Efficiency of Transformer

                                                Efficiency of Transformer

Output power and efficiency of the input transformer and power transformer ratio. Skills η in the directed

η = output / input

The Power input transformer and the output power of the percentages expressed as the percent efficiency.

% η = output / input × 100

= Output / output + Loss × 100

Or

% η = input - Loss / input × 100

                                                                                   = (1 - Loss / input) × 100

Transformer copper loss is that it all depends on the coil current flow caused by resistance and heating loss. The copper coil resistance and outlets are dependent attenuation. Pse = I²pRep = I²Res resistance cable size and length depends on the coil and the current load dependent. Load copper loss of the transformer depends on the specific outlets.

If you say that, on the transformer rating is KVA with kW  not write this? Or on the transformer rating is KVA you  ?

Answer: - Los outlets  on the transformer and copper and core loss depends on voltage phase differences, but is not dependent on them. The phase difference between voltage and current depends on the load type. It is the same for Watt or kW load current may be different, but KVA Voltage and current are the same on all the power factor. The rating transformer kW with KVA It is on.

transformer efficiency is a maximum when the core loss and copper loss is equal.

That is, the core loss = copper loss

Pc = Ph + Pe transformer core loss and copper loss Pse = I ² pRep = I ² Res

Core loss is not dependent on electrical outlets.

Considered the direction of the primary input VpIp Cosθp

η = VpIp Cosθp - Loss / VpIp Cosθp

= 1 - I ² pRep / VpIp Cosθp - Pc / VpIp Cosθp

= 1 - IpRep / VpIp Cosθp - Pc / VpIp Cosθp

Differencent  both parties relative to the Ip,

dη / dIp = 0 - Rep / Vp Cosθp + Pc / VpI ² p Cosθp

Ip variable. Therefore, the maximum efficiency dη / dIp = 0

For maximum efficiency

= -Rep/VpCosθp + Pc / VpI ² p Cosθp = 0

Or Rep = Pc / I ² p

I ² pRep = Pc

Psc = Pc

Therefore, the maximum efficiency of transformer core loss = copper loss.

For example of mathematics

1. A 20KVA, 2300/230V transformer and full load copper loss 200W  is calculated to -

a.  6KVA load of copper loss.

b. High Voltage in 4A load of copper loss.

c. 10KW 0.8 power factor load of copper in Loss.

Solution: -

a. Copper loss = (6/20) ² × 200

                          = 18 w.

b. Full load current IFL = 20 × 1000/2300

                                       = 8.69 A

             Copper loss = (4 / 8.69) ² × 200

                                   = 42.38 W.

c. KW load = 10/0.8

                     = 12.5 KVA

Copper loss = (12.5 / 20) ² × 200

                                   = 78.125 W. Ans.

2. A 5KVA, 2300/230V transformer and 24 hours of low load calculation forms -

1 .1 ½ load  power factor 0.8 to 1 hours.

3. 1 ¼ load  power factor 0.8 to 2 hours

4. Rated load power factor of 0.9 to 3 hours

5. ½ load  power factor unity  of 6 hours

6. ¼ load  power factor unity of 8 hours

7. No load power factor unity of 4 hours

Core Loss 40 Watt and Copper loss 112 Watt  now the skills of % η all day

Solution: -

24 Hour output

= (1 ½ × 5 × 0.8 × 1) + (1 ¼ × 5 × 0.8 × 2) + (5 × 0.9 × 3) + (½ × 5 × 1 × 6) + (¼ × 5 × 1 × 8)

= 54.5 KWh

24 hours in Los copper

= [(1 ½) ² × 112 × 6] + [(1 ¼) ² × 112 × 2] + [112 + 2] + [(½) ² × 112 × 6] + [(¼) ² × 112 × 8]

= 1162 Wh = 1.162 KWh

24 hour core loss

= 40 × 24 = 960Wh = 0.96 KWh

All day efficiency = 54.5 / 54.5 + 1.162 + 0.96 × 100%

= 96.25% Ans.

3. A 10KVA transformer core loss and full load copper loss 55W 130W., Which follows 24 hours of loading.

1 15KVA load  power factor 0.85 to 2 hours

2. Rated load power factor of 0.90 to 5 hours

3.7.5  load  power factor 0.95 to 6 hours

4.5KVA unity load  power factor of 7 hours

5. No load of 4 hours

Transformer and all-day skills that now.

Solution: -

    Rating = 10 KVA

Pc = 55W

Psc = 130 W

% η all day =  ?

24 hrs. output

15 × 0.85 × 2 = 25.5 KWh

10 × 0.9 × 5 = 45 KWh

7.5 × 0.96 × 6= 42.75 KWh

5 × 1 × 7 = 35 KWh

0          4     =      00

Total 24 hrs. = 148.25 KWh

Core loss = 55 × 24 = 1320 Wh = 1.32 KWh.

Copper loss = (15/10)² × 130 × 2 = 585 Wh.

                                         130 × 5 = 650 Wh.

                    (7.5/10)² × 130 × 6 = 438.75Wh

                    (5/10)² × 130 × 7 = 227.5 Wh

                                 Total = 1.9 KWh.

% η all day =  output / output + losses × 100%

                  = 148.25 / 148.25 + 1.32 +1.9 × 100

                   = 97.87% Ans.