Phase 6 science exam papers

sorry the power factor is normally a lagging power factor never leading and the phase are 120 degrees out of phase..

OK, so you understand that the neutral current is the phasor sum of the 3 phase currents?

Do you understand basic trigonometry (cos, tan, sin and Pythagoras theorem)?

Are you only unclear about how to calculate the phasor sum of the three currents?

kingden2006 wrote: »

sorry the power factor is normally a lagging power factor never leading and the phase are 120 degrees out of phase..

In "real life" power factor is generally lagging, but not always.
This is because most loads have a larger inductive component than capitative.

Remember anything with a coil (such as a motor or fluorescent light) is inductive.
Inductive loads cause the current to lag the voltage.

Capacitive loads on the other hand cause the current to lead the voltage.
That is why capacitor banks are connected to improve "bad power factor".

yes i understand basic trigonometry 2011 and i know that an inductor lags and capacitor leads the voltage.
yes thats it 2011 just unclear on how to calculate the phasor sums...you hit the nail on the head there.
your very good at all this,ever think about teaching?

kingden2006 wrote: »

doing phase 6 in sligo IT arse biscuits..

I teach in DIT, would have been happy to explain in person if you were there.

ah thanks arse biscuits you"ve been great answering all the questions i"ve put up and bruthal and 2011,i know were ill go if i get stuck in phase 6..

Have a look at this and this:

Do you think you can work out the angles A & B?

.

would i not need the length of 2 sides for that?

kingden2006 wrote: »

would i not need the length of 2 sides for that?

No.
You already have the information you require in post #56

Look at it this way:

From this it should be possible to work out angle A.
Using the same technique work out angle B.

Does this make sense?

no can"t seem to get my head around it 2011,you"d think i would remember stuff like this especially since i"ve only done phase 4 this year,and got all credits..um

kingden2006 wrote: »

no can"t seem to get my head around it 2011,you"d think i would remember stuff like this especially since i"ve only done phase 4 this year,and got all credits..um

I am not familiar with the syllabus for any of the apprenticeship phases.
Have you covered this type of question yet?

Can you understand the diagram in my last post?
I derived it from the diagram in post #56.

no i would not have covered this type of question yet,yes i understand the phasor diagram in 56 and can get as far as what you have done in that but i"m struggling after that for some reason...

kingden2006 wrote: »

no i would not have covered this type of question yet

Then don't worry.

yes i understand the phasor diagram in 56

In many ways that is the hardest part.
If you drew that diagram as I did you would get most of the marks for the question. The rest is just a little bit of maths.

Once you cover it in college it should make sense.

My advice is to revise what you have already covered to date.
That would be far more beneficial that trying to get ahead of the rest of your class.

yes and your advise is good advise 2011,thats exactly what ill be doing from tomorrow until sunday,ill keep you posted on how i"m getting on,and thanks for all the help it means a lot,think i"ll just relax now.
thanks 2011 and a happy new year to you.

kingden2006 wrote: »

thanks 2011 and a happy new year to you.

You are welcome, happy new year to you too.

BrianDug

Just to add to whats already been said, il explain how I look at these problems it may or may not be an easier way for you to visualise whats happening. I want to try avoid giving you the answer but just to hopefully point you in the right direction.

The diagram that 2011 has drawn is really the first thing you should sketch out as its much easier to visualise whats happening then.

My approach:

1. Choose a reference point for the angle start point, so all other angles are measured from this point. I choose L1 as my 0 degree reference. So from there L2 is 120 degrees and L3 is 240 degrees naturally enough.
2. Break each phase into a cos & sine component e.g ( 3.45 , -5.87). Then add up the cos column and sine column separately before taking the square root of the cos & sine component to get the neutral current.

So as and example say we broke the each phase into sine and cosine components and ended up with L1 = ( 3.45 , -5.87), L2 = ( 5.85 , 1.99), L3 = (6.40 , 2.22) we would add up all the cos components (3.45+ 5.85+6.40) = 15.7 and the sine components (-5.87+1.99+2.22) = -1.66.
Take the square root of this to get the neutral current .i.e sqrt(15.7^2 + 1.66^2) = N-current. [note:if you have a minus sign change it to positive, it wont effect the total neutral current]

2. To break down a phase into a sine and cosine component the first thing (as in step 1) is to find the distance from your reference point in degrees. So say we had phase L3 20A @ 0.95 lagging. If you draw the phasor as 2011 has done and mark your angle in (rough sketch) it would lag L3 by 18.19 degrees, but now here is the key part, how many degrees is this from your 0 degree REFERENCE point?? [ it will be 240+18.19 = 258.19 degrees.

