Caluculating Real Power

[andrew_JKTA]

Hi,

I am trying to measure the power consumption of different appliances connected to the mains. I have used the PCI-6024E to measure the voltage across the appliance and the current through it (using a current shunt).

Since this is connected to the mains, I should use the equation:

Real Power = (RMS Voltage) * (RMS Current) * power factor

where power factor is cos(phase angle)

How do I measure or calculate this phase angle because I have two unknowns in one equation at the moment.

Is there another method of finding the real power when the only known quantities I have are voltage and current?

Thank you very much,

Andrew

[Gabi1]

i guess this is not directly a Labview question 🙂

well, you need another instrument: the wattmeter.
this instrument will measure the instantaneous real power by measuring the induced magnetic field.
as you know, the value might be lower than directly calculated from VoltAmpere. the ratio is the powerfactor.

note: even if you dont saturate the wattmeter, it might overload and burn. this is specially the case where powerfactor is very bad (below 90%), and the wattmeter might still be loaded with very (!) high Current and Voltage, yet giving low amplitude results.

[andrew_JKTA]

Yeah, sorry, didn't know where else to post it

So does this wattmeter measure the power factor or the instantaneous real power?

I need to measure the real power consumed by the appliance over a period of time (about 10 hours) and not just instantaneously, hence I'm using the DAQ board.

Is there some other way to just measure the power factor or am I just misunderstanding you?

Thanks very much for your very quick reply!!

[Gabi1]

no problem 🙂

the wattmeter measure the instantaneous real power. you can record it over 10h, and you will get an array of 10h of measurement of real power consumption :). this is the same as recording any other value for 10 hours.

i do not know of another way to measure power factor, as you need both the real power (measured with a wattmeter) and the observed power (measured with ampermeter and voltmeter, then multiply VA).

a wattmeter, especially analog, is relatively cheap to purchase in any electronic shop. but you better purchase a digital one, with analog output, so you can connect to your card input. check that it is overpower protected (so that it wont burn your card which cost 100times more, in case you run in overpower consumption).

good luck!

[andrew_JKTA]

So what you're saying is, I don't need to measure voltage and current anymore, I can just directly measure the power using a wattmeter?

Then connect the wattmeter to my DAQ so it can automatically record the readings from the wattmeter?

Thanks

[andrew_JKTA]

Also, surely the accuracy from a wattmeter would be much lower than if I used my DAQ board?

Thanks

[Gabi1]

nono!

let me rephrase:

to measure the aparent power, you need to measure the current and the voltage on the line (using a voltmeter and an ampermeter). you probalby do so already using some instrumentation.

to measure the real power, you need a wattmeter.

to record all those values, you need communication with your computer. in your case you do it with the 6025 DAQ board, so i suggest your wattmeter has an analog output, to connect as a voltage reader to an analog input of your card.

note: if you use an analog wattmeter, you could (but not recommended!!) connect the leads of the wattmeter internal coil windings directly to the daq card (you would need to disconnect them from the needle coil). you would need also to calibrate the value in to reflect power. be aware: you will (most probably) burn your card!

[andrew_JKTA]

But I'm trying to measure the power consumption, so surely all I need is the real power and I don't need to know what the apparent power is?

[Gabi1]

in this case yes, a wattmeter would be enough 🙂

yet, if you already have ampere and current meters connected, you can just record the whoel bunch, as there are lots of cases where apparent power is required (and is sometimes more important than real power)

for example, electricity companies will charge you differently if your apparent power differe significantly from the real power. the reason being that the reactance involved induces quite a bit of losses on the line (and somebody needs to pay the price 🙂 ). so they charge in W(real power), but check on apparent power (AV).

[andrew_JKTA]

ok

thanks very much for your help!!