HELP with Transformer Calculation

[Ingenieur]

Please don't confuse real power (W) with apparent power (VA)
They are not equivalent.
Now, can we get back to the OP's question?

'active' power, real would imply imaginary power, trust me it's real (reactive) lol

no one is misunderstanding the concept except you
nope, you claim power is only watts, it is actually va
watt = J/sec = C V / sec = VA
s is in va, so the units of sqrt (P^2 + Q^2) must also be
a va ~ w, we call them different things for clarity of convention, not because they are different, all are v x a

now, what assumptions need made to answer the op's question?

 

[Besoeker]

'active' power, real would imply imaginary power, trust me it's real (reactive) lol

VAr is not power.
Go in peace.

 

[Ingenieur]

VAr is not power.
Go in peace.

In electric power transmission and distribution, volt-ampere reactive (var) is a unit by which reactive power is expressed in an AC electric power system. Reactive power exists in an AC circuit when the current and voltage are not in phase. Per EU directive 80/181/EEC (the "metric directive"), the correct symbol is lower-case "var",^[1] although the spellings "Var" and "VAr" are commonly seen, and "VAR" is widely used throughout the power industry. The term var was proposed by the Romanian electrical engineer Constantin Budeanu and introduced in 1930 by the IEC in Stockholm, which has adopted it as the unit for reactive power.


the reason you avoid the 'what is the assumption' question is you were in err making one because none need be made
man up

btw VAr is incorrect usage, should be var

 

[Besoeker]

In electric power transmission and distribution, volt-ampere reactive (var) is a unit by which reactive power is expressed in an AC electric power system. Reactive power exists in an AC circuit when the current and voltage are not in phase. Per EU directive 80/181/EEC (the "metric directive"), the correct symbol is lower-case "var",^[1] although the spellings "Var" and "VAr" are commonly seen, and "VAR" is widely used throughout the power industry. The term var was proposed by the Romanian electrical engineer Constantin Budeanu and introduced in 1930 by the IEC in Stockholm, which has adopted it as the unit for reactive power.


the reason you avoid the 'what is the assumption' question is you were in err making one because none need be made
man up

btw VAr is incorrect usage, should be var

OK. but var is not real power as I'm sure you know.
You pay for energy usually in kWh. Note the W in there. Not VA.
Now on yer bike.

 

[Ingenieur]

OK. but var is not real power as I'm sure you know.
You pay for energy usually in kWh. Note the W in there. Not VA.
Now on yer bike.

it is a form of power
glad you learned something
some pay a charge for var consumption

assumption?

lol

 

[petersonra]

VAR is an acronym. Acronyms are spelled out in capital letters.

 

[Ingenieur]

VAR is an acronym. Acronyms are spelled out in capital letters.

nope
"the correct symbol is lower-case "var",^[1]"



https://eur-lex.europa.eu/LexUriServ/LexUriServ.do?uri=OJ:L:1980:039:0040:0050:EN:PDF#page=6
not an acronym
it is a unit

 

[Besoeker]

it is a form of power
glad you learned something
some pay a charge for var consumption

Hence PFC.

 

[Ingenieur]

FACTS textbook
'var' not 'VAR'

 

[ActionDave]

no assumption required

S (va) = P (w) + jQ (var)
since they give P in watts you know pf = 1
they do not mention S or Q

if they gave S va then an assumption may be required: does eff include reactive 'losses'?
or is it all R losses?
you can still calc current but losses won't be accurate

if they only give Q var same as S

Power IS watts regardless of PF.

nope
total power S is va, hence xfmr POWER capacity rating in va
reactive in var
but they are equivilent

what assumption must be made?

watt = J/sec = C V / sec = VA

Please don't confuse real power (W) with apparent power (VA)
They are not equivalent.
Now, can we get back to the OP's question?

'active' power, real would imply imaginary power, trust me it's real (reactive) lol

no one is misunderstanding the concept except you
nope, you claim power is only watts, it is actually va
watt = J/sec = C V / sec = VA
s is in va, so the units of sqrt (P^2 + Q^2) must also be
a va ~ w, we call them different things for clarity of convention, not because they are different, all are v x a

now, what assumptions need made to answer the op's question?

