Unit I

INTRODUCTION

S. No. TOPIC Page No.

1 Introduction 1.1 Basic Structure of Electromagnetic rotating machine 1.1.1 Major considerations in Electrical Machine Design 1.1.2 Limitations in design 1.1.3 Classification of design problems 1.2 Materials for Electrical Machines 1.2.1 Electrical conducting materials 1.2.2 Magnetic materials 1.2.3 Insulating materials 1.3 Space factor 1.4 Magnetic and Electric Loadings 1.4.1 Total loadings 1.4.2 Specific Loadings 1.4.3 Choice of Specific Magnetic loadings 1.5 Thermal considerations 1.5.1 Heat flow 1.5.2 Newton’s law of Cooling 1.5.3 Internal temperatures (Hot temperatures) 1.5.3.1 Calculation of internal temperature 1.5.3.2 Temperature gradients in cores 1.5.3.3 Heat flow in two dimensions 1.5.3.4 Thermal gradients in conductors placed in slots 1.5.3.5 Heating of turbo alternators 1.5.4 Thermal state in electrical machines 1.5.4.1 Theory of solid body heating 1.5.4.2 Heating 1.5.4.3 Cooling 1.6 Rating of machines 1.6.1 Motor ratings based on duty 1.7 Standard specifications 1.8 Solved problems 1.9 Two mark Q;A 1.10 University Question 1.11 Additional Problems 1. INTRODUCTION

In General, designing of any machine bring about application of science and technology. It is very much essential to produce machines with more cost-effective, highly durable, better quality and efficient without compromising its standard specifications. It is understand that electrical machines such as generators, motors and transformers are used to convert and transfer energy. The rotating machines are made of field and armature windings which are influencing magnetic flux. Amongst the two windings of rotating machines, field winding produces the flux and armature winding supplies electrical power/mechanical power. In stationary machine, primary winding supplies the power which is demanded by the secondary winding.

The designing of an electrical machine requires the dimensions of magnetic and electrical circuits, insulation system, etc. The analytical methods are used to find out the dimensions of the above mentioned parameters.

In general, the designer of a machine deals with the different type and number of problems. For all the problems solution differs and for sometimes, ends without a solution or many solutions for a single problem. However, it is important that solution provided by the designer should satisfy the requirements with appreciable efficiency. The designer should also ensure the machine is highly reliable, durable, least weight, lower temperature rise and lower cost.

A practical designer not only concentrates on the specifications of the machines but also should ensure the adaptability of the machine under stock. This situation sometimes forces the designer of the machine to compromise between the ideal design and a design which comply with manufacturing conditions.

It is stated earlier that a machine should be designed by following the national and international standards. Therefore, the electrical designer must be familiar with the standards. The list of recommended national and international standards is,

British Standard (BS), England

Indian Standard (IS), Bureau of Indian Standard (BIS), India

International Electro technical Commission (IEC)

NEMA (The National Electrical Manufacturers Association).

Basic Structure of Electromagnetic rotating machine

The main parts of the rotating machine and its structure have been depicted through the Figure 1.1.

Mechanical parts: The frame, bearings and shaft are the main mechanical parts of the machine

Constructional elements of rotating machines: The main constructional elements of the machine are stator and rotor. The construction of these elements differs for the DC and AC machines.

DC Machine

Stator: Yoke, Field Pole, Pole Shoe, Field Winding, InterpoleRotor: Armature Core, Armature Winding, CommutatorOthers: Brush and Brush holder

AC Machines

Synchronous Machine

Stator: Yoke, Field Pole, Pole Shoe, Field Winding, InterpoleRotor: Armature Core, Armature Winding

Squirrel Cage Induction Motor

Stator: Frame, stator core and stator winding

Rotor: Rotor core, rotor bars and End rings

Slip Ring Induction Motor

Stator: Frame, stator core and stator winding

Cores

Frames

Bearing

Air gap

Rotor

Stator

Shaft

Windings

Rotor: Rotor core and rotor bars

Figure1.1 Basic Structure of Electromagnetic Rotating Machine

The machine functions four important circuits namely Magnetic, Electric, Dielectric and Thermal circuits.

Magnetic Circuit: This circuit provides the magnetic flux path which consists of air gap, stator and rotor teeth and stator and rotor cores (yoke).

Electric Circuit: The stator and rotor windings of the rotating machine are the main components of the electric circuit. The windings are used to transfer electrical energy between the two regions. It ensures the production of emf and development of magneto mechanical force (mmf) in the machine. In order to protect the electric circuit, the windings are suitably insulated.

Dielectric circuit: The insulation between the conductors, and core and windings are made through the dielectric circuit. The non-metallic materials preferably organic or inorganic, and natural or synthetic are recommended as the insulating materials.

Thermal Circuit: The thermal circuit is very much essential for heat dissipation which is produced inside the machine. The circuit is concerned with mode and media for dissipation of heat.

1.1.1 Major considerations in Electrical Machine Design

As mentioned above, the Electrical machine has different functions which are supported by dielectric or insulation, cooling system and mechanical parts. Therefore, it is advisable to put up extra care while designing the machine. The following list shows the important factors need to be considered for designing the machine,

Magnetic circuit or the flux path:

Magnetic circuit is the main source for core losses. It can be reduced by minimizing the usage of mmf without compromising the required amount of flux.

Electric circuit or windings:

Electric Circuit should ensure induction of required EMF in the machine with lesser complexity in winding arrangement. Additionally, the construction of electric circuit should produce lesser copper loss.

Dielectric circuit or insulation:

It should ensure the trouble free separation of machine parts which operating at different potential and restrict the current flow in the given paths of the circuit.

Cooling system or ventilation:

It should ensure operating temperature of the machine within the specified limit.

Machine parts:

It should ensure the robustness and accommodate the changes in size.

Other factors:

The machine design also involves optimization of manufacturing, operating and maintenance cost. The other factors requires consideration apart from the above mentioned are

Limitation in design (saturation, current density, insulation, temperature rise etc.,)

Customer’s needs

National and international standards

Convenience in production line and transportation

Maintenance and repairs

Environmental conditions etc.

1.1.2 Limitations in design

Many materials are involved in designing a machine and its supporting peripherals which imposes a restriction in design. These limitations due to the saturation of iron, current density of conductors, temperature change, insulation properties, mechanical properties, efficiency, power factor, commutation, etc.

Saturation: The volume of iron material decides the flux density of the field. The higher value of flex density drives the iron to operate beyond knee of the magnetization curve or in the region of saturation. This saturation of iron leads to the increase in core loss and performs under excitation and also produces harmonics. The excitation should be improved to meet the desired flux.

Current density: The current density can be improved by using greater volume of copper which results increase in the core losses and temperature.

Insulation (which is both mechanically and electrically weak): The most susceptible part of any machine is its insulation. The life of a machine depends upon the type of insulating materials used for its construction. It is most important to select the insulation material for the machine in order to withstand the electrical, mechanical and thermal stresses produced in it. In case of transformer, the mechanical strength of insulation material is one of the most important parameter. For instance, when a transformer is forced under short circuited secondary and primary is operating, it produces Large axial and radial forces. Therefore, while designing insulation for the transformer, it is essential to select the material to withstand large mechanical stresses. Further, the insulating material should be selected such a way to withstand higher operating temperature. The size of insulation is decided by the voltage stresses and mechanical stresses produced.

Temperature: As mentioned above insulating materials decides operating life of any machine. Notably, the life of the insulating materials in turn depends upon the withstand capacity of machine under higher temperature. The life of the insulating material gets reduced if it is operated more than its allowable temperature. This issue can be addressed by providing proper cooling and ventilation techniques. In the cooling system, coolant flows along proper paths absorbs the heat from the machine parts for dissipation to outside the machine. This helps to prevent the temperature rise of the machine and life time of the machine can be taken care.

Efficiency: The operating cost of any machine has been decided by its efficiency. The efficiency of the machine can be improved by using small magnetic and electrical loading with the support of large amount of conducting materials (both iron as well as copper/Aluminium). Though the capital cost of the machine is high but it reduces the operating cost.

Mechanical parts: It is necessary to address numerous technological requirements while constructing the electrical machine. The machine should be designed with lesser complexity without compromising on its technological concepts. The technology adapted for machine design should be unfailing with the performance requirements, reliability and durability. Most importantly, the design of mechanical parts for high speed machines is more critical. For example, while designing a turbo-alternator, the rotor slot dimensions are so selected that the mechanical stresses at the bottom of rotor teeth do not exceed the maximum allowable limit. In large machines, the size of the shaft is decided by considering the critical speed which depends of the deflection of the shaft.

Power factor: It is obvious that induction motors introduce poor power factor as the size and cost of induction motors are reduced with using a high value of flux density in the air gap. This meager power factor results in larger values of current for the same power demand and, therefore, conductor sizes are increased to the higher value. Further, the power factor also decides the value of flux density of the field. Hence, power factor is considered to be one of the limiting factors for machine design.

Commutation: Commutation of a relevant machine is important as it limits the output of the machine. For example, at present the maximum power output of a single unit d.c machine is approximately 10MW and this limitation is due to the commutation difficulties.

In addition to the above mentioned factors, Consumer, manufacturer or standard specifications may also create limitations in machine design.

1.1.3 Classification of design problems

The electrical machine design is a combination of most complex and diverse engineering problems. The engineering problems involved in designing rotating machines and Transformer is shown through the Table 1.1.

Table 1.1 Classification of design problems

Classification Rotating Machines Transformer

Electromagnetic Design Involves,

Stator and rotor core dimensions

Stator and rotor teeth dimensions

Air gap length

Stator and rotor windings Involves designing core and windings

Mechanical Design Involves design of frame, shaft and bearings Involves design of housing the core and winding assembly

Thermal Design Involves in the design of cooling ducts in core and cooling fans Involves in the design of cooling tubes and radiators

Dielectric Design Involves in the design of insulation between conductors Involves in the design of insulation between core and windings

Materials for Electrical Machines

Electrical Machine requires different varieties of materials for conducting, insulating and magnetic circuits. The selection of good conductors, insulators and permanent magnets are most essential for designing machine to bring out better performance.

1.2.1 Electrical conducting materials

The electrical conducting materials are divided into two main categories such as High conductivity materials and High resistivity materials. Amongst the two types, the later one is used for making resistances and heating devices.

All types of windings required for making electrical machines, apparatus and devices are made by high conductivity materials. The important properties of a good conductor are listed below in alphabetical order,

Allow brazing, soldering or welding so that the joints are reliable

Durable and cheap by cost

High melting point

High resistance to corrosion

High tensile strength

Highly malleable and ductile

Low value of resistivity or high conductivity

Low value of temperature coefficient of resistance

The above mentioned requirements differ with the intention. For instance, it is recommended to manufacture electrical windings with least possible resistivity for the cost of slight loss in tensile strength. Copper and Aluminum are the most commonly used high conductivity materials. The significant properties of copper and aluminum are shown in the Table 1.2.

Table 1.2 Properties of Copper and Aluminum

S.

