Help Needed With these 3 Phase 4 Electricial Science Questions!!!

RJ59

I really am getung it hard to answer these 3 ..they are the only1s outa 20 i cant answer so any help would be most greetful!!!

1.A 3phase delta connected inducatuib motor has a 15kw output 400v supply 80% efficiant, p.f. 0.85 Calculate Phae currents of winding???


2.The Voltage in a Balanced Delta load is 400v,line current is 10a @lagging.Calculate Totla Consumed??


3.The magnetic flux per pole in a dc machine is 2m wb and the poleforce diemssions are 0.2 x0.2 Calculate flux denisty at the poleface???


Any help would really be apprecited i cant find the answers at all have checked scaddens books for clues and just cant work them out i really need to no the answers and how to do them thanks for any replys:D

Bruthal

RJ59 wrote: »

I really am getung it hard to answer these 3 ..they are the only1s outa 20 i cant answer so any help would be most greetful!!!

1.A 3phase delta connected inducatuib motor has a 15kw output 400v supply 80% efficiant, p.f. 0.85 Calculate Phae currents of winding???


2.The Voltage in a Balanced Delta load is 400v,line current is 10a @lagging.Calculate Totla Consumed??


3.The magnetic flux per pole in a dc machine is 2m wb and the poleforce diemssions are 0.2 x0.2 Calculate flux denisty at the poleface???

Question 1
3 phase power is root 3 V x I x PF x efficiency
so I = power/(root3 x V x PF x efficiency)
So thats 15000/(1.73 x 400 x 0.85 x 0.8) = around 33 amps which is the line current of each phase.

Update as forgot the question asked for each winding current, (pointed out by mceebe) so-
In a delta motor the line currents split so divide the 33A by root 3 to get 19Amps approx.

The reason the 3 phase power is volts x amps x 1.73 (root 3) is the 3 phases currents at any one time will add up to 1.73 times the current measured in on of the phases because of the way the current splits into the 3 delta windings, if it was star the 3 windings would have the line current but the voltage across the star connected windings would now be split by line voltage/root 3. So the power will always add up to (Line voltage x line current x root 3 x power factor)KW`s . Leave out the power factor and you have the KVA required to run the motor, thats why a generator output is on KVA, its the volts x amps it can deliver, where as the motor kw will be its useful power output. Its useful power may be 15kw, but it will require 17 or 18kva of power to get 15kw output out of it.
I left out efficiency there, just included power factor.

Question 2
Delta load 400v, line current 10 amps,
As in first question, 3 phase power is root3 x V x I x PF. We are not told the PF but for total consumed we dont need to know, we only need to know that for the KW output, so what we are really calculating is the KVA needed to supply the load.
So its 1.73(root3) x 400v x 10A = 6.92kva

I cant remember the exact details to do question 3, and could be corrected on the first 2 here, but they look ok.

M cebee

question 1 robbie

33amp will be your line current approx won't it?

multiply by 0.58 for phase current?

i can't understand the rest with the typos:)

Bruthal

M cebee wrote: »

question 1 robbie

33amp will be your line current approx won't it?

multiply by 0.58 for phase current?

i can't understand the rest with the typos:)

O yea i missed out on the question asking for the phase current of each winding which in delta would be 33amps line current/root 3 which is about 19 amps.

Dividing by root 3 (1.73) would be the same as multiply by 0.58

2011

Qustion 1. I agree with Robbie except: If the efficiency is 80% you must devide by 0.8, not multiply by it. This is because the kW in will be 20% larger than the kW out, deviding the output by 0.8 will achieve this.

Question 2. I would assume that the question is for the total power consumed and I would guess that you were give the PF but did not give it by
mistake. With the information given Robbie is correct.

Question 3. Units for the dimensions of the pole face are required. The answer is weber per m^2 and the unit for this is tesla.

