Cooling Transformer

                                                        Cooling Transformer



Transformer core loss and copper loss effect that removes heat from the cold to keep transformer  refrigeration or cooling process is called transformer.
This is especially the cooling system that is fully insulated transformer and its temperature increases. If the temperature exceeds a certain limit than the insulation between the coil and cores are insulation and coil is lost. Especially through such measures do not increase the temperature of the transformer and heat quickly. The transformer is to increase ratings.
Transformer and cooling system is described.
1. Natural cooling or Natural air cooling.
2. Natural oil cooling or oil immersed self cooling.
3. Oil immersed forced oil cooling.
4. Oil immersed forced water cooling.
5. Oil immersed forced air cooling.
6. Forced air cooling.
Transformer Action tharmosyipoun what?
The natural oil tank pipe or radiator cooling transformer oil transformer uttape them together in the light of the above radiator pipe or pipe enters the mouth and cold tank back into the mouth. The oil that is normal currents flow in the tharmosyipoun action.
For example, two of the transformer oil, Winding coil and cores as the insulation The heat from the core and coil out of the help.
Power transformer oil
1. Transformer oil dioxide - electric power (Di-electric strength) should be high quality.
2. Transformer oil does not contain any acids or alkali.
3. The material does not contain sulfur transformer oil.
4. Viscosity and consolidate all of the transformer oil.
5. Transformer oil must be fireproof.
6. Flash point below 160°c Do not transformer oil.
7. Fire point of transformer oil will be below 200°c degrees.
8. The relative importance of transformer oil is not more than 0.85.
9. Easy farces or transformer oil sludge (sludge) is not forded.
Transformer and the necessary parts and materials and work are described.
Sludge: - transformer oil reactions with oxygen in the air when coming into contact with the farces of oil molecules are broken, it is Slugging  It hampered the flow of oil accumulated in the bottom of the transformer Tanks and tank pipe to the pipe mouth. The transformer does not work well.
Transformer tank: - Basic transformer cold to keep a large urn set in a steel container is filled by oil. This urns is transformer tank. Tanks for the heat to spread quickly and the air wave shape of her body in the air to increase.
Conservator : - high capacity transformer wire on top of the drum-shaped part, said he was Conservator.
Brither: - transformer Conservator can not let any stream with silica gel Conservator for admission into the cylindrical part of the brither called him. Transformer birthing brither with Conservator is used to prevent errors. Usually with brither into and out of the air. geli streams takes time to absorb into the air. transformer oil with the water can not related.
Bukaloss Relay: - It is usually used between the transformer tank and Conservator is added's Work on the transformer internal short circuit fault protection.

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.