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.

Voltage regulation of Transformer

                                                      Voltage regulation of  Transformer.

Transformer no - load voltage and full-load conditions of voltage and full load voltage difference is ratio Voltage Regulation. The percentage rate of the ratio. Such as: -

Regulation of the voltage transformer must be less than that amount, the lower the better.

The regulation voltage transformer coil voltage detection, both with full load terminal voltage drops to the no - load voltage is determined. The no - load voltage is determined as the secondary.

 In other words,

Vp / a = Vs + Zes

Here,

Vp / a = no - load voltage secondary

Vs = secondary full-load terminal Voltage.

Is Zes = transformer coil in both the total voltage drop (secondary side)

The Vp / a = VNL and Vs = VFL in

Transformer regulation,

Vreg = VNL - VFL ÷ VFL × 100%

Vs-a reference to the

Vp / a = √ (Vs Cos θ + IsRes) ² + √ (Vs Sin θ + Is Xes) ²

Or Vs-a reference to the

Vp / a = Vs + Is (Cos θ - j Sin θ) (Res + j Xes)

Methods and equations can be used.

When, Cos θ = load power factor

If the power factor leading θ-value is negative.

 Example of mathematics & solution :-

1. 25 KVA, 2300/230 V Transformers Rp = 0.8Ω Xp = 3.2Ω, Rs = 0.009Ω, Xs = 0.03Ω equal to transformer of regulation. When -

1. Power Factor unit.

2. Power factor 0.8 lagging

3. Power factor leading 0.866.

Ans: - a = 2300/230

              = 10

Is = 25 × 1000 ÷ 230 = 108.7 A

Secondary terms

Res = Rp / a ² + Rs = 0.8 / (10) ² + 0.009 = 0.017 Ω

Xes = Xp / a ² + Xs = 3.2 / (10) ² + 0.03 = 0.062 Ω

IsRes = 108.7 × 0.017 = 1.85 V

IsXes = 108.7 × 0.062 = 6.74 V

1. V / a = VNL = √ (VFL cosθ + IsRes) ² + √ (VFLsinθ + IsXes) ²

= √ (230 + 1.85) ² + √ (0 + 6.74) ²

= 231.95 V.

Vreg = VNL - VFL ÷ VFL × 100%

= 231.95 - 230 ÷ 230 × 100%

= 0.84%

2. Here Cos θ = 0.8, Sin θ = 0.6

V / a = VNL = √ (VFL cosθ + IsRes) ² + √ (VFLsinθ + IsXes) ²

= √ (230 × 0.8 + 1.85) ² + √ (230 × 0.6 + 6.74) ²

= 235.57 V.

Vreg = VNL - VFL ÷ VFL × 100%

= 235.57 - 230 ÷ 230 × 100%

= 2.82%

3. Here Cos θ = 0.866 Sin θ = - 0.5

(Θ-'s value because of the negative power factor leading)

V / a = VNL = √ (VFL cosθ + IsRes) ² + √ (VFLsinθ + IsXes) ²

= √ (230 × 0.8 + 1.88) ² + √ (230 × (- 0.5 + 6.74) ²

= 228.32 V.

Vreg = VNL - VFL ÷ VFL × 100%

= 228.32 - 230 ÷ 230 × 100%

= - 0.73%


2. Short-Circuit Test of a 10KVA 2300/230V Transformers were found to have the following information, Ese = 137V, Psc = 192 W, Isc = 4.34A, 0.8 power factor of percent voltage regulation  lagging to out now.

Ans: -

That have been invoked

Rating = 10KVA. % Reg =   ?

Ep = 2300V Pf = 0.8 lagging

Es = 230V

Esc = 137V

Psc = 192W

Isc = 4.34A

Re = Psc / I ² sc

 = 192 / (4.34) ² = 10.19 Ω

Ze = Esc / Isc

= 137 / 4.34 = 31.57 Ω

Xe = √ Ze - √ R ² e

= √ (31.57) - √ (10.19) ²

= 29.88 Ω

I = KVA × 1000 / Ep

= 10 × 1000/2300

= 4.35A

IRe = 4.35 × 10.19 = 44.3V

IXe = 4.35 × 29.98 = 129.98V

Cosθ = 0.8 & sinθ = 0.6

Vp = √ (V cosθ + IRe) ² + √ (Vsinθ + IXe) ²

= √ (2300 × 0.8 + 44.3) - √ (2300 × 0.6 + 129.98) ²

= 2414.8 V

Vreg = VNL - VFL ÷ VFL × 100%

= 2414.8 - 2300 ÷ 2300 × 100%

= 4.99%

Short Circuit Test of Transformer

                             Short Circuit Test of Transformer

While a transformer short circuit the other hand ammeters watt meter and voltmeter with Connected to supply transformer short circuit test is that the test will be called. The transformer equivalent Resistance, impedance   evaluate and determine the copper loss is short-circuit test.