No. Particulars Copper Aluminum

Resistivity at 200C 0.0172 ohm / m/ mm2 0.0269 ohm / m/ mm2

Conductivity at 200C 58.14 x 106 S/m 37.2 x 106 S/m

Density at 200C 8933 kg/m3 2689.9 m3

Temperature coefficient

(0-100oC) 0.393 % per 0C 0.4 % per 0C

Explanation: If the temperature increases by 1oC, the resistance increases by 0.4% in case of aluminum

Coefficient of linear expansion (0-100oC) 16.8×10-6 per oC23.5 x10-6 per oCTensile strength 25 to 40 kg / mm2 10 to 18 kg / mm2

Mechanical property

highly malleable and ductile not highly malleable and

ductile

Melting point 10830C 6660C

Thermal conductivity (0-100oC) 599 W/m0C 238 W/m0C

From the Table 1.2, it is understood that for the desired resistance and length, cross-sectional area of aluminum is 61% larger than that of the copper conductor and almost 50% lighter than copper. Aluminum materials are lesser expensive and easily available compared to Copper.

1.2.2 Magnetic materials

The orientation of the crystals of the material and technical specifications of a machine or equipment decides the magnetic properties of a magnetic material. The significant properties of the good magnetic materials are listed below,

A high curie point. (Above Curie point or temperature the material losses the magnetic property or becomes paramagnetic, that is effectively non-magnetic)

High electrical resistivity (to reduce the eddy current loss)

High saturation induction (to minimize weight and volume of iron parts)

Low reluctance or high value of relative permeability ?r.

Narrow hysteresis loop or low Coercivity (to minimize hysteresis loss and improve the operation efficiency)

Should have a high value of energy product (expressed in joules / m3).

Magnetic materials are broadly classified into Diamagnetic, Paramagnetic, Ferromagnetic, Antiferromagnetic and Ferrimagnetic materials. Amongst the all types, ferromagnetic materials are most preferred and suitable for electrical machines. Ferromagnetic properties are confined almost entirely to iron, nickel and cobalt and their alloys. The relative permeability (?r) of ferromagnetic material is greater than 1.0. The dipoles of the magnetic field made by ferromagnetic materials align themselves in the direction of the applied field and get strongly magnetized.

Further the Ferromagnetic materials can be classified as Hard or Permanent Magnetic materials and Soft Magnetic materials.

Hard or permanent magnetic materials:

This type of materials has large hysteresis loop (obviously hysteresis loss is more) and step by step rising magnetization curve. The examples are carbon steel, tungsten steal, cobalt steel, alnico, hard ferrite etc.

Soft magnetic materials:

This type of materials has small size hysteresis loop and a steep magnetization curve. Soft Magnetic materials are further classified as,

Solid Core Materials

Generally used for yokes poles of dc machines, rotors of turbo alternator etc., where steady or dc flux is involved. The examples are cast iron, cast steel, rolled steel, forged steel etc., (in the solid form).

Electrical Sheet and Strip materials

Silicon steel (Iron + 0.3 to 4.5% silicon) cab ne utilized in the laminated form. The ageing issue and core loss of a machine can be reduced by proper addition of silicon. Low silicon content steel or dynamo grade steel is used in rotating electrical machines and are operated at high flux density. High content silicon steel (4 to 5% silicon) or transformer grade steel (or high resistance steel) is used in transformers. Further sheet steel may be hot or cold rolled. Cold Rolled Grain Oriented Steel (CRGOS) is costlier and superior to hot rolled. CRGO steel is generally used in transformers.

Special purpose alloys

Nickel iron alloys have high permeability and addition of molybdenum or chromium leads to improved magnetic material. Nickel with iron in different proportion leads to

High nickel permalloy (iron +molybdenum +copper or chromium), used in current transformers, magnetic amplifiers etc.,

Low nickel Permalloy (iron +silicon +chromium or manganese), used in transformers, induction coils, chokes etc.

Perminvor (iron +nickel +cobalt)

Pemendur (iron +cobalt +vanadium), used for microphones, oscilloscopes, etc.

Mumetal (Copper + iron)

iv. Amorphous alloys (often called metallic glasses)

Amorphous alloys are produced by rapid solidification of the alloy at cooling rates of about a million degrees centigrade/sec. The alloys solidify with a glass-like atomic structure which is non-crystalline frozen liquid. The rapid cooling is achieved by causing the molten alloy to flow through an orifice onto a rapidly rotating water cooled drum. This can produce sheets as thin as 10?m and a meter or more wide.

These alloys can be classified as iron rich based group and cobalt based group.

Material Maximum permeability ? x 10-3 Saturation magnetization in teslaCoercivity A/m Curie temperature oCResistivity ?m x 108

3% Si grain oriented 90 2.0 6-7 745 48

2.5% Si grain non – oriented 8 2.0 40 74.5 44

<0.5% Si grain non oriented 8 2.1 50-100 770 1.2

Low carbon iron 3-10 2.1 50-120 770 12

78% Ni and iron 250-400 0.8 1.0 350 40

50% Ni and iron 100 1.5-1.6 10 530 60

Iron based Amorphous 35-600 1.3-1.8 1.0-1.6 310-415 120-140

1.2.3 Insulating materials

Insulating materials are used to isolate the electrical parts of a machine at different potentials in order to avoid malfunction of a machine. The properties of an ideal insulating material is listed below,

Flexible and cheap

Good moisture resistant

Good thermal conductivity

Good thermal oxidation capability

High dielectric strength.

High temperature with standing capacity.

High value of resistivity (like 1018 ?cm)

Liquid insulators should not evaporate or volatilize

Low dielectric loss angle ?

Should withstand stresses due to centrifugal forces (as in rotating machines), electro dynamic or mechanical forces (as in transformers)

Should withstand vibration, abrasion and bending

Insulating materials are classified as Solid, Liquid and Gas, and vacuum. The applicability of materials depends on the requirements.

i. Solid: This type of insulating materials is used with field, armature, and transformer windings. The examples of solid materials are listed below,

Ceramic : porcelain, steatite, alumina etc.,

Fibrous or inorganic animal or plant origin, natural or synthetic paper, wood, card board, cotton, jute, silk etc.,

Glass : soda lime glass, silica glass, lead glass, borosilicate glass

Mineral : mica, marble, slate, talc chloride etc.,

Non-resinous : mineral waxes, asphalt, bitumen, chlorinated naphthalene, enamel etc.,

Plastic or resins. Natural resins-lac, amber, shellac etc., Synthetic resins-phenol formaldehyde, melamine, polyesters, epoxy, silicon resins, bakelite, Teflon, PVC etc

Rubber : natural rubber, synthetic rubber-butadiene, silicone rubber, hypalon, etc.,

ii. Liquid: This material is used in transformers, circuit breakers, reactors, rheostats, cables, capacitors etc., & for impregnation. The examples are listed below,

Mineral oil (petroleum by product)

Synthetic oil askarels, pyranols etc.,

Varnish, French polish, lacquer epoxy resin etc.,

iii. Gaseous: Different gases are used for different purposes. The examples are listed below,

Air used in switches, air condensers, transmission and distribution lines etc.,

Halogens like fluorine, used under high pressure in cables

Hydrogen though not used as a dielectric, generally used as a coolant

Inert gases neon, argon, mercury and sodium vapors generally used for neon sign lamps.

Nitrogen use in capacitors, HV gas pressure cables etc.,

Out of the above mentioned list of insulating materials, any material can be utilized to meet all the desirable properties of equipment.

Insulating materials are further classified based on thermal consideration. The insulation system (also called insulation class) for wires used in generators, motors, transformers and other wire-wound electrical components is divided into different classes according the temperature that they can safely withstand.

As per Indian Standard ( Thermal evaluation and classification of Electrical Insulation,IS.No.1271,1985,first revision) and other international standard insulation is classified by letter grades A,E,B,F,H (previous Y,A,E,B,F,H,C).

Insulation class Maximum operating temperature in 0C Typical materials

Previous Present Y 90 Cotton, silk, paper, wood, cellulose, fiber etc., without impregnation or oil immersed

A A 105 The material of class Y impregnated with natural resins, cellulose esters, insulating oil etc., and also laminated wood, varnished paper etc.

E E 120 Synthetic resin enamels of vinyl acetate or nylon tapes, cotton and paper laminates with formaldehyde bonding etc.

B B 120 Mica, glass fiber, asbestos etc., with suitable bonding substances, built up mica, glass fiber and asbestos laminates.

F F 155 The material of class B with more thermal resistance bonding materials

H H 180 Glass fiber and asbestos materials and built up mica with appropriate silicon resins

C

C >180 Mica, quartz, ceramics, glass and asbestos with binders or resins of super thermal stability

1.3 Space factor

Space factor is defined as the ratio of copper area to the total winding area. It is used to select the size of conductor, air gap and insulation thickness. The expression for space factor for a field coil is given by,

Sf= afTfhfdf <1 where,

af = net cross section of each field conductor or turn, mm2

Tf = number of field turns

hf = height of field coil, mm

df = depth of field coil, mm

1.4 Magnetic and Electric Loadings

This section describes the loadings of the machine.

1.4.1 Total loadings

Total Magnetic Loading (TML): The total flux (Ø) around the armature (or stator) periphery at the air gap is called the total magnetic loading.

TML= pØ (ie. Total flux entering and leaving the armature)

Total Electric Loading (TEL): The total number of ampere conductors (Iz) around the armature (or stator) periphery is called total electric loadings.

TEL = IzZ

1.4.2 Specific Loadings

In the design of rotating electrical machines two types of loadings are specified namely magnetic loading and electric loading.

Specific magnetic loading :

It is defined as the average flux density over the air gap of a machine.

Bav= total flux around the air gaparea of flux path at the air gap= p??DL= ??LSpecific electric loading :

It is defined as the number of armature (or stator) ampere conductors per metre (or stator) periphery at the air gap.

Specific electric loading ac= Total armature ampere conductorsarmature periphery at the air gap= IzZ?D1.4.3 Choice of Specific Magnetic loadings

The choice of specific magnetic loading is influenced by definite factors. Some of these factors are general and applies to all types of machines and some of them are specific to individual machines. This section describes some of the general factors which influence the choice of specific magnetic loadings suitable for all types of electrical machines. Most of the specific loadings are determined by the following factors,

Maximum flux density and iron loss of the machine

Magnetizing current and

Core loss

Maximum flux density

The maximum flux density of the magnetic circuit should be maintained below the threshold value. The set value for the threshold depends on the material used for designing the circuit. The flux density in iron parts is directly proportional to the average flux density in the air gap which is expressed in equation 1.1. In a well constructed electrical machine, the maximum flux density occurs at the teeth of the machine. Therefore, the relationship has been derived between the flux density in the teeth and flux density in the air gap.

Let us consider a non-salient pole machine having S armature slots.

Flux over one slot pitch= p?S=p.Bav?DLp L . 1S= Bav?DSL= BavysL Where, ys= slot pitch = ?DS

Air gap

Slot

Tooth

Slot

WS

Wt

WS

YS

After neglecting the saturation, the entire flux over a slot pitch is carried out by the tooth which is shown through the Figure 1.2.