3. To break this into cos and sine components it would be : cos component = 20cos(258.19) = -4.0933
The sine component would = 20sin(258.19) = -19.5766
So once you have done this for all the phases sum them and take the sort to get the neutral current, as for the angle think about it.
note: be careful of leading angles, again draw a rough sketch to figure out the total degrees from your reference.

4.21 amps in the neutral,is this right?

bassey

Wondering if any one has any past/sample papers to work through for phase 6?

kingden2006 wrote: »

4.21 amps in the neutral,is this right?

No.

8.94A

kingden2006 wrote: »

4.21 amps in the neutral,is this right?

You may have gone wrong at the end, the total x value is -5.58
when you square this value, just leave the minus out as minus by minus is positive.

no went wrong on the second part arse biscuits,yes i got 8.93 amps...
cheers for the correction...

anyone tell me where im going wrong on this one?
I1=40 AMPS AT UNITY.
I2=30 AMPS LAG VOLTAGE BY 30 DEGREES
I3=30 AMPS LEAD VOLTAGE BY 30 DEGREES..
VERTICAL AND HORIZONTAL COMPONENTS.
I GOT...
VC1=40
HC1=0
VC2=-25.98
HC2=15
VC3=0
HC3=-30

=VC=+14.02 HC=-15
MY ANSWER IS 20.53,I KNOW THAT THIS IS WRONG.

THIS IS ANOTHER ONE IM GOING WRONG SOMEWHERE.

I1=20AMPS AT UNITY
I2=30 AMPS AT .866 LAGGING WHICH IS 30 DEGREES
I3=40 AMPS AT 30 DEGREES LEADING.

I AGAIN GOT

VC=20
HC=0
VC=- 25.98
HC=+ 15
VC=0
HC=-40...AND I GOT

VC=-5.98 HC=-25 AND AFTER I SQUARED THEM BOTH AND GOT THE ROUTE I GOT 25.70?

I DONE IT LIKE THIS VC1 20 COS 0=20
HC1 20 SIN 0=0
VC2 30 COS 150=25.98
HC2 30 SIN 150=15
VC3 40 COS 270=0
HC3 40 SIN 270=-40.
IM GETTING SOME OF THEM RIGHT MAINLY THE ONES THAT ARE LAGGING AND AT UNITY BUT SEEM TO BE GOING WRONG ON THESE TWO.
PROBABLY SOMETHING SIMPLE.

With this type of question most of the marks will be awarded for the method used and drawing the above diagram.
A mathematical error would result in only a small reduction in marks.
I would not get too hung up on simple maths.

BrianDug

HC1 20 SIN 0=0
VC2 30 COS 150=25.98
HC2 30 SIN 150=15
VC3 40 COS 270=0
HC3 40 SIN 270=-40.

Hey,

If you did a quick sketch (as we keep telling you to do :-D ) you would see your error immediately!!!!
If L3 is 30 degrees LEADING, the angle from your 0 degree reference is 210 degrees not 270.

2011 wrote: »

With this type of question most of the marks will be awarded for the method used and drawing the above diagram.
A mathematical error would result in only a small reduction in marks.
I would not get too hung up on simple maths.

The marking system is 1 or 0

kingden2006 wrote: »

anyone tell me where im going wrong on this one?
I1=40 AMPS AT UNITY.
I2=30 AMPS LAG VOLTAGE BY 30 DEGREES
I3=30 AMPS LEAD VOLTAGE BY 30 DEGREES..
VERTICAL AND HORIZONTAL COMPONENTS.
I GOT...
VC1=40
HC1=0
VC2=-25.98
HC2=15
VC3=0
HC3=-30

Angle 1 = 0
Angle 2 = -120-30=-150
Angle 3 = -240+30=-210

X1=40.cos0=40 Y1=40.sin0=0
X2=30.cos-150=-25.98 Y2=30.sin-150=-15
X3=30.cos-210=-25.98 Y3=30.sin-210=15
Xtotal =-11.96 Ytotal=0

In=Square root (11.96^2 + 0^2)=11.96A

arse..biscuits wrote: »

The marking system is 1 or 0

In what exam, phase 6 electrical science?
Have you a link to this?

I served my time in the early nineties.
This certainly was not the way that our trade exams were marked.
I am open to correction on this but I would be very surprised if marking systems have changed so radically.
You stated that you teach in DIT (apprentices I assume?) so I acknowledge that you would have a better insight into this than me.

To clarify:
Are you suggesting that if a someone makes a mathematical error but is 100% correct with everything else that they get a 1 or a 0?

Many thanks.