Thermal Design Aspects of Electrical Machine

Introduction

Heat transfer always occurs when there is a temperature difference in a system. The temperature
difference evens out naturally as heat transfers from the higher temperature to the lower
according to the second law of thermodynamics. In electrical machines, the design of heat
transfer is of equal importance as the electromagnetic design of the machine, because the
temperature rise of the machine eventually determines the maximum output power with which
the machine is allowed to be constantly loaded. As a matter of fact, accurate management of heat
and fluid transfer in an electrical machine is a more difficult and complicated issue than the
conventional electromagnetic design of an electrical machine.
The problem of temperature rise is twofold: first, in most motors, adequate heat removal is
ensured by convection in air, conduction through the fastening surfaces of the machine and
radiation to ambient. In machines with a high power density, direct cooling methods can also be
applied. Sometimes even the winding of the machine is made of copper pipe, through which the
coolant flows during operation of the machine. The heat transfer of electrical machines can be
analysed adequately with a fairly simple equation for heat and fluid transfer. The most important
factor in thermal design is, however, the temperature of ambient fluid, as it determines the
maximum temperature rise with the heat tolerance of the insulation.
Second, in addition to the question of heat removal, the distribution of heat in different parts of
the machine also has to be considered. This is a problem of heat diffusion, which is a
complicated three-dimensional problem involving numerous elements such as the question of
heat transfer from the conductors over the insulation to the stator frame.
The distribution of heat in the machine can be calculated when the distribution of losses in
different parts of the machine and the heat removal power are exactly known. In transients, the
heat is distributed completely differently than in the stationary state. For instance, it is possible to
overload the motor considerably for a short period of time by storing the excess heat in the heat
capacity of the machine.
Following effect might occur due to the overheating of the machine
 The lifetime of insulation can be estimated by statistical methods only. However, over a
wide temperature range, the lifetime shortens exponentially with the temperature rise Θ
of the machine. A rise of 10K cuts the lifetime of the insulation by as much as 50%
 A similar shortening of the lifetime applies also to the bearings of the motor, in which
heat-resistant grease can be employed. In critical drives, oil mist lubrication can be used,
in which case the oil is cooled elsewhere and then fed to the bearings Even ball bearings
can be used at elevated speeds if their effective cooling is ensured for instance by oil
lubrication.
 The temperature rise of the winding of an electrical machine increases the resistance of
the winding. A temperature rise of 50K above ambient (200C) increases the resistance by
20% and a temperature rise of 135K by 53%. If the current of the machine remains
 unchanged, the resistive losses increase accordingly. The average temperature of the
winding is usually determined by the measurement of the resistance of the winding.
So due to excess internal temperature hot spot may occurs in the inner part of the core and
conductors
Heat Removals
The heat is removed by convection, conduction and radiation. Usually, the convection through
air, liquid or steam is the most significant method of heat transfer. Forced convection is,
inevitably the most efficient cooling method if we do not take direct water cooling into account.
The cooling design for forced convective cooling is also straightforward: the designer has to
ensure that 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. If the
motor is flange mounted, a notable amount of heat can be transferred through the flange of the
machine to the device operated by the motor. The proportion of heat transfer by radiation is
usually moderate, yet not completely insignificant. A black surface of the machine in particular
promotes heat transfer by radiation. All of the following means of heat transfer will be found in
the electrical machine
 Conduction
 Convection
 Radiation

Conduction
The conduction is heat transformer by diffusion in stationary material due to temperature
gradient . The medium can be solid , liquid and gasses .. The diffusion processes is through
substance. The conduction phenomena occur in heat transfer from inner core to the surface of
core,inner part of the conductor to the surface of the conductor heat transfer in liquid and gasses
medium etc. t
S
2


1
Q

Fig.3.1 Heat flow due to conduction

So Heat flow by means of the conduction

 𝜃 −𝜃 𝑡 𝑡
(𝑄) = 𝑤ℎ𝑒𝑟𝑒 𝑅 =𝜌 =
𝑅 𝑆 𝜎 𝑆

Where th and th are thermal resistivity and thermal conductivity of the material and it
depends upon the material properties is number of the free electrons.

