 Open access

# Calculation Methods for Heating and Ventilation System of Electrical Machines

Written By

Otilia Nedelcu and Corneliu Ioan Sălişteanu

Submitted: December 7th, 2011 Published: October 24th, 2012

DOI: 10.5772/51962

From the Edited Volume

## Heat Transfer Phenomena and Applications

Edited by Salim N. Kazi

Chapter metrics overview

View Full Metrics

## 1. Introduction

To design as accurately electrical machines it is necessary as the cold machine to be efficient and to determine values and distribution of temperature in the machine (along the rotor). For this, must do a complete and accurate analysis of temperature field in steady state (or in transient regime) which requires the knowledge and location of the temperature loss amounts, the knowledge of the velocity distribution of the cooling fluid in the cooling channels different of the machine, the determination of thermal flux values and the distribution of these values in some component parts of the electric machine.

Considering the analogy between a ventilation network of cooling and a nonlinear resistive circuit, the flows of the cooling agent of an electric machine can be determined.

The aim of thermal computation of electric machines is necessary to establish heating, temperature overrunning or overheating in some parts of the machine from the environment temperature at a given operating condition of the machine.

The service life of the insulations used in electric machines is limited heating which greatly decreases in the same time with temperature increasing. Heating determined is stabilized, because the electrical machine which will be discussedhas a long-lasting operating regime. Inside the electric machine a heat quantity is produced which has to be exhausted to outside for limiting the excessive heating, heating due to the losses which occur in the active materials.

## 2. Heat transfer

The heat transfer is the science of spontaneous and irreversible processes of heat propagation in space and is the exchange of thermal energy between two objects, two regions of the same part and respectively two fluids, as a result of a difference of temperature between two fluids.

In technical, the heat transfer must be:

• more intense (e.g. steam boilers, the optimal exchange in heat exchangers);

• Stopped (e.g. steam by thermal insulation).

The heat is the transfer of energy between phisical-chemical systems or between different parts of the same systems, if there is a transformation where is not performed any mechanical work.

The heat transfer deals with the processes where the thermical energy at higher parameters is changed in thermal energy to lower parameters, usually, the parameter which appreciates the heat quality is the temperature, defined as a global measure of processes’ intensity which determines the internal energy of an object (thermal agitation of molecules in liquid and gases, atoms vibration and free electrons motion in metals, etc.).

The heat exchange follows thermodynamics principles: the first principleof thermodynamicsexpresses the energy conservation rule and it establishes the quantitative relationship between heat and mechanical work, and allows integral transformation of heat in mechanical work and the reverse; the second principle of thermodynamicsstates the necessary conditions for transforming heat in mechanical work and establishes natural meaning of heat propagation, always from the source with higher temperature to the source with lower temperature.

The main objectives of the heat transfer are: first, the determination or ensuring the amount of heat exchanged per time unit in fixed temperature conditions, and second, checking the compatibility of materials used with the regime of temperature faced by, in determining the temperature field. We can say that an exchange of heat device is an optimal solution from thermic, hydraulic, mechanic, economic and reliability points of view, mainly, the heat transfer being the determining factor. We should add to these, finding the enhancing methods and procedures or, in certain cases, heat transfer breaking. 

Conduction, radiation and convectionare distinct forms of heat transfer. As defined above, only conductionand radiationare heat exchange processes due solely to the difference in temperature, convectionis a more complex process, necessarily involving the mass transfer. Because, the convection makes energy transfer from regions with higher temperature in regions with lower temperature, it was generally accepted „heat transfer by convection” as being the third mode of heat exchange.

Basic sizes in heat transfer

The temperature field. The status scalar parameter, the temperature, in any point of space M(x, y, z) depends on position and time, namely it is a function of the following form.

t=f(x,y,z,τ)E1

The temperature field is such, the total of the temperature tvalues throughout all space, at some time τ, the expression (1) representing the equation of this field.

The temperature field may be constant (stationary or permanent) and transient (non-stationary or variable) as time τ is explicitly or nonexplicitly in equation (1) and namely:

• the temperature constant field has the equation:

t=f1(x,y,z),   tτ=0E2
• the temperature transient field is expressed by equation (1)

Regarding the number of coordinates, the temperature field can be one, two or tridirectional. So, in equations (1) and (2), the temperature field is tridirectional in transient regime, respectively, constant. If the temperature is depending on two coordinates and time, the field is bidirectional transient, with the equation:

t=f2(x,y,τ),   tz=0E3

and if it is expressed as a function of a coordinate and time, the field is onedirectional transient, with equation:

t=f3(x,τ),  ty= tz=0.E4

The equation of temperature in the onedirectional constant field is the most simple:

t=f4(x),  ty= tz=tτ=0.E5

All points of space considered which have the same temperature t at a moment in time τ form isothermal surface. Because, a point in an object can not concomitant has two different temperature values, the result is that isothermal surfaces are continuous surfaces which do not intersect one another. 

The temperature gradientis a measurement which expresses elementary increase of temperature to a point of a temperature field, at a given time τ.

The temperature gradient is a normal vector to isothermal surface and it is numerically equal with the limit of the report between the temperature variation Δtbetween two isothermal surfaces and the distance Δnof these two, measured on normal to surface, when Δntends to zero, namely:

The heat amount ΔQpassing through an object or from an object to another, by an isothermal surface S, in time unit Δτ, is called heat flow (thermal flow)Q:

Q.=ΔQΔτ   [W]E7

The unitary heat flowqsis the heat flow which crosses the surface unit in the time unit:

q.S=Q.S=ΔQSΔτ   [W/m2]E8

where Sis the area of heat exchange surface, in m2.