Parallel sided

Slot

Tapered tooth

Figure 1.2 Flux over a slot pitch Figure 1.3 Armature with tapered teeth

Area of flux path in each tooth = width of tooth x core length = Wt L

?Flux density in teeth Bt= flux in each tooth are of ach tooth= BavysLWtL= BavysWt= Bg.ysWt —- (1.1)

In a salient pole machine, the pole arc originates the flux and therefore the teeth which are under the pole arc carry whole of the flux and hardly any flux is carried by the teeth lying outside the pole arc.

Therefore, flux density in the teeth of salient pole machines is Bt= Bav?.ysWt —- (1.2)

= Bg.ysWt —- (1.3)

Where, Bg = maximum flux density in the air gap, and

? = ratio of pole arc to pole pitch

From the equations (1.2) and (1.3), it is evident that the flux density in the teeth/any part of the magnetic circuit is directly proportional to the specific magnetic loading. From the above equations (1.2) and (1.3), we get,

BtBav= ys?WtThus for a machine of given dimensions ratio BtBav is constant and therefore if Bt is not to exceed a maximum specified limit (2.0 to 2.2 wb/m2 in case of dc machines), specific magnetic loading has to be kept below a certain limiting value. For instance, let us consider a dc machine with tooth width equal to slot width and ? = 0.66,

Wt=Ys2 as Ys=Wt+ Ws and Wt= Ws

and therefore Bt=ys?WtBav=2?Bav=20.66Bav=3 Bavand if Bt is to be limited 2.1 wb/m2 then the value of Bav will not exceed 0.7 wb/m2.

Electrical machines with rectangular parallel sided slot will have tapered teeth and therefore the tooth width will not be same over the entire height of tooth which is shown through the Figure 1.3. As a result, it produces different values of flux density in teeth at different heights. The maximum value of flux density in teeth occurs where the tooth width is smallest i.e., at the root of the teeth in case of dc machines and at a section near the air gap for synchronous machines.

In large machines with larger diameter, the taper of teeth is least bothered and therefore the value of width of teeth will be same over their entire height. On the other hand, in small machines with smaller diameters, the taper of teeth is very prominent and consequently the ratio Bt/Bav is very large at the section where the teeth have the smallest width and hence for a given maximum value of Bt, the Bav must be reduced. Thus, it is concluded that specific magnetic loadings for small machines will be minimum.

Magnetizing current

The magnetizing current of electrical machine is directly proportional to the mmf required to strengthen the flow of flux through the air gap and iron parts of the machine. Moreover, the mmf required for air gap is directly proportional to the gap flux density i.e. the specific magnetic loading. With regards to the iron parts of the machine, it is discussed that the value of flux density of iron parts depends upon the value of specific magnetic loading. For the small value of specific magnetic loading, the flux density in the iron parts is low. It results on the linear or knee portion of the B-H curve which is shown in Figure 1.4. This requires a small or even negligible value of mmf for iron parts, as H, the mmf/m length is very small for flux densities on the linear and knee portions of the curve.

Nevertheless, for the large value of specific magnetic loading, the flux density in iron parts may be such as to work these parts in the saturation region of the B-H curve. It is evident from the Figure 1.4 that if iron parts are worked in the saturation region then the mmf/m length and consequently, the mmf required for iron parts is excessively large. Thus a large value of specific magnetic loading results in increased values of magnetizing mmf and hence more magnetizing current.

Saturation

Knee

Linear

H

B

Saturation

Knee

Linear

H

B

Figure1.4 B-H curve

The value of magnetizing current will not be significant while designing d.c machines as d.c machine has ample space on salient poles to accommodate the required number of field turns. On the other hand, for induction motors, it is very much essential to consider the magnetizing current as increase of magnetizing current leads for poor power factor. Therefore, it is recommended that specific magnetic loading for induction motors is lower than that of d.c machines. Whereas for synchronous machines, the magnetizing current will not be more critical and the suggested value of specific magnetic loading is intermediate between dc and induction machines.

Core loss

The area of cross section of iron parts of the magnetic circuit of an electrical machine is expressed as,

A= flux carried by the partflux density in the part?1B The core loss will be generated when subjecting this part to alternating magnetization. The specific core loss i.e. loss per unit volume or weight is approximately proportional to the square of the flux density or iron per unit volume ? B2.

Now total core loss = loss per unit volume x volume

= loss per unit volume x area x length ? B2 x (l/B) ? B (as the

length of flux path is constant though not strictly)

It is proven that the core loss in any part of the magnetic circuit is directly proportional to flux density for which it is going to be designed. (It should be noted that it is not true for existing machine for which the iron loss is proportional to B2). As the flux density in any part of magnetic circuit is proportional to the specific magnetic loading, it is concluded that the core loss in a machine varies with respect to the variation in specific magnetic loading. Thus a large value of specific magnetic loading indicates an increased core loss and consequently a decreased efficiency and an increased temperature rise.

With a desired specific magnetic loading, the core loss increases as the frequency of reversals is increased. This is because the hysteresis loss is directly proportional to the frequency and eddy current loss is proportional to the square of the frequency.

1.4.4 Choice of specific electric loading

The factors which influence the choice of specific electric loading are discussed in the following subsections.

Permissible Temperature rise

Figure 1.5 shows an armature of a rotating machine.

Let us consider,

Z= total number of armature conductors,

S = number of armature slots,

az = area of each conductor,

? = current density

? = resistivity of conductor material

YS

dSWS

L

Tooth

Slot

Figure 1.5 Armature of a rotating machine

Therefore, for a slot pitch, ampere conductors/metre for the portion is expressed as,

ac=ZIz?D =IzZS?DS=IzZsys where,

Zs=Z/S = number of conductors per slot

Resistance of slot portion of each conductor = ?LazI2R loss in slot portion of each conductor = Iz2 ?LZI2R loss in each slot = Zs Iz2 ?LZHeat produced in a slot is dissipated over the surface over one slot pitch. Considering only the cylindrical surface,

Heat dissipating surface S = ysL

Loss dissipated per unit area of armature surface

q = loss/ surface = ZsIz2 ?LazysL = ZsIz ys .?. Iz azwe know,ac = ZsIz ys and ?= Iz azq= ac .? .?—- (1.4)

From the above derived equation (1.4), it is understood that the heat dissipated per unit area of armature surface is proportional to specific electric loading.

It is important to be noted that the heat dissipating surface of the overhang has been not considered for expression as it is not used quantitatively.

Temperature rise is expressed as, ?=QcS

Where,

Q = the loss to be dissipated

S =the dissipation surface

c =the cooling co-efficient

Also,

Loss dissipated per unit area q = Q/S

Therefore, temperature rise ?=qc=ac ? ? cFrom the above expression, the specific electric loading (ac) can be expressed as,

ac = ?? c?—- (1.5)

From equation 1.5, it can be concluded that the limiting value of specific electric loading ac is fixed by maximum allowable temperature rise and the cooling co-efficient. The effects of different factors are discussed in details below.

Temperature rise :

It is evident from equation (1.5) that a high value of specific electric loading is used for a high temperature. With the type of insulating materials used for insulation the maximum allowable temperature rise of a machine can be determined. The temperature rise of organic and inorganic materials is different. For example, organic materials like cotton, paper and many varnishes may be worked up to a maximum temperature (not temperature rise) of 1500C while inorganic materials like mica, asbestos and glass fibre bounded with silicone withstand temperature rise of 1800C without deterioration.

Cooling co-efficient:

From the equation (1.5), it is understood that the cooling co-efficient of the machine is inversely proportional to the specific electric loading. Therefore, for a small value of cooling co-efficient, a high value of specific electric loading is used in the machine. The value of cooling co-efficient can be decided by the provided ventilation conditions in the machine. A machine with better ventilation has a lower of cooling co-efficient and therefore, a high value of specific electric loading may be used in it.

Current density:

From the equation (1.4), it is clear that specific electric loading and current density are inversely proportional each other. Therefore, a higher value of specific electric loading requires conductors with lower current density. A perfect conductor usually have current density within a range of 2 – 5 A/mm2, the temperature rise is usually 400C for normal applications and cooling co-efficient is between 0.02 to 0.035 oC W-m2. The ambient temperature, the conductor resistivity (?) at operating temperature and, a compatible set values for current density (?) and specific electric loading (ac) can be derived from the equation (1.4) once knowing the insulating material and the withstand temperature.

Voltage:

The influence of voltage to the specific electrical loading is derived below,

Area of each slot = height of slot * width of the slot = dsWsTotal area of all slots = SdsWs= ?Dys. dsWs=?D ds(wsys)

Total area of conductors in slots = ?D ds(wsys) Sf —- (1.6)

Sf=space factor for slotsAlso,

Total area of conduction = Zas=ZIz?=?D ac?—- (1.7)

Where, az=Iz? and IzZ= ?D acEquating (1.6) and (1.7) we get,

?D ac?=?D ds(wsys) Sfac= ds= wsys? Sf —- (1.8)

The equation (1.8) describes that for a fixed ratio of slot width to slot pitch and fixed values of depth of slot and current density, the specific electric loading is directly dependent upon space factor Sf and the ratio of bare conductor area to total slot area. The high voltage machines require greater insulation thickness and lower space factor for this machine. Therefore, increase in voltage reduces specific electric loading of a machine. However, the shape of the conductors decides the value of space factor.

Size of machine:

For the design purposes, it is assumed that depth of slot, ratio of width of slot to slot pitch, current density and slot space factor are the same for all machines, and the specific electric loading is constant. Practically, the above said assumptions are insignificant and it is advised to use the slightly varying values within a range of sizes. In fact, if the current density and the slot space factor are assumed constant then specific electric loading is proportional to the diameter, referring equation (1.8), as slot depth usually depends upon the diameter.

1.5 Thermal considerations

1.5.1 Heat flow

The energy transfer by transformers and electromechanical energy conversion by the rotating electrical machines absorbs currents in the conductors and fluxes in the ferromagnetic parts. The current flow in conductors produces I2 R losses in the windings and core losses in the ferromagnetic cores. Further, the additional losses occur in the tank walls, end plates and covers on account of leakage flux. The above mentioned losses appear as heat which increases temperature above the ambient medium in every affected part of the machine. The dissipation of heat by the parts of an electrical machine has been carried out by conduction, and convection assisted by radiation.

Amongst the two methods of heat dissipation, the convection through air, liquid or steam is the most significant method. Forced convection through the coolant system is used for heat dissipation. However, the design for forced convective cooling is a straightforward and the designer should ensure for a large enough amount of coolant flows through the machine. This means that the cooling channels have to be large enough. If a machine with open-circuit cooling is of IP class higher than IP 20, using heat exchangers to cool the coolant may close the coolant flow. On the other hand, the proportion of heat transfer by radiation is usually not appreciable, yet not completely insignificant. Heat transfer by radiation can be done by a providing black surface to the machine.