(𝜃 − 𝜃 ) 𝑆
(𝑄) = 𝑜𝑟 𝑡𝑒𝑚𝑒𝑝𝑟𝑎𝑡𝑢𝑟𝑒 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑑𝜃 = (𝑄) 𝑅
𝜌 𝑡
( )
Hence heat dissipation per unit surface area (𝑞) = 𝑊𝑎𝑡𝑡/𝑚

The thermal resistivity depends on the materials as shown in below table

Material Thermal resistivity -m

Air 20
Micanite 8
Transformer oil 0.5
Transformer sheet along lamination 0.02
Transformer sheet across lamination 0.05-0.1
Aluminium 0.005
Copper 0.0026
Thermal conductivity of the coolant is much lesser than metal . So in practice the heat
conducted in interface is negligible

(𝑞) =𝜎 (𝜃 − 𝜃 ) 𝑊𝑎𝑡𝑡/𝑚

𝑤ℎ𝑒𝑟𝑒 𝜎 = (50 − 1500) 𝑓𝑜𝑟 𝑜𝑖𝑙 𝑎𝑛𝑑 𝜎 = (10 − 50) 𝑓𝑜𝑟 𝑎𝑖𝑟

Convection
Convection is heat transfer between either hot surface and cool moving fluid or cold
surface and hot moving fluid . Convection occurs in Liquid and gaseous

t
S
2

2

1
Q

Fig.3.2 Heat flow due to convection

 (𝑄) = 𝛼 (𝜃 − 𝜃 )𝑆 𝑤𝑎𝑡𝑡 (𝑞) = (𝜃 − 𝜃 ) 𝑤𝑎𝑡𝑡/𝑚

Liquid and gas particles near heated body will heated and becomes lighter and moves giving
space for the cool particles which in turn get heated and moves again. This process is called
natural convection . The convection is complicated phenomenon depends on velocity of coolant
, pressure , viscosity , gravitational height orientation configuration and condition of heated
surface , temperature difference , density so on .

For the transformer oil and vertical plain of height less than 1m
. . .
(𝑞) = 2𝑃 𝜃 𝑊/𝑚 or (𝑞) = 1.3𝑑 − 0.25 𝜃 𝑊/𝑚

Where P is pressure ,  is temperature difference and d is the diameter of the vertical tubes
Similarly for the forced cooling

= (1 + 𝐾 𝑉) 𝑤𝑎𝑡𝑡/𝑚
Where Kv depends on blast and V is the velocity of the coolants

Radiation
Radiation is the process of heat transfer from the heated surface of area S , temperature 2
and ambient temperature 1 via electromagnetic radiation . . The heat dissipation due to
radiation is Stefan and Boltzman law
(𝑞) = 5.7𝑥10 𝑒(𝑇 − 𝑇 ) 𝑤𝑎𝑡𝑡/𝑚

Where e , T2 and T1 are emissivity surface temperature in Kelvin and Ambient Temperature
in Kelvin . The value of emissivity is 1 for the perfect black body

(𝑞) = (𝜃 − 𝜃 ) 𝑤𝑎𝑡𝑡/𝑚

Where  is specific heat dissipation due to radiation and it depends upon the surface
nature of the heated body .

Newton’s Law of cooling

When the machine is loaded continuously loaded heat loss occurs continuously and
temperature rises . After some interval it attains its final steady state temperature . To
maintain the final steady state temperature after attaining final value the total heat loss
must be balanced by heat dissipation by means of convection and radiation
𝑄 = ( +  )𝜃𝑆 𝑤𝑎𝑡𝑡
Specific heat dissipation due to both convection and radiation depends upon the surface
Surface  + = watt/m2
 Polished Metal 8.2
Aluminium Paint 10.8
Oil Paint 30

Calculation of the internal Temperature rise

Lets q be the heat flow per unit volume . The path of heat flow is only in the direction of
x axis there will be no heat flow form the bottom , top and front and back sides of the
rectangular conductor as shown above figure .

