I1/I2 = N2/N1Φmax = Bmax * AcSingle Phase Transformer Calculations

[oxon88]

Homework Statement

A single phase transformer has the following rating: 120 kVA, 2000 V/100 V, 60 Hz with 1000 primary turns.

Determine:

a) the secondary turns
b) the rated primary and secondary currents
c) the maximum flux
d) given a maximum flux density of 0.25 T, the cross sectional area of the core.

Homework Equations

V₂/V₁ = N₂/N₁

 

[oxon88]

trying to work out the answer for part a

N₂ = N₁*V₂ / V₁


= (1000 * 100) / 2000


N₂= 50 turns


Is this correct?

 

[NascentOxygen]

trying to work out the answer for part a

N₂ = N₁*V₂ / V₁


= (1000 * 100) / 2000


N₂= 50 turns

1000 turns : 50 turns ≡ 2000V : 100V ✔

 

[oxon88]

Thanks, can you offer any advice on solving part b?

Im stuck trying to work out the current in either N₁ or N₂

once i have one of these values I guess i can use V₂/V₁ = N₂/N₁ = I₂/I₁

 

[oxon88]

ok here is an attempt at b

I₁=120000/2000 = 60A

I₂ = 60/0.05 = 1200A
how does this look?

 

[NascentOxygen]

What does it mean when it says a single phase transformer has the following rating: 120 kVA?

 

[oxon88]

120kVA is the maximum power the transformer can output. its a measure of volt-amps.

so... 120kVA / 100v = 1200A on the secondary?

 

[NascentOxygen]

120kVA is the maximum power the transformer can output. its a measure of volt-amps.

Close enough.

so... 120kVA / 100v = 1200A on the secondary?

Yes. And the same idea applies to the primary, too.

 

[oxon88]

I'm a bit stuck with part c. Any advice on this one?

 

[NascentOxygen]

I'm a bit stuck with part c. Any advice on this one?

What equations do you know that involve magnetic flux?

 

[oxon88]

Would this be appropriate to use?

Magnetic Flux = B.A.cosθ

 

[NascentOxygen]

Would this be appropriate to use?

Magnetic Flux = B.A.cosθ

Φ=B.A.cosθ is a good equation to memorize, and we'll need that for part (d) which relates flux to flux density and area.

For part (c) we need an equation relating flux, voltage and number of turns for a transformer.

 

[oxon88]

ok I see. Would this be more appropriate?

Φ = V × T / N

Φ = Flux (weber)
V = Voltage
T = Time
N = No. of turns

 

[NascentOxygen]

In an operating transformer the flux is constantly changing, and in a sinusoidal manner. It's the changing flux linking the secondary that induces the secondary voltage.

E1 = N1 . ΔΦ/Δt

You're getting close. It's based on that, but others have done all the necessary fiddle-factoring and you've almost certainly been presented with the final equation in lectures or textbook reading.

Hint: there's a 4.44 factor in it.

 

[oxon88]

right then I think I may have got it...

V_p = 4.44 * f * N_p * Φ_max2000 = 4.44 * 60 * 1000 * Φ_max

2000 = 266400 * Φ_max2000 / 266400 = Φ_maxΦ_max = 7.51 mWb

 

[NascentOxygen]

Φmax = 7.51 mWb ✔

 

[oxon88]

Great thanks for you help and advice. Its making more sense now. Last part to go...

Φ = B * A * cosθ

would θ = 0 ?



7.51x10^-3 = 0.25 * A * cos(0)

7.51x10^-3 / 0.25 = A

A = 0.03003mm²

 

[NascentOxygen]

Here's a reference for this topic, might be useful to others. http://books.google.com.au/books?id...v=onepage&q=4.44 transformer equation&f=false

 

[NascentOxygen]

Great thanks for you help and advice. Its making more sense now. Last part to go...

Φ = B * A * cosθ

would θ = 0 ?

EDIT: Oops. Misread the Greek. :tongue:
Yes, windings are perpendicular to the core.

