ac/dc power supply efficiency

[[email protected]]

Hello to all,
making measurements on an ac/dc power supply (230Vac -> 24Vdc) I noticed that the cos phi is 0.56.
The active power is 65W and the reactive power is about the same value.
Now, I'm sure my load for power supply is 59W at 24Vdc (resistive load)
How would you compute the efficiency of the power supply, i.e. power of the load / active power or power of the load / (active+reactive power)?
Obviously there is a big difference. The current absorbed by the network is the sum of the two components related to active and reactive power, then I would suggest the latter calculation . Thank you. Greetings. Andrea

 

[Phil Allison]

Hello to all,

making measurements on an ac/dc power supply (230Vac -> 24Vdc) I noticed
that the cos phi is 0.56.

** No it is not !!!

The ratio of watts to "VA" may well be 0.56 though - aka the "power
factor".

The active power is 65W and the reactive power is about the same value.

** There is simply no "reactive power" involved.

Now, I'm sure my load for power supply is 59W at 24Vdc (resistive load)

** Easy enough to compute.

How would you compute the efficiency of the power supply, i.e. power of
the load / active power or power of the load / (active+reactive power)?

** Where do folk get these whacky ideas ??????

Efficiency has NOTHING to do with any of this nonsense.

Obviously there is a big difference.

** No such conclusion is "obvious" at all.

Efficiency = power input / power output.

The DC power supply you allude to could easily be close to 100% efficient.

Many are.

Think again.




..... Phil

 

[[email protected]]

Phil I'm not agree with you: Efficiency = power input / power output.
If so efficiency > 100%

Efficiency is power output/power input.
My question is if I have to compute power input like the sum of active and reactive power

Andrea

 

[[email protected]]

Efficiency = power input / power output.

no, it isn't. it is power output/power input
otherwise efficiency would be more than 100% because input power is always greater than output power

My question is if I have to consider input power as the sum of active plus reactive power
Andrea

 

[Phil Allison]

no, it isn't. it is power output/power input

** Slip of the tongue................

My question is if I have to consider input power as the sum of active plus
reactive power

** That is YOUR delusion.

Cos "reactive power " is imaginary and totally non existent in the example
you are considering.



.... Phil

 

[Phil Allison]

Phil I'm not agree with you: Efficiency = power input / power output.
If so efficiency > 100%

Efficiency is power output/power input.
My question is if I have to compute power input like the sum of active and
reactive power

Andrea

** Now you sound like a fucking, nut case troll.

Piss off.

 

[John S]

Hello to all,
making measurements on an ac/dc power supply (230Vac -> 24Vdc) I noticed that the cos phi is 0.56.
The active power is 65W and the reactive power is about the same value.
Now, I'm sure my load for power supply is 59W at 24Vdc (resistive load)
How would you compute the efficiency of the power supply, i.e. power of the load / active power or power of the load / (active+reactive power)?
Obviously there is a big difference. The current absorbed by the network is the sum of the two components related to active and reactive power, then I would suggest the latter calculation . Thank you. Greetings. Andrea

Using your numbers, your efficiency is 59/65 or about 90.8 %.

 

[Phil Allison]

"John S"

Using your numbers, your efficiency is 59/65 or about 90.8 %.

** Correct.

Andrea is another fucking TROLL


.... Phil

 

[amdx]

Hello to all,
making measurements on an ac/dc power supply (230Vac -> 24Vdc) I noticed that the cos phi is 0.56.
The active power is 65W and the reactive power is about the same value.
Now, I'm sure my load for power supply is 59W at 24Vdc (resistive load)
How would you compute the efficiency of the power supply, i.e. power of the load / active power or power of the load / (active+reactive power)?
Obviously there is a big difference. The current absorbed by the network is the sum of the two components related to active and reactive power, then I would suggest the latter calculation . Thank you. Greetings. Andrea

We interrupt this SED thread to apologize for the rantings of Phil.
He lives to show he is right without teaching the OP why.

Nut Case Acknowledgement Committee

 

[Phil Allison]

"amdx = fuckhead & troll "

We .......

** There is no " 'we" - just one, anonymous, smug, egomaniac ****.

He lives to show he is right without teaching the OP why.

** One cannot teach a pig to sing - nor a trolling fool.

And YOU are both.

You shit for brains porker.



..... Phil

 

[amdx]

"amdx = fuckhead & troll"



** There is no " 'we" - just one, anonymous, smug, egomaniac ****.




** One cannot teach a pig to sing - nor a trolling fool.

And YOU are both.

You shit for brains porker.



.... Phil

You must recall my previous request that you be creative in you
responses. Most of this was new, very good on you.
Thanks, Mikek

porker? lol

 

[whit3rd]

Phil I'm not agree with you: Efficiency = power input / power output.

If so efficiency > 100%



Efficiency is power output/power input.

My question is if I have to compute power input like the sum of active and reactive power

What is the computation FOR? If you're sizing a fuse or circuit breaker, it
might be useful to know the input in VA terms, as well as in real-power terms.

 

[[email protected]]

What is the computation FOR? If you're sizing a fuse or circuit breaker, it

might be useful to know the input in VA terms, as well as in real-power terms.

I have to deal with a certain number of loads feeded by that kind of power supply, so the amount of reactive power affects the choice of circuit breakers and power consumption from the wall.

I'm seriously thinking to replace the power supplies with others that have a better cos fi (0,96 vs. 0,56) to deal with less current from the wall (and also less harmonics)

 

[John S]

I have to deal with a certain number of loads feeded by that kind of
power supply, so the amount of reactive power affects the choice of
circuit breakers and power consumption from the wall.

Forget "reactive power". Measure the input current with a true RMS meter
and size your breakers and wiring based on that.