Fig For internal heat flow

The heat encountered to the strip of thickness dx at x distance from the centre of the bar is
𝑄 = 𝑞𝑥𝐿𝑊

The thermal resistance of the elementary heat flow path 𝑅 =𝜌

Hence temperature difference of this elementary path

𝑑𝑥
𝑑𝜃 = 𝑄 𝑅 = 𝑞𝑥𝐿𝑊𝜌 = 𝑞𝜌𝑥𝑑𝑥
𝐿𝑊

Since the direction of heat flow is bidirectional 𝜃 = ∫ 𝑑𝜃 = 𝑞𝜌 ∫ 𝑥𝑑𝑥 = 𝑞𝜌

Hence the temperature variation from the inner part to the surface is parabolic and the

inner temperature =𝜃 + 𝑞𝜌
If the core is made of laminated sheet then the heat dissipation will be different along the
lamination and the across the lamination for the small height the dissipation along the
lamination will be better but for large height heat flow across the the lamination will be
better
 Figure representing heat flow along and across the lamination

Here 𝜃 = 𝑞𝜌 𝑎𝑛𝑑 𝜃 = 𝑞𝜌

Thermal resistivity along the lamination is less than across the lamination so chances of
formation of Hot spot will be less if the height of sheet is less . But if the height of sheet
will be more than the temperature difference between surface and the centre along the
lamination will be more there will be chances of the formation of the hot spots . So for the
larger machine axial ventilation isn’t enough and radial ventilation is also preferred

Thermal resistance of winding

When the current carrying conductor is coated with insulation then the effective thermal
resistivity of the conductor and winding will be

𝜌 = 𝜌 (1 − 𝑆 ) where S is Space Factor which is the ration of Copper area to total

winding area .𝜌 is the thermal resistivity of the insulation

Temperature gradient of the conductor placed in slot

Fig. for the heat flow in a conductor placed in slot

Let Iz is the current flowing in the conductor of Cross-sectional area az placed in slot .
The axial length of the slot is L . Assume that the heat flows along the length of the
material from its centre to end ring . Lets consider the elementary heat flow path of
thickness dx at x distance from the middle if the conductor.
Total Heat Loss =Iz2R =Iz2 x/az
The thermal resistance of the elementary path 𝑑𝑅 = 𝜌
Temperature Gradient 𝑑𝜃 = 𝐻𝑒𝑎𝑡 𝑙𝑜𝑠𝑠 𝑥 𝑇ℎ𝑒𝑟𝑚𝑎𝑙 𝑅𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 𝐼 𝜌
Assume the insulation thickness is very high and no heat transfer via insulation and heat
conducted through conductor upto end ring can be computed by integrating above equation

𝜃 = ∫ 𝑑𝜃 = 𝜌𝜌 𝛿 ∫ 𝑥𝑑𝑥 = 𝜌𝜌 𝛿
If the end rings are heated then the heat flows via insulation to armature yoke

Heat Flow Via Insulation

Let ds is the depth of slot which is covered by conductor and Ws is width of the slot .
Since the air has high thermal resistivity so there will no heat flow via open side slot .
Area for the heat flow path A=L(2ds+Ws)
Thermal Resistance Ri=i ti/L(2ds+Ws) where ti is thickness of the insulation
Temperature Gradient 𝑑𝜃 = 𝐻𝑒𝑎𝑡 𝑙𝑜𝑠𝑠 𝑥 𝑇ℎ𝑒𝑟𝑚𝑎𝑙 𝑅𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 𝐼 𝜌
( )