Thermal Resistance: The thermal resistance is defined as the thermal resistance which causes a drop of 1oC per watt of heat flow. The thermal resistance, like electrical resistance, can be written as:

Re= ?tS—- (1.8)

= t?S—- (1.9)

? = thermal resistivity of material, ? (thermal) m or oC-m/W

? = 1/? = thermal conductivity, W/oC – m

t = length of medium, m

S = area of surface separated by the medium, m2

The heat flow equation between two conducting surfaces is expressed as,

Qcon= ?1-?2Rc W .

The above equation can be written as,

Qcon= S(?1-?2)?t W—- (1.10)

Heat dissipated per unit surface area by conduction is qcon= ?1-?2?t W/m2 —- (1.11)

The temperature difference across the conducting medium, ?= ?1-?2

= QconRc —- (1.12)

= Qcon?tS —- (1.13)

From the equation (1.13), it is understood that a material with a high value of thermal resistivity will dissipate less amount of heat or alternatively for the dissipation of same heat the temperature rise will be higher.

1.5.2 Newton’s law of Cooling

It is described previously that the machine parts temperature rises by the increase in losses. After sometime, the parts of a machine attain a steady temperature rise. The produced heat is dissipated by radiation and convection. At this temperature, the heat produced in the machine parts equal the heat dissipation.

The equation relating the heat dissipation by radiation is given by,

Qrad=?rad ?S wattThe above expression can be modified if temperature rise varies between moderate limits.

The heat dissipated by convection is given by,

Qrad=?con?S wattTherefore, the sum of heat dissipation by radiation and convection results the total heat dissipation. It is expressed as,

Q=Qrad+Qrad= ?rad?S watt+?con?S =(?rad+?con ) ?S= ??S watt —- (1.14)

where, ?=?rad+?con —- (1.15)

= specific heat dissipation or emissivity due to radiation plus convection

Equation (1.14) represents the Newton’s law of cooling. Newton’s law of cooling varies where the body is acted upon by a uniform current of air. Therefore, this law applicable to the natural process of cooling for restricted range of temperature.

1.5.3 INTERNAL TEMPERATURES (HOT SPOT TEMPERATURES):

The loss in electrical machines occurs inside the iron cores and windings. The produced loss is dissipated to the surface by the cooling medium. The internal flow of heat, from the parts in which it is actually generated to the cooling surface from which it is transferred to the coolant, is important in determining the hot spot temperatures and the temperatures to which the insulating materials would we subjected.

If the cross-section of a coil, which produces electrical losses, is very large or if the insulation around the coil or the core is very thick, then there is always danger of exceptionally high internal temperatures developing, even when the temperature of external surface is below the maximum specified limit. In order that there should not be any injury to the insulating materials we must determine the temperature of the hottest spot, a place where the local temperature is the highest. Hence, the problem is to find out the difference in temperature between the outside surface from which heat is carried away and the hottest spot inside the winding, from which the heat must travel through the conductors and insulation before it can be dissipated away.

1.5.3.1 Calculation of internal temperature

The Figure 1.6 shown below has a very large plate of thickness ‘t’, consisting of a homogeneous material. It is assumed that the length and width of the plate are very large as compared with its thickness. The direction of heat flow across its thickness is shown through the Figure 1.6. The two surfaces are assumed to be at the same temperature.

Heat flow

dxxQ

Q0

xQS

tQmt

Fig. 1.6 Temperature gradient

Let us consider,

l=length of the plate, m

w=width of the plate, m

?=thermal resistivity of material along the direction of heat flow, m

q=heat produced per unit volume, W/m3

?=temperature rise, oCd? = the temperature difference across the walls of this strip

dx =elementary strip thickness

x =distance from the centre

We get,

Qx = per unit volume x volume = q x lwxThe temperature difference between the walls of this strip is,

d?= heat conducted x thermal resistance of the strip

= q lwx x ?dxlw=q?x dx ? The difference in temperature between the centre and any point at a distance x from the centre along the path of heat flow,

?= 0x?x dx=q?x22 —- (1.16)

Thus the temperature difference curve is a parabola shown in Figure1.6.

From equation (1.16), the difference in temperature between the centre of the plate and the outer surface (i.e. x=t/2) is,

??=q?t28?—- (1.17)

Hence the temperature of the hottest spot (centre of plate in this case)

?m=??+?s=q?t28+?s —- (1.18)

where,

?S = temperature the surface.

Temperature gradients in cores

The built up area of laminated core in electrical machine is shown in Figure1.7. It consists of steel laminations insulated from each other by varnish. The heat is dissipated duo to the iron loss at the core when it is subjected to alternating magnetization.

The point O shown in the Figure 1.7 is the hottest part of the laminated core and the generated heat is to be conducted to surfaces A and B. The path of heat flow along X axis is across the laminations while the heat flow along Y axis is along the laminations.

Y

B

B

A

xA

X

yO

Figure 1.7 Built up laminated core

Consider that all the heat flows alongside the direction OX.

Therefore, from equation (1.16), temperature difference between O and A is given by

?OA=q?xx22?Where, ?x is the thermal resistivity across the laminations.

Considering the total heat to flow alongside the laminations (along OY), temperature difference between O and B is given by

?OB=q?yy22?Where, ?y is the thermal resistivity along the laminations.

When comparing the two laminations, the value of thermal resistivity along the laminations is low as compared with that across the laminations. Therefore, at first sight it would indicate that all the heat should be taken along the laminations in order to keep down the internal temperature or indicates that axial ventilation wherein air is blown across the laminations would be most effective.

1.5.3.3 Heat flow in two dimensions

In practical, the heat does not flow along parallel paths and the dissipating surface as discussed in the previous sections. Actually, it flows in different directions and the windings and the cores have insulation in addition to copper and iron respectively. The thermal resistivity of built up windings and cores depends upon relative thickness of insulation to copper or iron.

Let us consider a coil or a core which has a large axial length comparing its width or thickness. Figure 1.8 shows the cross section through the coil. It is obvious that the entire outer surface must be maintained at constant temperature.

W

bB

bC

taO

xwtdxwtA

D

a

Figure 1.8 Heat flow in two dimensions

Let, l=length of the coil, m

w= width of the coil, m

t=thickness of coil, m

?y =thermal resistivity along aa, ?m?x =thermal resistivity along aa, ?mQ=heat produced per unit volume, W/m3

It is assumed that the heat flows outward through walls of successive imaginary spaces of rectangular. The boundary of the rectangular space is specified as ABCD, the thickness of the wall is dx in the direction bb and dx w/t in the direction aa.

Thermal resistance of horizontal elementary strips,

=?ydx w/tlAB+CD=?ywdx4lxtThermal resistance of vertical elementary strips,

=?x dxlAD+BC=?x t dx4 lwxTherefore, the total thermal resistance of walls is expressed as, (as the above mentioned two paths are in parallel),

dR?=?ywdx4 lxt . ?xtdx4 lxw?ywdx4 lxt+?xtdx4 lxw=?x?ytw dx4lx?yw2+?xt2qx be the heat produced in ABCD,

qx=ql2x.2xwt=4qlx2wtTemperature difference between inner and outer walls of ABCD

d?=heat loss in space ABCD x thermal resistance of walls=4qlx2wt. ?x?ytw dx4lx?yw2+?xt2=qw2?x?yx dx?yw2+?xt2Difference of temperature between O and outer surface

?=qw2?x?y?yw2+?xt20t2x dx=qw2t2?x?y8?yw2+?xt2 —- (1.19)

Let Q be the total heat produced in the coil,

Q = ql wt

Substituting this value of Q in equation (1.19) we get,

?=Q8lwt?x?y?yw2+?xt2=Q8l wt?x+tw?y —- (1.20)

1.5.3.4 Thermal gradients in conductors placed in slots

The conductors placed in slots carry current which produces heat in them on account of copper loss. The temperature in the centre of the core is assessed by considering the following two cases.

Case i: Considering the slot insulation to be very thick as compared with that on the end connections, the heat produced in the embedded portions of the conductor is conducted along its length to the end windings. Figure 1.9 shows a conductor placed in a slot. It is preferred to find the temperature difference between its centre O and the overhang.

Embedded

Conductor

Overhang

Core

L

xdx

Fig. 1.9 Temperature gradient across the length of embedded conductor

Let Iz=current carried by the conductor, A

L=length of the embedded conductor, m

az=area of conductor in m2

?=current density in the conductor, A/m2

?c=thermal resistivity of the conductor, ?m

? =electrical resistivity of the conductor, ?mConsider a strip of width dx at a distance x from O

Heat conducted through the strip,Qx=I2R loss between O and strip=Iz2?xazThermal resistance of the strip dR?=?cdxazTemperature across the strip,

d ?=Iz2?xaz. ?cdxaz= ?Iz2az2?cxdxTemperature difference between O and overhang

?=0L2? x Iz2az2 ?c x dx=?Iz2az2 ?cL28?—- (1.21)

=? ?2?cL28? —- (1.22)

Case ii: Consider that the overhang is considerably hot and therefore the heat produced in the embedded portion of the conductor is to be conducted through the slot insulation to the iron core. we have now to compute the difference between copper and surrounding iron.

Let,

Ws=width of the slot, m

ds=depth of the slot iron, m

t=thickness of insulation, m

?i=thermal resistivity of insulation, ?mConductor

Insulation

WS

dS

Fig. 1.10 Temperature gradient across the insulation of conductor embedded in a slot.

Heat produced in the conductor Q=Iz2 ?LazArea to the path of heat flow= L (2 ds + Ws ) Resistance of thermal insulation R?=?itL2ds+WsTemperature gradient across the insulation ?=Iz2?Laz . ?itL2ds+WsFrom equation (1.12), temperature across the strip is expressed as,

?=Iz2?Laz x ?itL2ds+Ws ?=Iz2?az x ?it2ds+Ws?—- (1.23)

putting Iz?az , we have

?=?2az??itL2ds+Ws?—- (1.24)

1.5.3.5 Heating of turbo-alternator rotors:

The temperature difference between the conductors in a slot and its iron walls is

?=Iz2?az . ?itL2ds+Ws? (see from eqn.1.23)

We can neglect the term Ws for the case of strip copper laid flat in the rotor slots of turbo-alternators because the heat will not travel down the layers of insulation.

? For turbo alternator rotors

?=Iz2?az . ?it2ds—- (1.25)

1.5.4 THERMAL STATE IN ELECTRICAL MACHINES:

1.5.4.1 THEORY OF SOLID BODY HEATING

The temperature of a machine rises when it is run under steady load condition starting loading conditions. The temperature at first increases at a rate determined by power wasted. As the temperature rises, the active part of the machine dissipates heat partly by conduction, partly by radiation and in most cases, largely by means of air cooling. Therefore as the temperature rises, its rate of increase falls off owing to better heat dissipating conditions .the temperature -time curve is exponential in nature.

The temperature of any part of a machine, not only depends on the heat produced in itself but also on heat produced in other parts. This is because there is always a heat flow from one part to another. For example, the heat produced in the part of the winding embedded in the slot flows partially through the insulation to the laminations partially to the end windings.