Electrical analogy of heat transfer

Two systems are analogous when they have different natures, but respect similar equations with similar boundary conditions. This implies that equations describing the action of a system can be transformed into the equations of the other system by simply exchanging the variable symbols. So, Ohm’s law for Electrotechnics expressing the link between DC I, voltage (potential difference) ΔUand the electrical resistance Re,has an analog form in heat transfer, by the relationship between unitary thermal flow q, temperature difference (thermal potential) Δtand a size called thermal resistanceRt,,namely:

I=ΔURe ;  q.S=ΔtRt E9

In equation (9), when qis measured in W/m2 and Δtis measured in °C, the thermal resistance Rtis expressed in (m2 °C)/W.

Based on this analogy, a series of concepts of DC circuits can be applied to the heat transfer problems (for example, electrical circuit has an equivalent thermal circuit and reverse) and in dynamic regime (for example, electrical modeling of some transient thermal processes). The electrical analogy of heat transfer can be used as a calculation tool and as viewing heat transfer equations by linking them to the electrical engineering domain.

Corresponding to the three fundamental types of heat transfer, we can establish the expressions for calculating the thermal resistance to conduction, convection and respectively, radiation, which can take place in the complex processes of heat exchange, equivalent electrical diagrams serially or mixed connected.

Inverse thermal resistance is called thermal conduction. 

## 3. The calculation methods for heating system of electrical machines

The heat flow Q (the amount of thermical energy which pass through a given surface S in time unit) expressed in J/s is numerically equal with losses of power P, expressed in W, which determine heat in any part of the machine. So, instead of heat flow, in relations used, will use losses P, directly. The object characteristics and the difference between object temperature to the environment that transfers heat determines heat transfer on the mass and on the surface of that object. For thermal calculation, the analogy is made between electrical resistance from the power of a circuit and thermal resistance R from heat flow Q.

Based on this analogy, the temperature drop in mass or surface of an object corresponding to heat flow Q, (or losses P) is determined by multiplying between thermical resistance of the object opposed to heat flow in its mass, respective its surface, R, and losses P, relation (10):

θ=PR[°C],E10

Where:θ= ν12 – is the temperature drop, over temperature or object heating, in °C.

On the relation (10) we may work with equivalentthermal diagramswith means which, for certain values of thermal resistance and losses, the heatθis resulting.

The way to solve the thermal schemes, depending on the thermal resistances connection (series, parallel or mixed) is identical to the electrical schemes. [2, 3]

Next we will present the thermal resistances expressions depending on the types of heat transmission and the structure encountered in building electrical machines.

Heat transmission by radiation and convection in the case of heat transmission in axial channels and radial channels of electrical machines

In heat transmission by radiation and convection in the case of heat transmission in axial channels and radial channels of electrical machines, the transmission coefficients of heat depend on the air velocity which blows on the surface. For certain lengths of the channel, transmission coefficients of heat will be taken from the diagrams shown in „figure 1” for axial channels and „figure 2” for radial channels.

## 4. Common thermal schemes used by electrical machines

Operating winding of electrical machines Figure 3.Transversal section through a winding pole of a DC machine.

In “figure 3”, it is presented the transversal section through a winding pole of an electrical DC machine, for that the temperature is considered constant around the coil mass, calculating the average temperature value, the temperature drop is practically null, as well as in the coil mass and in the machine housing, for the part of housing which contributes to cooling of a magnetizing coil and corresponds an angleα=π2p, this is the angle between symmetry axes from a main pole and the auxiliary neighboring poles.

The real thermal diagramme of heat transmission to a coil from a pole ofDC machine is indicated in „figure 4”, where there are indicated losses by Joule effect of one coil Pcub,thermal resistance from coil to housing R1corresponding the domainof „figure 3”, thermal resistance from coil to pole by pole piece R2corresponding the domainof „figure 3”, thermal resistance from coil to pole by the pole body R3corresponding the domainto figure 3, thermal resistance from pole to housing R4considering the pole-housing jonction 0,1 mm corresponding the domain, thermal resistance from pole to air by air-gap R5corresponding the domain, thermal resistance from pole to the space air between poles R6corresponding the domain, thermal resistance from housing to air inside machine R7corresponding the domain, thermal resistance from housing to environmental medium R8corresponding the domain, thermal resistance from coil to air R9corresponding the domain. All resistances are measured in 0C/W. Figure 4.Main diagram of heat transmission to a coil from a pole of DC machines

For learning the total equivalent resistance we need to reproduce intermediate thermal diagrams, „figure 5.a, b”.

Thermal resistances for determining the heating of magnetizing coil from „figure 5.a” correspond to relations:

RA=R2R3R2+R3;RB=R1;RC=R;RD=R5R6R5+R6;RE=R7R8R7+R8E11

And for „figure 5.b” after transforming the triangle RD,RCand REfrom „figure 5.a” into star Rx, Ryand Rzthe relations for thermal resistances are:

Rx=RCRDRC+RD+RE;Ry=RCRERC+RD+RE;Rz=RDRERC+RD+RE;RI=RA+Rx;  RII=RB+RyE12

According to (12) and the equivalent diagram from „figure 5.b” we can determine the total equivalent resistance:

RT=(RIRIIRI+RII+Rz)R9RIRIIRI+RII+Rz+R9E13

After having known the total equivalent resistance RT, we can determine the heating of magnetizing coil towards the cooling air in the machine:

θcub=RTPcubE14 Figure 5.Intermediate thermal diagrams for determining the heating of magnetizing coil:a – real diagram; b – equivalent diagram.

