Skip to main content

Transformer Interview Questions Part 2

Q. Why the transformer ratings are in kva?

A. Since the power factor of transformer is dependent on load we only define VA rating and does not include power factor .In case of motors, power factor depend on construction and hence rating of motors is in KWatts and include power factor.

Q. Is copper loss affected by change in power factor?

A. Yes. Copper loss varies inversely with power factor.It depends on current in primary and secondary windings. It is known that current required is higher when power factor is lower

Q. How do we minimize eddy current loss?

A. By laminating the core we can minimize eddy current loss.

Q. Does the transformer draw any current when its secondary is open?

A. Yes. It draws No load primary current.

Q. What is the difference between delta-delta, delta-star transformer?

A. Delta-delta transformer is used at generating station or a receiving station for Change of Voltage (i,e) generally it is used where the Voltage is high & Current is low whereas Delta-star is a distribution kind of transformer where from secondary star neutral is taken as a return path and this configuration is used for Step down voltage phenomena.

Q. What is the function of transformer oil?

A. Transformer Oil serves mainly two purposes
1. It acts as liquid insulation in electrical power transformer
2. It dissipates heat of the transformer i.e acts as coolant
.          In addition to these, this it serves other two other purposes, it helps to preserve the core and winding as these are fully immersed inside oil and also prevents direct contact of atmospheric oxygen with cellulose made paper insulation of windings, which is susceptible to oxidation.

During 1970s, polychlorinated biphenyls (PCB)s were often used as a dielectric fluid since they are not flammable. But, They are toxic, and under incomplete combustion, can form highly toxic products such as furan.
Presently, non-toxic, stable silicon-based or fluoridated hydrocarbons are used. Combustion-resistant vegetable oil-based dielectric coolants and synthetic pentaerythritol tetra fatty acid (C7, C8) esters are also becoming increasingly common as alternatives to naphthenic mineral oil.

Q. If we give 2334 A, 540V on Primary side of 1.125 MVA step up transformer, then what will be the Secondary Current, If Secondary Voltage=11 KV?

A. As we know the Voltage & current relation for transformer-V1/V2 = I2/I1
We Know, VI= 540 V; V2=11KV or 11000 V; I1= 2334 Amps.
By putting these value on Relation-
540/11000= I2/2334
So,I2 = 114.5 Amps

Q. Explain the operation of variable frequency transformer?

A. A variable frequency transformer is used to transmit electricity between two asynchronous alternating current domains. It is a double fed electric machine resembling a vertical shaft hydroelectric generator with a three-phase wound rotor, connected by slip rings to one external ac power circuit. A direct-current torque motor is mounted on the same shaft. Changing the direction of torque applied to the shaft changes the direction of power flow; with no applied torque, the shaft rotates due to the difference in frequency between the networks connected to the rotor and stator. Thus it acts as a continuously adjustable phase shifting transformer. It allows control of the power flow between two networks.

Q. What will happen if DC supply is given on the primary of a transformer?

A. Mainly transformer has high inductance and low resistance. In case of DC supply there is no inductance, only resistance will act in the electrical circuit. So, high electrical current will flow through primary side of the transformer. Due to this the coil and insulation will burn out.

Q. Why Delta Star Transformers are used for Lighting Loads?

A. For lighting loads, neutral conductor is must and hence the secondary must be star winding.  This lighting load is always unbalanced in all three phases. To minimize the current unbalance in the primary we use delta winding in the primary. So delta / star transformer is used for lighting loads.

Q. What are the types of cooling systems it transformers?

A. 1. ONAN (oil natural, air natural)
     2. ONAF (oil natural, air forced)
     3. OFAF (oil forced, air forced)
     4. ODWF (oil direct, water forced)
     5. OFAN (oil forced, air forced)

Comments

Popular posts from this blog

Transformer multiple choice questions part 1

Hello Engineer's Q.[1] A transformer transforms (a) frequency (b) voltage (c) current (d) voltage and current Ans : D Q.[2] Which of the following is not a basic element of a transformer ? (a) core (b) primary winding (c) secondary winding (d) mutual flux. Ans : D Q.[3] In an ideal transformer, (a) windings have no resistance (b) core has no losses (c) core has infinite permeability (d) all of the above. Ans : D Q.[4] The main purpose of using core in a transformer is to (a) decrease iron losses (b) prevent eddy current loss (c) eliminate magnetic hysteresis (d) decrease reluctance of the common magnetic circuit. Ans :D Q.[5] Transformer cores are laminated in order to (a) simplify its construction (b) minimize eddy current loss (c) reduce cost (d) reduce hysteresis loss. Ans : B Q.[6] A transformer having 1000 primary turns is connected to a 250-V a.c. supply. For a secondary voltage of 400 V, the number of secondary turns should be (a) 1600 (b) 250 (c) 400 (d) 1250 A

Condition for Maximum Power Developed In Synchronous Motor

The value of δ for which the mechanical power developed is maximum can be obtained as, Note : Thus when R a is negligible, θ = 90 o for maximum power developed. The corresponding torque is called pull out torque. 1.1 The Value of Maximum Power Developed        The value of maximum power developed can be obtained by substituting θ = δ in the equation of P m .        When R a is negligible,     θ = 90 o  and cos (θ) = 0 hence, . . .               R a = Z s cosθ   and X s = Z s sinθ        Substituting   cosθ = R a /Z s in equation (6b) we get,         Solving the above quadratic in E b we get,        As E b is completely dependent on excitation, the equation (8) gives the excitation limits for any load for a synchronous motor. If the excitation exceeds this limit, the motor falls out of step. 1.2 Condition for Excitation When Motor Develops ( P m ) R max        Let us find excitation condition for maximum power developed. The excitation

Effect of Slip on Rotor Parameters : Part2

Effect of Slip on Rotor Parameters 2. Effect of Slip on Magnitude of Rotor Induced E.M.F        We have seen that when rotor is standstill, s  = 1, relative speed is maximum and maximum e.m.f. gets induced in the rotor. Let this e.m.f. be,                 E 2 = Rotor induced e.m.f. per phase on standstill condition         As rotor gains speed, the relative speed between rotor and rotating magnetic field decreases and hence induced e.m.f. in rotor also decreases as it is proportional to the relative speed N s - N. Let this e.m.f. be,                E 2r = Rotor induced e.m.f. per phase in running condition  Now        E 2r α N s while E 2r α N s - N        Dividing the two proportionality equations,               E 2r /E 2 = ( N s - N)/N s    but (N s - N)/N = slip s               E 2r /E 2 = s               E 2r = s E 2        The magnitude of the induced e.m.f in the rotor also reduces by slip times the