Math problem set Grade 12 Math
Craymoor Pump Station: Horsepower, Efficiency, and Energy Cost
A grade 12 problem set on pump math: total dynamic head, water and brake horsepower, wire-to-water efficiency, energy cost in kWh, and the pump affinity laws.
Objectives
- Students will calculate total dynamic head, water horsepower, and brake horsepower for a pump.
- Students will convert pump power to energy use and cost and apply the affinity laws to a speed change.
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: water horsepower, brake horsepower, wire-to-water efficiency, affinity laws
Teacher overview
A pump does not just lift water to a height; it also pushes against the friction of the pipe, and it loses some energy to inefficiency in the pump and the motor. This set follows the fictional Craymoor high-service pump station through the horsepower, energy, and affinity-law math a distribution operator uses to judge pumping cost. A common misconception to correct is treating the vertical lift alone as the head the pump works against; the pump must overcome total dynamic head, which adds friction loss to the static lift, and the motor must supply more power still because neither the pump nor the motor is perfectly efficient.
Problems
Craymoor high-service pump station is a fictional station. Use these values for every problem. Water horsepower equals flow in gpm times head in feet divided by 3960; one horsepower equals 0.746 kilowatts.
- Flow: 1,400 gpm
- Static head (vertical lift): 180 feet
- Friction head loss: 25 feet
- Pump efficiency: 0.80
- Motor efficiency: 0.92
- Electricity rate: $0.11 per kWh
- Daily run time: 18 hours
- Calculate the total dynamic head in feet, adding the static head and the friction head loss.
- Calculate the water horsepower using flow times total dynamic head divided by 3960. Round to the nearest tenth.
- Calculate the brake horsepower by dividing water horsepower by the pump efficiency of 0.80. Round to the nearest tenth.
- Calculate the motor input horsepower by dividing brake horsepower by the motor efficiency of 0.92. Round to the nearest tenth.
- Convert the motor input horsepower to kilowatts, using 1 horsepower equals 0.746 kilowatts. Round to the nearest tenth.
- Calculate the station's wire-to-water efficiency as a percent, the product of pump efficiency and motor efficiency. Round to the nearest tenth.
- Calculate the energy used in one day, in kWh, using the unrounded input power from problem 5 and the 18-hour run time. Round to the nearest whole kWh.
- Calculate the daily energy cost at $0.11 per kWh. Round to the nearest dollar.
- A variable-frequency drive slows the pump to 90 percent of its speed. Using the affinity law that flow is proportional to speed, find the new flow in gpm.
- Using the affinity law that pump power is proportional to the cube of speed, find the new input power in kW at 90 percent speed, starting from your problem 5 value. Round to the nearest tenth.
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: 205 feet. Total dynamic head equals 180 feet plus 25 feet, which is 205 feet.
- Answer: 72.5 horsepower. Water horsepower equals 1,400 times 205 divided by 3960: 1,400 times 205 is 287,000; divided by 3960 is 72.47, or about 72.5 horsepower.
- Answer: 90.6 horsepower. Brake horsepower equals 72.47 divided by 0.80, which is 90.59, or about 90.6 horsepower.
- Answer: 98.5 horsepower. Motor input equals 90.59 divided by 0.92, which is 98.47, or about 98.5 horsepower.
- Answer: 73.5 kilowatts. 98.47 horsepower times 0.746 kilowatts per horsepower equals 73.46, or about 73.5 kilowatts.
- Answer: 73.6 percent. Wire-to-water efficiency equals 0.80 times 0.92, which is 0.736, or 73.6 percent.
- Answer: 1,322 kWh. Daily energy equals 73.46 kilowatts times 18 hours, which is 1,322.3, or about 1,322 kWh.
- Answer: $145 per day. 1,322.3 kWh times $0.11 per kWh equals $145.45, or about $145 per day.
- Answer: 1,260 gpm. Flow is proportional to speed, so the new flow is 0.90 times 1,400, which is 1,260 gpm.
- Answer: 53.6 kW. Power scales with the cube of speed: 0.90 cubed is 0.729; 0.729 times 73.46 kilowatts is 53.55, or about 53.6 kW, a large drop for a small speed reduction.
Differentiation
Support
Give students the water-horsepower and horsepower-to-kilowatt formulas on a reference card with problem 2 fully worked. Let them reuse the same chain in problems 3 through 5 with only the divisor changing.
Extension
Ask students to find how much the fictional station would save per day by running at 90 percent speed, using the problem 10 power, then explain why a 10 percent speed cut saves far more than 10 percent of the energy.
ELL supports
Pre-teach "head," "horsepower," "efficiency," and "speed" with a labeled pump-and-pipe diagram showing static lift, friction, and the motor. Provide the sentence frame "The motor needs more power than the water horsepower because ___" for problems 3 and 4.
Extensions
Have students compare the fictional Craymoor cost per million gallons pumped with a published pumping-energy figure for a real utility, discussing how head, efficiency, and electricity price each move that cost.