Hi Everyone,
As a first time poster, I'd first like to say that this website has been invaluable throughout my university studies.
Now onto my question.
For the system in the included picture you are to calculate the currents in the secondary windings.
The Loads are
Single phase loads: 30kW each
Motor M1: 50kVA, cosθ=0.5 lagging
Motor M2: 160kVA, cosθ=0.8 lagging
Now the answer given is 777A and the reason i'm skeptic that it's right, is i know it came from a companion textbook we are using that has had a plethora of mistakes in it.
So, here is my attempt at this solution, hopefully someone can let me know if i'm on the right track or not.
To me the easiest way is to investigate a single phase and since the three phase loads are balanced this should be fine to do so. Starting with the single phase load
I = P/E = 30e3/120 = 250A, so each single phase load will have a current of 250A
Now my understanding regarding the three phase loads is that the actual power i.e with power factor must be determined and the divided by three, the number of phases. So the values become.
M1=(50e3*0.5)/3 = 25e3/3 = 8.33e3
M2=(160e3*0.8)/3 = 128e3/3 = 42.67e3
then using the equation S =√3 E I, I can be found. But since the Δ connected I is the line current divided by √3, the found ( I ) will need to be multiplied by √3, or as i have done, cancle √ 3 from the equation.
So I = S/E, where E is the line voltage. Applying this results in.
M1: I = 8.33e3/120 = 69.42 A
M2: i = 42.67e3/120 = 355.58 A
The total line current per phase is then the sum of all of these, which is
I = 250+69.42+355.58 = 675 A.
This answer is off by 102A and I personally know three phase is a real weak point for me. I'm a radio tech by trade and I simply don't have much to do with it. Unfortunately it's not even worth contacting my lecturer( i'm a correspondance student) as he is always unavailable(still waiting for a reply on a question i asked 1 month ago.
I have tried asking local electricians about it, but they are practical sorts of people and have no idea about the math ( which is sort of scary when you think about it).
Hopefully someone out there can help me out with this, as i really need to know what's what before i attend my final exam next friday.
Many thanks in advance
Dylan Whittaker
As a first time poster, I'd first like to say that this website has been invaluable throughout my university studies.
Now onto my question.
For the system in the included picture you are to calculate the currents in the secondary windings.
The Loads are
Single phase loads: 30kW each
Motor M1: 50kVA, cosθ=0.5 lagging
Motor M2: 160kVA, cosθ=0.8 lagging
Now the answer given is 777A and the reason i'm skeptic that it's right, is i know it came from a companion textbook we are using that has had a plethora of mistakes in it.
So, here is my attempt at this solution, hopefully someone can let me know if i'm on the right track or not.
To me the easiest way is to investigate a single phase and since the three phase loads are balanced this should be fine to do so. Starting with the single phase load
I = P/E = 30e3/120 = 250A, so each single phase load will have a current of 250A
Now my understanding regarding the three phase loads is that the actual power i.e with power factor must be determined and the divided by three, the number of phases. So the values become.
M1=(50e3*0.5)/3 = 25e3/3 = 8.33e3
M2=(160e3*0.8)/3 = 128e3/3 = 42.67e3
then using the equation S =√3 E I, I can be found. But since the Δ connected I is the line current divided by √3, the found ( I ) will need to be multiplied by √3, or as i have done, cancle √ 3 from the equation.
So I = S/E, where E is the line voltage. Applying this results in.
M1: I = 8.33e3/120 = 69.42 A
M2: i = 42.67e3/120 = 355.58 A
The total line current per phase is then the sum of all of these, which is
I = 250+69.42+355.58 = 675 A.
This answer is off by 102A and I personally know three phase is a real weak point for me. I'm a radio tech by trade and I simply don't have much to do with it. Unfortunately it's not even worth contacting my lecturer( i'm a correspondance student) as he is always unavailable(still waiting for a reply on a question i asked 1 month ago.
I have tried asking local electricians about it, but they are practical sorts of people and have no idea about the math ( which is sort of scary when you think about it).
Hopefully someone out there can help me out with this, as i really need to know what's what before i attend my final exam next friday.
Many thanks in advance
Dylan Whittaker
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