MAGNETIC CIRCUIT AND COOLING OF ELECTRICAL MACHINES (2marks)

SHORT QUESTIONS AND ANSWERS.
1.Write any two similarities and differences between Electric and Magnetic Circuits.
Similarities-
Electric Circuits
1.Emf circulates current in a closed path.
2.Flow of Current is opposed by Resistance
Differences:
1.When current flows,energy is spent continuously
2.Current actually flows in Electric Circuits
3.Ohms Law:emf=Current XResistance
Magnetic Circuits
1.Mmf creates flux in a closed path.
2. Creation of flux is opposed by reluctance.
1.Energy is needed to create the flux,but not to maintain it.
2.Flux does not flow in a magnetic circuit.
3. Ohms Law:mmf=Flux XReluctance
2.What is magnetization curve?
*Curve – Y axis-Magnetic field intensity(H) Vs X axis-Flux density(B).of Magnetic material.
*To estimate mmf required for flux path in the magnetic material.
3.Define gap contraction factor for slots and ducts and gap contraction factor..
Kgs=Gap contraction factor for slots.=Reluctance of AG for slotted arm/Rel.Of AG for smooth arm.
Kgd=GCF for ducts.=Rel. of AG for armature of a machine with ducts/Rel.Of AG without ducts .
K g = Rel. of AG for arm. in m/c with slotted arm& ducts/Rel.Of AG of smooth arm. and without ducts .
K g =Total gap contraction factor for slots and ducts=KgsxKgd.≈1.2
Note:The slots and ducts in the armature (or stator or rotor) of electrical m/c increases the reluctance of air-gap ,which in turn increases the mmf required for air-gap.Kg represents the increase in reluctance as an increase in air-gap length.With the knowledge of Kg,the mmf required for air-gap can be estimated without calculating the increase in reluctance due to slots &ducts.
4.What are ventilating ducts?
The radial ventilating ducts are small gaps of width wd in between the stacks of armature core.They are provided for improving cooling of the core when the length of the core is greater than 0.1m.

5.What is Cater’s coefficient?What is its significance?.
It is a parameter used to estimate the contracted or effective slot pitch in case of armature with open or semi enclosed slots .It is a ratio of wo/lg,where wo=slot opening,lg=length of AG.
It is also used to estimate the effective length of armature when ducts are employed. It is a ratio of wd/lg,where wd=width of duct,lg=length of AG.
6.What is the effect of salient poles on the air-gap mmf?
In salient pole machines, the length of air-gap is not constant over the whole pole pitch.
Hence the effective air-gap is not constant over the whole pole pitch. Hence the effective air-gap length is kglg,where kg is the gap contraction factor. Also for calculating mmf, the maximum gap density Bg,at the centre of the pole is considered instead of average gap density.The field form factor kf, relates the average gap density over the pole pitch to maximum flux density in the air-gap ,given by,
Field form factor,kf=Bav/Bg;Also,kf≈ψ=Pole arc/Pole pitch.
7.What are the problems encountered in estimating mmf for teeth?List the methods of estimating them
Problems:(i).The flux density in different section of a tooth is not uniform.(ii).The slots provides another parallel path for the flux.
Estimation of mmf for teeth: (i).Graphical method.(ii)Three ordinate method.(iii).B1/3 method.
8.Distinguish between real and apparent flux densities in the tooth section of slot.
The real flux density is due to the actual flux through a tooth.The apparent flux density is due to total flux that has to pass through the tooths.Since some of the flux passes through slot,the real flux density is always less than the apparent or total flux density.
Bapp=Total flux in a slot pitch/Tooth area. Breal=Actual flux in a tooth/Tooth area.
9.What is leakage flux and leakage co-efficient?How will you minimize the magnetic leakage?
The leakage flux is the flux passing through unwanted path. It will not help either for transfer or conversion of energy. Leakage co-efficient,Cl=Total flux/Useful flux.The leakage flux affects the excitation demand,regulation,forces on the winding under short circuit conditions, commutation etc.
Leakage flux flows through air-gap of the m/c.If the air-gap of the m/c is kept as low as possible ,LF↓.
10.What is fringing flux?What are the differences between leakage flux and fringing flux?
The bulging of magnetic path at the air-gap is called fringing.The fluxes in the bulged portion are called fringing flux. The leakage flux is not useful for energy transfer or conversion and flows in unwanted path. But the fringing flux is useful flux and flows in the magnetic path.The effect of leakage flux on the m/c performace is accounted by leakage reactance whereas fringing flux increases the slot reactance.
11.List the various types of armature leakage flux.
Slot LF:flux crosses the slot from one tooth to the next and returning through iron.
Tooth topLF:Flux flowing from top of one tooth to the top of another tooth.
Zig-Zag LF:Flux passing form one tooth to another in a Zig-Zag fashion across the air-gap.
Overhang LF:Flux produced by overhang portion of the armature winding.
Harmonic or Differential LF:Flux produced due to difference in stator and rotor harmonic content.
Skew LF:Reduction in mutual flux due to skewing of rotor in induction motor.