 

[[email protected]]

Forget "reactive power". Measure the input current with a true RMS meter

and size your breakers and wiring based on that.

I've done this, I obtain 0,72A @ 230Vac as input current.
I'll use this data to size breakers and wiring.
Using a different power supply, with a good active pfc, the current drops to 0,37A @ 230Vac for the same system - the second power supply is obviouslymore expensive but I'm wondering if reactive power will be figured out as a cost by the customer that know current consumption

 

[[email protected]]

I have to deal with a certain number of loads feeded by that kind of power supply,
so the amount of reactive power affects the choice of circuit breakers and
power consumption from the wall.

Fuses are actual rated with the I²t parameter, so you really want to
know the power supply peak current.

In a simple capacitor input power supply, current is drawn only just
before the positive or negative voltage peak. The half cycle is 180
degrees long, but the actual conduction angle might be as short as 18
degrees i.e 1 ms at 50 Hz.

However, the total charge for this half cycle must be transferred
during the conduction angle, i.e in this case 10 times the average
current.

Assuming a resistive load drawing 10 Arms (2.3 kW@230 V), the fuse
heating would be 1 A²s during each half cycle. However, if the same
charge needs to be transferred during 18 degrees (1 ms), the peak
current would be 100 A and the fuse heating 10 A²s, thus blowing the
fuse long before 2.3 kW power.

Putting such peaky loads on all three phases would put the _sum_ of
all three phases into the neutral wire, since the current peaks do not
overlap. The neutral wire heating would be up to 9 times the heating
of an individual phase conductor.

Only with overlapping currents will some (or all) phase currents
cancel out in the neutral conductor.

The conduction angle depends on the source impedance of the feeding
electric network and the size of the capacitor in the power supply. In
practice, conduction angles would be slightly longer than 1/10 of the
half cycle time.

You really need to measure the peak current or conduction angle using
a current transformer and oscilloscope.

making measurements on an ac/dc power supply (230Vac -> 24Vdc) I noticed that the cos phi is 0.56.
The active power is 65W and the reactive power is about the same value.
Now, I'm sure my load for power supply is 59W at 24Vdc (resistive load)

In the EU, power supplies greater than 75 W needs to have a good power
factor, requiring some PFC circuitry. It appears that your supply has
been designed to be just below that limit and hence does not need a
PFC.

I'm seriously thinking to replace the power supplies with others that have a
better cos fi (0,96 vs. 0,56) to deal with less current from the wall
(and also less harmonics)

At least in EU, buy one that is rated above 75 W and the power factor
is OK.

 

[John S]

I've done this, I obtain 0,72A @ 230Vac as input current. I'll use
this data to size breakers and wiring. Using a different power
supply, with a good active pfc, the current drops to 0,37A @ 230Vac
for the same system - the second power supply is obviously more
expensive but I'm wondering if reactive power will be figured out as
a cost by the customer that know current consumption

I think you have an instrumentation problem. Read Phil Allison's early
response to you for understanding.

If your input power is 65W and your input VA is 230*.72, then your power
factor is 65/166 or about .4 which does not agree with your initial post
of .56 (Cosine whatever).

Additionally, using the PFC, the output power is 65W and your input
power is .37*230 (85W) so the efficiency is now 76% (maybe).

One or more of your instruments may not be capable of measuring what you
need. Use caution when making measurements. Check your instruments'
specifications to see if they can handle the waveform factors. Do you
have an oscilloscope?

Cheers,
John S

 

[Phil Allison]

I have to deal with a certain number of loads feeded by that kind of
power supply, so the amount of reactive power affects the choice
of circuit breakers and power consumption from the wall.

** FORGET the term "reactive power " - it is meaningless for your
situation where there is NO reactive current.

I'm seriously thinking to replace the power supplies with others that have
a better cos fi (0,96 vs. 0,56) to deal with less > current from the wall
(and also less harmonics)

** You are obsessed with serious miscomprehensions.

If you have 65W load with a *power factor* of 0.56 that load is 65/0.56 =
116VA.

So the RMS current draw is just under 1 amp at 120VAC.

Inrush surge current will be the thing that gets you with such multiple
loads and you do not even know what it is.



.... Phil

 

[Phil Allison]

Fuses are actual rated with the I²t parameter,

** Only in relation to short surges.

so you really want to
know the power supply peak current.

** Which happens only at switch on !!!!

Assuming a resistive load drawing 10 Arms (2.3 kW@230 V), the fuse
heating would be 1 A²s during each half cycle. However, if the same
charge needs to be transferred during 18 degrees (1 ms), the peak
current would be 100 A and the fuse heating 10 A²s, thus blowing the
fuse long before 2.3 kW power.

** Absolute CRAP !!!!!!!!!!!

Fuses and breakers are rated with the RMS value of the current and that is
all you need to know.

Putting such peaky loads on all three phases would put the _sum_ of
all three phases into the neutral wire, since the current peaks do not
overlap. The neutral wire heating would be up to 9 times the heating
of an individual phase conductor.

** More absolute CRAP !!

Piss off FOOL !!!!!!


... Phil

 

[Jasen Betts]

Assuming a resistive load drawing 10 Arms (2.3 kW@230 V), the fuse
heating would be 1 A²s during each half cycle. However, if the same
charge needs to be transferred during 18 degrees (1 ms), the peak
current would be 100 A and the fuse heating 10 A²s, thus blowing the
fuse long before 2.3 kW power.

Putting such peaky loads on all three phases would put the _sum_ of
all three phases into the neutral wire, since the current peaks do not
overlap. The neutral wire heating would be up to 9 times the heating
of an individual phase conductor.

no. becauae they don't overlap, only three times the phase wire heating,