Phase voltage in AC Delta connection

[jacklam2048]

Hi, my customer asked us to develop a three phase AC deepfryer, the input voltage is 230/380V Main voltage, Power 7000W. My colleague created a drawing in Wye connection as shown.

I try to create the same but in Delta connection, but I am unable to figure out the phase voltage (i.e. the voltage across each heating element). Can anyone help or give me some clues ?

Thanks a lot !

 

[Harald Kapp]

see e.g. here.

 

[Martaine2005]

This site also has all the calculations.. Here.
It's way over my head but the information is there.

Martin

 

[Bluejets]

If your elements are the original star connected, you'll blow the guts out of them on delta.
Reason being the original star has 2 elements per phase with any imbalance on the neutral.
Delta has one element across across the two phases.
Why you would need to change to delta is beyond me.

 

[jacklam2048]

If your elements are the original star connected, you'll blow the guts out of them on delta.
Reason being the original star has 2 elements per phase with any imbalance on the neutral.
Delta has one element across across the two phases.
Why you would need to change to delta is beyond me.

Hi Bluejets, Thanks your reply.
I think in Delta connection, the current across each heating element is much smaller comparing to the phase current in Star, thus the heating element connections are not so easy to burn out. Sometimes the burn out of the terminals at the heating elements is a headache and cause a lot of maintenance cost.

 

[Harald Kapp]

I think in Delta connection, the current across each heating element is much smaller comparing to the phase current in Star

Only if the impedance of the heating element is increased accordingly.
As the voltage increases by sqrt(3), impedance needs to be tripled to maintain the same power:

Star: P_star = V_star² / R_star

Delta: P_Delta= V_Delta² / R_Delta = (V_star × sqrt(3))² / R_Delta = 3 × V_star² / R_delta

With the requirement P_star = P_delta you get

P_Delta= 3 × V_star² / R_delta = V_star²/ R_star

Cancelling V_star out of the equation gives:

3 / R_delta = 1 / R_star or equivalently: R_delta = 3 × R_star

 

[Bluejets]

I think in Delta connection, the current across each heating element is much smaller comparing to the phase current in Star,

Nope....

 

[Tha fios agaibh]

I think the confusion is between supply and load.

The phase current inside the windings of a Delta connected transformer will draw less current than a Wye.

But, you are showing your deepfryer load wired with a Delta configuration.

If you apply 230v 3p to the Delta load, you are changing the impedance with the extra windings.

Rough math shows your amperage increasing from 10 to 15 amps per leg and wattage increasing from 2.34kw to 3.45kw per element.

I agree with Bluejets that you'd blow it up if hooked up this way.

 

[jacklam2048]

I think the confusion is between supply and load.

The phase current inside the windings of a Delta connected transformer will draw less current than a Wye.

But, you are showing your deepfryer load wired with a Delta configuration.

If you apply 230v 3p to the Delta load, you are changing the impedance with the extra windings.

Rough math shows your amperage increasing from 10 to 15 amps per leg and wattage increasing from 2.34kw to 3.45kw per element.

I agree with Bluejets that you'd blow it up if hooked up this way.

 

[jacklam2048]

Yes， I have calculated:
In Wye connection, the resistance of each heating element is 22.6 Ohm.
If I convert it to Delta connection, the resistance of each heating element has to be 68.48 Ohm.
Thanks all people's reply !
You are helpful !

 

[jacklam2048]

Only if the impedance of the heating element is increased accordingly.
As the voltage increases by sqrt(3), impedance needs to be tripled to maintain the same power:

Star: P_star = V_star² / R_star

Delta: P_Delta= V_Delta² / R_Delta = (V_star × sqrt(3))² / R_Delta = 3 × V_star² / R_delta

With the requirement P_star = P_delta you get

P_Delta= 3 × V_star² / R_delta = V_star²/ R_star

Cancelling V_star out of the equation gives:

3 / R_delta = 1 / R_star or equivalently: R_delta = 3 × R_star

Thanks Harald, Your calculation is a great help !