Temperature Gradient 𝑑𝜃 = 𝜌𝛿 𝑎 𝐿𝜌 == 𝜌𝛿 𝑎 𝜌
( ) ( )
For the Turbo alternator ds  Ws
𝑡
𝑑𝜃 = 𝜌𝛿 𝑎 𝜌
2𝑑
Heating and Cooling of Machine
Heating
Heating of machine relates with how fast the machine reaches its final steady state
temperature and the factors affect the it. If the f and I are the initial and final temperature
then (t) will be described by below equation

𝜃(𝑡) = 𝜃 1−𝑒 +𝜃𝑒

Where 𝜏 is the heating time constant of the machine
Where 𝜏 =

G = weight of the active part of the machine in Kg
.h= Specific heat J/Kg 0C
= Specific heat dissipation W/m2
S= Cooling surface m2

If the Machine is start from cold then 𝜃(𝑡) = 𝜃 1−𝑒

Cooling
When the machine is running with full load then it attains its final steady state temperature ,
now if the machine is stops then the temperature exponentially starts to decrease .
 𝜃(𝑡) = 𝜃 𝑒
Cooling
A processes by means of which heat resulting from losses occurring in the machine is given
up to a primary coolant by increasing its temperature. The heated primary coolant will be
light in density so it is replaced by cool coolant or it will cooled by secondary coolant
in some cooling chamber or heat exchanger . A method of the cooling in which the coolant
is drawn from the surrounding then passed to machine and then return to the surrounding is
called Open circuit cooling
Whereas in closed circuit cooling the primary coolant is circulated in the different parts of the
machine if necessary with heat exchanger . The heat transferred from primary coolant to
secondary coolant either via structural parts or heat exchanger .
Methods of cooling
The factors which determines the size of the machine for the given duty cycle is the
temperature rise of the machine due to various losses . The maximum allowable temperature rise
in the various parts of the machine is standardize by the IEC and accordingly the insulation
will be chosen considering the operating voltage and temperature . Hence the after the
development of the new insulation materials the suitable for the higher temperature the
standards can be revised . Another approach to operate the machine in the higher temperature
is having better cooling system .

Small machine ie having fractional horse power rating the natural cooling is employed
where as for the large machine the artificial cooling system are found to regulate the
temperature of various parts of machine . In the most cases the cooling of the electrical machine
is done by air stream and this cooling is called ventilation where as for the speed machine
such as turbo alternator hydrogen is used for the cooling .
Ventilating System
The ventilating system of the electrical machine is either induced ventilation or forced
ventilating system
Induced ventilating system
The ventilation of the machine is induced if the fan produce decreased air pressure inside the
machine causing the air to be sucked into the machine under the external atmospheric pressure,
then the air is pushed out by the fan into atmosphere. The following figure show the induced
ventilation using the external and internal fan. The self or induced ventilation is most common in
small machine
 Fig. Self or induced ventilation
Forced ventilation
The ventilation of the machine is said to be forced if the fan sucks the air from the atmosphere
and forced it into the machine and finally pushed out the atmosphere . The forced ventilation
system is used in large machine . The air also carries the heat of fan so to limit the temperature
the volume per sec requirement is more . Below figure shows the scheme for external and
internal fan based forced ventilating system

Radial and Axial Ventilating system

Fig. forced ventilation

The ventilating system can be divided on the basis how the air passed inside the machines . It
will be mainly radial , axial and combined type .
Radial Ventilating system
This system is most commonly employed because the movement of rotor induced a natural
centrifugal movement of air which may be augmented by provision of fans if required. Below
figure show the radial ventilating system for the small core length.The end shield are shaped to
guide air on the overhang and then on to the back of core .The scheme is suitable for the
machine upto 20KW .For the large core length the core is subdivided in order to have radial
ventilating ducts a shown below figure .The core is normally divided to 40 to 89mm thick with
ventilating duct of 10mm between them .The advantage of this system is minimum energy losses
in the ventilation and uniform temperature rise in axial direction . The main disadvantage of this
scheme is it increase the length of core
 Fig for Radial Ventilation

