Energy metering

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#1
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Total votes: 0

Dear friends 

Hi and good day

I have got a lamp which is written that is 50W . I have connected it to my ampermeter and voltmeter and saw that the voltage is around 220V and current is 155ma . so the Lamp must be 220*0.15=33w (Am i right ?)

how much energy does this lamp consume during :

a) one second ?

b)1 hour ?

I'm working on an energy meter project using Ade7758 . I have connected this lamp to the energymeter IC and i can see the voltage and the current are like those my multimeter . then I read the active and apparent energy from the Ic . which is calculated by the ic itself . in a 0.5 second timer , i read them and I got the number  331 which shows that is related to the product of I and V. then i read the energy during one second , and i got 662 - As expected - then I read this values during an hour and the sum of these values was about 2240000 -as expected again-

But I cant understand the relation between these values and the real consumed energy .

please kindly help me to understand what I must do 

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Total votes: 1

20 samples/second, 3600 seconds per hour = 72,000 samples/hour. 2240000 / 72000 = 31 watt-hours.

 

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Total votes: 0

Thank you very much for your kind reply 

A new question is where did you find the "20samples per second" ?

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Total votes: 1

I expect it has more to do with the effective gain configured by the analog front-end and the ADE7758 gain registers.

"Experience is what enables you to recognise a mistake the second time you make it."

"Good judgement comes from experience.  Experience comes from bad judgement."

"Wisdom is always wont to arrive late, and to be a little approximate on first possession."

"When you hear hoofbeats, think horses, not unicorns."

"Fast.  Cheap.  Good.  Pick two."

"We see a lot of arses on handlebars around here." - [J Ekdahl]

 

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Total votes: 1

Given the sparsity of information provided , I made some simple assumptions about the system (voltage division ratio , current transformer ratio, register settings, communication rate).

 

You measured 33 watts instantaneous power with ammeter and voltmeter. Assuming this is a steady state system, the energy consumed in 1 hour would be 33 watt-hour.  Assuming the system gain is 1.0 , the results provided (331, 662, 2240000) indicate 20 samples/second. 

 

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Total votes: 1

The OP rounded 0.155 down to 0.15, so the reported calculation of 33 W should actually be 34.1 W.

 

The OP also has already stated that measurements at 1 second intervals give accumulated reading of 662.  The real (metered) energy over the same period is 34.1 joules (1 joule = 1 watt second), this implies about a 19.41:1 ratio between readings and joules, i.e. gain of 19.41.  A watt hour is 3600 joules, so the expected accumulated readings should be about 19.41 * 34.1 * 3600 = 2,383,200 which agrees fairly well with the reported value of 2,240,000 (6.4% off).

"Experience is what enables you to recognise a mistake the second time you make it."

"Good judgement comes from experience.  Experience comes from bad judgement."

"Wisdom is always wont to arrive late, and to be a little approximate on first possession."

"When you hear hoofbeats, think horses, not unicorns."

"Fast.  Cheap.  Good.  Pick two."

"We see a lot of arses on handlebars around here." - [J Ekdahl]