Power factor question

[DT1878]

revising for my 301 on Thursday and Ive come stuck on a question, any ideas?



A 30KW 3 phase delta connected motor has a power factor of 0.86 lagging. Calculate the line current.

answer is 50.35a but i dont know how to get there





thanks in advance

 

[Guest123]

Ok

30000/(400x1.732x0.86) = 50.35A.

 

[pmac10]

rearrange the equation
power=volts*amps*SQ RT 3*power factor
to make amps the subject
30000=400*50.35*root3*0.86

 

[pmac10]

that was quick lenny

 

[DT1878]

thanks lads thats great

 

[Guest123]

that was quick lenny

Story of my life....

 

[DT1878]

erm another quick problem.......


A medium sized factory has a three phase load of 50kVA and 30kVAr. Calculate the power factor?

a. 0.8



again im struggling with the formula, i must be arranging it in the wrong order, sorry to be a pain but anybody help us out?

 

[pmac10]

post the formulae, what is 30kvAr, should that read 30kvA, is 50kvA apparent power and 30kvA true power?

 

[DT1878]

well it will be this wont it?

The power factor for a three-phase electric motor can be expressed as:
PF = P / [(3)1/2 U I] (2)
where
PF = power factor
P = power applied (W, watts)
U = voltage (V)
I = current (A, amps)

 

[pmac10]

where did the motor come from???
can you type the question as it reads??

 

[DT1878]

sorry i must be using the wrong forumula then, this is exactly how the question reads:


A medium sized factory has a three phase load of 50 kVA and 30 kVAr. Calculate the power factor

 

[pmac10]

to me, and im pretty sure im right(99.9%), you have a factory with apparent power 50kvA and true power of 30kvA, power factor=true power/apparent power, 30kvA/50kvA=0.6 which means your answer is wrong. text books with incorrect answers are not uncommon. does this sound plausible?? also i'm not sure what kvAr is?

 

[DT1878]

to me, and im pretty sure im right(99.9%), you have a factory with apparent power 50kvA and true power of 30kvA, power factor=true power/apparent power, 30kvA/50kvA=0.6 which means your answer is wrong. text books with incorrect answers are not uncommon. does this sound plausible?? also i'm not sure what kvAr is?

kvar is Reactive Power so i think the question is right im just not getting it.

or you could be right and the paper is wrong lol

 

[pmac10]

you are right, i didnt think you would that at level 3, i cant cope with these questions after a few beers so ill get back to you tomorrow nite if thats ok

 

[DT1878]

haha ok pmac no probs, cheers anyway

 

[pushrod]

erm another quick problem.......


A medium sized factory has a three phase load of 50kVA and 30kVAr. Calculate the power factor?

a. 0.8



again im struggling with the formula, i must be arranging it in the wrong order, sorry to be a pain but anybody help us out?

pythagorus gets you true power at 40kw.
then 40/50 = 0.8


edit; just to put a bit more meat on the bones of it - its your power triangle that you need to know , with kVA on the hypotenuse , kW on the adjacent and kVAr on the opposite.

(notice it is another 3, 4, 5 triangle)

 

[pmac10]

had a look through my HNC notes and icant give you an answer from those two figures, i would need the phase angle between the supply voltage and the supply current or use as pushrod says pythgorus with impedance triangle of the reactive component with impedance (Z) on the hypotenuse,resistance (R) on the adjacent and reactance(X) on the opposite,
that make any sense???

 

[pmac10]

pythagorus gets you true power at 40kw.
then 40/50 = 0.8


edit; just to put a bit more meat on the bones of it - its your power triangle that you need to know , with kVA on the hypotenuse , kW on the adjacent and kVAr on the opposite.

(notice it is another 3, 4, 5 triangle)

actually that makes far more sense, im over complicating it, but the cosine of the phase angle equals the power factor

 

[DT1878]

cheers anyway lads still a bit confused on that one, if it does come up in the exam al just have to guess lol

 

[pushrod]

This is what you need to know.( It is very similar to the impedance triangle)

VA² = W² + VAr²
so in your example W² = VA² - VAr²
W = sq rt ( VA² - VAr² )
W = sq rt (50² - 30²)
W = 40 k watts

Power factor = real power ÷ apparent power
= 40 ÷ 50
= 0.8

hope that is a bit clearer - if it isn't at least try and remember that since the hypotenuse part of the triangle is the longest that means that the VA figure will be bigger than the W.
So if the question says the apparent power is 15 VA what is the true power in watts you at least know you are looking for a smaller answer

 

[DT1878]

ahhhh brilliant pushrod thats much clearer thanks a lot.

 

[JUD]

This is what you need to know.( It is very similar to the impedance triangle)

VA² = W² + VAr²
so in your example W² = VA² - VAr²
W = sq rt ( VA² - VAr² )
W = sq rt (50² - 30²)
W = 40 k watts

Power factor = real power ÷ apparent power
= 40 ÷ 50
= 0.8

hope that is a bit clearer - if it isn't at least try and remember that since the hypotenuse part of the triangle is the longest that means that the VA figure will be bigger than the W.
So if the question says the apparent power is 15 VA what is the true power in watts you at least know you are looking for a smaller answer

Glad to see you're making use of the ² & ÷ symbols, but what about poor old √

Really good stuff by the way

 

[pushrod]

Glad to see you're making use of the ² & ÷ symbols, but what about poor old √

Really good stuff by the way

Cheers, tried it but couldn't get the 251 code to work on my ' pooter

 

[JUD]

You get it using HTML code &# 8730; (without space)

 

[pushrod]

You get it using HTML code &# 8730; (without space)

i couldn't find it in the full list of html codes , but now i see you actually included that one in your actual post - DOH!

 

[JUD]

i couldn't find it in the full list of html codes , but now i see you actually included that one in your actual post - DOH!

Yer just had a look at list. Not as big as I thought it was. I must have got them from a different website. If I find it again I'll post it up.

 

[JUD]

This is what you need to know.( It is very similar to the impedance triangle)

VA² = W² + VAr²
so in your example W² = VA² - VAr²
W = sq rt ( VA² - VAr² )
W = sq rt (50² - 30²)
W = 40 k watts

Power factor = real power ÷ apparent power
= 40 ÷ 50
= 0.8

hope that is a bit clearer - if it isn't at least try and remember that since the hypotenuse part of the triangle is the longest that means that the VA figure will be bigger than the W.
So if the question says the apparent power is 15 VA what is the true power in watts you at least know you are looking for a smaller answer

Would I be right in saying that

VA = √(W² + VAr²)
W = √(VA² - VAr²)
VAr = √(VA² - W²)

Seems to work with the 30, 40, 50 from this thread.

 

[pushrod]

Would I be right in saying that

VA = √(W² + VAr²)
W = √(VA² - VAr²)
VAr = √(VA² - W²)

Seems to work with the 30, 40, 50 from this thread.

Certainly would

 

[JUD]

Here's another good way of remembering it (for me anyway )

S = Apparent Power (VA), P = True Power (W), Q = Reactive Power (VAr)

S = √(P² + Q²)
P = √(S² - Q²)
Q = √(S² - P²)

PF (Power Factor) = P ÷ S

 

[DT1878]

thanks for the explanations