The calculation of thermal resistances from the real diagram(figure 4)

The thermal resistance from coil to machine housing

Thermal resistance from coil to machine housing R1is calculated according to total thickness βt1(thickness coil insulation βb, insulation thickness to the housing βc, thickness of protecting lac from impregnation βl0,025 cm), of equivalent thermal conductivity λ1(thermal conductivity when coil is separately isolated λb, thermal conductibility for insulation from housing λc, thermal conductivity for protecting lac λl) and from the corresponding surface S1by which the heat is transmitted, considering only the side parts. [2, 3]

βt1=βb+βc+βl[cm];λ1=βt1βbλb+βcλc+βlλlS1= 2b1(lm+bm+2b1), [cm2]E15

Where:

b1- the thickness of the coil to the upper part, in cm;

bm– the polar core width, in cm;

lm– the polar core length, in cm.

These are obtained by:

R1=βt1λ1S1[oC/W]E16
.

The thermal resistance from the coil to the polar piece.

The thermal resistance from the coil to the polar piece, R2, is calculated as the thermal resistance from coil to machine housing for same insulation, but get calculating only the isolated frame bottom and the transmission surface of heat S2.

R2=βt2λ2S2[oC/W]βt2=βt1andλ2=λ1S2= 2b2(lm+bm+2b2), [cm2]E17

where b2is the thicknessthe coil to the below part, in cm.

The thermal resistance from the coil to the pole object

The thermal resistance from the coil to pole object R3, is calculated according to the thickness of the insulated coil βbonly when the coil is isolated separately, βm, the thickness of the insulation toward the body of the pole, the impregnation layer βl0,025 cm, the thermal conductivity of the insulated coil λb, thermal conductivity of insulation from the pole body λm, the thermal conduction of the protecting lac λland of the corresponding surface S3.

βt3=βb+βm+βl[cm]λ3=βt3βbλb+βmλm+βlλlS3= 2b3(lm+bm+2b3), [cm2]E18
R3=βt3λ3S3[oC/W]E19

The thermal resistance from the pole core to the housing

The thermal resistance from the pole core to the housing R4 is calculated according to the thickness of the parasite airgap to the pole-housing jonction at about 0,1 mm, for thermal conductivity in electrical machines whose stator is not impregnated after windingλ4= 0,00025 W/cm°C, or for electrical machines with the impregnated stator λ4= λl, and, also according to the corresponding calculated surface with bmthepolar core width and lmthe polar core length.

R4=0,01λ4S4[oC/W]E20

The thermal resistance to the heat transmission from the core pole to airgap

 α0=1,42 x 10-3W/cm2 oC Iron and steel surfaces, grouted (housing surfaces) α0=1,67 x 10-3W/cm2 oC Iron and steel surfaces, nongrouted α0=1,33 x 10-3W/cm2 oC Cooper or brass surfaces, and coil impregnated or varnished surfaces

### Table 1.

The coefficient values of heat transmissionα0, on thenon-blast surfaces

Considering the corresponding surface S5 calculated according to the length and width of the polar piece lprespectively bp, the thermal resistance to the heat transmission from the core pole to airgap is determinate by relation:

R5=1αt5S5[oC/W]E21

The coefficient of heat transmission on the object surface αt5is determined admitting that θaer'= 20°C is half of the total heating of the air in the machine, k = 0,8 chosen from „table 2”, the coefficient of heat transmission chosen from „table 1” is α0= 1,67 x 10-3W/cm2 OC and air velocity in the airgapvδ=12vr.

αt5=α0(1+kvδ)(10,5a)[W/cm2°C]E22
a=θaer'θaE23
a= 20o70o= 0,28 class of insulation B, a= 20o90o= 0,22class of insulation F

S5 =bplp[cm2].

 The name of the surface and the type of blast k The blast surface in the most perfect way 1,3 The surface of the front linkage of induced winding 1,0 The surface of the active part of the induced 0,8 The surface of the magnetizing coils 0,8 The collector surface 0,6 The outer surface of the housing 0,5

### Table 2.

The coefficient values of blast intensities k

The thermal resistance to heat transmission from polar piece to the space between poles

In the heat transmission to space between poles, by polar pieces, the thermal resistance is determined regarding the coefficient of heat transmission αt6and the surface located on the polar width of the object bclmpolar core length:

R6=1αt6S6[oC/W]S6= 2bclm[cm2]αt6= 1,6.7 x103(+ 0,8v6)( 0,5a) [W/cm2°C], cuv6=vpE24

The thermal resistance to heat transmission from the housing to interior

When heat transmission is done from housing to interior, the thermal resistance R7 is determined the same as in heat transmission to the space between poles, by polar pieces with αt7αt6, v7 = v6, and the corresponding surface S7 is calculated regarding the interior diameter of the housing Dc, the number of pole pairs pand the housing length lc.

R7=1αt7S7[oC/W]S7πDc4plc(bm+2b1)(lm+2b1)[cm2]E25

The thermal resistance to heat transmission from housing to exterior (quiet air)

The heat transfer from the housing to outer where the air is quiet, thermal resistance is calculated taking into account the heat transmission coefficient α0= l,42x10-3 W/cm20C determined in „table 1”, by the appropriate surface S8, at which involved, the height of stator yoke hjs, the inner diameter of the housing Dc, the number of pole pairs pand the housing length lc.

R8=1α0S8[oC/W]S8=π(Dc+2hjs)4plcE26

The thermal resistance to heat transmission from coil directly to cooling air in the machine

When the heat transmission is done from coil directly to cooling air in the machine, the thermal resistance depends on the thickness and the thermal conductivity of coil insulation when the coil is isolated separately βbandλb, by the coefficient of heat transmission αt9, and by the corresponding surface S9.