7.51x10^-3 = 0.25 * A * cos(0)

7.51x10^-3 / 0.25 = A

A = 0.03003mm²

Check those units.

 

[oxon88]

would it be m²? (standard SI base unit)

 

[NascentOxygen]

would it be m²? (standard SI base unit)

It should be.

0.03 m² sounds better. 120kVA would be a heavy industrial transformer. If the core were to have a circular cross-section it would be 20cm. diameter, without the windings, so I guess that sounds in the ballpark.

So, you have the transformer equations all connected up now?

 

[oxon88]

so that would mean if the core was circular, the diameter would be roughly 19.5cm? That is huge.

yes its all a lot clearer now. thank you for your help and time.

 

[terrytibbs]

Hi could somebody expand on 4.44 factor if you don't mind.

 

[braceman]

hmmm

I'm guessing that the equation used to find the max flux is the universal EMF equation :-

E = 4.44 x f x N x a x B

where
E = rms voltage of winding
f = frequency
N= number of turns
a = cross=-sectional area of core
B = peak flux density

Just wondering how we can actually use this to find 'B' as we don't know what 'a' is either? It looks like it was just left-out. Surely this would give a wrong answer then??

I've reached this question too, and my notes have no mention of this universal equation so I'm struggling a bit to work it out..

Could anyone please explain why it's been left out and yet the answer is still right..

 

[NascentOxygen]

I'm guessing that the equation used to find the max flux is the universal EMF equation :-

E = 4.44 x f x N x a x B

where
E = rms voltage of winding
f = frequency
N= number of turns
a = cross=-sectional area of core
B = peak flux density

Just wondering how we can actually use this to find 'B' as we don't know what 'a' is either?

If you are asked to find maximum flux, then you are to find flux, Φ, not flux density, B. https://www.physicsforums.com/showpost.php?p=3871394

 

[braceman]

bugger

thanks for the reply Nascent,

yeah, I misread it to read just flux and not flux density like it was.

Going back and trying to find the equation just for flux now...

 

[Luk-e-2012]

I'm getting this for d)

Bm = 0.25T
Φm = Bm*A
Or , A = Φm/Bm
= 0.0075/0.25 m2
= (0.0075/0.25)*10000 cm2
= 300 cm2

Thoughts

 

[Jason-Li]

Hi NascentOxygen,

I understand this is an old thread but just wondering if you clarify one thing. Would I be correct in saying the 4.44 figure comes from 2π/√2=4.44, and that the √2 is there for the fact it is single phase? I was doing some reading and found this online, why has the RMS value been used on this rather than the max since the question asks for maximum flux?

Thanks,
Jason

https://www.electronics-tutorials.ws/transformer/trans6.gif

 

[jim hardy]

Would I be correct in saying the 4.44 figure comes from 2π/√2=4.44, and that the √2 is there for the fact it is single phase?

I would say you're exactly right.
for sine waves
ω = 2πf
and √2 is the ratio of peak to rms

Hats off to you for digging into understand the equations.
It's a lot easier than trying to remember them, isn't it ?
4.44 is slide rule accuracy.
With today's calculators you can postpone rounding off until the last step.
Sometimes textbook authors arrange their example problems so that pi's and √'s cancel out , and angles have convenient cosines.
That's when a little bit of such 'procrastination' just might pay off .

old jim
.

 

[Jason-Li]

I would say you're exactly right.
for sine waves
ω = 2πf
and √2 is the ratio of peak to rms

View attachment 236282

Hats off to you for digging into understand the equations.
It's a lot easier than trying to remember them, isn't it ?
4.44 is slide rule accuracy.
With today's calculators you can postpone rounding off until the last step.
Sometimes textbook authors arrange their example problems so that pi's and √'s cancel out , and angles have convenient cosines.
That's when a little bit of such 'procrastination' just might pay off .

old jim
.

Yeah precisely the reason I don't just copy people who have done the question previous, I want to understand it so I can apply it more easily in future! Thanks for confirming that with the good explanation and picture also.

An old Jim is a wise Jim !