Tested for transformer short circuit coils full load current to flow in the equivalent electrical outlets.

Tested for transformer short circuit transformer and low - voltage of the circuit is short because of the -

1. Less than the full load current of high voltage electrical outlets are used for low rang ammeters and watt meters.

The. Supply only 4% of the rated voltage is 10% of the supply is not a problem.

3. It is the correct voltage for full load outlets.

Transformers short circuit test method is described

Normally the transformer and low - voltage short circuit in the high voltage side of the meter and the watt meter ammeters volts is connected to the supply. Variable voltage supply source or from the series resistance 2% of rated voltage connection of the supply voltage with a gradual increase in the full-load current to flow until the transformer. High voltage to low voltage at full load current to flow when a particular certificate will not be short of the full load current to flow. The ammeters watt meter and volts meters are read into the record.

Received watt meter reading = Psc

Ammeter text = Isc

And voltmeter readings = Esc

Psc = transformer copper loss,

 Re = Psc / I ² sc

Ze = Esc / Isc

Xe = √ Z ² e - √ R ² e

Mentioned that, for Re, Ze and Xe equivalent to the value of the high voltage.

Transformer short circuit test is when you are careful.

1. Transformer low fat side of the terminal, in short by resistance .

The. High side of transformer and the appropriate ratings -'s watt-meter, Volt meter, you will ammeter connection.

3. Ammeter range of outlets than the little ones to be loaded.

4. Transformer and voltage rating in volts meters range is about 10%.

5. 4% or 10% of the high-side voltage is applied. The voltage It is so full load current flow is ammeter.

The low core loss is needed voltage is inconsiderable.

Both coil full load current is flowing because this is the time to get Transformer that it accepts full load copper loss.

For example of mathematics

1. A 100 KVA 2400/240 V distribution Transformer short circuit test results are as follows -

Esc = 60 V, Isc = 34.67 A, Psc = 819.4 W transformer full load copper loss determined now.

Answer: -

Transformer IFL = 100,000 / 2400 = 41.6 A

Isc = 41.6 A.

Here,

I'sc = 34.67 A

P'sc = 819.4 W

Psc = (Isc / I'sc) ² × 819.4

   = 1180 Watt.

2. A 100 KVA 2400/240 V distribution Transformer short circuit test results are as follows -

Esc = 60 V, Isc = 34.67 A, Psc = 819.4 W load, the transformer

1. 125 KVA,

2. 75 KVA

3. 0.772 power factor 85 KW to determine if each of the copper loss.

Answer: -

Transformer IFL = 100 × 1000/2400 = 41.6 A

 Full load copper loss Psc = 1180A.

Here,

1. PscL = (125/100) ² × 1180 = 1845 W

2. PscL = (75/100) ² × 1180 = 663 W

3. KVAL = 85 / 0.772

               = 1180 Watt.

PscL = (85/100 × 0.772) ² × 1180 = 1430 W

Open Circuit / No Load Test on transformer.

                                  Open Circuit / No Load Test on transformer.

                                    

One side the other side of the transformer and open circuit watt meter and voltmeter Ammeter  connection with the supply voltage to the full test is called the open circuit test.

Transformer and four parameters - 1.resistance equivalent(Re). Reactance equivalent (Xe) 3. Core Loss, Conductance (Go) 4. Magnetizing  or reactance transformers tests for the diagnosis of (Xo). The core loss, copper loss and specific test for the diagnosis of polarity.

The two most widely used test of transformer -

1. No - load test or open circuit test.

2. Short circuit test.

The back-to-back test for the diagnosis and polarity Los polarity test for the diagnosis. Open circuit test another no - load test. Transformer no - load current and power factor angle determine the core loss of the open circuit test.