Electrical machines are not homogeneous bodies .their parts are made up of different materials like copper, iron and insulation. these materials having different thermal resistivity’s and due to this, it is rather difficult to calculate the temperature part of the machine .the results obtained from such a theory are applicable to a certain degree, to the different parts of the machine as a whole.

Q=power loss or heat developed, J/s or W

G=weight of the active parts of machine, kg

H=specific heat J/kg-?S=cooling surface, m2

?=specific heat dissipation W/m2 – oCC=1/? = cooling co-efficient oC – m2/W

? = temperature rise at any time t, oC?m = final steady temperature raise while heating , ? ?n = final steady temperature raise while cooling , ??i = initial temperature rise over ambient medium,?Th = heating time constant ,sTc = cooling time constant ,st=time, s.

1.5.4.2 HEATING:

Considering the conditions at any time t from start, heat energy developed in the body during an infinitely small time dt,

=heat energy developed per second x dt = Qdt —- (1.26)

If during this period dt the temperature of the body raises by d?, the heat energy stored in the body=weight of the body x specific heat x difference in temperature

=G h d? —- (1.27)

If in the process of heating, the temperature of the surface rises by ? over the ambient medium, at the instant considered, the heat energy dissipated by the body into the ambient medium due to radiation, conduction and convection

= specific heat dissipation x surface x temperature raise x time

=? x S x ? x dt=S ? ? dt —- (1.28)

As the heat developed in the machine is equal to the stored in the parts plus the heat dissipated, we have from equations 1.26, 1.27 & 1.28

Q dt=G h d?+S ? ? dt—- (1.29)

dt(Q-S ? ? ) =G h d?dt=d?QGh-S?Gh? —- (1.30)

Solving the differential equation 1.30

t=-GhS?log e(QGh-S?Gh?)+K —- (1.31)

where K is the constant of integration,

The value of K is found by applying the boundary condition, when t = 0, we have ? = ?i .

Putting this in eqn.1.31

0 = – GhS?logeQGh-S?Gh?i+K or K=GhS?loge(QGh-S?Gh?i)Substituting this value of K in eqn. 1.31

t= – GhS?logeQGh-S?Gh?+GhS?loge(QGh-S?Gh?i)=- GhS?log eQS? -?QS? – ?i—- (1.32)

The machine reaches a final steady temperature rise ?m when t = ?. Under this condition there is no further temperature rise and the rates of heat production and dissipation are equal.

This means d?=0 or Ghd?=0when the machine attains final steady temperature rise.

From eqn. 1.29, for ?=?m, Q dt=S ? ?m dt?m=QS? —- (1.33)

?m=QS?=QCS —- (1.34)

Putting?m=QS? in eqn. 1.32, we have

t= GhS?loge?m -??m -?i —- (1.35)

The term GhS? has the dimensions of time and is called the heating time constant Th.

? Th=GhS? —- (1.36)

Putting this value, Th=GhS? in eqn. 1.35

t= Thloge?m -??m -?i —- (1.37)

or ?m -??m -?i=e-tTh

or0=?m1-e-tTh+?ie-tTh—- (1.38)

If the machine starts from cold conditions,

?i=0 (No temperature rise over the ambient medium)

?=?m1-e-tTh —- (1.39)

Equation 1.39 is the temperature rise with time. The temperature rise curve is exponential in nature as shown in fig.1.11.

Heating time constant:

Consider equation 1.39, ?=?m1-e-tThPutting t=Th in above expression, we have,

?=?m1-e-1=0.632?mThus we can define the heating time as the time taken by the machine to attain 0.632?m of its final steady temperature rise. The heating time constant of a machine is the index of time taken by the machine to attain its final steady temperature rise.

ThTime, t

?m0.632 ?mTemperature risTemperature rise, ?

Fig.1.11 Heating curve

Considering the relationshipTh=GhS? we conclude that the time constant is inversely proportional to ? (specific heat dissipation). ? has a large value for well ventilated machines and thus the heating constant is small. The value of heating time constant is large for poor ventilated machines.

Since, the volume of machine and hence its weight increases in proportion to the third power, and the surface area in proportion to the second power of its linear dimensions, the heating time constant of a machine increases at first power of linear dimensions, because of this, large sized machines have large heating time constants.

The heating time constant of a well ventilated induction motor of about 20 kW rating may be out of order of minutes while that of large or totally enclosed machines may reach several hours or even days. The heating time constant of conventional electrical machines is usually within the range of ½ to 3-4 hours.

Final Steady temperature rise: Considering an expression ?m=QS?, it is clear, other things being equal, the final steady temperature rise is directly proportional to the losses. It is also evident that the final steady temperature rise is inversely proportional to surface area and specific heat dissipation. Thus for the same loss, the machine would attain a higher temperature rise if its dissipating surface is small or, if its ventilation is poor.

Cooling

The equation for cooling can be derived by considering eqn.1.29

Q dt = Gh d? + S?? dtThe solution for the above equation is:

t= GhS?logeQGh-S?Gh?i+KThe value of K is obtained by putting boundary conditions,

When t = 0, ? = ?i. From this we have,

0= GhS?logeQGh-S?Gh?i+KOr K= GhS?logeQGh-S?Gh?iSubstituting this value of K and proceeding as in the case of heating, we get,

?=?n1-e-tTc+?ie-tTc—- (1.40)

Where ?n = QS? —- (1.41)

= QcS —- (1.42)

and Tc=GhS?—- (1.43)

The value of ? under cooling conditions is usually different from that under heating conditions and so the heating and cooling time constants of a machine may have different values.

If the machine is shut down, no heat is produced and its final steady temperature rises when cooling is zero or ?n = 0. Under these conditions eqn. 1.40 thus reduces to

?= ?ie-tTc—- (1.44)

Temperature rise, ?0.368 ?iTemperature ris?= ?i(e-tTC)ThTime, t

fig.1.12 Cooling curve

It is clear from eqn.1.44, that the cooling curve is also exponential in nature as shown in fig.1.12. Eqn.1.44 is applicable to machines which are shut down while eqn.1.40 is applied to machines allowed to cool owing to partial removal of load.

Cooling Time Constant: Consider the relation,

?= ?i e-tTc

Putting t = Tc, we have

?= ?i e-1=0.368 ?iThus we can define the cooling time constant as the time taken by the machine for its temperature rise to fall to 0.368 of its initial value.

As stated earlier the cooling time constant may be different from the heating time constant for the same machine, as the ventilation conditions in the two cases may not be the same. The cooling time constant is usually larger owing to poorer ventilation conditions when the machine cools. In self cooled motors the cooling time constant is about 2-3 times greater than the heating time constant because cooling conditions are worse at stand still.

Final Steady temperature rise: Considering an expression ?n=QS?, the final steady temperature rise when the machine is cooling is directly proportional to the losses. When the machine is shut down, Q = 0 and so the final temperature rise is zero, i.e. the machine finally comes to ambient temperature conditions.

1.6 Rating of machines

The rating of machine refers to the whole of the numerical values of electrical and mechanical quantities with their duration and sequences as designed to the machines by the manufactures and stated on the rating plate, the machine complying with the specified conditions. The duration of the sequence may be indicated by the qualifying term.

In most of the situations, motors are selected on the basis of their heating since temperature rise of the motor is the prime factor in determining its life. The capacity of the motor is then checked for overload capacity. Motor power capacity is selected in accordance with the work of the motor is desired to perform simultaneously ensuring that the motor will operate with permissible limits of mechanical loading.

The rating of a motor is the power output or the designated operating power limit based upon certain definite conditions assigned to it by the manufacturer.

1.6.1 Motor Ratings Based on Duty

As per IS: 4722-1968 Specification the types of duties are classified for rotating electric machines as:

S1 – continuous duty

S2 – short time duty

S3 – Intermittent periodic duty

S4 – Intermittent periodic duty with starting

S5 – Intermittent periodic duty with starting and braking

S6 – continuous duty with intermittent periodic loading

S7- continuous duty with starting and braking

S8- continuous duty with periodic speed changes.

N

Time, t

Temperature

?mElectric Losses

Load

P

Continuous duty (S1):-

On this duty the machine operate with constant load for the duration long enough such that all the parts of the motor to attain its final steady temperature.

Therefore the continuous rating of a motor is defined as the load carried by the machine for a long time without the temperature rise of any part exceeding the maximum permissible value.

Eg: fans, pumps, compressors, conveyors, centrifugal pumps, paper mill drives and other equipment which operate for several hours and even days at a time.

The simplified load diagram for this duty is a horizontal straight line shown in fig.1.13.

N – operation under rated conditions, S; Fig.1.13 Continuous duty (S1)

?m – maximum temperature, oC

Short Time Duty (Duty type S2):-

For this duty cycle 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 final steady temperature rise. But the period for rest is so long that the motor temperature drops to the ambient value.

The short time rating is defined as the output of a motor at which it may be operated for a certain period without exceeding the maximum permissible value of temperature rise.

Eg: Machine tool drive for position control, railway turntable, navigation lock gates are some examples of the drives which operate on short time duty.

The simplified diagram for short time duty is shown in fig.1.14.

N

Temperature

?mElectric Losses

Load

P

Time, t

N

Time, t

Temperature

?mR

Period of One Cycle

Load

P

Electric Losses

Fig.1.14 Short Time Duty (S2) Fig. 1.15 Intermittent Periodic duty ( S3)

N – operation under rated conditions, S;N – Operation under rated conditions, S;

?m – maximum temperature attained R – Operation at rest and de-energized, S

during duty cycle, oC?m – maximum temperature attained

during duty cycle, oCThe period of operation is so short that the temperature rise of the motor does not reach its final steady value and the period of rest is so long that the motor returns to cold conditions.

Standard short time ratings are: 10, 30, 60 and 90 minutes.

Intermittent Periodic duty (S3):-

For this duty cycle the periods of constant load (N) and rest (R) with machine de-energized alternatively. The load periods are too short to allow the motor to reach its final steady state value while period of rest are also too small to allow the motor to cool down to the ambient temperature.

Eg: Cranes, lifts and certain metal cutting machine tool drive.

The simplified load for this type of duty is shown in fig.1.15.

Duty factor is used to evaluate the intensity of heating due to intermittent period loads. The duty factor is generally defined as the ratio of the heating (working) period to the period of whole cycle.

Duty factor ? = thth+tc—- (1.45)

Where, th= heating period, tc = period of rest.

The duty factor is determined on the basis of a cycle 10 minutes long.

The Intermittent Rating of a motor applied to an operating condition during which short time load periods alternate with periods of rest or no load without the motor reaching the thermal equilibrium and without the maximum temperature rising above the temperature rise.

The duty factor ? = NN+R —- (1.46)

D

Time, t

Temperature

?mElectric Losses

Load

P

R

Period of One Cycle

N

Intermittent Periodic duty with starting (S4):-

This type of duty consists of a sequence of identical duty cycles each consisting of a period of starting (D), a period of operation at constant load (N) and a rest period (R).

The operating and rest periods are too short to reach its final steady state value.

In this duty the stopping of the motor is obtained either by natural deceleration after disconnection of the electric supply or by means of braking such as mechanical brake which does not cause additional heating of windings.