R9=βbλ9S9+1αt9S9[oC/W]αt9=α0(+ 0,8v9)( 0,5a) [W/cm2°C],α0= 1,33 103. W/cm2°C, from table 1; v9= v6;S92(bm+lm+4b1)1,1hb[cm2]E27

## 5. The losses by Joule effect, a single coil

The losses by Joule effect Pcubare the total losses in the magnetizing reeling Pcucreported to the number of poles of the electrical machine 2p.

Pcub=Pcuc2p[W]E28

## 6. The total heat of magnetizing reeling

Heating the magnetizing coil to the air in the machine, is determined by the relation (14) together with heating the air to the contact with the coil (θaer20°C – from rotor to stator) determining heating of the magnetizing reeling towards the external environment.

## 7. The diagrams of ventilation and calculation of aerodynamic resistances

In the construction of rotating electrical machines, the cooling air is most frequently used. Depending on the construction type and the type of protection of the electrical machine the best ventilation system is chosen.

Calculating ventilation of rotating electrical machine is done as follows:

• establishing the ventilation diagram andcalculating aerodynamic resistances;

• calculating the pressure determined by means of air entrainment;

• calculating air flow at different sections.

Preparing the ventilation diagram for a rotating electrical machine requires prior execution of the diagram to the scale of general assembly (longitudinal section and cross section) and establishing the routes where cooling air circulates. For calculation aerodynamic resistances we need to know longitudinal and transversal dimensions of all pipes that form the routes for air circulation.

When we solve an aerodynamic diagram, we aim to find the equivalent resistance Rae.. Between aerodynamic diagrams used for air circuits and electrical diagrams may be an analogy, the aerodynamic resistances act as passive elements.

Determining the equivalent resistance depends on how aerodynamic resistances are connected:

• to series arrangement Raeis the sum of the aerodynamic resistances and it is determined by relation:

Rae=i=1nRaiE30
• to shunt connectionRaeis determined by relation:

1Rae=i=1n1RaiE31
• to mixed connection, „figure 6”, Raeis calculated according to relations of calculation for the series arrangement and shunt connection and we obtain:

Rae=Ra1+Ra2Ra3(Ra2+Ra3)+Ra4E32

Bilateral axial-radial ventilation for an electrical machine in protected construction

In „figure 7.a” are represented the longitudinal section and the frontal view for an induction motor with the rotor in short-circuit, in protected construction, and with arrows we indicate routes that circulate the cooling air.

Typically, in asynchronous machine the role of inductor is accomplished by the stator, while the rotor is the induced machine. The machine excitation is done in AC. The collector body is of „collector rings” type.

The asynchronous machine can not develop a couple unless the speed „n” of the rotor is different by the speed „n1 ofthe stator rotating field. [2, 3] Figure 7.Ventilation diagram of asynchronous electrical machine with bilateral radial ventilation: a) physical diagram; b) main diagram.

Because depression created in the rotor bars ends and the rotor ribs (the electrical machine has not its own ventilator), the air is absorbed through the windows of the two fore-shields, washes the active elements and it is thrown out through the windows of its middle, by the bottom of the housing.

Air flow is symmetrical to the middle of the machine length, and is, therefore, sufficient to calculate ventilation only on one side, which is half of the machine.

The sources of pressure are denoted by „H”, the aerodynamic resistance by „Ra” and are divided in sections. [3, 11, 12]

## 8. Section I

We established I section, which contains aerodynamic resistances Ra1, Ra2, Ra3, andRa4, which are connected in series and is calculated the total aerodynamic resistance for section I, according the relation:

RatrI=i=14RaiE33

For determining total aerodynamic resistance of the I section we calculate, in turns, each aerodynamic resistance which is part of the I section regarding the coefficients of aerodynamic resistances and the sections of the channels by which the air circulates.

 Conditions of entry ζ Entry with sharp edges 62 x 10-2 Entry with edges at angle of 90o 30 x 10-2 Entry with rounded edges (12,5...0) x 10-2

### Table 3.

The values of the resistance coefficient to the air admission pipes

 The angle of deviation ζ 135o 32 x 10-2 90o 70 x 10-2

### Table 4.

The values of the resistance coefficient to deviation in different angles

The air enters through the shield windows under sharp edges (blinds) by narrowing from the external environment to the input section. From „table 3”, for sharp edgesζas=62102, and for narrowingS1Sext0, as in „figure 8ζing32102.

Aerodynamic resistance Ra1is calculated according to the net input surface in the shield windows S1, to the coefficient of aerodynamic resistance to narrowing ζingand to the coefficient for sharp edges ζas.

Ra1=ζas+ζingS12=(62+32)102S12E34

Where:-S1= kiSi[m2]

• Siis the total surface ofthe shield windows, in m2;

• kiis the coefficient of reduction in section because the coatings required by the machine protection: ki=0,6÷0,7

There is a relaxation of the air coming out of the shield and to the input in the machine and a deviation to 135o by deflector (figure 7.a)

S2=π4(D12D22)[m2]E35
ForS1S2=...ζlarg=...., and for the deviation to 135o from „table 4” ζr=32102

The aerodynamic resistance Ra2is calculates according to the net output surface of windows shield S1, to the coefficient of aerodynamic resistance to enlargement ζlargand to the coefficient to the air relaxation of shield output ζas.

Ra2=ζlarg+ζrS12E36

Next, there is a relaxation to the output of deflector under sharp edges.

S3=π4(D32D42)[m2]E37
ForS2S3=...ζlarg=...., and for sharp edges ζas=62102

We calculate the aerodynamic resistance Ra3regarding the net input surface in the machine S2, the coefficient of aerodynamic resistance to widening ζlargand the coefficient for sharp edgesζas.

Ra3=ζlarg+ζasS22E38

At the rotor winding input occurs a narrowing and there are sharp edges for the short-circuited rotor.

S4=π4(D52D42)[m2]E39
ForS4S3=...ζing=...., and for sharp edgesζas=62102.

