Class 12 Physics Alternating Current Energy losses in actual transformers

Energy losses in actual transformers

Flux leakage:-

  •  There are air gaps between the primary and the secondary coils  because of which the change of flux which is associated with the primary coil is not completely transferred to secondary coil.
  • In order to reduce the loss secondary coil can be wound over the primary coil.
  • For example:- In case of toroidal transformer cores, over the primary coil secondary coil is wound above it. As a result there is no air gap in between them.

 

Resistance of windings:-

  • The wire used for the windings has some resistance and so, energy is lost due to heat produced in the wire (I2R).
  •  If the area of the cross section of the wire is increased then the resistance will be reduced considerably.
  • So the thick wires are used in the windings of primary and secondary coils as a result resistance will be less .
  • The amount of heat lost because of wires will be less as resistance is minimal.

Eddy currents:-

  • Soft Iron core also gets heated up because of magnetic flux as a result eddy currents are developed in the soft iron core.
  • Core gets heated up because of eddy currents. This will harm the transformer core.
  • In order to prevent this laminated core can be used. Because of insulated covering eddy  currents are not able to produce the heating effects.

 

Hysteresis:-

  • There is energy loss involved during the magnetisation of the material of the core.
  • Always those materials are to be chosen for which hysteresis loss is minimum.
  • That is why Soft Iron core is used instead of permanent magnets.

Problem:- A small town with a demand of 800 kW of electric power at 220 V is situated 15 km away from an electric plant generating power at 440 V. The resistance of the two wire line carrying power is 0.5 Ω per km. The town gets power from the line through a 4000 - 220 V step-down transformers at a sub-station in the town.

(a) Estimate the line power loss in the form of heat.

(b) How much power must the plant supply, assuming there is negligible power loss due to leakage?

(c) Characterise the step up transformer at the plant.

Answer:-

Total electric power required, P = 800 kW = 800 × 103 W

Supply voltage, V = 220 V

Voltage at which electric plant is generating power, V' = 440 V

Distance between the town and power generating station, d = 15 km

Resistance of the two wire lines carrying power = 0.5 Ω/km

Total resistance of the wires, R = (15 + 15)0.5 = 15 Ω

A step-down transformer of rating 4000 − 220 V is used in the sub-station.

Input voltage, V1 = 4000 V

Output voltage, V2 = 220 V

RMS current in the wire lines is given as:

I  = (P/V1)

= (800 x 103) / (4000)

I = 200A

(a) Line power loss = I2R

= (200)2× 15

= 600 × 103 W

= 600 kW

(b) Assuming that the power loss is negligible due to the leakage of the current:

Total power supplied by the plant = 800 kW + 600 kW

= 1400 kW

(c) Voltage drop in the power line = IR = 200 × 15 = 3000 V

Hence, total voltage transmitted from the plant = 3000 + 4000

= 7000 V

Also, the power generated is 440 V.

Hence, the rating of the step-up transformer situated at the power plant is

440 V − 7000 V.

Problem:- Do the same exercise as above with the replacement of the earlier transformer by a 40,000-220 V step-down transformer (Neglect, as before, leakage losses though this may not be a good assumption any longer because of the very high voltage transmission involved). Hence, explain why high voltage transmission is preferred?

 

Answer:-

The rating of a step-down transformer is 40000 V−220 V.

Input voltage, V1 = 40000 V

Output voltage, V2 = 220 V

Total electric power required, P = 800 kW = 800 × 103 W

Source potential, V = 220 V

Voltage at which the electric plant generates power, V' = 440 V

Distance between the town and power generating station, d = 15 km

Resistance of the two wire lines carrying power = 0.5 Ω/km

Total resistance of the wire lines, R = (15 + 15)0.5 = 15 Ω

P = V1I

RMS current in the wire line is given as:

I = (P/V1)

= (8000 x 103) / (40000)

=20 A

(a) Line power loss = I2R

= (20)2 × 15

= 6 kW

 

(b) Assuming that the power loss is negligible due to the leakage of current.

Hence, power supplied by the plant = 800 kW + 6kW = 806 kW

(c) Voltage drop in the power line = IR = 20 × 15 = 300 V

Hence, voltage that is transmitted by the power plant = 300 + 40000 = 40300 V

The power is being generated in the plant at 440 V.

Hence, the rating of the step-up transformer needed at the plant is 440 V − 40300 V.

Hence, power loss during transmission = (600/1400) x 100% = 42.8%

In the previous exercise, the power loss due to the same reason is:-

(6/806) x 100 = 0.744%

Since the power loss is less for a high voltage transmission, high voltage transmissions are preferred for this purpose.

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