Energy Cost from a Utility Rate

Ce=EpeC_e = E \, p_e

Enter your known values, leave one input blank, and solves for the missing one. Try different units for next level excitement!

Learning zone

Water is not the only meter a cooling system spins. Tower fans, condenser-water pumps and the compressor itself all draw power, and boilers burn gas — so the same product, energy times a rate, prices both. A tower's fans and pumps drawing 250,000 kWh a year at $0.11/kWh cost $27,500; a boiler burning 20,000 MMBTU of gas at $8.00/MMBTU costs $160,000. North American electricity sits around $0.08–0.15/kWh commercial and natural gas around $6–10/MMBTU, but demand charges, ratchets and time-of-use blocks mean the effective rate on a bill is often well above the headline commodity rate — take it from the bill, dividing total dollars by total kilowatt-hours, rather than from the tariff sheet.

This calculation is what makes the water-treatment argument financial rather than technical. Scale is an insulator: a 0.6 mm (1/64 in) carbonate film on condenser tubes lifts compressor power by roughly 20%, and on a plant with a six-figure electricity bill that dwarfs the entire chemical budget. The same arithmetic prices the other direction too — boiler blowdown leaves at saturation temperature, so every percent of continuous blowdown costs a fraction of a percent of fuel, and a blowdown heat exchanger's payback is nothing more than this equation applied to recovered energy. Energy is entered and answered in kilowatt-hours and the rate in dollars per kilowatt-hour regardless of the metric/imperial toggle, because that is how every electricity meter on earth reads; and as with every money answer here, the currency is whatever currency you typed the rate in.

Energy Cost from a Utility Rate
Ce=EpeC_e = E \, p_e
Where
  • CeC_e= Energy cost
  • EE= Energy consumed
  • pep_e= Energy rate
Missing one of these? Work it out first, then come back