We calculate the aerodynamic resistance Ra4regarding the net input surface under rotor winding S4, the coefficient of aerodynamic resistance to narrowing ζingand the coefficient for sharp edges ζas.

Ra4=ζing+ζasS42E40

After we calculated the aerodynamic resistances Ra1, Ra2, Ra3,Ra4,regarding the narrowing or widening ventilation channels by relation (31) we calculate the total aerodynamic resistance of the section I.

## 9. Section II

The II section contains aerodynamic resistances Ra5, Ra6, Ra7, Ra8, Ra9, Ra10, Ra11, Ra12,connected in series and we calculate the total aerodynamic resistance to the section II with the relation:

RatrII=i=512RaiE41

To calculate the aerodynamic total resistance by section II, we calculate, as for section I, each aerodynamic resistance, in turns, considering the widening and the narrowing of the ventilation channels.

When the air begins to cross the section II, it enters under the rotor core which occurs in narrowing, and the input is done under sharp edges (figure 9).

S5=π4(Dir2Dax2)nnbnhn[m2]E42

Where:

• Dir, is the interior diameter of the armature core disc, in m;

• Daxis the diameter of the shaft, in m;

• bnis the ridge width, in m;

• hn=DtrDax2
• nnis the ridges number.

For S5S4=...ζing=....and for sharp edgesζas=62102.

We calculate the aerodynamic resistance Ra5, regarding the net surface to the output under the rotor core S5, the coefficient of aerodynamic resistance to the narrowing ζingand the coefficient for sharp edges ζas.

Ra5=ζing+ζasS52E43

For the input of radial ventilation channels of the rotor under sharp edges occurs a deviation to 90o,ζc=70102, from „table 3(ζas=62102)and a narrowing.

S6=0,9πDirbvnv2[m2]E44

Where:

• bvis the width of the radial channel for ventilation, in m;

• nvis the number of radial channels for ventilation;

• 0,9 is the coefficient which takes into consideration the presence of spacers.

For

S6S5=...ζing=....E45

The aerodynamic resistance Ra6is calculated, using the relation (42) which takes into consideration the net surface to the input of radial channels of the ventilator S6, the coefficient of aerodynamic resistance to narrowing ζingand the coefficient for sharp edges ζas.

After the air entered in the channels of ventilation the relaxation occurs (figure 9)

S7=0,9πDicbvnv2[m2]E46

Where:

• Dic is the diameter of the rotor slots base, in m.

For

S6S7=...ζlarg=....E47

The aerodynamic resistance Ra7is calculated, taking into consideration the relaxation in the channels of ventilation:

Ra7=ζlargS62E48

A new narrowing occurs to the entry in the region of rotor coils under sharp edges, for the short-circuited rotor or the rounded edges of rotors and widening to the outside rotor.

S8'=S7z2bc2bvnv2[m2]E49
S8"=0,9πDrbvnv2Z2bc2bvnv2[m2]E50

Where:

• Dr is the exterior diameter of the rotor, in m;

• z2 is the number of rotor slots;

• bc2 is the width of the rotor slot, in m.

For S8'S7=...ζing=....and for sharp edgesζas=62102.For
S8'S8"=...ζlarg=....E51

We obtained the aerodynamic resistance Ra8, calculated to the input of the coils region by narrowing the ventilation channel and the output of rotor by widening the ventilation channels for short-circuited rotor where the edges are sharp:

Ra8=ζing+ζas+ζlargS8'2E52

And Ra8, for phase-wound rotor where the resistance of coils rounded edges are neglected is:

Ra8=ζing+ζlargS8'2E53

At the transition from the rotor coils zone to the stator coils zone and the input under rounded edges a widening (or a narrowing) occurs(ζr=12,5102).

S9'=0,9πDbvnv2z1bc1bvnv2[m2]E54

Where:

• D is interior diameter of the stator, in m;

• z1 is the number of slots of the stator;

• bc1is the width of stator slots, in m.

For the rapport S8"S9'orS9'S8"(the lower section is used to the numerator) we determine ζlargor ζing. For determining the resistance we put at the denominator the smallest section.

The resistance Ra9is obtained, taking into account the net surface in the stator coils zone S9, the aerodynamic resistance coefficient to narrowing/widening ζing/ζlargand the coefficient for the input under sharp edges ζr.

Ra9=ζlarg(ing)+ζrS8(9)"2E55

In the stator coils zone and to the output between coils, the net surface increases and a relaxation occur.

S10=0,9π(D+2hcl)bvnv2[m2]E56
S9'=S10z1bc1bvnv2[m2]E57

Where:

• hc1is the height of the stator slot, in m.

ForS9'S9"=...ζlarg'=....; For

S9"S10=...ζlarg"=....E58

The resistance Ra10, is calculated taking into account the net surface in the stator coils zone S9and the net surface to the output between coils S10, the aerodynamic resistance coefficient to widening ζlargin both zones.

Ra10=ζlarg'S9'2+ζlarg"S9"2E59

To the exterior armature core disks in the radial channels of the stator a widening is produced, again, the surface S11is bigger than the surface S10.

S11=0,9πDebvnv2[m2]E60

Where:

• Deis the exterior diameter of the stator core disk, in m.

For

S10S11=...ζlarg'=....E61

The result is R a11:

Ra11=ζlargS102E62

To the output of radial channels of the stator, the air suffers a relaxation again.

S12=πDeL12[m2]E63

Where:

• L1is the geometric length of the machine plus the thickness pressure rings, in m.

