Part 3 – Transformer Calculations
Transformer Efficiency
The two components that makeup transformer losses are: Copper losses – (load loss) which depends on the amount of load connected to a transformer and Iron (core) losses – (no-load) which is independent of the load.
Transformer efficiency is:
Efficiency = 
x 100
= 
x 100
=
x 100
Copper Loss (I2R Loss)
Results from the resistance of the primary winding and secondary winding. Resistance is a function of is conductor material used for the windings, cross-sectional area of the conductor, and the length of the conductor.
Copper is the natural choice for a conductor because of:
- availability
- strength
- cost
- low resistance
Iron Loss (Core Loss)
Energy is required to sustain the magnetic field in the steel core. Results from the constant magnetizing and de-magnetizing of the core does not depend on the load for a given transformer. Energy is required to align and realign the molecular particles of the magnetic material and this is called hysteresis loss.
Various core materials have different permeability, which is the ability to increase flux density within the material when electric current flows through a conductor wrapped around the magnetic materials. For example, silicon steel has fewer losses than carbon steel. A relatively new metal alloy called amorphous iron has fewer losses than silicon steel.
Eddy current loss is eddy currents are circulating currents that are induced by the magnetic material and heat is generated as the current circulates. Eddy current losses are reduced by making the core out of laminated sheets and varnishing each sheet so that it is electrically insulated. Laminations do not affect the permeability of the core.
Primary and Secondary Loading
Primary Load – Connected to AC power source: Primary loading (Pp) = Vp x Ip
| Pp = Primary Loading (kVA) |
| Ip = Primary winding current (A) |
| Vp = Applied Voltage (kV) |
Secondary Load – Connected to an external load: Secondary loading (Ps) = Vo x Is
| Ps = Secondary Loading (kVA) |
| Vo = Secondary winding current (A) |
| Is = Output Voltage (kV) |
Calculating Transformer Efficiency
Formulas to use:
Power = VA cosθ
Input Power = Primary Loading (Pp) x Power Factor (PF)
Output Power = Secondary Loading (Ps) x Power Factor (PF)
Efficiency = Power Output (kW) x 100
Power Input (kW)
Efficiency = 
x 100
= 
x 100
Input Power = Output Power + Copper Loss + Iron Loss (Core loss)
An example of a transformer’s primary windings has a current flow of 30 A. The secondary windings have a current flow of 80 A. The applied voltage is 1.8 kV and the output voltage is 660 V. The power factor is 0.89. The transformer’s measured copper loss is 0.9 kW. Calculate:
- primary loading in kVA
- secondary loading in kVA
- input power
- output power
- efficiency
- iron loss
The solution to the example:
Primary winding current (Ip) = 30 A
Secondary winding current (Is) = 80A
Applied Voltage (Vp) = 1.8 kV
Output Voltage(Vs) = 660 v
Power Factor = 0.89
Copper loss = 0.9 kW
Primary loading (Pp) = Vp x Ip
=1.8 kV x 30 A
=54 kVA
Secondary loading (Ps) = Vo x Is
=660 V x 80 A
=52,800 VA
= 52.8 kVA
Solutions
Solution 1:
Primary winding current (Ip) = 30 A
Secondary winding current (Is) = 80A
Applied Voltage (Vp) = 1.8 kV
Output Voltage(Vs) = 660 v
Power Factor = 0.89
Copper loss = 0.9 kW++
Primary loading (Pp) = Vp x
=1.8 kV x 30 A
=54 kVA
Secondary loading (Ps) = Vo x Is
=660 V x 80 A
=52,800 VA
=52.8 kVA
Solution 2:
Primary winding current (Ip) = 30 A
Secondary winding current (Is) = 80A
Applied Voltage (Vp) = 1.8 kV
Output Voltage(Vo) = 660 v
Power Factor = 0.89
Copper loss = 0.9 kW
Input power = VA cosΦ
=Primary Loading (Pp) x PF
=54 kVA x 0.89
=48.1 kW
Output power = VA cos Φ
=Secondary Loading (Ps) x PF
=52.8 kVA x 0.89
=47.0 kW
Solution 3:
Primary winding current (Ip) = 30 A
Secondary winding current (Is) = 80A
Applied Voltage (Vp) = 1.8 kV
Output Voltage(Vo) = 660 v
Power Factor = 0.89
Copper loss = 0.9 kW
Efficiency = Output VA (Secondary Loading) x 100
Input VA (Primary loading)
=54.0 kVA x 100
52.8 kVA
= 97.7%
Or
Efficiency =Output power x 100
Input power
=47 kW x 100
48.1 kW
=97.7%
Solution 4:
Primary winding current (Ip) = 30 A
Secondary winding current (Is) = 80A
Applied Voltage (Vp) = 1.8 kV
Output Voltage(Vo) = 660 v
Power Factor = 0.89
Copper loss = 0.9 kW
Input power = Output power + Copper loss + Iron loss
48.1 kW = 47 kW + 0.9 kW + Iron loss
Iron loss = (48.1 – 47 – 0.9) kW
=0.2 kW