Please verify my answer!!!

[Le7316]

This is one of my hw and just wondering if someone can verify my answer,

Q: A single phase load at 480 V draws 33.7 A at 15degree lagging. What is the power factor and apparent power of the load?

A:

R = V/I = 480/33.7= 14.24 ohm
True power (P) = I² x R = (33.7)^2 x 14.24 = 16.17 Kw
cos 15 = 0.96 = pf
Apparent power (S) = VI = P/pf = 16.17Kw/cos 15 = 16.75 KVA

Is there anything wrong with my calculation ?

Thanks

 

[Besoeker]

This is one of my hw and just wondering if someone can verify my answer,

Q: A single phase load at 480 V draws 33.7 A at 15degree lagging. What is the power factor and apparent power of the load?

A:

R = V/I = 480/33.7= 14.24 ohm
True power (P) = I² x R = (33.7)^2 x 14.24 = 16.17 Kw
cos 15 = 0.96 = pf
Apparent power (S) = VI = P/pf = 16.17Kw/cos 15 = 16.75 KVA

Is there anything wrong with my calculation ?

Thanks

Yes.
Apparent power is just V*I. You don't need to calculate resistance, actually it's impedance anyway since the PF isn't unity.
Real power is V*I*cos phi.

Try that.

 

[Le7316]

how about this ?

Apparent power = 480v x 33.7 A = 16176 VA
Real Power = (480v x 33.7A) cos 15 = 15624.8W

PF = Real Power/ Apparent power = 0.9

 

[Besoeker]

how about this ?

Apparent power = 480v x 33.7 A = 16176 VA
Real Power = (480v x 33.7A) cos 15 = 15624.8W

PF = Real Power/ Apparent power = 0.9

OK. You got the real and apparent.

But where did the 0.9 come from?
You had already (correctly) determined PF in post#1.

I'll help if I can but you need to put the effort in to learn.

 

[charlie b]

PF = Real Power/ Apparent power = 0.9

This is a round off error. If you divide the calculated real power by the calculated apparent power, and carry it out to at least three significant digits, you will get the same answer you had in your first post.

 

[charlie b]

Let?s talk about simplification of effort. In your first post, you first calculated R as V/I, then you used that value in the formula (P) = I² x R. But if you put those two together, you get

• (P) = I² x R
• (P) = I² x (V/I)
• (P) = I x V

So as Besoeker pointed out, your first effort included an unnecessary step.

In your second post, you did something similar.

• Apparent power = 480v x 33.7 A = 16176 VA
• Real Power = (480v x 33.7A) cos 15 = 15624.8W
• PF = Real Power/ Apparent power

But looking at that third formula, you can get,

• PF = Real Power/ Apparent power
• PF = (Apparent Power times cos 15) / Apparent power
• PF = cos 15

Here again, you could have gone directly to the PF = cos 15, without the first two steps. Indeed, as Besoeker also pointed out, you did that in your first post.

We engineers like to be lazy. :happyyes: We don't like to do extra work. :happyno:

 

[Le7316]

Let?s talk about simplification of effort. In your first post, you first calculated R as V/I, then you used that value in the formula (P) = I² x R. But if you put those two together, you get

• (P) = I² x R
• (P) = I² x (V/I)
• (P) = I x V

So as Besoeker pointed out, your first effort included an unnecessary step.

In your second post, you did something similar.

• Apparent power = 480v x 33.7 A = 16176 VA
• Real Power = (480v x 33.7A) cos 15 = 15624.8W
• PF = Real Power/ Apparent power

But looking at that third formula, you can get,

• PF = Real Power/ Apparent power
• PF = (Apparent Power times cos 15) / Apparent power
• PF = cos 15

Here again, you could have gone directly to the PF = cos 15, without the first two steps. Indeed, as Besoeker also pointed out, you did that in your first post.

We engineers like to be lazy. :happyyes: We don't like to do extra work. :happyno:

Thank you for all the explanation guys. I appreciate it !!! Anyways, the extra work (PF = 15624.8W/16176 VA = 0.9) that i did was just to verify i got the right answer thats all and looks like i made a right decision to join this forum :thumbsup:. Now, lets go further if its 3 phase then the equation going to be like this correct ?

Apparent power = 1.73 x 480v x 33.7 A = 27984.5 VA
Real Power = 1.73 (480v x 33.7A) cos 15 = 27030.9W

 

[Besoeker]

Thank you for all the explanation guys. I appreciate it !!! Anyways, the extra work (PF = 15624.8W/16176 VA = 0.9) that i did was just to verify i got the right answer thats all and looks like i made a right decision to join this forum :thumbsup:. Now, lets go further if its 3 phase then the equation going to be like this correct ?

Apparent power = 1.73 x 480v x 33.7 A = 27984.5 VA

Well, I'd have used sqrt(3) in the calculation.
Most calculators, even the most basic, can do that.
So you get 28018 VA.
Not such a big error - but an easily avoidable one.

Real Power = 1.73 (480v x 33.7A) cos 15 = 27030.9W

More simply, since you have already calculated it, VA*cos15.
So 27063 W.

 

[Besoeker]

We engineers like to be lazy.

Nah!!!
Just efficient.

 

[Le7316]

Well, I'd have used sqrt(3) in the calculation.
Most calculators, even the most basic, can do that.
So you get 28018 VA.
Not such a big error - but an easily avoidable one.


More simply, since you have already calculated it, VA*cos15.
So 27063 W.

Ok. Thanks but the answer is still right if i dont use sqrt(3) but 1.73

 

[Besoeker]

Ok. Thanks but the answer is still right if i dont use sqrt(3) but 1.73

Yes.
Near enough for most practical purposes.

 

[Le7316]

Thank you sir !!!