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

 

[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