For

S11S12=...ζlarg=....E64

Because there is a widening of ventilation channels to the output of the stator radial channels Ra12is calculated regarding the aerodynamic resistance coefficient to widening and S11:

Ra12=ζlargS112E65

## 10. Section III

The aerodynamic resistances Ra13, Ra14, Ra15, Ra16, Ra17, Ra18constitute the section III, they are serially connected. Taking into account the equation (28) the total aerodynamic resistance is calculated by the section III with equation:

RatrIII=i=1318RaiE66

After the air got out of the stator radial channels, it will cross the rotor ends coils by deviation to 135o for which ζc=32102(figure 7 and figure 10).

S13=πD3lbr[m2]E67

Where:

• lbris the axial length of the rotor coil end, in m.

Because there is no widening or narrowing, the resistance Ra13is calculated regarding the crossing surface to the rotor coil ends and the coefficient for rounded edges:

Ra13=ζcS132E68

After that a narrowing to input follows between rotor bars ends and sharp edges. ζas=62 x 10-2 for short-circuited rotors from bars with sharp edges. For the rotor with bars with rounded edges ζas=0. Figure 10.Partial longitudinal section through a rotor cage induction motor with ribbed shafts and bars ends having a ventilator effect (the rotor fromfigure 7)
S14=πD3(lbrlbi)z2br2(lbrlbi)[m2]E69

Where:

For

S14S13=...ζing=....E70

Resulting the resistance Ra13calculated using the net surface to the input between the rotor bars ends and the coefficients of aerodynamic resistances to narrowing under sharp edges:

Ra14=ζas+ζingS142orRa14=ζingS142E71

The isolated support of the bandages does not allow air to enter through the coils and as a result in the right front end, to winding motors only a narrowing occurs.

To the ends output (the bars) of the rotor coils there is a relaxation of the air, a relaxation caused by widening the channel (figure 7)

S15=πDld[m2]E72

For

S14S15=...ζlarg=....E73

Is obtained:

Ra15=ζingS142E74

There is a passing through the spaces between the stator coils ends which can be of two types.

• prefabricated coils

• winding to the round conductor.

The prefabricated coils(from profiled conductor or bars) are represented in „figure 11”.

We determine the surface S16under the front part to which we consider the narrowing of the lower layer.

S16=πDlf[m2]E75

We admit then, that the distances d1and kbetween coil ends, to lower part, are equal to those at the upper part (d1and kare constant on front ends height).

Narrowing at the lower layer (figure 11)

S16'=[lbsd1+(lmibc)a+kcbdd1nd]z[m2]E76

Where:

• ndis the number of distances between coil adjacent ends;

• lmiis the dental corresponding to the mean diameter of the coil lower side, in m. Figure 11.End of prefabricated stator coilForS16'S16=...ζing=...., and for rounded edges ζr=12,5 x 10-2

The resistance Ra16is:

Ra16'=ζr+ζingS16'2E77

For the output relaxation from lower layer S16is:

S16"π(D+hc1)lf[m2]E78

Where:

hc1is height of stator slot, in m.

For

S16'S16"=...ζlarg=....E79

We obtained:

Ra16"=ζlargS16'2E80

Narrowing in the upper layer and the input under the slightly rounded edges.

S16"'S16';for
S16'S16"=...ζing=....E81

For the slightly rounded edges from „table 3ζr=12,5 x 10-2

The resistance R”’a16is:

Ra16"'=ζr+ζingS16'2E82

The total resistance to airflow between the ends of stator winding is the sum of calculated resistances for narrowing in the lower layer, the relaxation to the output from lower layer and the narrowing in the upper layer and the input under the slightly rounded edges:

Ra16=Ra16'+Ra16"+Ra16"'E83

In the case of winding from round conductor (from wire),the air can not penetrate through the coil ends, because spaces between coils are obstructed of isolated strips between phases, so the air avoid them. The distance Xbetween the end of stator coil and deflector, must be large enough not to exclude the air (figure 7.a)

The narrowing in the section X.

S16π(D+hc1)X[m2]E84

For S16S15=...ζing=....and the resistance Ra16is:

Ra16=ζlargS162E85

After passing by front ends of stator winding there is a relaxation of air after widening the ventilation channel.

S17π(D+2hc1)(lf+X)[m2]E86

For S16S17=...ζing=....and aerodynamic resistance Ra17is:

Ra17=ζlargS162E87

The air flows in a narrow channel between housing ribs to a deviation of 135o (ζc=62 x 10-2).

S18=πDcpbncpnnc[m2]E88

Where:

• bncis the housing rib width, in m;

• nncis the number of the housing ribs;

• Dc, p(from figure 7.a) in m.

For S18S17=...ζing=....and so:

Ra18=ζc+ζas+ζingS182E89
(77)

## 11. Section IV

The total resistance of section IV is calculated as in the other sections with the equation (26) for the series connection:

RatrIV=i=1920RaiE90

A deviation at 90o occurs (ζc=70 x 10-2) to the output sided windows (figure 7.a) without the narrowing or widening of ventilation channels:

S19=LnBn2[m2]E91

And the result is:

Ra19=ζcS192E92

The output of the air from the housing is done to the deviation at 90o(ζc=70 x 10-2) for the relaxation of the outside air (figure 7.a).

S20=12Lphpki[m2]E93

For this kiwe must take into account the calculation ofRa1.

For S20Sex0ζlarg=60102and so:
Ra20=ζc+ζlargS202E94

In the scheme from „figure 7.b”, Hbis the pressure produced by the ends of moving rotor bars, Hnis the pressure produced by rotor ribs, and Hdis the pressure produced by rotor spacers.

## 12. The calculation air flow

The calculation of air flow is done after we solved the ventilation scheme regarding the air flows of different sections and after we calculated the aerodynamic resistances values and pressures values created by sources.

