.
Objectives
- Students will calculate total dynamic head as the sum of static lift and friction head.
- Students will convert between feet of head and psi using the 0.433 psi-per-foot rate.
Vocabulary
- Total dynamic head (TDH)
- Total dynamic head is the total amount of lift a pump has to provide to move water where it needs to go, adding up the height it must lift the water, the friction it must overcome in the pipes, and the pressure needed at the far end. Engineers calculate it to pick a pump that is powerful enough for the job.
Also: static lift, friction head, psi
Teacher overview
A pump that lifts water uphill has to overcome two things: the height it raises the water, called static lift, and the resistance of the pipe, called friction head. Their sum is the total dynamic head, the real load the pump works against. This lesson uses a fictional pump station to make that idea concrete, then connects head in feet to pressure in psi with the constant every operator memorizes, 1 foot of head equals 0.433 psi. A common misconception is treating the pump's rated pressure as if it must only match the static lift; students learn to add friction head first so the pump is not undersized.
Procedure
Hook (5 min): Show a photo of a water tank on a hilltop above a river. Ask students how the water "knows" to push back against a pump, and record two or three ideas without judging them. Steer toward the idea that both the climb and the pipe fight the pump.
Direct instruction (15 min): Define static lift, friction head, and total dynamic head (TDH = static lift + friction head). Introduce the conversions 1 foot of head = 0.433 psi and 1 psi = 2.31 feet. Model one fictional warm-up example on the board: a pump with 150 feet of static lift and 15 feet of friction has a TDH of 165 feet, which is 165 x 0.433 = 71.4 psi. Emphasize that this example is instruction only, not a graded problem, and that friction was added before converting.
Guided practice (15 min): Hand out the student page. In pairs, students complete handout questions 1 through 4, which build the Ridgeport TDH and convert it to psi. Circulate and check that pairs add friction to static lift before converting.
Independent practice (10 min): Students complete handout questions 5 through 8 on their own, working through a higher friction case and writing a general equation for required pressure.
Closure / exit ticket (5 min): Students answer handout question 9 on a half sheet: a pump must overcome 210 feet of TDH, so what pressure in psi must it produce? Collect and scan for the 0.433 conversion. If more than a quarter of the class multiplies by 2.31 instead, reteach the direction of the conversion the next day.
Student Handout
Ridgeport's pump station (a fictional utility) lifts water from a river intake up to the Beacon Heights tank. The static lift is 180 feet. At the design flow, friction in the pipe adds another 20 feet of head. Use 1 foot of head = 0.433 psi and 1 psi = 2.31 feet. Round pressures to one decimal place and heights to the nearest foot.
- Find the total dynamic head (TDH), in feet, as static lift plus friction head.
- Convert the TDH from question 1 to psi.
- The pump's data plate says it can produce 95 psi at the design flow. Convert 95 psi to feet of head.
- Does the pump's head from question 3 exceed the required TDH from question 1? By how many feet?
- In summer the friction head rises to 35 feet. Find the new TDH in feet.
- Convert the new TDH from question 5 to psi.
- Compare the new required pressure from question 6 to the pump's 95 psi rating. Does the pump still keep up, and by how much?
- Write an equation for the required pump pressure P (in psi) as a function of friction head f (in feet), keeping static lift at 180 feet. Then find P when f = 25 feet.
- Exit ticket: A different pump must overcome 210 feet of TDH. What pressure in psi must it produce?
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: 200 ft. TDH = 180 + 20 = 200 feet.
- Answer: 86.6 psi. 200 feet x 0.433 psi per foot = 86.6 psi.
- Answer: 219 ft. 95 psi x 2.31 feet per psi = 219.45, which rounds to 219 feet of head.
- Answer: Yes, by about 19 ft. The pump can make 219 feet of head against a 200-foot requirement, so it clears the lift by 219 - 200 = 19 feet.
- Answer: 215 ft. TDH = 180 + 35 = 215 feet.
- Answer: 93.1 psi. 215 feet x 0.433 psi per foot = 93.095, which rounds to 93.1 psi.
- Answer: Yes, by about 1.9 psi. The pump is rated for 95 psi and now needs 93.1 psi, a margin of only 95 - 93.1 = 1.9 psi, much tighter than before.
- Answer: 88.8 psi. The equation is P = (180 + f) x 0.433. At f = 25, P = 205 x 0.433 = 88.765, about 88.8 psi.
- Answer: 90.9 psi. 210 feet x 0.433 psi per foot = 90.93, which rounds to 90.9 psi.
Differentiation
Support
Give students a filled-in first line for question 1 (TDH = static + friction) and a labeled arrow card showing "feet to psi: multiply by 0.433" and "psi to feet: multiply by 2.31." Let pairs finish questions 1 and 2 before moving on, and allow calculators throughout.
Extension
Ask students to find the maximum friction head, in feet, the 95 psi pump could tolerate while still meeting the 180-foot static lift, working backward from the pump's rated head. Have them explain what a shrinking friction-head margin would mean for the pump as pipes age.
ELL supports
Provide a labeled side-view diagram showing the river intake, the vertical static lift, and the hilltop tank, with friction shown as small arrows along the pipe. Pre-teach "lift," "friction," and "head," and post a two-line conversion reference beside the handout.