The simplified load diagram for intermittent periodic

duty with starting (S4) is shown in fig.1.16.

The duty factor ? = D+ND+N+R —- (1.47) Fig.1.16 Intermittent Periodic duty

with starting, (S4)

D – Starting period, S, N – Operation under rated conditions, S,

R – at rest and de-energized, S, ?m – maximum temperature attained during duty cycle, oCIntermittent periodic duty with starting and braking (S5):-

This type of duty consists of a sequence of identical duty cycles each consisting of a period of starting (D), a period of operation at constant load (N), a period of braking (F) and a rest period (R). The operating and rest periods are too short to obtain thermal equilibrium during one duty cycle. In this duty braking is rapid and is carried out by electrical means.

The duty factor ? = D+N+FD+N+R+F—- (1.47)

The simplified load diagram for this duty type S5 is shown in fig.1.17.

D

Time, t

Temperature

?mElectric Losses

Load

P

R

Period of One Cycle

N

F

Time, t

Temperature

?mElectric Losses

Load

P

Period of One Cycle

N

Y

Fig. 1.17 Intermittent Periodic duty with Fig.1.18 Continuous duty with

starting and braking duty (S5) intermittent periodic duty (S6)

D – Starting period, S; N – Operation under rated conditions, S;

N – Operation under rated conditions, S;V – Operation on no load, S;

F – Breaking period, S?m – maximum temperature attained

R –Rest and de-energized, S during duty cycle, oC?m – maximum temperature attained

during duty cycle, oC

Continuous duty with intermittent periodic duty (S6):-

This type of duty consists of a sequence of identical duty cycles each consisting of a period of operation (N) at constant load and period of operation at no load. The operation and no load periods are too short to attain thermal equilibrium during one cycle.

The duty factor ? = NN+V —- (1.48)

The simplified load diagram for this duty type S6 is shown in fig.1.18.

Unless and otherwise specified the duty cycle is 10 minutes. The recommended duty factors are 15, 25, 40 and 60 percent.

Continuous duty with starting and braking (S7):-

This type of duty consists of a sequence of identical duty cycles each having a period of starting, a period of operation at constant load and a period of electric braking. There is no rest or de-energized period.

The simplified load diagram for this duty type S7 is shown in fig.1.19.

Speed

D

Time, t

Temp.

?mElectric Losses

Load

P

Period of One Cycle

N1

N2

F1

F2

N3

The duty factor for this duty cycle is 1.

D

Time, t

Temperature

?mElectric Losses

Load

P

Period of One Cycle

N

F

Fig.1.19 Continuous duty with starting and Fig.1.20 Continuous duty with periodic

electric braking (S7) speed changes Duty type (S8)

D – starting period, S F1, F2 – braking periods, S

N – operation under normal conditions, S D – acceleration period, S

F – period of electric braking, S N1, N2, N3 – operation under rated

conditions, S

?m – maximum temperature attained during duty cycle, oC

Continuous duty with periodic speed changes (S8):-

This type of duty consists of a sequence of identical duty cycles each consisting of a period of operation at constant load corresponding to determined speed of rotation, followed immediately by a period of operation at another load corresponding to a different speed of operation. The operating time is too short to attain thermal equilibrium during one duty cycle there is being no rest and de-energized period.

The simplified load diagram for this duty type S8 is shown in fig.1.20.

1.7 Standard specifications:

The standard specifications shall be marked with the appropriate items as follows:

The type of machine, whether motor or generator.

The class of rating : Continuous or Short time

The rated Output : kW (for generators), kW or HP (for motors)

The rated voltage : V volt

The rated Speed : N rpm

Type of current (d.c or a.c), for a.c machines the rated frequency and number of phases

The class of insulation or permissible temperature rise (?0 C)

Cooling : Natural or forced cooling

Type: Generator or motor, separately excited or self-excited-shunt, series, or compound, if compound type of connection – long or short shunt, type of compounding – cumulative or differential, degree of compounding – over, under or level. With or without inter poles, with or without compensating windings, with or without equalizer rings in case of lap winding.

Voltage regulation ( in case of generators) : Range and method

Speed control ( in case of motors ) : range and method of control

Efficiency: must be as for as possible high (As the efficiency increases, cost of the machine also increases).

Type of enclosure: based on the field of application – totally enclosed, screen protected, drip proof, flame proof, etc.,

The manufacturer’s name

The manufacturer’s serial number or identification mark

1.8 Indian Standard specifications for electrical machines

IS 325-1966: Specifications for three phase induction motor

IS 4029-1967: Guide for testing 3ph induction motor

IS 1231-1967: Dimensions of three phase foot mounted induction motors

IS 996 – 1979: Specifications foe single phase ac and universal motor

IS12615-1986: Specifications for energy efficient induction motor

IS13555-1993: Guide for selection & application of 3ph induction motor for different types of driven equipment

IS 8789-1996: Values of performance characteristics for three phase induction motor

IS 4722-1992: Specifications for rotating electrical machines

IS 12802-1989:Temperature rise measurement of rotating electrical machines

IS 4889-1968: Method of determination of efficiency of rotating electrical machines

IS 7231-1973: Guide for testing synchronous machines

IS 1180-1989: Specifications for outdoor three phase distribution transformer upto 100 kVA. (sealed non – sealed type)

IS 2026-1994: Specifications of power transformers

IS 13956-1994: Testing of transformers

IS 5142-1969: Continuously variable voltage auto transformers

IS 2705- 964: Specifications for Current Transformers

1.8 Solved Problems

A 350 kW, 500 V, 450 rpm, 6-pole, dc generator is built with an armature diameter of 0.87 m and core length of 0.32 m. The lap Wound armature has 660 conductors. Calculate the specific electric and magnetic loadings. ( May/June 2012 – 8 mark)

Given Data:

P = 350 kW D = 0.87m V = 500V L = 0.32m

n = 450/60 rps Z = 660 p = 6 Lap Wound

Solution:

Specific electric loading, ac= IZ Z?DSpecific Magnetic Loading, Bav=p??DL

The power output of the generator, P= VI x 10-3 in kW

Therefore Full load current I= PV× 10-3=350500×10-3 = 700 amps

Neglecting field current, Ia?I

Current through each armature conductor IZ= Armature Current No. of Parallelpaths=Iaa=7006

= 116.67 amps ( ?a=p in lap wound )Specific electric loading, ac= IZ Z?D =116.67 × 660?×0.87= 28173 amp.cond./m Induced emf in dc generator, E= ? Z n pa= = ? Z n , for lap winding (?p=a )Hence flux pole, ?= EZn? VZn= 500660×450/60= 0.101 wb

Specific magnetic loading, Bav= P??DL= 6×0.101?×0.87×0.32= 0.6929 wb/m2

Result:

Specific electric loading, ac = 28,173 amp.cond./m

Specific magnetic loading, Bav = 0.6929 wb/m2A 600 rpm, 50 Hz, 10,000 V, 3 phase, synchronous generator has the following design data. Bav = 0.48 wb/m2, ? = 2.7A/mm2, slot space factor = 0.35, number of slots = 144, slot size =120×20mm, D = 1.92m and L = 0.4m. Determine the kVA rating of the machine. ( May 2012 – 8 mark)

Given data

B av = 0.48wb/m2 D = 1.92mslot space factor = 0.35

? = 2.7A/mm2L = 0.4m3 phase

slots = 144 slot size = 120×20mm600 rpm, 10000V

Solution

Specific electric loading , ac= IzZ?D —– (i)

Let az = area of cross –section of each conductor in mm2

The current density ,?=Izaz ; ?Iz= ?az —– (ii)

Conductors in a slot = slot area × slot space factor = 120 × 20 × 0.35 = 840mm2

Also, conductorsarea in a slot = Number of conductersper slot x area of cross section of each conductor = Total number of armature conductorsnumber of slots x area of cross section of each conductor Total number of armature conductorsZ= Conductors area in a slot × Number of slotsArea of cross-section of each conductor

= 840×144az—– (iii)

Specific electric loading, ac= IzZ?D

From equations (i), (ii) and (iii) we get,

ac= ?.az840×144az?D=?×840×144?D= 2.7×840×144?×1.92= 54144 amp.cond/m kVA rating of the machine Q=C0D2Lns and C0=11 Bav ac Kws x 10-3

?Q=11 Bav ac Kws x 10-3D2Lns = 11×0.48×54144×0.955×10-3×(1.92)2×0.42×60060 =4025 kVAResult

kVA rating of the machine = 4025 kVAA 20-HP, 440V, 4Pole, 50Hz, 3phase induction motor is built with a stator bore of 0.25m and core length of 0.16m.The specific electric loading is 23000 ampere conductors per metre. Find the specific magnetic loading of the machine .Assume full load efficiency of 84% and a power factor of 0.82. Using the data of the above machine determine the main dimensions for a 15HP, 460 V, 6 pole , 50 Hz motor .

Given data

MACHINE-IMACHINE-II

20 HP 4 Pole 15 HP

440 V 50Hz 460 V

D = 0.25 m L = 0.16 m 50 Hz

ac = 23000 amp.cond./m. 6 poleSolution

Machine – I

Input kVA, Q1=HP ×0.7460.84 ×0.82=21.66 kVASynchronous speed ,ns=2fp=2 ×504= 25 rps

Also , the input kVA , Q1= = C0 D2 L ns , Where Co = 11 Bav ac Kws ×10-3

? Q1 = 11Bav ac KWS × 10-3D2 L ns

Bav = Q111 ac KWS × 10-3 D2 Lns = 21.6611×23000×0.955×10-3 ×(0.25)2 ×0.16×25= 0.3586 wb/m2

MACHINE-II

Input kVA, Q2=HP×0.746? × pf=15×0.7460.84×0.82=16.246 kVA

Synchronous speed, ns=2fp = 2×506 = 16.667 rpsThe value of ac (specific electric loading) decreases when voltage rating is increased. Hence the ratio of specific electric loading can be expressed as shown below .

ac2ac1= V2V1 ? ac2 = ac1× V1V2 = ac1V2V1The ratio of voltage rating = V2V1= 460440 = 1.0455

ac2 = ac11.0455 = 230001.0455 = 21999.044 ? 22000 amp.cond./m

Let us assume same L? Ratio for both the machines.

