Induced EMF Calculations

[ROHM]

I would like to know how much emf is induced by a 15 KVA 480D/240V single phase transformer. Can anyone tell me how to calculate the emf in volts for that transformer at full capacity?

Please include steps on how to calculate it.

 

[Julius Right]

If it is a single-phase transformer no D[delta] is here, as for 3 phases.
Maybe you need Emf for one phase of a three-phase transformer.
Let's say Vp=15 kV in star connection phase-to-phase voltage.
The attached sketch represents a schematic diagram of a two winding transformer: primary and secondary windings.
One phase voltage it is V1=V3/sqrt(3)=15/sqrt(3)=8.66 kV
Let's say uk[short-circuit voltage]=5%
Then total Ztot=15^2/(15/1000)*5%=750 ohm[KV^2/MVA*uk%]
Emf=Vp-Zp*Ip. Neglecting Io Ip=Is’ [Is referred to primary]=MVA/sqrt(3)/KV=0.577 A
Let’s say Zp=Zs’ [Zs referred to primary] then Zp=Ztot/2=375 ohm
Efm=15/sqrt(3)-0.577*375/1000=8.44 kV
However, the impedance is Zp= Rp+jXp and the Ip= Irat*(cosφ-jsinφ) usually.
Let's say Xp/Rp=5 and cosφ=0.85 then sinφ=sqrt(1-0.85^2)=0.527
Xp=Zp/sqrt(1+1/5^2)=375/1.04=360.6 and Rp=360.6/5=72.1
Efm=15/sqrt(3)-0.577/1000*(0.85-j0.527)*(72.1+j360.6)=8.515 kV.

 

[ptonsparky]

If it is a single-phase transformer no D[delta] is here, as for 3 phases.
Maybe you need Emf for one phase of a three-phase transformer.
Let's say Vp=15 kV in star connection phase-to-phase voltage.
The attached sketch represents a schematic diagram of a two winding transformer: primary and secondary windings.
One phase voltage it is V1=V3/sqrt(3)=15/sqrt(3)=8.66 kV
Let's say uk[short-circuit voltage]=5%
Then total Ztot=15^2/(15/1000)*5%=750 ohm[KV^2/MVA*uk%]
Emf=Vp-Zp*Ip. Neglecting Io Ip=Is’ [Is referred to primary]=MVA/sqrt(3)/KV=0.577 A
Let’s say Zp=Zs’ [Zs referred to primary] then Zp=Ztot/2=375 ohm
Efm=15/sqrt(3)-0.577*375/1000=8.44 kV
However, the impedance is Zp= Rp+jXp and the Ip= Irat*(cosφ-jsinφ) usually.
Let's say Xp/Rp=5 and cosφ=0.85 then sinφ=sqrt(1-0.85^2)=0.527
Xp=Zp/sqrt(1+1/5^2)=375/1.04=360.6 and Rp=360.6/5=72.1
Efm=15/sqrt(3)-0.577/1000*(0.85-j0.527)*(72.1+j360.6)=8.515 kV.

Whooshing sound over my head..


Did you get 5 volts?

 

[jim dungar]

If it is a single-phase transformer no D[delta] is here, as for 3 phases.
Maybe you need Emf for one phase of a three-phase transformer.
Let's say Vp=15 kV in star connection phase-to-phase voltage.
The attached sketch represents a schematic diagram of a two winding transformer: primary and secondary windings.
One phase voltage it is V1=V3/sqrt(3)=15/sqrt(3)=8.66 kV
Let's say uk[short-circuit voltage]=5%
Then total Ztot=15^2/(15/1000)*5%=750 ohm[KV^2/MVA*uk%]
Emf=Vp-Zp*Ip. Neglecting Io Ip=Is’ [Is referred to primary]=MVA/sqrt(3)/KV=0.577 A
Let’s say Zp=Zs’ [Zs referred to primary] then Zp=Ztot/2=375 ohm
Efm=15/sqrt(3)-0.577*375/1000=8.44 kV
However, the impedance is Zp= Rp+jXp and the Ip= Irat*(cosφ-jsinφ) usually.
Let's say Xp/Rp=5 and cosφ=0.85 then sinφ=sqrt(1-0.85^2)=0.527
Xp=Zp/sqrt(1+1/5^2)=375/1.04=360.6 and Rp=360.6/5=72.1
Efm=15/sqrt(3)-0.577/1000*(0.85-j0.527)*(72.1+j360.6)=8.515 kV.

