Solve Transmission Losses: 50kW @ 230V, PF 0.6 Lagging

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The discussion centers on calculating transmission losses for a load of 50kW at 230V with a power factor of 0.6 lagging, where the supply system has a resistance of 0.1 ohms. The user initially struggles to connect the formulas for real power, apparent power, and transmission loss, eventually realizing that using the apparent power leads to the correct loss calculation of 13.13kW. Another user shares a similar problem involving a 2MW load at 6.35kV and a power factor of 0.9, ultimately finding the power loss to be approximately 1.22kW after correcting their approach. The discussion highlights common challenges in power engineering calculations and the importance of understanding the relationships between power, current, and resistance. Overall, the thread serves as a collaborative space for troubleshooting power loss calculations in electrical engineering.
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I need help with this tutorial question I cannot solve!

a consumer demands 50kW at an effective voltage of 230V. supply system can be represented by a 0.1ohm resistance. Find the transmission losses in the event of the load being supplied with a power factor of 0.6 lagging.

I know the answer works out to be 13.13kW, but can't figure out the solution!
 
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Welcome to PF!

Hi SGL18! Welcome to PF! :wink:

Show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
 
i've used the power factor=real power/apparent power (P/S) and from the power triangle i can get the real, reactive and apparent power. Can also get the impedance angle from the power factor using inverse cos function! (cos-1 0.6 = 53.13) and i know other formulas such as real power=p=V.I.cos (angle) but I'm just not sure how to put it together to get transmission loss!?
 
i Just mananged to solve this using Apparent power instead of real power! P=IV so I=P/V or rather I=S/V then P loss= I (squared)R! Thanks for replying anyway tiny-tim! just started a new power engineering class soe rest assured I'll have plenty more questions! :)
 
Hi SGL18 (and/or anyone else who reads this),

I have a similar question but can't seem to work it out from what you've posted.

Question:

A 6.35kV single phase transmission line has a series resistance of 0.01 ohms supplies a 2MW load of 0.9 lagging power factor. Find the power loss in the tansmission line.

Ans: 1.22kW

I know its not particularly difficult but I have tried everything I can and am close to suicide :/

Please help.
 
Welcome to PF!

Hi MBgigs! Welcome to PF! :wink:

Show us what you've done so far. :smile:
 
Hi and thanks for replying,

Well I figured find the Power and subtract the given power of 2MW. P = VI* so I need I. Therefore V = IR or IZ.

6350 = (0.01)(I)

I = 635000 (Already I see something is wrong)

I then subbed in for P = VI* and get a ridiculous figure not worth mentioning. I've thrown away (or recycled :) ) my scrap paper working which I won't do from now on but I've tried many other methods using the many available formula's but I am not getting anywhere.

Its been 3 years since I've done anything like this and I've jumped right into the deep end as this is just the basics for 3 phase power.

MBgigs
 
Hi MBgigs! :smile:
MBgigs said:
A 6.35kV single phase transmission line has a series resistance of 0.01 ohms supplies a 2MW load of 0.9 lagging power factor. Find the power loss in the tansmission line.

Ans: 1.22kW
MBgigs said:
Well I figured find the Power and subtract the given power of 2MW. P = VI* so I need I. Therefore V = IR or IZ.

6350 = (0.01)(I)

Oh I see, you're trying to find I from V and R. :redface:

No, you must find I from V and W (and then use power loss = I2R, as SGL18 did) …

what do you get? :smile:
 
I gave that a go but guess I don't know what to do next:

2000000 = (6350)(I)

I = 314.96A

314.96^2 x 0.01 = 992W
 
  • #10
MBgigs said:
I gave that a go but guess I don't know what to do next:

2000000 = (6350)(I)

What about the power factor? :wink:
 
  • #11
2000000 / 6350 x 0.9 = 349.96I

349.96^2 x 0.01 = 1.224kW (1.2kW) :)

Thanks again for your help. I may be in serious trouble this year :/

All the best.
 

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