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Math problem set Grade 11 Science

Harwick's Pumping Bill: Energy, Efficiency, and Cost per Gallon

A grade 11 problem set where students turn a fictional utility's flow, head, and pump efficiency into daily energy, annual cost, and the savings from an upgrade.

Objectives

  1. Students will calculate a pump's water horsepower, electrical power, and daily energy use from flow, head, and efficiency.
  2. Students will convert energy use into annual cost and unit cost and evaluate the savings from a more efficient pump.

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: wire-to-water efficiency, water horsepower, kilowatt-hour, unit cost

Teacher overview

Pumping is usually the largest item on a water utility's electricity bill, and the size of that bill depends as much on efficiency as on how much water moves. This set follows a fictional district's high-service pumps from raw flow and head down to the dollars per year and dollars per million gallons that a manager actually reports to the board. A common misconception is to bill the water horsepower directly and forget efficiency; because pumps and motors are not perfectly efficient, the electric meter always records far more energy than the water receives, and the utility pays for the full metered amount.

Harwick's Pumping Bill: Energy, Efficiency, and Cost per Gallon — Grade 11 Science
Name: ____________________________ Date: ______________

Problems

The Harwick Water District is a fictional utility. Use these stated values for every problem. Its high-service pumps deliver an average of 3 MGD against a total dynamic head of 300 feet. The combined wire-to-water efficiency (pump and motor together) is 65 percent. Electricity costs $0.14 per kWh. Use 1 MGD = 694 gpm, water horsepower = (flow in gpm x head in feet) / 3960, and 1 horsepower = 0.746 kilowatt. Round sensibly and carry your rounded values forward.

  1. Convert the average flow of 3 MGD into gpm.

  2. Find the water horsepower using WHP = (flow in gpm x 300) / 3960.

  3. The pumps are only 65 percent efficient wire-to-water, so the electrical input is larger. Find the electrical input horsepower: water horsepower / 0.65.

  4. Convert the electrical input horsepower from item 3 into kilowatts, using 1 horsepower = 0.746 kilowatt.

  5. If the pumps run continuously, find the energy used in one day, in kWh (kilowatts x 24 hours).

  6. Find the daily electricity cost at $0.14 per kWh.

  7. Find the annual electricity cost (use 365 days).

  8. Find the unit cost of pumping, in dollars per million gallons (daily cost divided by 3 MG).

  9. The district could replace the pumps with units that reach 78 percent wire-to-water efficiency. Recompute the daily energy and daily cost at 78 percent, and find the annual savings compared with the 65 percent pumps.

  10. Besides efficiency, name two factors that would raise this pumping bill, and explain how each one increases the energy used.

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: 2,082 gpm. 3 MGD x 694 gpm per MGD = 2,082 gpm.

  2. Answer: about 157.7 hp. WHP = (2,082 x 300) / 3960 = 624,600 / 3960 = 157.73, about 157.7 water horsepower.

  3. Answer: about 242.7 hp. Electrical input = 157.73 / 0.65 = 242.66, about 242.7 horsepower.

  4. Answer: about 181.0 kW. 242.66 horsepower x 0.746 = 181.02, about 181.0 kilowatts.

  5. Answer: about 4,344 kWh per day. 181.0 kilowatts x 24 hours = 4,344 kWh in one day.

  6. Answer: about $608 per day. 4,344 kWh x $0.14 per kWh = $608.16, about $608 per day.

  7. Answer: about $222,000 per year. $608.16 per day x 365 days = $221,978, about $222,000 per year.

  8. Answer: about $203 per million gallons. $608.16 per day / 3 million gallons = $202.72, about $203 per million gallons.

  9. Answer: about $37,000 per year saved. At 78 percent, electrical input = 157.73 / 0.78 = 202.2 horsepower, so 202.2 x 0.746 = 150.8 kilowatts, giving 150.8 x 24 = 3,619 kWh per day and 3,619 x $0.14 = $506.66 per day. Daily saving is $608.16 - $506.66 = $101.50, so annual saving is about $101.50 x 365 = about $37,000 per year.

  10. Answer: Any two reasonable factors, such as a higher total dynamic head (the pumps lift or push against more pressure), a larger volume pumped (more MGD means more running time), or throttled valves adding artificial head. Each raises the energy because energy use scales with head and with the volume pumped.

Differentiation

Support

Provide a worked one-line example for a different pump (say 1 MGD at 100 feet) showing the full chain from gpm to daily cost, and give the three constants (694, 3960, 0.746) on a labeled reference card so students focus on the sequence of steps.

Extension

Suppose the 78 percent pumps in problem 9 cost $180,000 more to install than a like-for-like replacement. Have students find the payback time using the annual savings, and explain why a utility board would still weigh reliability alongside payback.

ELL supports

Provide a labeled flow chart of the calculation (flow to power to energy to cost) with the unit of each box written in, and allow students to record answers as numbers with units on the chart rather than in full sentences. Pre-teach "efficiency," "head," and "kilowatt-hour."