Science Buddies: "Ask an Expert"

Hello,
My son is trying to do an experiment in which he wants to find out 'vampire load' or 'standby power' of various household equipment. We bought Kill-a-Watt and plugged it into various appliances/equipment. Now our dilemma is this
One of the laptop consumes 0.4 watt when it is turned off. Kill-a-Watt shows readings upto two decimal points when it shows Kwh measurement. We left that laptop connected to Kill-a-Watt for about 12 hours and still the reading of KWH was 0.00. From what I understand to find the energy consumption formula is
(Wattage × Hours Used Per Day) ÷ 1000 = Daily Kilowatt-hour (kWh) consumption
(0.4 * 12) / 1000 will be 0.0048 kwH . This will not show in the Kill-a-Watt monitor as it only shows readings upto two decimal places.

Now, knowing this I have two choices
1. Plug in this laptop for 30 hours for it to show a reading. The cons of this is that we will not be able to test too many appliances and take multiple readings as his project is due on Jan 17th 2012.
2. Take a wattage reading when appliance is turned off and if it is greater than zero, then use the above formula to find the KWH.

My question is this...Is option 2 a valid option for a science fair ? Or only empirical reading in Kill-A-Watt would make his experiment credible. What are practical options for him given the time constraint ?
Thanks again for your support.

anuami - I wasn't familiar with the instrument you were using, so I downloaded the instruction manual for the Kill-A-Watt P4400 which seemed to be a low cost device, possibly identical, or close to the unit you have. The problem you are having is just the magnitude issue of the units you are dealing with. A device consuming 4 watts in standby mode would have to be in that mode for 250 hours to read 1.0 kW. Much too long for you to get meaningful results for your time budget. You got a better reading most likely when you measured the instantaneous value of 4 watts in standby mode using this equipment. (I assume you also had two decimal resolution capability from the meter and it could have read 4.03 watts or 3.91 watts). This would be the way I would get around the kW magnitude issue. So I definitely vote for your option 2. If the standby power doesn't change over time, just use the formula you correctly stated.

Rick Marz

Rick,
Thanks for your help. I have one more question.
We took measurements of standby power for some devices in our home. Now for the experiment we had thought that
Volt and Frequency is a control variable as it comes from the wall outlet and it is a constant. However, when we took readings the voltage and frequency differed slightly for each reading. So, do the appliances make any difference or is it just a fluctuation of electricity?
So, my question to you is 'Is Volt and Frequency a control variable or dependant variable' ?
Thanks
Ami

Anuami - it is possible that the voltage in your home may read differently at different outlets due to differing loads on the circuit. If you turn on a high power consumption appliance like a toaster or hair dryer you may measure a few volts lower than a measurement on a different outlet in the house. These are 'copper losses' and amounts to the voltage drop across your wiring in that circuit. The 60 Hz frequency of your electrical service is constant and closely controlled by the power company. No $20 meter would be able to measure any deviation from the 60 Hz supply (50 Hz in Europe and many other countries of the world).

If I understand the operation of the Kill-A-Watt unit, you can measure instantaneous power and voltage. Use that voltage if you want to calculate current. I think you are only interested in measuring watts, so that calculation is not relevant. You are correct to imply that fluctuating voltage is a variable, so you may want to account for it. Consider this. At a supply voltage of 110 VAC you will measure a standby (or operating power) that is about 10% lower than a measurement made if the voltage went up to 120 VAC.

Rick Marz