VAr is not power.
Go in peace.

In electric power transmission and distribution, volt-ampere reactive (var) is a unit by which reactive power is expressed in an AC electric power system. Reactive power exists in an AC circuit when the current and voltage are not in phase. Per EU directive 80/181/EEC (the "metric directive"), the correct symbol is lower-case "var",^[1] although the spellings "Var" and "VAr" are commonly seen, and "VAR" is widely used throughout the power industry. The term var was proposed by the Romanian electrical engineer Constantin Budeanu and introduced in 1930 by the IEC in Stockholm, which has adopted it as the unit for reactive power.


the reason you avoid the 'what is the assumption' question is you were in err making one because none need be made
man up

btw VAr is incorrect usage, should be var

OK. but var is not real power as I'm sure you know.
You pay for energy usually in kWh. Note the W in there. Not VA.
Now on yer bike.

it is a form of power
glad you learned something
some pay a charge for var consumption

assumption?

lol

VAR is an acronym. Acronyms are spelled out in capital letters.

nope
"the correct symbol is lower-case "var",^[1]"



https://eur-lex.europa.eu/LexUriServ/LexUriServ.do?uri=OJ:L:1980:039:0040:0050:EN:PDF#page=6
not an acronym
it is a unit

Hence PFC.

FACTS textbook
'var' not 'VAR'

There was a time when I regretted not pursuing a chance to be an engineer. You two have convinced me I did not miss an opportunity but rather dodged a bullet.

 

[Ingenieur]

Simple.
You can't calculate the current without knowing the power factor.
Since it is not given, you have to make an assumption about it.

If I were answering the question I'd state that I had made that assumption.

Anyway, the OP has had his question answered, analysed, and disected.
Shall we leave it at that, please?

if so simple why not grasp it?
if they give you P (and no pf, S, Q etc) then the pf =1, not an assumption, but fact, you are assuming in err otherwise

simple
pf = cos(ang V to I) = P (w) / S (va) = pf is unit less, ie, w and va cancel meaning they are equivalent
P = 1.732 ( ) x V (v) x I (A) x pf ( ) = units va, not w

you can leave it, I'll do as I please

 

[Ingenieur]

There was a time when I regretted not pursuing a chance to be an engineer. You two have convinced me I did not miss an opportunity but rather dodged a bullet.

and the profession is better off for it

 

[kingpb]

if so simple why not grasp it?
if they give you P (and no pf, S, Q etc) then the pf =1, not an assumption, but fact, you are assuming in err otherwise

It is just as correct, and in the real world IMO more accurate to assume a power factor and as long as the assumption is stated, it is just as correct as your assumption that it is 1.0. My reason; if I size a circuit and assume power factor is 1.0, then my current is 6.8A. If I assume a power factor of 0.85, which as you are aware is typical in the power industry for that assumption, my current 8.0A. If I have capacity for 8 I have capacity for 6.8; but the same does not hold true the other way around.

Yes, you can argue that there isn't any difference between the two, and using significant figures there probably is not, but every case may not be so trivial.

The point being, my stated assumption and reasonable logic is just as good as yours. IYO, you may think yours is better and that's fine, but no reason to get cheeky about it.

 

[Besoeker]

It is just as correct, and in the real world IMO more accurate to assume a power factor and as long as the assumption is stated, it is just as correct as your assumption that it is 1.0. My reason; if I size a circuit and assume power factor is 1.0, then my current is 6.8A. If I assume a power factor of 0.85, which as you are aware is typical in the power industry for that assumption, my current 8.0A. If I have capacity for 8 I have capacity for 6.8; but the same does not hold true the other way around.

Yes, you can argue that there isn't any difference between the two, and using significant figures there probably is not, but every case may not be so trivial.

The point being, my stated assumption and reasonable logic is just as good as yours. IYO, you may think yours is better and that's fine, but no reason to get cheeky about it.

Well put that man. I agree. It is an assumption.