At what distance are you calculating this EMF? Doesn't EMF drop off at something like the inverse of the square of the distance?
The OP is likely using a customer owned dry type transformer with a 480V primary and 120V secondary.

 

[ROHM]

I need to know at what safe distance I can place it away from a DC circuit cabinet. I don't want any induced EMF onto those DC circuits by this transformer. What emf would be at 4 feet? And is there a formula I can calculate the emf based on distance, say 1 feet, 2 feet, 3 feet.

 

[don_resqcapt19]

I need to know at what safe distance I can place it away from a DC circuit cabinet. I don't want any induced EMF onto those DC circuits by this transformer. What emf would be at 4 feet? And is there a formula I can calculate the emf based on distance, say 1 feet, 2 feet, 3 feet.

There won't be any issue no matter how close together they are, assuming that they are in separate enclosures.

 

[tortuga]

I kinda followed along till here:

Efm=15/sqrt(3)-0.577/1000*(0.85-j0.527)*(72.1+j360.6)=8.515 kV.

what does the j mean?
using your number I get

15/1.732 = 8.66
0.577/1000 = 0.000577
0.85-0.527 = 0.323
74.1+360.6 = 343.7

0.000577 * 0.323 * 434.7 = 0.081

8.66 - .081 = 8.579

EMF = 8.579

 

[Julius Right]

Sorry, I did not understand what about emf of a transformer you intended. So, it is about Magnetic field near transformer.
An American institute- EPRI - conducted research on some pole and pad mounted transformers.
For a 50 kVA transformer the magnetic field was as per attached figure.
The emf resulted it depends on the area exposed to this magnetic field.
However, as you can see, the only 3 or 4 feet is important where it is an average total of 150*4 mG.
So, for an area between two conductors 1 foot distance and a length of 100 feet [total 12*1200=14400 square inches] the average total magnetic flux will be only as for the first 4 feet closed to transformer.
Total magnetic flux is 200*4*12=9600 mG.
emf=4.44*f*mag.flux[mG]/10^7=0.256 V
emf=4.44*60*9600/10^7=0.256V



























































Sorry, I did not understand what about emf of a

transformer you intended. So, it is about Magnetic field near transformer.

An American institute- EPRI -

conducted research on some pole and pad mounted transformers.

For a 50 kVA transformer the magnetic field was as per attached figure.

The emf resulted it depends on the area exposed to this magnetic field.

However, as you can see, the only 3 or 4 feet is important where it is an average total of 150*4 mG.

So, for an area between two conductors 1 foot distance and a length of 100 feet [total 12*1200=14400 square inches] the average total magnetic flux will be only as for the first 4 feet closed to transformer.

Total magnetic flux is 200*4*12=9600 mG.

emf=4.44*f*mag.flux[mG]/10^7=0.256 V

emf=4.44*60*9600/10^7=0.256V





























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.

 

[winnie]

I need to know at what safe distance I can place it away from a DC circuit cabinet. I don't want any induced EMF onto those DC circuits by this transformer. What emf would be at 4 feet? And is there a formula I can calculate the emf based on distance, say 1 feet, 2 feet, 3 feet.

You need to specify the sensitivity of the DC circuits. With the requirement 'I don't want any induced EMF...' the answer is easy: you must have infinite separation between the transformer and the DC circuit cabinet.

At 60Hz and normal building distance you are concerned about the magnetic field leaking out of the transformer. The EPRI study @Julius Right mentioned gives you an idea of how strong the field will be, but the actual number depends on the specific design of the transformer. You may need to measure the field around your transformer.

Once you know the field strength you can calculate the voltage induced in a loop of wire at a given distance. A key factor here is the 'loop area' of the target circuit; the bigger the open loop the more flux it intercepts and the greater the voltage induced in the loop. The internal layout of your DC circuit cabinet will impact how much voltage gets induced.

Then you need a spec for how much induced voltage your system can tolerate. Are you talking about 48V on/off control circuits, telephone (voice) circuits, or some esoteric high sensitivity instrumentation system?

For just about any industrial system my guess is that the enclosures can be adjacent with minimal spacing for mechanical access and ventilation. But without more details there can be no certain answer.

Jonathan

 

[Julius Right]

Sorry once again! I forgot to multiply by 12 the 4 feet in the magnetic flux calculation.
Instead of: Total magnetic flux is 200*4*12=9600 mG. and
emf=4.44*60*9600/10^7=0.256V
It has to be:
Total magnetic flux is 200*4*12*12=115200 mG. And the voltage [emf] should be:
emf=4.44*60*115200/10^7=3.07 V