So if the area is .2 x .2m the the flux density is:

B =0.002 x 0.2 x 0.2 = 80 x 10^-6 tesla

Bruthal

2011 wrote: »

Qustion 1. I agree with Robbie except: If the efficiency is 80% you must devide by 0.8, not multiply by it. This is because the kW in will be 20% larger than the kW out, deviding the output by 0.8 will achieve this.

Question 2. I would assume that the question is for the total power consumed and I would guess that you were give the PF but did not give it by
mistake. With the information given Robbie is correct.

Question 3. Units for the dimensions of the pole face are required. The answer is weber per m^2 and the unit for this is tesla.

So if the area is .2 x .2m the the flux density is:

B =0.002 x 0.2 x 0.2 = 80 x 10^-6 tesla

Well kw = 1.73 x 400 x I x 0.85 x 0.8 i would of thought.
so to get I you divide both sides of equation by 1.73 x 400 x 0.85 x 0.8
to be left with 15000/(1.73 x 400 x 0.85 x 0.8) = Line current.
Divide again by 1.73 to get phase current or in the original equation above use 3 instead of 1.73 for phase current in a delta.


I you look at it again, i believe i am dividing by 0.8 because
its 15000/(1.73 x 400 x 0.85 x 0.8)=33 line amps. The 0.8 in the brackets is making the result within brackets smaller, therefor the lower the efficiency the higher the total result. This result is the input current. You are still dividing the 15000 by each figure within brackets.

It is the same as 15000 divided by 1.73 divided by 400 divided by 0.85 divided by 0.8



If you divide 100 by 10 and then divided by 0.8 as example, then you are doing the same as 100/(10 x 0.8) =12.5
The 0.8 has made the answer bigger, as it has been multiplied by the numbers after the division. Do the same calculation i did in question 1 but replace the 0.8 with 0.4 and you get 63 or so amps, an answer i would expect when the motor is now 40 percent efficient.

And in the end the phase current was shown to be 19 amps or there abouts. So 19 x 400 x 3 is 22.8kva
Multiply this KVA result by 0.85 pf and then by 0.8 efficiency to give you around the 15kw output of motor.

Or use the line current 33A. Kva is root 3 x 400 x 33amps is 22.8kva

As a matter of interest, i did the whole calculation again but divided by the 0.8 within the brackets and the answer for line current is then 20 amps, clearly not correct.

The answers above may be 1 or 2 amps out as i did rough calculations.

Question 2 asks for the total consumed rather than the output, which i would assume is the KVA. So the PF is not required in the question.

Question 3 you will have to give me some grinds on:)
Although i know its the pole face area divided by the flux density probably to give you tesla per meter square in your calculation or something like that.

Bruthal

2011 wrote: »

Question 2. I would assume that the question is for the total power consumed and I would guess that you were give the PF but did not give it by mistake.

I suppose it depends how you look at this, the total power output would not be the same as the total consumed i dont think. I think the total consumed would be the amount a generator would need to be able to provide, as to run a load requires enough power to provide for its output plus its losses, which for a balanced 3 phase load is 1.73 x V x I (KVA). If the PF is taken account of then we are calculating the loads output power. The total the load consumes will be higher than its output and is just the KVA value.

Maybe im looking at it wrong though.

M cebee

yes-i'm sure robbie was correct

he was dividing by the 0.8


how are motors normally quoted -input or output kilowatts? i'm gone confused

obviously if it's horspower it's output

Bruthal

M cebee wrote: »

yes-i'm sure robbie was correct

he was dividing by the 0.8


how are motors normally quoted -input or output kilowatts? i'm gone confused

obviously if it's horspower it's output

The KW is the motors useful output rating, so the motor in this case can output 15KW. But a generator to power it would have to be able to deliver 23KVA which is the input to the motor, which is the 33amps, probalby 31 in actual fact as i roulghly calculated. But of the 33amps anyway only about 22amps is actually doing any work, the rest is lost in the efficiency and PF losses. But the generator or power supply needs to deliver 33 amps to get the 15kw from the motor. So the KVA input would be 23 or so for the 15kw motor power output

Horse power is the same thing, the motors output, just a different unit.
15/0.746 = 20HP

So to answer, the motor is rated by its KW output.

