Calculating the Watts by Energy Meter Reading?

Your example is not wrong. But isn't it 410 rev/kWh? It will also turn 410 times for 10 watts in 100 hours. So how big is my fan? Sorry - Your Fan: How is the relation (ship) between the 42 seconds and the hour? And 1 turn to 410? You put that all in 1 pot, stir and see what you cook. I have no calculator and don't find my pencil anymore. Will you tell us? Thanks

I hope this isn't a homework problem.

Units shown to verify proper calculation.

410 [rev/kW-hr] / 3600 [s/hr] = 0.114 [rev/kW-s]
1 / 0.114 [rev/kW-s]] = 8.78 [kW-s/rev]
8.78 [kW-s/rev] / 42 [s/rev] = 0.209 [kW] = 209 [W]

It is good answer. You can also use simply this formula:

T = (3600 x 1000) / (W x R) or

W = (3600 x 1000) / (T x R)

Where,

T = Time taken per revolution (in seconds)

W = Load in Watts

R = Rev/kWH from meter dial

W = (3600 x 1000) / (42 x 410) = 209 Watts

This is the watts drawn by the load. The name plate load will be larger than this because the voltage applied to the load is generally less than the rated voltage of the load due to the voltage drop in the system.

Keep in mind it is not a recommended way to measure the load, but you can get your approximate load amount.

Assumptions:

(1) The load is constant during the measured time.

(2) The energy meter is accurate (not fast or not slow)

See also http://cr4.globalspec.com/comment/453657 if you want to test the meter (either slow or fast).

- MS

You say "This is the watts drawn by the load." No, it's the watts input to the motor. The input to the load (the fan) is lower due to motor efficiency. If it were an ordinary motor and running at rated load, the rated load on the nameplate is the motor output (shaft) power.

The reason I say "If it were an ordinary motor...." is because some small appliances with built-in motors e.g. power tools give a rating in watts equal to the electrical input power (probably cos it looks better) which might apply to this fan.

Cheers..........Codey