Bilateral radial-axial ventilation for an electrical machine in protected construction

We calculate the flows for the asynchronous electrical machine with bilateral radial ventilation from „figure 7.a”, the cooling air is driven only by the effect of ventilator of ends of rotor bars and of ribs and spacers on rotor.

The flows Qb, Qrand Qby sections III, II, and V, respectively I and IV from „figure 7.b”, are determined by equations:

Qb=y[m3/s]E95
Qr=1RaIIRaIIIQb2Hb+Hr[m3/s]E96
Q=Qb+QrE97

When the flow which enters and exit from the machine is less than the flow calculated for half of the machine it is necessary to adopt a solution by decreasing the equivalent resistances on the sections by sections increasing, or giving up the constructive solution adopted and finding other solutions.

The example of calculation for asynchronous electrical machine with bilateral radial ventilation:

• We calculate the aerodynamic resistance on every section, starting from the nominal data.

• Equivalent resistance by section I is:RatrI=i=14Rai

• Equivalent resistance by section II is:RatrII=i=512Rai

• Equivalent resistance by section III is:RatrIII=i=1318Rai

• Equivalent resistance by section IV is:RatrIV=i=1920Rai

By means of the computer program PANCIA (Program Analysis of Analog Circuits) we obtain the system solution. [7, 8, 9, 10]

 The unknowns of the systemV1 V2 V3 V4 V5 V6 V7 V8 V9 V10V11V12V13V14V15V16V17V18V19V20V21I21I22I23 The system of equations+ (+G1+G2)*V1+ (-G2)*V2= 0+ (-G2)*V1+ (+G2+G3)*V2+ (-G3)*V3= 0+ (-G3)*V2+ (+G3+G4)*V3+ (-G4)*V4= 0+ (-G4)*V3+ (+G4+G5+G13)*V4+ (-G5)*V5+ (-G13)*V15 = 0+ (-G5)*V4+ (+G5)*V5+ (+1)*I21 = 0+ (+G6)*V6+ (-G6)*V7+ (-1)*I21 = 0+ (-G6)*V6+ (+G6+G7)*V7+ (-G7)*V8= 0+ (-G7)*V7+ (+G7)*V8+ (+1)*I22 = 0+ (+G8)*V9+ (-G8)*V10 + (-1)*I22 = 0+ (-G8)*V9+ (+G8+G9)*V10 + (-G9)*V11 = 0+ (-G9)*V10 + (+G9+G10)*V11 + (-G10)*V12 = 0+ (-G10)*V11 + (+G10+G11)*V12 + (-G11)*V13 = 0+ (-G11)*V12 + (+G11+G12)*V13 + (-G12)*V14 = 0+ (-G12)*V13 + (+G12+G18+G19)*V14 + (-G18)*V20 + (-G19)*V21 = 0+ (-G13)*V4+ (+G13)*V15 + (+1)*I23 = 0+ (+G14)*V16 + (-G14)*V17 + (-1)*I23 = 0+ (-G14)*V16 + (+G14+G15)*V17 + (-G15)*V18 = 0+ (-G15)*V17 + (+G15+G16)*V18 + (-G16)*V19 = 0+ (-G16)*V18 + (+G16+G17)*V19 + (-G17)*V20 = 0+ (-G18)*V14 + (-G17)*V19 + (+G17+G18)*V20 = 0+ (-G19)*V14 + (+G19+G20)*V21 = 0+ (+1)*V5+ (-1)*V6= -E21+ (+1)*V8+ (-1)*V9= -E22+ (+1)*V15 + (-1)*V16 = -E23 Potential at nodesV1=-31.752799V2=-43.259396V3=-57.299053V4=-58.125131V5=-335.922113V6=864.077887V7=330.162981V8=293.853281V9=2393.853281V10=2357.543581V11=2193.074695V12=1471.273161V13=1469.399393V14=1425.531437V15=-58.851254V16=1441.148746V17=1439.724722V18=1439.473272V19=1428.010008V20=1427.879152V21=913.802203V22=0 Currents and voltages sides U1=31.752799U2=11.506597U3=14.039657U4=0.826077U5=277.796983U6=533.914906U7=36.309700U8=36.309700U9=164.468886U10=721.801534U11=1.873768U12=43.867956U13=0.726124U14=1.424024U15=0.251449U16=11.463264U17=0.130856U18=2.347715U19=511.729234U20=913.802203U21=-1200.000000U22=-2100.000000U23=-1500.000000 I1=0.365521I2=0.365521I3=0.365521I4=0.365521I5=0.108940I6=0.108940I7=0.108940I8=0.108940I9=0.108940I10=0.108940I11=0.108940I12=0.108940I13=0.256581I14=0.256581I15=0.256581I16=0.256581I17=0.256581I18=0.256581I19=0.365521I20=0.365521I21=0.108940I22=0.108940I23=0.256581 Balance of powersThe power generated:744.373309The power consumption: 744.373309 Nodal analysis resultsV1 = -31.752799V2 = -43.259396V3 = -57.299053V4 = -58.125131V5 = -335.922113V6 = 864.077887V7 = 330.162981V8 = 293.853281V9 = 2393.853281V10 = 2357.543581V11 = 2193.074695V12 = 1471.273161V13 = 1469.399393V14 = 1425.531437V15 = -58.851254V16 = 1441.148746V17 = 1439.724722V18 = 1439.473272V19 = 1428.010008V20 = 1427.879152V21 = 913.802203I21 = 0.108940I22 = 0.108940I23 = 0.256581

ANCIA is based on the modified nodal method (MNM) which is, due to its flexibility, the most used and the most efficient method of simulating the electrical circuits on computer. [7, 8] The system of the equations that describes the equivalent circuit as ventilation scheme is a nonlinear algebraic system because the constants of pressure have the expressions:
ΔHK=RaKQK2(UK=RKIK2)E98

For every nonlinear resistance, we considered the linear section of the characteristic U-I, so the circuit obtained is a linear resistive circuit.

