From high school science I recall that W = VA. (power = current * voltage) 

 

So why is the output of my old UPS rated at 3200 VA or 2000 W? 

Power factor would be my Guess

Maybe a PF of 0.62 .

Jay

What's that when it's at home? 

1 hour ago, Bobster said:

From high school science I recall that W = VA. (power = current * voltage) 

Only in special instances, is this true.

Namely, with a DC source or a purely resistive or a resonant load.

In practice, inductive and capacitive loads cause a phase shift between the current and the voltage.

And power actually equals the amount that the voltage sine wave and current sine wave are in phase.

The actual relationship is Power = V x I x Cos (Phi), where Phi represents the angle that the Voltage and current are out of phase ( the power factor).

Cos(Phi) = 1, when the voltage is in phase with the current, in that special instance VA and W will be equal.

 

 

9 hours ago, phil.g00 said:

Only in special instances, is this true.

Namely, with a DC source or a purely resistive or a resonant load

 

What? Next I'll find out there's another 3 letters in the alphabet. Modern education, I tell you...

Seriously, it did occur to me that my puzzlement might have something to do with all the electrical stuff I did at school being DC and not AC.

I recall I've written essay lengths about this in the past on this very forum.

So yeah, the first thing to understand is the AC voltage as we use it daily is an average, a so-called RMS value. If you put a 230VAC waveform on an oscilloscope, you will see that the peaks are actually around +-325V. The equivalent DC voltage (if you were to rectify it, and then dump the peaks into the valleys to get a flat line) would average out at 230V. So that's the first thing.

The second thing to understand is that while I=V/R according to your physics class, the reality is actually that I = V/(R+X), where X is the impedance caused by inductive components. An inductive component is basically something that has a magnetic component to it, so in the beginning as the voltage ramps up and there is no magnetic flux in this component, it will accept a large inrush of current , but then as the magnetic flux builds up it will start to resist the flow of current and the impedance (X) will increase. This new-found resistance to current always lags behind the voltage, and what this means is that if you were to put a current clamp and put that on the same oscilloscope as your voltage waveform, you will see that the two waveforms don't line up.

The amount of shift between the two waveforms can be measured as an angle, and that angle is usually called by the greek letter theta. The more X in the system, the larger your theta will be.

(DC systems don't care about X, once the initial inrush current is over, X is essentially zero because the voltage is constant).

So now I get to the third thing you need to understand. While P=VI, you can only use the RMS values (aka the averages) if their wavecorms line up perfectly. If there is a phase shift, then it stops working. This is because you have to multiply the two waveforms you get, using the instantaneous voltage and current values at infinitely many points and add it all up (known in mathematical circles as integration). If the two are exactly in phase, P=VI.

It turns out that as long as the V and I waveforms are sinusiodal, then P = VI*cos(theta).

But in a modern home... the current is not necessarily sinusiodal. Enter non-linear loads such as cheap SMPSes...

I will end the story here 🙂

Edited July 26, 20196 yr by plonkster

Something interesting to keep in mind, that will help your overall understanding: Irrigation valves. Those little 24V jobbies you burry in a box somewhere in your garden. They are meant to be driven by 24VAC. If you use 24VDC, you will burn them out in the long run. And here is why: The solenoid in those valves have an inductive component, so under AC they have a higher impedance than on DC. They literally have a higher resistance on AC than on DC and therefore pass more current on DC than on AC.

That is also why 12VDC is enough to switch a 24VAC irrigation solenoid 🙂

Thanks all for all the info. I did realise during all the reading that my training was a little more up to date than 70s high school - IE I worked for what is not Telkom in the late 70s and then fixed computer power supplies in 1980. Clearly all that experience doesn't make me an engineer 🙂 

Edited July 26, 20196 yr by Bobster
inclusivity

Yeah technically my explanation still dumbed it down. It's more like P(t) = V(t)^2/R+X(t), that is instantaneous power at time t is the voltage (at the time) squared divided by the (constant) resistance plus the additional impedance X at the time. Because the X(t) function does not necessarily create a sine-wave (where dealing with non-linear loads), nor is it necessarily in phase with V(t), all bets are off. All you can say with certainty is that the real power is going to be less than perfect. So inverter makers (and generator salespeople) tend to assume your power factor will be at least 0.8 or better, and hence you get the two values, with the power almost always 80% of the VA value.

VAs are better for selling stuff of course, because they are mooaar. Also see PMPO for audio stuff.

22 hours ago, Bobster said:

From high school science I recall that W = VA. (power = current * voltage) 

 

So why is the output of my old UPS rated at 3200 VA or 2000 W? 

In short and simple layman's terms; VA is what power you put in and W is what power you get out. A VA rating will always be higher than the W rating due to losses in the conversion. It is often indicative of a systems efficiency regarding AC.

22 hours ago, KevinM said:

In short and simple layman's terms; VA is what power you put in and W is what power you get out. A VA rating will always be higher than the W rating due to losses in the conversion. It is often indicative of a systems efficiency regarding AC.

No, you've misunderstood the explanations above, it is to describe the capacity of the machine for different types of loads, not the efficiency of the machine.

On 2019/07/26 at 11:43 AM, plonkster said:

Also see PMPO for audio stuff.

That one I do know about.

On 2019/07/26 at 7:04 PM, KevinM said:

In short and simple layman's terms; VA is what power you put in and W is what power you get out. A VA rating will always be higher than the W rating due to losses in the conversion. It is often indicative of a systems efficiency regarding AC.

I don't think so. The input voltage and current are stated separately. The 2000W/3200VA is stated as output.

Edited July 28, 20196 yr by Bobster
grammar

On 2019/07/26 at 7:04 PM, KevinM said:

VA is what power you put in and W is what power you get out

The VA vs P of a system actually has nothing to do with the efficiency of the inverter. It depends solely on the loads.

I have a bunch of old iron transformers for MR16 downlights, and I drive LED lamps with them. The LED lamps draw about 10% of what the transformers were designed for, so this means the magnetic flux is not properly discharged through the secondary of the transformer, which means the residual magnetic flux in the core ends up fighting the current change on the primary side as the voltage swings up and down. This causes a rather horrible power factor of around 0.2, which literally means 200W of real power looks like 1000VA to the inverter.

That means that 200W of lighting takes up a third of the 3000VA inverter's capacity... but if you measure on the DC side you will see only 200W of actual power is used from the batteries.

On 2019/07/25 at 9:05 PM, Bobster. said:

From high school science I recall that W = VA. (power = current * voltage) 

 

So why is the output of my old UPS rated at 3200 VA or 2000 W? 

@KevinM @Bobster. according to my understanding 

 

The volt-ampere (VA) is the power that the inverter provides, while the watt (W) is the power that gets converted into useful work.

 

It’s like this: I’m the air conditioner, and you’re the inverter. If I need 1,000 dollars from you, you have to give me 1,200 in total, and I will return 200 of it because I won’t use it.

 

The 1,200 represents the volt-amperes.

The 1,000 represents the watts that I actually need.

 

Now, to complete the transaction, you, as the power source, must provide the full 1,200, but I only use 1,000 and return the remaining 200.

I don’t use them.

 

If there’s another load at the same time, it might use that 200.

 

So, when calculating consumption, whether on an electricity bill or from a battery, you count the watts.

 

But when you’re checking if a generator or inverter can handle the load, you look at the KVA.