1. The document discusses heating and cooling of electric motors, with heating caused by losses in the steel core, windings, and bearings, and cooling facilitated by ventilation and a motor's heat dissipation capacity. 2. It provides equations for calculating a motor's heating and cooling time constants based on its heat production, heat capacity, and heat transmission coefficient. 3. Operating a motor above its permissible temperature rise rating can cause damage from thermal breakdown of insulators or deterioration of insulating materials.

1. Heating and Cooling of Electric Motors for Electric Drives

2. INTRODU
CTIONOD
UCTION

3.  Losses constitute heat
 Losses in the steel core
 Losses in the motor's windings
 Losses due to bearing friction
 Heating is a function of losses while cooling is function of heat dissipation
facilities such as:
 Ventilation to air, liquids and solids

4.  Under steady state condition the temperature rise is the reached when the rate of
heat production and dissipation are equal.
 Electric machines have a permissible temperature rise rating which must not be
exceeded
 The rating is a function of the thermal strength of the insulators used

5.  Damage of the motor
 Thermal breakdown of insulating material (short circuit)
 Fire outbreaking
 Deterioration of Insulating materials

6. 1. Ambient air density.
2. Motor enclosure.
3. Airflow over motor.
4. Frame surface area.

7.  Heat Balance Equation for an Electric Motor under Constant Load
Mathematically the heat balance equation for an electric motor under constant load
is given by the expression below:
𝑄𝑑𝑡 = 𝐴τ𝑑𝑡 + 𝐶𝑑τ (joules)
Where Q = total heat evolved by the motor per unit time (joules/sec)
C = Heat capacity of the electric motor (joule/oC)
A = Coefficient of heat transmission (joule/sec.oC)
τ= Rise in temperature of the motor above the ambient temperature.
𝜏 = 𝑄
𝐴 1 − 𝑒−
𝐴𝑡
𝐶 + τ𝑜 𝑒−
𝐴𝑡
𝐶

8.  Heating time constant is the time it takes for the motor to reach 0.623 of its final
steady temperature rise.
 Heating time constant = C/A
 Cooling time constant is defined as the time taken to cool the motor down to 0.368
times its initial temperature rise above ambient temperature.

10. 1. The machine is considered to be a homogeneous body having a uniform
temperature gradient.
2. Heat dissipation taking place is proportional to the difference of temperature of
the body and surrounding medium.
3. The rate of dissipation of heat is constant at all temperatures.

11. The heat developed = p.dt
The heat stored = Gh.dΘ
The heat dissipated = SΘλ.dt
Total heat developed = ?
Θm =
𝑝
𝑆λ
and 𝜏 =
𝐺ℎ
𝑆λ
Θ = Θm 1 − 𝑒−
𝑡
𝜏
 P =heat developed, joules/sec or watts
 G =weight of active parts of machine, kg
 h =specific heat per kg per oC
 S = cooling surface, m2
 λ = specific heat dissipation (or) emissivity, J
per sec per m2 of surface per oC difference
between surface and ambient cooling medium
 Θ = temperature rise,
 Θm =final steady temperature rise, oC
 t = time, sec
 𝜏 = heating time constant, seconds
 𝜏 ' = cooling time constant, seconds

12. The enclosure of a l0kW motor is equivalent to a cylinder of 70 cm
diameter and 100 cm length. The motor weighs 500 kg. Assuming
that the specific heat is 700 J /kg/°C and that the peripheral surface
of the enclosure of the motor alone is capable of heat dissipation of
12.5 W/sq. m/°C, calculate the heating time constant of the motor
and its final temperature rise. Efficiency of motor is 90 per cent.
A 10-kW motor has a heating time constant and cooling time
constant of 45 and 70 min, respectively. The final temperature
attained is 60°C. Find the temperature of motor after 45 min full-
load run and then switched of for 30 min.

13. The heating time constant of a 80-kW motor is 60 min. The
temperature raise is 65°C when runs continuously on full load. Find
the half-hour rating of motor for the same temperature raise.
Assume that the losses are proportional to the square of the load
and the motor cools to ambient temperature between each load
cycle