Transformer and open circuit test / no - load test method described as follows: -

High open-circuit voltage of the transformer and the low - voltage side ammeter, volts watt meter and Frequency meter connection with the rated supply voltage is reached. The watt meter that measures the power of the transformer core loss and the current size of the ammeter no - load current. Now, the core loss -

Pc = EocIocCosΦoc

When,

Eoc = Open Circuit Voltage

Ioc = Open Circuit Current

Φoc = open circuit voltage and the phase difference between outlets.

Φoc = Cos ֿ ¹ Pc / Eoc Ioc

The Ioc = IN and Φoc = ΦN

If  Iw = IN Cos ΦN

       Im = IN Sin ΦN

The open-circuit test transformer or no - load test that shows all the information -

. Core Loss

. No - load current.

. Transformer and power factor.

. Loss conductance core,

. Magnetizing reactance.

Open circuit test transformer / No - only no-load transformer test of time - full load current flowing in the load outlets that only 2% of 3% and it is the only one Coil  flow. The copper loss is very minor. As a result of the full supply voltage is generated disables full Flux. It disables the hysterics  and Eddie Current full marks loss. The reason watt transformer and open circuit test meters only shows the core loss and loss of full core.

For example of mathematics :-

1. A 50 KVA 4600/230 V, 60 Hz transformer and open circuit test results are as follows -

Eoc = 230 V, Ioc = 12.5 A.

Pc = 285 W.

Working Current transformer or magnetic current component and component tracking now.

Solution: -

Φoc =Cos ֿ ¹  Pc / Eoc Ioc

= Cosֿ¹ Pc / Eoc Ioc

Cosֿ¹ 285 /230 × 12.5

= 84.3º

Im = IN Sin ΦN = 12.5 Sin 84.3º

= 12.44 A

Iw = IN  Cos ΦN

= 12.5 Cos 84.3º

= 1.24 A.

Transformation ratio

                                                     Transformation ratio

Transformer primary voltage and secondary voltage ratio is transformation ratio. Is expressed by ratio a = Ep / Es

The symbol K to K = Es / Ep

Voltage transformer ratio

Transformer primary voltage and the voltage transformer secondary voltage ratio called. ratio a = Ep / Es are expressed.

The symbol K to K = Es / Ep

Transformer turn ratio

Transformer primary coil and secondary coil of the screw of the screw is ratio turn ratio. Turn ratio a = Np / Ns

The symbol K to K = Ns / Np

Transformer voltage is equal to ratio turn ratio and why?

Transformer can be seen e.m.f. Equation E = 4.44ƒNΦ

Primary voltage Ep and Np is the number of primary turn or screw

Ep = 4.44ƒNpΦ ................... (1)

The secondary voltage Es and Ns the number of secondary turn or screw

Es = 4.44ƒNsΦ ........................ (2)

Equation (1) and (2) divided by the -

Ep / Es = Np / Ns

Frequency is equal to the flux of each coil, and both have the same voltage and the voltage turn , turn ratio ratio and the same.

280 × 224 - ... to have a turns ratio of 2:3 rather than 0.667 or 100:150.

For example of mathematics

1. A 4600/230 V Transformer Secondary to the primary coil 36 of the screw of the screw and how?

Answer: - 

that is,

 Ep = 4600 V

Es = 230 V

Ns = 36

Np =?

Ep / Es = Np / Ns

Np = Ep / Es × Ns

      = 4600/230 × 36

     = 720 turns.

2. A 25Hz 2400/230 Volt single phase voltage transformer 8V per piece to determine: -

1. The number of primary and secondary screw,

2. Maximum flux cores.

Answer: -

Reply: -

Are,

ƒ = 25Hz

Ep = 2400 V

Es = 230 V

E / turn = 8 V

Np =  ?

Ns =   ?

Φm =  ?

1. Np = 2400 / 8 = 300 turns

Ns = 230/8 = 29 turns

2. Ep = 4.44ƒNpΦ

Φ = Ep / 4.44ƒNp

 = 2400 / 4.44 × 25 × 300

= 72.07 m wab.

3. A 50KVA 440/110 Volt Transformer transformer has a primary coils turn 1000. The transformer Ip, Is and Ns to determine.

Answer: -

 that is,

Rating = 50 KVA

Ep = 440 V

Es = 110 V

Np = 1000 turn

Ip =?

Is =?

Ns =?

Ip = 50 × 1000 ÷ 440 = 113.64 A

Is = Ep ÷ Es × Ip

    = 440 ÷ 110 × 113.64 = 454.54 A

Ns = Es÷ Ep × Np

    = 110 ÷ 440 × 1000

    = 250 turns.