View Full Version : Measuring power
mtlman
Aug 13, 2009, 05:01 AM
How do I measure the electric power being used in my shop, for lights , machines and accessories. I need to know how much each item uses and the cost. How can I measure each to give me a daily total cost.?
KISS
Aug 13, 2009, 08:03 AM
This is an interesting queston, but I need to ask/eduacate as well.
Energy Kilo-Watt-hours. Depends on Current, voltage and time.
AC power is difficult to measure. For this discussion, I'll assume sine wave voltage, current and a phase angle. This isn't true for a few items such as electronic ballasts, computer power supplies.
Power = voltage * Current;; for resistive loads (e.g. a light bulb at steady state)
Power (watts) is also voltage (Volts) * Current (Amps) * PF
Power for 3 phase is a little more complicated. I'll add that if necessary. Nameplate data is OK.
The PF is the power factor and is defined as cos(theta) and varies from -1 to +1; generally 0 to 1. Motors reduce the power factor. The PF is usually on the nameplate.
Motors draw a large start-up current and the load varies based on what's it's cutting etc.
So, you nned to know how long the machine is operating during the day and at what load. A tall order.
You can estimate using nameplate data.
A device called Killawatt. Kill-A-Watt Electric Usage Monitor Review ? The Gadgeteer (http://the-gadgeteer.com/2003/12/23/kill_a_watt_electric_usage_monitor_review/)
This device can monitor any corder appliance and compute a cost based on the cost per KWH. If you have electric rates that are baseed on say 15 minute peak demands all bets are off.
This won't help you with the lights or the 240 V devices or things that draw more than 15 A at 120 V, but it's a start and it's cheap. Note that a reviewer said, it won't work on a modified sine wave inverter.
I bought a 3 phase energy monitor and it cost about $600, if I remember right. We used it to measure single phase power of a non-sinusoidal voltage of continually varying voltages around 30 VAC nominal. It was made by this company. Schneider Electric United States - Solutions, Products and Services in Electrical Distribution and Automation and Control (http://www.powerlogic.com/index.cfm) before they were gobbled up. The current transformer used selects the current range.
tkrussell has a better link with various products.
Generally these take current transformers to measure current (you pass a wire through the core and it's non-contact) to measure current and the then attach power to the monitor.
The monitor I had was able to be powered by 120 VAC, but measure much lower voltages of non sine wave power.
In light of thie, there will be inaccuraces based on what technique you use:
Some possibilites:
1. Nameplate data
2. Measure at the panel; sub-panel for the shop and get a nice average.
3. Measure individual devices or group of devices at the panel.
4. Use Kill a watt is the loads are not very odd
5. Measure V and I of motor loads and estimate the PF of the motor at 0.8 while cutting, not unloaded.
There are plenty of other devices on the market. Most will use a clamp on meter or torroid coil. The current can be measured using a hall effect sensor or a current transformer.
If you measure currents, it's important to use a Trms meter. It will remove some of the effects of waveform distortion.
Some idea of what you want and why? Cost per widget and overhead expenses. And what you have: 460 V equipment, 3 phase, 120 V? All part of a single sub-panel or a shop attached to a house.
I leave you with this link for now:
Power meters, watt meters, kWh meters - PowerMeterStore.com (http://www.powermeterstore.com/index.php?s=g&gclid=CLins6b2oJwCFVhB5godZ2Ohiw)
Newton1Law
Aug 17, 2009, 12:37 PM
KISS has given you the answer but I propose a second alternative. List the devices that you want to measure the energy usage of. Note if they are single phase or three phase. Gather the nameplate information for each device.
For example starting with single phase type loads:
10 - 120 Volt, 100 Watt lamps. These are on 8 hours per day.
1 - 240 Volt, 5 horsepower motor. This runs a total of 4 hours per day.
1 - 120 volt, 1500 watt refrigerator. This runs a total of 6 hours per day.
1) Lights -> 10, 100 watt lights equals (10X100) = 1000 watts. 1000 watts on for 8 hours gives you an energy usage of 8000 watt-hours. We divide this by 1000 to get kilowatt-hours so 8000 watt-hours divides by 1000 = 8 kilowatt-Hours.(or kWH)
2) Motor -> 1 Horsepower = 746 watts. 5 horsepower = 3730 watts, on for 4 hours gives 14,920 watt-hours or 14.95 kWH.
3) Refrigerator -> 1500 watts on for 6 hours = 9000 watt-hours or 9 kWH.
4) Total energy used by these devices for one day then is (8 + 14.95 + 9) = 31.9 kWHs per day.
5) If your billing period is for thirty days then you would multiply 31.9 kWH per day by 30 days to get = 957 kWH for the month. Assuming your pay about 12 cents per kWH, your cost would be 957 kWH times $0.12/kWH = $114.84 cents.
All this is of course an approximation, but if you plug in your numbers you should get a reasonable number for your cost per month.
I have made a couple of exceptions here in that I assumed your power company does not charge for power factor (or VAR loading) and that you still have to add taxes.
For three phase loads then nameplates should give you the wattage of the device or the VA (volt-amps) or kVA (kilo volt-amps) and power factor. If given in watts, just use watts times time in hours to get watt-hours of energy use. If given in kVA with a power factor, then kVA X power factor = Watts. Again I am assuming no cost for Var loading. If you can give me the local power company that supplies your power, I can give you a better calculation.
Hope this helps.
InfoJunkie4Life
Oct 4, 2009, 04:04 AM
I think I have an easier solution. Write down the current numbers on your meter and wait thirty days and write it down again. Subtract the two to find how much you used that month. Repeat, except don't use the shop for that thirty days... say vacation or something. Then subtract your final numbers to get the amount you use per month. You can do it over a week and just multiply your final difference by 4 if you want. I wouldn't go any shorter than a week because after that you become unreliable, unless you use exactly the same tools the exact amount of time. That'll give you an average though.