 

[Ingenieur]

It is just as correct, and in the real world IMO more accurate to assume a power factor and as long as the assumption is stated, it is just as correct as your assumption that it is 1.0. My reason; if I size a circuit and assume power factor is 1.0, then my current is 6.8A. If I assume a power factor of 0.85, which as you are aware is typical in the power industry for that assumption, my current 8.0A. If I have capacity for 8 I have capacity for 6.8; but the same does not hold true the other way around.

Yes, you can argue that there isn't any difference between the two, and using significant figures there probably is not, but every case may not be so trivial.

The point being, my stated assumption and reasonable logic is just as good as yours. IYO, you may think yours is better and that's fine, but no reason to get cheeky about it.

it is a basic, elementary level test question to test basic concepts and math skills, not a trick question requiring assumptions
they give P
they do NOT mention pf, S or Q, and would do so if relevant, they are not, so the only logical deduction is the load is 1500 va = 1500 w + 0j var
don't get personal with the insults, no one is getting 'cheeky' with the possible exception of you

 

[kingpb]

it is a basic, elementary level test question to test basic concepts and math skills, not a trick question requiring assumptions

Unless it was you taking the test, or have a copy in front of you, that sounds like a lot of assumptions; just say'in.:blink:

they give P they do NOT mention pf, S or Q, and would do so if relevant, they are not, so the only logical deduction is the load is 1500 va = 1500 w + 0j var

So, what it seems is you are asserting is that they want the person to only ASSUME that the pf=1 and thus KW=KVA :?

Just checking, cause it appears that it's ok to ASSUME as long as they agree with your position, and not ok to make assumptions that are counter to your position.

 

[Ingenieur]

It is just as correct, and in the real world IMO more accurate to assume a power factor and as long as the assumption is stated, it is just as correct as your assumption that it is 1.0. My reason; if I size a circuit and assume power factor is 1.0, then my current is 6.8A. If I assume a power factor of 0.85, which as you are aware is typical in the power industry for that assumption, my current 8.0A. If I have capacity for 8 I have capacity for 6.8; but the same does not hold true the other way around.

Yes, you can argue that there isn't any difference between the two, and using significant figures there probably is not, but every case may not be so trivial.

The point being, my stated assumption and reasonable logic is just as good as yours. IYO, you may think yours is better and that's fine, but no reason to get cheeky about it.

it is a test question, to test basic understanding and math skills
they give power P
they do mention pf, S or Q, so only logical to assume they DO NOT factor in
it is not I getting 'cheeky'

 

[Ingenieur]

Unless it was you taking the test, or have a copy in front of you, that sounds like a lot of assumptions; just say'in.:blink:


So, what it seems is you are asserting is that they want the person to only ASSUME that the pf=1 and thus KW=KVA :?

Just checking, cause it appears that it's ok to ASSUME as long as they agree with your position, and not ok to make assumptions that are counter to your position.

he gave us the numbers, I used them with 0 assumptions

they want you to know the load is 1500 W, not va, not var...1500 W
you do not need to assume, they tell you the load is 1500 W, not 1500 va, not 1500 W + 100j var, etc...simply 1500 W (and everyone knows W = va
otherwise dividing P/v would not = Amps, but Watt/Volts

I assumed nothing, the stated load is 1500 W

 

[kingpb]

he gave us the numbers, I used them with 0 assumptions

they want you to know the load is 1500 W, not va, not var...1500 W

I assumed nothing, the stated load is 1500 W

Zero assumptions? Since by vector math: S(va)=P(w)+jQ(var), then only way to get S(va) = P(w) is to assert that jQ(var) = 0; i.e. pf=1
which means you had to Assume pf=1 since none was given.

I can honestly say, where I went to school, you would have just gotten at least 10pts marked off for not stating the assumption that pf=1 and then gotten a zero for arguing with the professor that there were no assumptions, or he would have laughed and ridiculed you in front of the class. :thumbsup:

 

[Besoeker]

they do mention pf, S or Q, so only logical to assume

not a trick question requiring assumptions.

To assume or not to assume to paraphrase Willie Wiggledagger.