For machine – I, L1?1 = L1?D1P1= 0.16?(0.25)4 = 0.8149

For Machine -II, L2?2= 0.8149

Hence,L2 = 0.8149 ?2= 0.8149 × ?D2P2= 0.8149×?6 D2 = 0.4267 D2

The KVA input of machine -2, Q2= C02 D22L2ns2 Where , C02=11 Bav ac2 Kws ×10-3? Q2 = 11 Bavac2Kws ×10-3ns2

D22 L2= Q211 Bavac2Kws ×10-3 ns2 =16.24611×0.3586×22000×0.955×10-3×16.667=0.0118 m3On substitution for L2 = 0.4267D2 in the equation for D22 L2

We get,

D22L2 = 0.0118m3;

D22(0.4267 D2) = 0.0118

D2 =(0.0118/0.4267)1/3 = 0.3024m

L2 = 0.4267×0.3024 = 0.129m

Result

Machine-I Specific magnetic loading, Bav = 0.3586 wb/m2

Machine-II Diameter of stator bore, D2 = 0.3024m

Length of stator core, L2 = 0.129m

Determine the air –gap length of a dc machine from the following particulars: gross-length of core = 0.12m, number of ducts = one and is 10mm wide, slot pitch = 25mm, slot width = 10mm, carter’s coefficient for slots and ducts = 0.32, gap density at pole centre = 0.7wb/m2 field mmf/pole = 3900AT, mmf required for iron parts of magnetic circuit = 800 AT. ( April /May 2015 – 8 mark )

Given data

L = 0.12 m ys=25 mm Bg=0.7 wb/m2nd=1 wt =10 mm mmf per pole = 3900 AT

wd=10 mm Kcs=Kcd=0.32 mmf for iron = 800 AT

Solution

Mmf for air-gap, ATg = mmf per pole – mmf for iron parts

= 3900 – 800 = 3100 AT

Gap contraction factor for slots, Kgs = ysys-Kcswt= 2525-0.32 ×10=1.1468

Gap contraction factor for ducts, Kgd= LL- Kcd nd Wd = 0.120.12-0.32 ×1 ×10 ×10-3= 1.0274

Gap contraction factor Kg= Kgs × Kgd= 1.1468 ×1.0274 = 1.1782

Mmf for air-gap , ATg= 800,000 Kg Bg lg? Length of air-gap, lg=A Tg800,000 Bg Kg = 3100800,000 ×0.7 ×1.1782 = 4.7 ×10-3 = 4.7 mm

Result

Length of air-gap, lg= 4.7 mm

5) A field coil has a cross-section of 100 x 50 mm2 and its length of mean turn is 1m. Estimate the hot spot temperature above that of the outer surface of the coil, if the total loss in the coil is 120 W. Assume space factor = 0.56. Thermal resistivity of insulating material = 8 ?m. (Nov/dec-2012 – 6mark)

Given data

?i = 8 ?mSf = 0.56

a = 100 x 50 mm2

Volume = a x l = 5 x 10-3 m3

Solution;

?e= ?i1-Sf12= 8 1-0.561/2=2 ?mVolume of coil = 100 ×10-3×50 ×10-3×1=5 ×10-3m3 Q = 1205 ×10-3 = 24 ×103 wb/m3Assuming equal in ward and outward heat flow, hot spot temperature is,

?0 = q?et28q=heat produced / unit volume =1205 ×10-3=24 ×103 W/m3

?0 = q?et28= 24 ×103×2 ×50 ×10-328 =150CResult

Hot spot temperature ?0 = 15 oC6) A field coil has a heat dissipating surface of 0.15 m2 and a length of mean turn of 1 m. it dissipates loss of 150 W, the emissivity being 34 W/m2 oC. Estimate the final steady state temperature rise of the coil and its time constant if the cross section of the coil is 100 × 50 mm2. Specific heat of copper is 390 J/kg oC. The space factor is 0.56. Copper weighs 8900 kg/m3. (April /May 2011 -16 mark)

Given data

Q = power loss or heat dissipated = 150W

S = cooling surface = 0.15 m2

? = specific heat dissipation = 34 W/m2 oC

G = weight of active part of machine

h = specific heat = 390 J/kg oC

Solution

Final steady temperature rise,

?m=Q?S= 1500.15 ×34 = 29.40C

Heating time constant, Th=Gh?SVolume of copper =1×100 ×50 ×0.56 ×10-6=2.8 ×10-3m3.

Weight of copper G = 2.8 ×10-3× 8900 = 24.92 kg.Heating time constant,

Th=Gh?S= 24.92 × 3900.15 × 34 = 1906

Result

Heating time constant Th = 1906

7) The exciting coil of an electromagnet has a cross-section of 120 × 50 mm2 and a length of mean turn 0f 0.8 m. it dissipates 150 W continuously. Its cooling surface is 0.125m2and specific heat is 30 W/m2-0C.calculate the final steady temperature rise of the coil surface. Also calculate the hot spot temperature rise of the coil if the thermal resistivity of insulating material used is 8 ?m. The space factor is 0.56. (April/May 2015 – 8 mark)

Given data

Q = power loss or heat dissipated = 150W

S = cooling surface = 0.125 m2

? = specific heat dissipation = 30 W/m2 oC

?i = thermal resistivity of insulating material = 8 ?mSf = space factor = 0.56

G = weight of active part of machine

h = specific heat = 390 J/kg oC

Solution

Final steady temperature rise of coil surface,

?m=Q ?S= 150 0.125 ×30=400C

Effective thermal resistivity

?e= ?i1-Sf12= 8 1-0.561/2=2?m Power loss or heat dissipated q =Total lossvolume=150 120 ×( 50 × 10-3)2 x 0.8=31250 W/ m3Temperature difference between coil surface and hot spot ?0 = q?et28?0 = 31250 ×2 ×( 50 × 10-3)28= 19.5oC

? Temperature rise of hot spot = ?m+?0 = 40 + 19.5 = 59.50C

Result

Temperature rise of hot spot = 59.50C

8) The temperature rise of a transformer is 250C after one hour and 37.50C after two hour of starting from cold conditions. Calculate its final steady temperature rise and the heating time constant. If its temperature falls from the final steady value to 400C in 2.5 hour when disconnected. Calculate its cooling time constant. The ambient temperature is 300C.

(April /May 2015 – 8 mark)

Solution

When heating:

Since the transformer starts from cold conditions therefore its temperature rise is given by?=?m1-etThNow we have ?= 250C at t = 1 hour and

?=37.5 0C at t = 2 hour

25 = ?m1-etTh ….. (1)

37.5= ?m1-etTh …… (2)

From (1) and (2)

1 – e -2Th 1-e -1Th = 37.525= 1.5

1 – e -2Th 1-e -1Th x 1+e -1Th1+e -1Th =1 – e -2Th 1-e -2Th x 1+e -1Th =1.5

1+e -1Th = 1.5Or e -1Th = 0.5

-1Th=ln0.5= -0.693 Therefore, heat time constant Th=1.44 hourFrom eqn(1) , 25= ?m1-e11.44= ?m 1-0.5?Final steady temperature rise ?m=25/0.5= 500C

When cooling:

Temperature rise after 2.5 hour, ?=40-30= 100C. Since the transformer is disconnected its final steady temperature rise when cooling ?n=0.Initial temperature rise is ?i=?m=500CWe have, ?=?i e-tTc 10=50e-2.5Tce-2.5Tc =0.2-2.5Tc=ln0.2= -1.609 Therefore, Cooling time constant Tc=1.55 hour.Result

Final steady temperature rise ?m=500CCooling time constant Tc=1.55 hour9) The initial temperature of a machine is 400C. Calculate the temperature of the machine after 1hour if its final steady temperature rise is 800C and the heating time constant is 2 hours. The ambient temperature is 300C.

Solution

Initial temperature rise ?i=40-30= 100C.

?=?m1-e-tTh + ?ie-tTh?Temperature rise after one hour

?=80(1- e-t2)+10e-12 = 37.50C.? Temperature rise of machine after one hour = 37.54 +30= 67.540C.

1.9 Two mark Q;A

1. What are the considerations to be made while designing a electrical machines?

•Cost

•Durability

•Compliance with the performance specification and consumer requirement

2. List some limitation of the design

•Magnetic Saturation

•Temperature rise

•Efficiency

•Standard specifications

3. Define total magnetic loading.

The total magnetic load is defined as the total flux around the armature periphery and is given by p? Weber’s

4. Define total electric loading

The total armature ampere conductors around the armature periphery is known as the total electric loading and is given by IzZ.

5. Define specific magnetic loading

The specific magnetic loading is defined as the total flux per unit area over the surface of the armature periphery and is denoted by Bav also known as average flux density.

6. Define specific electric loading

It is defined as the number of armature conductors per meter of armature periphery at the air gap.

Specific electric loading=total number ampere conductors/armature periphery at air gap.

7. What are the factors that decide the choice of specific magnetic loading?

•Maximum flux density in iron parts of machine

•Magnetizing current

•Core losses

What is the factors that decide the choice of specific electric loading?

•Permissible temperature rise

•Voltage rating of machine

•Size of machine

•Current density.

9. How the design problems of electrical machines can be classified?

•Electromagnetic design

•Mechanical design

•Thermal Design

•Dielectric design

10. What are the major considerations to evolve a good design of electrical machine?

The major considerations to evolve a good electrical machine are the specific magnetic loading, specific electric loading, temperature rise, efficiency, length of air gap and power factor.

11. Write short notes on standard specifications.

The standard specifications are the specifications issued by the standards organization of a country. The standard specification serves as guidelines for the manufacturers to produce quality products at economical prices. The standard specifications for the electrical machines include Ratings, Types of Enclosure, Dimensions of the conductors, Name plate details, performance indices, permissible temperature rise, permissible loss, efficiency etc.,

12. What is a magnetic circuit?

The magnetic circuit is the path of magnetic flux. The mmf of the circuit creates flux in the path against the reluctance of the path. The equation which relates flux, mmf and the reluctance is given by,

Flux = mmf/reluctance

13. What are the constituents of magnetic circuit in rotating machine?

The various elements in the flux path of the rotating machine are poles, pole shoe, air gap, rotor teeth and rotor core.

14. Write ant two similarities between magnetic and electric circuits.

In electric circuit the emf circulates current in a closed path. Similarly in a magnetic circuit the mmf creates the flux in a closed path.

In electric circuit the flow of current is opposed by resistance of the circuit. Similarly in magnetic circuit the creation of flux is opposed by reluctance of the circuit.

15. Write any two essential differences between magnetic and electric circuit.

•When the current flows in electric circuit the energy is spent continuously, whereas in magnetic circuit the energy in needed only to create the flux but not to maintain it.

•Current actually flows in the circuit, whereas the flux does not flow in a magnetic circuit but is only assumed to flow.

16. What is magnetization curve?

The curve shows the relation between the magnetic field intensity (H) and the flux density (B) of a magnetic material. It is used to estimate the mmf required for the flux path in the magnetic material and it is supplied by the manufacturer of stampings or laminations

17. What is meant by magnetic circuit calculations?

The calculations of reluctance, flux density and mmf for various sections of magnetic circuit are commonly referred as magnetic circuit calculations.

18. How the mmf of a magnetic circuit is determined?

The magnetic circuit split into convenient parts (Sections) which may be connected in series or parallel. Then the reluctance, flux density and mmf for every section of the magnetic circuit is estimated. The summation of mmf of all sections in series gives the total mmf for the magnetic circuit.

19. Define gap contraction factor for the slots.

The gap contraction factor for slots Kgs is defined as the ratio of reluctance of air gap in machine with slotted armature to the reluctance of air gap in machines with smooth armature.

20. Define gap contraction factor for the ducts.

The gap contraction factor for the ducts Kgd is defined as the ratio of reluctance of air gap in machines with ducts to reluctance of air gap in machine without ducts.