2011

3 phase power is root 3 V x I x PF x efficiency

This is the line I think is wrong.

This it why!

The output =15kW

The input is always greater than the output.

The motor is 80% efficient so the input must be greater than 15kW

Therefore the input is:

15/0.8= 18.75kW

If you were to multiply by 0.8 then you would get 14.4 kW. There is no motor that has an input of 14.4kW and an output of 15kW

2011

Input Power = √3 x V(line) x I(line) x Cos θ

I(line) = Input Power /{√3 x V(line) x Cos θ}

I(line) = 18750 / {√3 x 400 x 0.85}

I(line) = 18750 / 588.9 =31.84 A

The current in each winding will be the phase current when connected in delta.
Therefore:

I(phase) = I(line) / √3 = 31.84 / √3 = 18.38 A

Bruthal

2011 wrote: »

Look it is really simple!

The output =15kW

The input is always greater than the output.

The motor is 80% efficient so the input must be greater than 15kW

Therefore the input is:

15/0.8= 18.75kW

If you were to multiply by 0.8 then you would get 14.4 kW. There is no motor that has an input of 14.4kW and an output of 15kW

Your right it is simple, but tell me where am i multiplying by 0.8. I am dividing 15000 by 4 numbers, and i believe your missing that. And im surprised i must say

Id take a closer look, we are looking for the input current,
15000/(1.73 x 400 x 0.85pf x 0.8eff) = 31.87 line amps imput.


Now lets recalculate with 40 percent efficiency.
15000/(1.73 x 400 x 0.85pf x 0.4eff) = 63.75 line amps imput.
Now tell me where the mistake is there. Get the calculator out and try it.


Now even more simple, the output is 15KW as per question
Input is 1.73 x 31.87 amps x 400v = 21.5KVA
Now unless im mistaken 21.5 is greater than 15 no?

And just to top it all off, multiply 21.5 x 0.85pf and what do you get? = Just over 18 kw input power if you want to look at it that way

2011

yes-it's not complicated but robbie divided
to correctly calculate the line current from output power

He did later, but when I saw the line below I stopped reading:

3 phase power is root 3 V x I x PF x efficiency

M cebee

yes-it's not complicated but robbie divided
to correctly calculate the line current from output power

quote "So thats 15000/(1.73 x 400 x 0.85 x 0.8) = around 33 amps which is the line current of each phase"

M cebee

"3 phase power is root 3 V x I x PF x efficiency"

this formula is output power

nevermind anyhow:)

Bruthal

At the end of the day, the question was whats the phase current and i calculated the line current, as did`t see the phase current part but then put that in,

Power in motor (KW) = 1.73 x V x I x PF x Efficiency.
Therefor I = KW/(1.73 x 400 x 0.85pf x 0.8eff)

So where you are getting me multiplying by 0.8 to get a bigger answer i fail to see. The smaller that eff number in the above formula the bigger the I answer will be. Same as the PF, the lower the PF decimal the higher the I we are looking for in the question, so its right beyond much doubt. Change the 0.8 to 0.4 above and re calculate, the answer is almost double. So how is the 0.8 is making the input smaller is a mystery.

Bruthal

2011 wrote: »

He did later, but when I saw the line below I stopped reading:

What do you mean i did later? I had the 33 amps line current in very first post, a simple 33/1.73 = 19 amps,

Now you got 18.3 amps

Now 2010, was i just lucky?

Bruthal

2011 wrote: »

He did later, but when I saw the line below I stopped reading:

Its neither here or there, the OP asked a question, i answered, no point arguing again over words, if my answer was wrong to the 2 questions then let me know,

You said you stopped reading after i said
power = 1.73 x V x I x PF x efficiency.

