energy used when voltage drops
I listened to all the threads so I thought I'd put my 2 cents in here concerning power consumption and decreased voltages, etc.
Let's ignore power meter inaccuracies at reduced or increased voltages and look at motors and then resistive elements such as water heaters:
For a motor: if you want to do "X" amount of work, I like to invoke what's called the "nth law of thermodynamics" that "there are no free lunches". In the case of a motor, if you want to do "X" amount of work it takes you the same amount of energy in kilowatt hours regardless of how the voltage or current may change. Again, assume here for the sake of discussion that the motor is not damaged at the lower voltage and we assume it still works as efficiently.
Side note: They call the meter on the side of your house a "power meter" but it's really an "energy meter". "Power" is measured in watts or Kilowatts but the meter on your house is an "energy meter" which measures "kilowatt hours consumed" which is a measure of "energy", NOT "power".
Back to the motor: If, for example, you would normally consume 120 Volts and 8 amps for 2 hours for a task, this means you used 960 Watts of power (120 Volts x 8 amps = 960 watts) but you used 1920 watt-hours of energy (120 Volts x 8 amps x 2 hours = 1920 watt-hours = 1.92 Kilowatt hours) and Kilowatt hours is what you get billed on.
Now, let's say the voltage now drops to 80 volts. You must still consume this same 1920 watt-hours of energy to do the same task and you can do this by drawing more amps for a shorter time or less amps for a longer time but in the end, to do the same task, you will consume the same amount of kilowatt hours of energy (Volts x Amps x hours) for this same task. And, since power companies bill us by the ENERGY we consume (kilowatt hours used), the cost to the consumer is the same regardless of fluctuations in voltage.
Note: we can argue if a motor or device will properly operate at these low/high voltage changes and what its efficiency is at these voltages and meter inaccuracies at voltages other than 120 volts, but, ignoring these, the main point here is that it takes a certain amount of energy in kilowatt-hours to do a given task and your bill will be the same to do the same task regardless of small fluctuations in voltage!!!
For a resistive heater element: Let's go back to the water heater scenario where the 1500W element running at 120 volts is then subjected to 80 Volts. At 120 Volts, we can calculate the resistance which is a FIXED value, regardless of applied voltage.
Watts = (V) x (I)
since I = V/R we can plug this into the above and arrive at this:
Watts = [(V) x (V)] / R
1500 = [ (120) x (120)] / R
Resistance = 9.6 Ohms for this element.
NOTE: for this element, the resistance is the SAME REGARDLESS OF THE VOLTAGE WE APPLY TO IT!!!! Apply 2000 volts across it and the resistance is still 9.6 Ohms. Apply 80 volts across it and the resistance is still 9.6 Ohms!
Now, using this fixed resistance of 9.6 Ohms, let's calculate the new watts (power) consumed when the voltage drops to 80 Volts:
Watts = [(V) x (V)] / R
Watts = [ (80) x (80)] / 9.6
Watts = 667 watts power consumed
Let's assume the heater is 40 gallons (about 332 lb of water) and the inlet water temp is 60F and we want to raise it to 186F. Thermodynamics tells us that (neglecting heat loss from the water heater jacket) that it takes 8.2 hours to raise 332 lb of water by 126 degrees F (186F - 60F inlet temp = 126F temperature rise). Here's the calculations (in the calculation, a BTU is a British Thermal Unit and one BTU raises 1 lb of water 1 deg F at 60 deg...it changes slightly when the temperature is different from 60F but we ignore it here...also, the 1500watt heater is equal to 1.5 Kilowatt:
3412 BTU/(Kilowatt hour) x 1.5 Kilowatt x (Time in hrs) = 332 lb water x 1 BTU/(lb water x deg F) x (186F - 60 F)
solve for Time and we get 8.17 hrs
Note: you might say, "well...it doesn't take a normal water heater 8 hrs to heat water"...and you're right but the average water heater has a heater element that is 4 times as powerful as this 1500 watt heater AND the average water heater only raises water to 120 F, NOT 186F. In the above example, raising water to 120F with this 1500 watt heater would only take 3.9hours and the typical water heater with a 6000 watt heater element would only require 58 minutes to heat this 60F water to 120F.
Back to the example:
Now, calculate the time required to raise the water to 186F using the reduced voltage (80 volts) where the watt consumption is 667 watts and you'll get 18.3 hrs, which, by the way is inversely proportional to the SQUARE of the drop in voltage (i.e., drop voltage by one-half and power consumed drops by a factor of four, etc.). BUT again (and we'll ignore heat loss here from the water heater jacket) you're billed by kilowatt hours and it takes the same amount of energy in Kilowatt hours to raise water temp by 1 deg F regardless of what the voltage is!! So, even if voltage fluctuates and wattage drops, you consume the same energy to heat your water but it just takes you a longer time to do so but THE COST IS THE SAME!!!
Bottom line in both examples: ignoring meter inaccuracies due to voltage fluctuations and heat loss, and whether or not a device will work without being destroyed at lower/higher voltages, it takes the SAME amount of energy (and therefore it will cost the same) to do the SAME task.....only difference is it may take longer to do it at reduced voltages.
Caveat: you could reduce voltages so low such that power consumed is so low that a task would never get done and then energy costs would rise dramatically. Example: put 10 watts into a 1,000 watt toaster and you'd never generate sufficient infrared energy to brown your toast and the device would run forever and consume infinite energy and your toast would never get done. Same if you tried to run a 1500 watt iron on say, 50 watts, etc.
BUT, don't lose sight of the big picture: small drops/rises in voltage mean you still will consume the same amount of energy to do the same task and your energy bill will be the SAME!
hope this helps