Axial Ventilating system

The axial ventilating system is adopted in small power and high speed induction machine Axial
ducts are arranged in rotor or in both rotor or stator , Generally cooling system holes are
punched where the considerable temperature rise occurs .The main disadvantage of this scheme
is non uniform heat transfer ie the temperature of the air will different in the cooling ducts

Fig for axial ventilation

Quantity of the cooling medium employed

Let
Q= Losses to be dissipated in KW
i=Inlet Temperature of the coolant
= Temperature rise of the cooling medium ‘
H= Barometric Height mm of Mercury
P = Pressure and Hear to be carried away =Q KW=Qx103
cP= Specific heat of air at constant pressure J/kg0C
Heat =Weight x Specific heat x Temperature
Heat / sec = Weight(Kg/sec) x Specific heat x Temperature
𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑎𝑖𝑟 𝑄𝑥10
= 𝐾𝑔/𝑠𝑒𝑐
𝑆𝑒𝑐 𝐶 𝑥𝜃
Let V be the volume of one Kg air at NTP
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑎𝑖𝑟 𝑎𝑡 𝑁𝑇𝑃 𝑄𝑥10
= 𝑥𝑉 𝑚 /𝑠
𝑆𝑒𝑐 𝐶 𝑥𝜃
From Kelvin law =
( )
Hence = 𝑉 𝑚 /𝑠𝑒𝑐
Cp = 0-995 and V = 0.775 m3 assumed; then
𝑉 0.78 𝑄 (𝜃 + 273) 760
= 𝑚 /𝑠𝑒𝑐
𝑆𝑒𝑐 𝜃 273 𝐻
The capacity of the fan required = Pfan = P Va /ηfan, W
Similar calculations can be made for the volume of hydrogen or air or oil used for cooling the
machine.
For Hydrogen cooling
𝑣 = 0.8 ( ) where H for the hydrogen is 2000 -2500mm
.
For water cooling 𝑣 = = 𝑙𝑖𝑡𝑒𝑟 /𝑠𝑒𝑐
.
.
For oil cooled𝑣 = 𝑙𝑖𝑡𝑒𝑟 /𝑠𝑒𝑐
( . . )
Problem 1
A copper bar of 12 mm diameter is insulated with micanite tube which fits tightly around the bar
and into the rotor slot of induction motor. The micanite tube is 1.5 mm thick and its thermal
resistivity is 8m calculate the losses that will pass copper bar to iron if the temperature difference
allowed is 250C. The axial length of the bar is 0.2m
Solution
Average Area for the path of the heat flow
S=(d+t)l=(12+1.5)x10-3x0.2=8.48x10-3 m2
Thermal resistance for the heat flow path
Rth=t/S=8x1.5x10-3/8.48x10-3 =1.415
Hence the permitted heat flow
Q=d/ Rth= 25/1.415=17.67 Watt

Problem 2
The thermal resistivity of a stack of laminations measured along the lamination is 0.02.The
iron loss is 40Kw/m3.Calulate the temperature difference between hot spot and outside surface
for following cases Length of stack measured along lamination =0.2m , Length of stack
measured across lamination =0.1m (ii) Length of stack measured along lamination = 1m ,Length
of stack measured across lamination =0.1m .The thermal resistivity across the lamination is 40
time more than along the lamination
Solution
We have temperature difference across and along
laminations are 𝜃 = 𝑞𝜌 𝑎𝑛𝑑 𝜃 = 𝑞𝜌
For Case1
Temperature rise along lamination
𝑦 40000𝑥0.02𝑥0.1
𝜃 = 𝑞𝜌 = =4 𝐶
2 2
Temperature difference across the lamination
𝑥 40000𝑥0.8𝑥0.05 Comments
𝜃 = 𝑞𝜌 = = 40 𝐶 For small height of the sheet the
2 2
For Case2 temperature different between hot
Temperature rise along lamination spot and outer surface isn’t
𝑦 40000𝑥0.02𝑥0.5 excessive but for sheet of large
𝜃 = 𝑞𝜌 = = 100 𝐶
2 2 height temperature different
Temperature difference across the lamination between hot spot and outer surface
𝑥 40000𝑥0.8𝑥0.05 is excessive so in this situation
𝜃 = 𝑞𝜌 = = 40 𝐶
2 2 radial ventilating duct is essential
Problem 4
The total losses in a 40MVA transformer are 200KW .The oil for the cooling is circulated by
the pump. .The oil after taking up heat from transformer tanks goes to the radiator where it is
cooled by the circulation of water . Calculate the amount of the oil if the temperature raise of oil
is 200C .Also calculate the amount of the water required for the cooling oil in the radiators if the
temperature difference of water in radiator is 100C.Assume the 20% of the heat will dissipated
from the wall of tank.