In the example given, we found that the method used is good for calculating the ventilation systems of different types of electrical machines.

## 13. The thermal calculation. The heat transfer

Our thermal model is based on the analyses of the characteristic circuit providing almost instant calculation speeds. This allows the user to perform any calculation on time. Alternative numerical methods, for example, the analysis of finite elements [FEA] and calculable dynamic fluids [CFD], normally require some days/weeks to put together a three-dimensional model. The numeric mode may then take some hours (days for a complex problem of transit) to calculate a solution. 

The characteristic diagram (the characteristic circuit) is like an electrical scheme:

• The thermal resistance instead the electrical resistance

• Sources of power instead of current sources

• Thermal capacity instead electrical capacity

• Power flow through the resistance instead the current.

All thermal resistances and thermal capacity from the model Motor-CAD are automatically calculated from geometric dimensions and materials properties. The user must be familiar with complex phenomena of heat transfer such as the dimensional analysis of conversion. Motor-CAD automatically selects and solves most appropriate formulation for a given surface and selected cooling time. Motor-CAD configures efficient, precise and robust mathematical algorithms for forced and natural convection, for cooling liquid, for radiation and conduction. A vast library of lamination correlations and demonstrated turbulent convections is used to give precise models for all internal and external surfaces. The air gap model includes laminar, vortex and turbulent convection.

The calculation in stationary regime

Motor-CAD simplifies the analyses of output dates of the stationary regime by using schematic diagrams, simple and easy to read. 

In “figure 14” we observe the solved main scheme. This scheme is used very much by post processing calculation in stationary regime. The resistances and the power sources indicated in scheme are color coded as machine components shown in editors of cross sections. The vertically oriented resistances represent radial heat transfer and horizontally oriented resistances represent axial heat transfer. The resistances which contain the letter “C” are resistances of convection and resistances which contain the letter “R” are resistances of radiation. Two colors resistances represent interface resistances between two components.

The diagram also shows labels for each of its components (resistances and nodes). With a single click of a button the user can change the scheme to indicate the nodal temperature and the thermal resistance or power values, so will avoid elaborate images. The main scheme is very useful for visualizing the place where there may be restrictions on the heat transfer and what can be done to improve cooling.

The calculation in transient regime

The Motor-CAD software allows the thermal transition of the motor to be calculated and the outputs to be viewed in graphical format. The opposite graph shows how the test data is important to emphasize on the precision of calculation. Also, Motor-CAD includes facility analyses of the operation cycle. A major advantage of this is that it allows the motor to be directed to its full potential without overheating.

The transient regime can be modeled, so in any case the thermal capacities are automatically added to circuit. It is essential to consider the transient thermal analysis when using operational complex loads if the motor will be used to its full potential. The graph below shows a typical transient regime. In this case we have a repetitive duty cycle (three overloads followed by half loads).

The Motor-CAD software incorporates mathematical efficient and suitable algorithms for the cooling methods. All these, and facility of prescribe miscalculation, if is necessary, provides customization calculations so as to suit the unique construction methods.

## 14. Conclusions

Determining or providing the amount of exchanged heat per unit in time in given conditions of temperature is a problem to be taken into account for calculating heating and ventilation for the electrical machines. Were presented the basic sizes of the heat transfer studied in the literature and the distinct forms of heat transfer, the conduction, the radiation and the convection, the basic equations and rules of conduction, radiation and convection. Between the thermal sizes and the electrical sizes we showed a correspondence using the method of thermal resistances and concentrated losses.

By analogy between a network of cooling ventilation and a nonlinear resistive circuits were determined the values of the cooling fluid flows of an electrical machine.

Also by analogy, the temperature drop in the mass or on the surface of an object, corresponding the caloric flow (or loses), was determined by the product between the thermal resistance of the object opposed to the caloric flow in the mass or on surface and loses.

Solving the thermal schemes, were done analogous with solving the electrical schemes, as in electrical circuits of conexions between thermal resistances are series, parallel or mixed.

As we said the exhaust heat from the electric machine is as important as and phenomena related to the proper functioning of the machine. Choosing the most appropriate ventilation is done even by the assignation of electromagnetic stresses.

Were showed the equivalent resistances,andthe aerodynamic schemes used were solved according to the conexions between aerodynamic resistances (series, parallel or mixed).

The discussed example for axial-radial bilateral ventilation in an electrical machineunder protected construction for which the air circulation is symmetrical towards the mid of machine and it is sufficient to calculate ventilation only for one part, which is half of the machine.

After calculating the aerodynamic resistances, were calculated the pressures and air flows using the computer program PACEN (Program of Analysis of Nonlinear Electrical Circuits), and were obtained the system solutions.

Motor-CAD is an example for the analytical analysis of the characteristic circuit, is a program dedicated to electrical motors and generators and has all the indications on limiting design effort.

• The problem of defining time (minutes to hours/days/weeks)

• The calculating rate (instantly to hours/days)

• The postprocesing time (instantly to hours)

• Easy to use.

Motor-CAD allows optimization of the cooling system and identifies vulnerable points of design and/or problems of fabrication; it checks if a provided motor is suitable for a particular application.

The final conclusion for using Motor-CAD is that the program is a quick methoddesigning for motors and electrical generators of thermal point of view, taking into account the fact that the thermal analogous circuit of the electrical machines is as important as designing the electromagnetical, mechanical.

Written By

Otilia Nedelcu and Corneliu Ioan Sălişteanu

Submitted: December 7th, 2011 Published: October 24th, 2012