21. Define total gap contraction factor, Kg

The total gap contraction factor Kg, is defined as the ratio of reluctance of air gap of machines with slotted armature & ducts to the reluctance of air gap in machines with smooth armature and without ducts. The total gap contraction factor is equal to the product of gap contraction factors for slots and ducts.

22. What is carter’s coefficient?

The carter’s coefficient is a parameter that can be used to estimate the contracted or effective slot pitch in case of armature with open or semi enclosed slots. It is the function of the ratio w0/lg where w0 is slot opening and lg is air gap length.

23. Write the expression for the gap contraction factor for slots and ducts

Gap contraction factor for slots, Kgs = ys / (ys – KcsWs) Gap contraction factor for ducts , Kgd = L / (L – Kcdndwd)

24. Write down the formula for computing the mmf for the air gap length.

Mmf for the air gap = 800000 B Kg lg in AT

25. Write the expressions for reluctance of air gap in machines with smooth armature and slotted armature.

Reluctance of air gap in machines with smooth armature and without ducts = lg / µ0Lys

Reluctance of air gap in machines with open armature slots and ducts = lg / µ0L’ys’

26. Define field form factor.

The filed form factor Kf is defined as the ratio of average gap density over the pole pitch to maximum flux density in the air gap.

Kf = Bav / Bg Kf ? ? = pole arc/pole pitch

27. List the methods used for estimating the mmf for the teeth (tapered teeth)

•Graphical method

•Three ordinate method (Simpson’s rule)

• Bt1/3 method

28. What is real flux density and apparent flux density?

The real flux density is due to actual flux through a tooth. The apparent flux density is due to total flux that has to be passed through the tooth. Since some of the flux passes through slot, the real flux density is always less than the apparent flux density

29. Define real flux density.

The real flux density is defined as the ratio of actual flux in the teeth to the area of the teeth.

30. Define apparent flux density

The apparent flux density is defined as the ratio of the total flux in the slot pitch to the area of the teeth.

31. State the relation between real and apparent flux density.

Breal = Bapp – µ0 atreal (Ks – 1)

32. Define leakage coefficient

The leakage coefficient is defined as the ratio of total flux to the useful flux.

33. What is fringing flux?

The bulging of magnetic path at the air gap is called fringing. The fluxes in the bulged portion are called fringing flux.

34. List some leakage fluxes available in the rotating machine.

•Slot leakage flux

•Zig-zag leakage flux

•Harmonic or differential leakage flux

•Peripheral leakage flux

•Tooth to leakage flux

•Skew leakage flux

35. Define permeance.

Permeance is the inverse of reluctance. The reluctance of magnetic path is given by the reluctance S = l/Aµ.

36. Define specific permeance of a slot

Specific permeance of a slot is defined as the permeance per unit length of slot or depth of field.

37. What is unbalanced magnetic pull?

The unbalanced magnetic pull is the radial force acting on the rotor due to non uniform air gap around the armature periphery.

38. What do you understand by slot pitch?

The slot pitch is defined as the distance between centres of two adjacent slots measured in linear scale.

39. Define slot space factor or slot insulation factor.

The slot space factor is defined as the ratio of conductor area to slot area.

40. List the different types of slots that are used in rotating machines.

•Parallel sided slots with flat bottom

•Tapered slots with flat bottom

•Parallel sided slots with circular bottom

•Tapered sided slots with circular bottom

•Circular slot

1.10 University Questions

PART-A

Define specific electric loading. (Apr/May2015, Nov/Dec2014, May/June 2013, April /May 2011, April /May 2010)

What are the factors that affect the size of rotating machine? (Apr/May2015, Nov/Dec2013)

What is gap contraction factor for slots? (Nov/Dec2014)

Define stacking factor. (Nov/Dec2014)

Write down the classification of magnetic materials. (May/June2014, Nov/Dec2011)

What is peripheral speed? Write the expression for peripheral speed of a rotating machine. (Nov/Dec2013, May/June 2012, April /May 2010)

What are the major considerations in Electrical Machine Design?( May/June 2013, Nov/Dec2011)

Mention the main areas of design of electrical machines.( Nov/Dec2012)

What is meant by magnetic loading?( May/June 2012)

How materials are classified according to their degree of magnetism? (April /May 2011)

List the advantage of Hydrogen Cooling. (April /May 2010)

What is meant by Specific Magnetic Loading?

PART B:

Discuss the advantages of hydrogen cooling. (8 marks – Apr/May2015)

The temperature rise of a transformer is 30 0C after one hour and 42 0C after two hours of starting from cold conditions. Calculate its final steady temperature rise and heating time constant. If its temperature falls from the final steady value to 48 0C in two hours when disconnected from the main. Calculate its cooling time constant. Assume the ambient temperature is 300C. (8 marks – Apr/May2015).

Describe the classification of insulating materials used for electrical machines. (8 marks – Apr/May2015).

The exciting coil of an electromagnet has a cross section of 120 x 50 mm2 and a length of mean turn 0.8m. I t dissipates 150W continuously. Its cooling surface is 0.125 m2 and specific heat dissipation is 30 W/m2-0C. Calculate the final steady temperature rise of the coil surface. Also calculate the hot spot temperature rise of the coil if the thermal resistivity of insulating material used is 8?m. (8 marks – Apr/May2015).

Derive the temperature rise of a machine when it is run under steady load conditions starting from cold conditions. (8 marks – Apr/May2015, (Nov/Dec2014)).

What are the limitations in design of electrical apparatus? Explain them. (8 marks – Nov/Dec2014))

Calculated the mmf required for air gap of a machine having core length = 0.32 m including 4 ducts of 10 mm each, pole arc= 0.19 m, slot pitch = 65.4 mm, slot opening = 5 mm, air gap length = 5 mm, flux / pole = 52 mwb. Given Carter’s coefficient is 0.18 for opening/gap = 1 and is 0.28 for opening/gap = 2. (8 marks – Nov/Dec2014))

Explain the modern trends in design of electrical machines. (8 marks – Nov/Dec2014))

What are the main groups of electrical conducting materials? Describe the properties and applications of those materials. (16 marks – May/June2014)

Explain in detail the various cooling methods of electrical machines. (16 marks – May/June2014)

Calculate the diameter and length of armature for a 7.5kW, 4 pole 1000 rpm, 220V shunt motor. Given: full load efficiency = 0.83, maximum gap flux density = 0.9 wb/m2, specific electric loading = 30000, armpere conductor per metre, field form factor = 0.7. Assuming that the maximum efficiency occurs at full load and the field current is 2.5% of rated current. The pole face is square. UNIT-II (8 marks – May/June2014)

Discuss the choice of specific magnetic loading. (8marks-Nov/Dec2013)

Discuss the choice of specific electric loading. (8marks-Nov/Dec2013)

Calculate the specific magnetic loading of 100HP, 300V, 3 phase 50Hz, 8 pole star connected flame proof induction motor having stator core length = 0.5m and stator bore = 0.66m, turns/phase = 286. Assume full load efficiency as 0.938 and power factor is 0.86. (8marks-Nov/Dec2013)

1.11 Additional Problems:

1. A laminated steel tooth of armature of a d.c. machine is 30 mm long’ and has a .taper such the maximum width is 1.4. times the minimum. Estimate the mmf required for a mean flux density of 1.9 wb/m2 in the tooth. B-H characteristics of steel is given below:

B wb/m2 1.6 1.8 1.9 2.0 2.1 2.2 2.3

H A/m 8700 0 000 7 000 7 000 1 000 0000 0 900

2. Determine the apparent flux density in the teeth of a d.c. machine when the real flux density is 2.15wb/m2. Slot pitch is 28 mm, slot width is 10 mm and the gross core length 0.35 metre. The number of ventilating ducts is 4. Each duct is 10 mm wide. The magnetizing force for a flux density of 2.15 wb/m2,is 55000 H/m. The iron stacking factor is 0.9.

3. Compute the main dimensions of a 2500 kVA, 187.5 rpm, 50 Hz. Three phase, 3 kV salient pole synchronous generator. The specific magnetic loading is 0.6 wb/m2 and the specific electric loading is 3400 ac/m. The ratio of core length to pole pitch= 0.65.

4. State and explain the main factors which influence the choice of specific magnetic loading and specific electric loading in a synchronous machine. Explain the role of digital computes in the design of electrical machines.

5. A 15 kW, 230 V, 4 pole d.c machine has the following data:

Armature diameter = 0.25 m ; armature core length = 0.125 m ; length of air gap at pole centre = 2.5 mm ; flux per pole = 11.7 x 10-3 wb, ratio Pole arc/ pole pitch=0.66. Calculate the mmf required for air gap.

6. Derive the expression for the specific permeance of slots with double layer windings and for special purpose induction motors.

7. Compute apparent magnetic flux density in the teeth of a dc machine when the real flux density is 2.15 wb/m2. Slot pitch is 28 mm. Slot width is 10mm and the gross core length is 0.35 metre. The number of ventilating ducts is 4, each 10 mm wide. The magnetizing force for a flux density of 2.15 wb/m2 is 55000 A/m. The iron stacking factor is 0.9.

350kW, 500V, 450rpm, 6-pole, dc generator is built with an armature diameter of 0.87m and core length 0.32m. The lap wound armature has 660 conductors . Calculate the specific electric and magnetic loadings.

The output co-efficient of 1250kVA, 300rpm, synchronous generator is 200 kVA/m-rps. (a) Find the values of the main dimensions (D,L)of the machine if the ratio of length to diameter is 0.2. Also calculate the value of main dimensions if(b)specific loadings are decreased by 10%each with speed remaining the same as in part (a).(c)speed is decreased to 150rpm with specific loadings remaining the same as in part(a)Assume the same ratio of length to diameter .comment upon the result

A20-HP , 440V, 4-pole ,50HZ , 3 phase induction motor is built with a stator bore of 0.25m and core length of 0.16m.the specific electric loading is 23000 ampere conductors per metre. find specific magnetic loading of the machine. Assume full load efficiency of 84percent and a power factor of 0.82. using the data of the above machine determine the main dimensions for a 15HP, 460v ,6 pole , 50Hz motor .

Calculate the main dimensions of 20kW, 1000rpm, dc motor. given that b=0.37tesla and ac=16000amp.cond./m. make necessary assumptions.

Determine suitable values for the number of poles, D and L for a 1000kW, 500v, 300 rpm dc generator. Assume average gap density =1 tesla and specific electric loading as 400 amp. cond./cm.

Calculate the specific electric and magnetic loading of 100HP, 300v , 3-phase 50HZ 8 pole, star connected, the flame proof induction motor having stator core length =0.5m and bore =0.66m. turn/phase = 286.Assume full load efficiency as 0.938 and pf as 0.86.

A 600 rpm, 50HZ, 10000 V ,3-phase, synchronous generator has the following design data. Bav=0.48 wb/m2, ?=2.7A/mm2 and space factor =0.35, number of slots =144, slot size=120*20mm, D=1.92m and L=0.4m. Determine kVA rating of a machine.

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