Solution
For Oil
Heat taken away by the oil Q= 0.8x200KW= 160KW
Temperature difference of oil =200C Assume Cp for oil=0.4
. .
The volume of oil required 𝑣 = = = 4.8
( . . ) .
For water
Heat taken away by the oil Q= 0.8x200KW= 160KW
Temperature difference of water =100C
. .
The volume of water required 𝑣 = = = 3.84

Problem 4
A 400 KVA transformer has its maximum efficiency at 80% of the full load .During short full
load heat run the temperature rise after one and two hour is observed to be 24 0C and 340C
respectively .Find the thermal time constant and steady state temperature rise .
If by the use of the fan specific heat dissipation ( watt/m20C is improved by 15% ,Find the
new KVA rating possible (i) for the same final steady state temperature as before (ii)if
allowable temperature raise is considered 50 0C

Solution

We have equation for the temperature rise 𝜃(𝑡) = 𝜃 1−𝑒

S0 24 = 𝜃 1−𝑒 and 34 = 𝜃 1−𝑒 and solving these two equation 𝜃 = 41.2 𝐶

Maximum efficiency occur at 80% of the full load
Let Pc= copper loss at full load =x2I2R =x2Pc
Pi= Iron loss at full load
Where x is percentage of full load
0.82Pc=Pi
Pc=1.563Pi
Total Full load loss =Pc+Pi=2.563Pi
 If the fan is used heat dissipation will improved by 15% for the same final steady state
temperature
Allowable full load loss will 15% more =1.15x 2.563Pi =2.947Pi
Allowable full load Cu Loss Pc =2.947Pi-Pi=1.947Pi
Increase in Cu loss =1.947Pi/1.563Pi=1.246
Let x be the overloading x2Pc=1.246Pc x=1.116

New KVA = 1.116X400=446.4KVA

If allowable temperature is 500C and fan is also used then due to fan the heat dissipation area
will increased by 15 %
New allowable loss =(50/41.2)x1.15xallowable losses in 400 KVA
=1.4x2.563Pi
New copper loss Pc =3.577Pi-Pi=2.577Pi
Increase in copper loss =2.577Pi/1.563Pi=1.65
Let x be the overloading x2Pc=1.65Pc x=1.26
New KVA = 1.28X400=512KVA

Problem 5
A 15 MVA Transformer has an iron loss of 80KW and full load copper loss 120KW .The tank
dimension are 3.5x3x1.4m .The transformer oil is cooled by water of 3liter per sec in cooling
coil .Estimate the average temperature raise of the transformer tank if the difference in
temperature of inlet and outlet of water is 150C.The specific loss dissipation of the tank wall is
10W/m20C.
Solution
Total Loss at full load = Q=Pi + Pc =80+120=200KW
Out of this total loss some losses will dissipated by tank and some will be carried away by
oil
Temperature difference of water in cooling tower =150C
.
The volume of water required 𝑣 =

Heat taken away by water =𝑄 = = = 187.5𝐾𝑊

. .
So the heat dissipated by the tank =200-187.5=12.5KW
The effective tank area for the heat dissipation neglecting the top and bottom
=2x3.5(3+1.4)= 30.8 m2
Hence temperature raise of the tank
𝑄 12.5𝑥1000
𝜃= = = 40.5 C
𝑆 30.8𝑥10