Why was this? Suddenly im clueless and not worthy is it.
Maybe i should of said this is the output, but still.
Then you claim i later did divide by the 0.8 instead of multiply like you claimed i incorrectly did. Total rubbish, its exact same formula,

I guess i was lucky getting the 33 line and 19 phase , although the 33 should be 31 or so:pac:

2011

You said you stopped reading after i said
power = 1.73 x V x I x PF x efficiency.

Why was this? Suddenly im clueless and not worthy is it.

Relax, life is too short. It has been a long day at work!
I stopped reading because I thought I saw an error and I tried to address it.
As it happens I was trying to cook the dinner, talk on the phone and feed the dog at the same time, end of!

I had the 33 amps line current in very first post, a simple 33/1.73 = 19 amps,

Now you got 18.3 amps

Now 2010, was i just lucky?

Hands up, you are 100% correct!!!

Sorry, I did the same sum in a different way and as you correctly point out I got the same answer. I just did it in a different order as this is the way that makes the most sense to me.

Anyway which ever way the OP does it it is correct getting the same answer.

2011

how are motors normally quoted -input or output kilowatts? i'm gone confused

Motors are normally quoted in terms of output. So a 15kW motor will have a 15kW output

2011

Regarding question 2 Robbie, I agree that your kVA calculation is 100% correct. It is not possible to do any further calculations with the information given.

My point was that I think the question given to the OP should have included the PF. As you know this is required to calculate the power as power is measured in watts as opposed to VA.


The OP said:

2.The Voltage in a Balanced Delta load is 400v,line current is 10a @lagging.Calculate Totla Consumed??

I think it should have said something like:

2.The Voltage in a Balanced Delta load is 400v,line current is 10a @ 0.8 lagging.Calculate Total Power Consumed??

Perhaps I am wrong, but clearly there are a few typos and it is a little strange not to state if power or VA are required in the answer!

Bruthal

2011 wrote: »

Regarding question 2 Robbie, I agree that your kVA calculation is 100% correct. It is not possible to do any further calculations with the information given.

My point was that I think the question given to the OP should have included the PF. As you know this is required to calculate the power as power is measured in watts as opposed to VA.

Yes i think your right, as mceebe said, lots of typos by the OP,
Im on here not to correct anyone, but to give my assesment as best i can, its not as good as it used to be as i am out of the game a few years now.

Dont forget, everyone can be wrong, and when we post we have to expect we may be corected, and also when we correct, our corrections may be wrong, and i actually remember mcebee and yourself saying its a constant learning curve which i completely agree with.

The OP asked for 3 questions to be answered and now he is probably worse off than he was before he asked them.

It should not be a competition about who is right, but a discussion for everyone to benefit, especially the OP. But along the way maybe the people answering can be assisted too.

I myself got carried away when i thought others were wrong in other threads, but again its a learning curve in every aspect.

I have been reminded of some aspects of the 3 phase stuff from this thread already, and other threads before. Everyones opinion should be respected regardless if they have 1 post or 10,000. But i know you would agree on that anyway.

So anyway, as you said, life is too short, and the main reason im happy about you saying that is your thinking the exact same thing i was thinking.

Its great, may the debates long continue, and the agreements also.

Bruthal

2011 wrote: »

My point was that I think the question (2) given to the OP should have included the PF. As you know this is required to calculate the power as power is measured in watts as opposed to VA.

Yes i agree, looking more carefully now your right, although question 2 asked for the total input and did not mention power. So we are left in the dark about that. But i think you are likely to be right and there are some missing elements in the question.

Bruthal

2011 wrote: »

This is the line I think is wrong.

3 phase power is root 3 V x I x PF x efficiency

This it why!

The output =15kW

The input is always greater than the output.

The motor is 80% efficient so the input must be greater than 15kW

Therefore the input is:

15/0.8= 18.75kW

If you were to multiply by 0.8 then you would get 14.4 kW. There is no motor that has an input of 14.4kW and an output of 15kW

Your correct here too, that line is wrong, the efficiency is only included to show the motor output rather than the power in the 3 phase circuit to the motor, the electrical imput power is the root 3 x V x I x PF.

Obviously enough too.