"SLC22/WK1: Advanced Electrical Transformers and the Power Factor."

dwings -

Home work on an Engagement Challenge.

A) Figure- 220/415V depicts a step-up transformer as the number of turns of the coil on the secondary core(Ns = 8) is more than the number of turns of coil on the primary core(Np = 4). Also, based on the voltage rating, the transformer's output voltage at the secondary side of the transformer(HL) is 415V which is more than the the transformer's input voltage of 220V at the primary side of the transformer(HT).

Figure - 415/220V depicts a step-down transformer as the number of turns of coil on the secondary core(Ns = 4) is less than the number of turns of coil on the primary core(Np = 8). Also, based on the voltage rating, the transformer's output voltage at the secondary side of the transformer(HL) is 220V which is less than the the transformer's input voltage of 415V at the primary side of the transformer(HT).

The reasons above strongly justify that Figures - 220/415V and 415/220V are step-up and step-down transformers respectively.

B)

A labelled Power Transformer

Name of labelled part and it's function:
• Radiator(A): As the transformer gets loaded, its temperature begins to increase with time causing the release of oil from the conservator to cool the transformer but with time the oil gets heated . The radiator helps in the cooling of the heated oil.

• Conservator(B): This is an essential part of a transformer though not all transformer has it based on their mode of operation. It is mostly cylindrical in shape and it helps in the storing of the transformer's oil.

• Buchholz relay(C): This is the part of the transformer functions like a circuit breaker by stopping the functioning of the transformer when there's a detected fault.
•Breather(D): Air is needed by the transformer, the breather helps in the inflow of air. It also helps to dehumidify the air getting into the transformer since it contains a silica gel.

• HT Brushing(E): This is the part of the transformer that determines the input voltage. It is also referred to the primary side of the transformer.

• Temperature gauge(F): The temperature of the oil and coil of the transformer is measured using this device. It also helps in switching-off the transformer when there's overheating.

• LT Brushing(G): This is the part of the transformer that determines the output voltage. It is also referred to the secondary side of the transformer. It is at this side of the brushing that the load is been connected.

C)


sourced
The type of transformer that is been used in my home's electricity is the step-down transformer. It's a type of transformer that is proficient in converting from a high voltage to a low voltage. The secondary coil has a less number of turns compare to that of the primary coil that has a greeter number of turns.

D) Using the conversion formula;
KVA = KW/PF
Where;
Kilovolt- ampere (KVA) = ?
Kilowatts (KW) = 5000KW
Power Factor(PF) = 0.9

Therefore,
KVA = KW/0.9
KVA =5000/0.9
KVA =5555.55KVA
KVA ~6000KVA
From the calculation, 6000KVA transformer will be required or needed for a 5000KW load.

E) Transformer name= Step-down transformer.
Transformer rating = 2500KVA
Primary Voltage(HV) = 33000KV
Secondary Voltage(LV) = 415KV
Current HV = 43.74A
Current HL = 3478.11A

F) Given:
Power of the transformer = 2.5MVA
Primary Voltage(HV) = 33KV
Secondary Voltage(LV) = 0.415KV
Current in HT = ?
Current in LT = ?

Solution:
Power of the transformer = 2.5MVA = 2.5 * 10^6VA
Primary Voltage(HV) = 33KV = 3310^3V
Secondary Voltage(LV) = 0.415KV = 0.41510^3V

Using the Formula:
P = √3 *V *I *Cos∅; where Cos∅ = 1

Current in HV will be calculated this;
2.510^6VA = √3 * 3310^3V * I
I = 2.510^6VA / √3 * 3310^3V
I = (2.510^6 / 1.73 * 3310^3)A
I = (2.5*10^6 / 57090)A
I = 43.79A
Therefore, the current in HV (HI) is 43.79A

Current in LV:
Recall that:
P = √3 *V *I Cos∅; where Cos∅ = 1
But V(LV) = 0.415KV = 0.41510^3V

Current in HV will be calculated this;
2.510^6VA = √3 * 0.41510^3V * I
I = 2.510^6VA / √3 * 0.41510^3V
I = (2.510^6 / 1.73 * 0.41510^3)A
I = (2.5*10^6 / 717.95)A
I = 2.5485A
I~2.55A
Therefore, the current in LV (LI) is 2.55A

G) Power factor based on my understanding is a ratio of active power(the fraction of the transformer's power that is been used or consumed) to the apparent power(the total amount of power generated by the transformer).
The power factor of my country is 0.8

H) The three(3) differences noticed in the two given transformers are:

Energy Pac

  1. The power of the transformer is 2500KVA
  2. Its Impedance is 6.6%
  3. The mass/weight of oil is 2266Kg

General Electronics Manufacturing Co. Ltd

  1. The power of the transformer is 200KVA
  2. Its Impedance is 4.01%
  3. The mass/weight of oil is 244Kg

Determine True/False:
• The name of the oil in a transformer is Pyranol- (True)
• The core loss of the transformer is in the winding -(False)
• The efficiency of the transformer is less than that of other electrical devices -(False)
• The transformer rating is in KW -(False)
• The transformer's insulation test is done with a Megger meter -(True)

I will like @bossj23 @basil20 @nsijoro @ahsansharif and @Cindy to share their views on the topic discussed.