Peripheral LF:Flux flowing circumferencially round the air-gap without linking with any of the winding.
12.Define leakage reactance. What is reactance voltage? Is it beneficial?.
The leakage reactance is the reactance offered by leakage flux.The induced voltage due to LF is reactance voltage. It is not beneficial as it affects commutation, and produce sparking in dc machines.
Leakage reactance,Xs=2πfZs2Lλs;L=Length of Arm.λs=Specific slot permeance,Zs=Conductors/slot.
Leakage reactance for poly phase m/c,Xs=8πfTph2Lλs/pq.
13.What is permeance and specific permeance?
Permeance is inverse of reluctance. S=Reluctance=l/μoμrA;Δ=Permeance(=1/S).
Specific permeance: Permeance per unit length of slot.
14.What is stacking factor?
The cores of magnetic circuits are built up with laminated steel plates wherever required. These laminations are insulated from each other by paper, stuck one side of the lamination.Also ventilating ducts are provided along the length of the armature.Hence ,it is clear that the whole length of the armature is not occupied by iron;some part of length is taken up by ventilating ducts and some part by lamination.It is usual to define iron space factor or stacking factor that relates the length of the armature interms of the non-iron portion as Stacking factor=Length of Iron in a stack of assembled core plates to total axial length.
∴Gross iron length,Ls=core length-length of ventilating ducts=L-ndwd;Net iron length,Li=Ki(L-ndwd)
15.On what factors Eddy current and Hysteresis loss depends?
Hysteresis Lossα fBm1.5;Eddy current lossαf2Bm2---wattss
16.What are the various methods of cooling of Turbo-alternator?
(a).Air Cooled Turbo-alternators:
(i).One sided axial ventilation :(upto 3MW).Machine is supplied with air by propeller fan and the air enters the machine from one side and leaves from the other.
(ii).Two sided axial Ventilation:(12MW).Air is forced through the machine from both sides.
(iii).Multiple inlet system:(60MW)Useful for machines having longer core lengths.Outer stator casing is divided into number of compartments, with alternative inlet(air is directed radially inwards) and outlet chambers(air is directed radially outwards).The air is drawn from the outlet chamber and is sent to the coolers where it is cooled and recirculated.
(b). Hydrogen Cooled Turbo-alternators:(more than 60MW).Hydrogen when mixed with air forms an explosive mixture over avery wide range.∴The frame of hydrogen cooled machines has to be made strong enough to withstand possible internal explosion without suffering serious damage.All joints in cooling circuits are made gas tight and oil film shaft seals are used to prevent leakage of hydrogen.
Intially,a pressure of 105kN/m2 was used .For modern conventionally cooled turbo-alternators,the pressure is about 200-300 kN/m2.Fans mounted on the rotor circulate hydrogen through the ventilating ducts and internally arranged coolers.The gas pressure is maintained by an automatic regulating and reducing valve controlling the supply from gas cylinders.
17.Define heating and cooling time constant.
Heating time constant: Time taken by the machine to attain 0.632 of its final steady temperature rise.It is an index of the machine to attain its final steady temperature rise.
θ=θm(1-e-t/Th);If t =Th ;θ=0.632θm;T=Gh/Sλ;λ=specific heat dissipation in W/m2-°C;G=weight of active part of machine inkg,h=specific heat in J/kg-°C,S=Cooling surface in m2.
Cooling time constant:θ=θie-t/Tc; If t =Tc ;θ=0.368θi;∴Cooling time constant is the time taken by the machine for its temperature rise to fall to 0.368 of its initial value.
18.Define rating of electrical machine.
The rating of machines refers to the whole of the numerical values of electrical and mechanical quantities with their duration and sequences assigned to the machines by the manufacturers and stated on the rating plate, the machine complying with the specified conditions.
19.Define Continuous, short time and intermittent short time duty of electrical machine.
(i).Continuous Duty:On this duty,the duration of load is for a sufficiently long time such that all parts of the motor attain thermal equilibrium,ie the motor attains maximum final steady temperature rise.
Continuous rating of a motor is defined as the load that may be carried by the machine for an indefinite time without the temperature rise of any part exceeding maximum permissible value.(eg)Fan,pumps.
(ii).Short time duty:The motor operates at a constant load for some specified time which is then followed by a period of rest.The period for load is so short that the machine cannot reach its thermal equilibrium,ie steady temperature rise while the period for rest is so long that the motor temperature drops to the ambient temperature.(eg).Railway turntable,navigation lock gates.
(iii).Intermittent periodic duty:On intermittent periodic duty,the periods of constant load and rest with machine de-energised alternate.The load periods are too short to allow the motor to reach its final steady state value while periods of rest are also too small to allow the motor to cool down to the ambient temperature.(e.g)Cranes,lifts,metal cutting machine.
20.Distinguish between conventionally cooled and direct cooling of Turbo-alternator.
Conventional cooling:Here machines dissipate their losses to a coolant which is entirely outside the coil.
Direct cooling:Process of dissipating the winding losses to a cooling medium circulating within the winding insulation wall.Here the Coolant is in direct contact with conductor.