help with calculation

[hidroela]

i got this homework problem that can not figure out

any help would be appreciated

Balance a 55-kVA-power load at 240 volt 3-phase a power factor of 90% and a 70 kW load at 120/240 volt single-phase at a power factor of 1.0 on a 120/240-volt three-phase, 4-wire service.

A = _______ ampere
B = _______ ampere
C = _______ ampere
N = _______ ampere

Best regards
HP

 

[skeshesh]

I think the forum's policy is that you post whatever attempt you made and help will be provided to reach the solution.

A couple of hints to get started: consider the power factor at the beginning so you only have to deal with VA, volts and amps; keep in mind that you have a 120/240 system which is a bit different from the 120/208 you may deal with more often.

 

[charlie b]

Please give us some idea of what you have tried to do on your own, before we offer any corrections or suggestions. We don't do homework for those who have not tried.

But let me ask you to confirm the statement of the problem. I am not certain what you would mean by a "120/240 volt, three phase system," or by a 70KW load having a 120/240V rating.

If you have a three phase delta secondary with 240 volts line to line, and if you center tap and ground one of the secondary windings, you could get both 120 and 240 out of the same transformer. But is that what you are dealing with? I don't think the description gives us enough information about the 70 KW load.

 

[hidroela]

this is what i got so far
-------------------------------------phase a--------------phase b-----------phase c------------ total
55 kva 3 phase------------------ 18333 va------------18333 va---------18333 va-----------55000 va
70 kva 1 phase------------------ 35000 va------------35000 va--------- ---------------------70000 va
------------------------------------53333 va------------53333 va---------18333 va------------125000 va

-------------------------------------phase a--------------phase b-----------phase c
55 kva 3 phase------------------ 104.65 A------------104.65 A-----------104.65 A
70 kva 1 phase------------------ 145.83 A------------145.83 A--------- ---------
------------------------------------250.48 A------------250.48 A-----------104.65 A


A phase I = 250.48 A
B phase I = 250.48 A
C phase I = 104.65 A
Neutral I = 0 A

 

[david luchini]

A phase I = 250.48 A
B phase I = 250.48 A
C phase I = 104.65 A
Neutral I = 0 A

I don't see the step from your VAs to your currents, so I'm not sure how you arrived at these values. But a balanced 55kVA, 3phase, 240V load would give you line currents of 132.3Amps, so having one current of 104.65 Amps is unlikely.

I see you followed skeshesh's advice about converting kW to kVA, but you also must pay attention to the factor that the power factors for the loads are different.

I did the calculation quickly, so it may be wrong, but I come up with currents of:

Ia = 132.3A
Ib = 381.9A
Ic = 423.4A
In = 0.0A

(based on a balanced 55kVA, 240V, 3ph load with a 0.9 lagging pf, and a 70kVA, 240V, 1ph load between B and C.)

 

[hidroela]

my bad i don't know neither
-------------------------------------phase a--------------phase b-----------phase c
55 kva 3 phase------------------ 132.31 A------------132.31 A-----------132.31 A
70 kva 1 phase------------------ 291.66 A------------291.66 A
-------------------------------------423.97 A------------423.97 A-----------132.31 A


A phase I = 423.97 A
B phase I = 423.97 A
C phase I = 132.31 A
Neutral I = 0 A

what i dont get is why you have Ib = 381.9A

 

[Dennis Alwon]

If the load is 70 kw single phase 120/240 are we to assume the load is balanced on 3 phases otherwise how could the neutral current be 0.

 

[hidroela]

If the load is 70 kw single phase 120/240 are we to assume the load is balanced on 3 phases otherwise how could the neutral current be 0.

i assumed the load is 240 volt

 

[Dennis Alwon]

i assumed the load is 240 volt

120/240 doesn't mean 240v to me. Bad question

 

[hidroela]

120/240 doesn't mean 240v to me. Bad question

i agree with you

 

[david luchini]

what i dont get is why you have Ib = 381.9A

...but you also must pay attention to the fact that the power factors for the loads are different.

This is where the importance of the loads having different power factors come into play. You cannot just add up the currents, but also must look at the phase relationships between the currents.

I you have a 240V delta system, you have 240V between each of the phases separated by 120 degrees. So you can assign system voltages of Vab=240<0, Vbc=240<-120 and Vca=240<-240.

You have (1) 240V, 3ph load and (1) 240V single phase load. In my mind, I simplify the three phase load into (3) single phase loads balanced between the phases. So, you have 1 load between A-B, two loads between B-C and one load between C-A (that's my configuration, yours has two loads between A-B, but that doesn't change anything.

We know that the 3 phase load has a power factor of 0.9. From this we know that the load current lags (or leads) the load voltage by an angle of 25.84 deg. So for the load current of the load between A-B, we have Iab=S/Vab = 18.333kVA/240V = 76.39A. But the load current is 25.84 degrees out of phase from the voltage, so we would note this as Iab=76.39<-25.84. Likewise, the load currents for the other two parts of the 3 phase load would be Ibc=76.39<-145.84 and Ica=73.39<-265.84.

For the single phase load (connected between B-C in my calc.) we have a 70kVA, 240V load with a unity power factor. The current for this load Ibc2=70Kva/240V=291.67A. Since it is unity, the angle of the load current Ibc2 is the same as the angle of Vbc, so Ibc2=291.67<-120.

To solve the line currents Ia, Ib and Ic, you would use KCL (with vector math):

Ia = Iab - Ica
Ib = Ibc + Ibc2 - Iab
Ic = Ica - Ibc - Ibc2.

I think this will give you the values I have indicated.

 

[skeshesh]

David beat me to it. I apologize if my "hint" was misleading - I did not mean to imply you can forget about the pf after considering KW vs. KVA, just that I like approaching that step first.

I still have to agree with the others that it's a badly worded question. It asks the reader/student to "balance" the loads to start off - in the case of this problem there is no way to balance the load assuming there's one single-phase load as the question implies.

Question for the OP: How close is what you stated to the actually problem statement? Is it possible that the actual question is worded differently and you're paraphrasing?

 

[hidroela]

This is where the importance of the loads having different power factors come into play. You cannot just add up the currents, but also must look at the phase relationships between the currents.

I you have a 240V delta system, you have 240V between each of the phases separated by 120 degrees. So you can assign system voltages of Vab=240<0, Vbc=240<-120 and Vca=240<-240.

You have (1) 240V, 3ph load and (1) 240V single phase load. In my mind, I simplify the three phase load into (3) single phase loads balanced between the phases. So, you have 1 load between A-B, two loads between B-C and one load between C-A (that's my configuration, yours has two loads between A-B, but that doesn't change anything.

We know that the 3 phase load has a power factor of 0.9. From this we know that the load current lags (or leads) the load voltage by an angle of 25.84 deg. So for the load current of the load between A-B, we have Iab=S/Vab = 18.333kVA/240V = 76.39A. But the load current is 25.84 degrees out of phase from the voltage, so we would note this as Iab=76.39<-25.84. Likewise, the load currents for the other two parts of the 3 phase load would be Ibc=76.39<-145.84 and Ica=73.39<-265.84.

For the single phase load (connected between B-C in my calc.) we have a 70kVA, 240V load with a unity power factor. The current for this load Ibc2=70Kva/240V=291.67A. Since it is unity, the angle of the load current Ibc2 is the same as the angle of Vbc, so Ibc2=291.67<-120.

To solve the line currents Ia, Ib and Ic, you would use KCL (with vector math):

Ia = Iab - Ica
Ib = Ibc + Ibc2 - Iab
Ic = Ica - Ibc - Ibc2.

I think this will give you the values I have indicated.

thanks for so detail explanation
i know i can digest all this theory with more time (i learn all this almost 30 year ago) an never used again. now my question is, there is a pre-elaborated formula
to do this at a no engineering knowledge level

 

[hidroela]

David beat me to it. I apologize if my "hint" was misleading - I did not mean to imply you can forget about the pf after considering KW vs. KVA, just that I like approaching that step first.

I still have to agree with the others that it's a badly worded question. It asks the reader/student to "balance" the loads to start off - in the case of this problem there is no way to balance the load assuming there's one single-phase load as the question implies.

Question for the OP: How close is what you stated to the actually problem statement? Is it possible that the actual question is worded differently and you're paraphrasing?

what i posted is a word by word of the original question

have to go to work

have all of you a nice day