Lecture 4: Losses and Heat Transfer

1 / 26 Lecture 4: Losses and Heat Transfer ELEC-E845 Electric Drives (5 ECTS) Marko Hinkkanen Aalto University School of Electrical Engineering Autumn 215

2 / 26 Learning Outcomes After this lecture and exercises you will be able to: Explain the dependency between the losses and the load Sketch the waveform of the motor temperature rise, when the load varies (and the thermal time constant is known) Choose a motor for a periodic duty cycle based on the average temperature rise

3 / 26 Outline Introduction Elementary Heat Transfer Model Motor Selection Based on the Average Temperature Rise

4 / 26 Rated Values Rated values will be marked with the subscript N Machine can operate continuously in the rated operating point Rated voltage U N is determined by insulation Rated current I N corresponds to the maximum acceptable resistive losses Rated field (and therefore k fn ) is determined by magnetic saturation Rated speed is obtained using the previous rated values ω N = U N R a I N k fn U N k fn Rated power refers to the output (mechanical) power, P N = T N ω N

5 / 26 About Thermal Limitations and Thermal Modelling Temperature rise is limited in motors (and power converters) Rated current values inherently include information of thermal limits Almost all motors are rated for continuous duty (S1 duty), while the load varies periodically in many applications Periodic duty is transformed into thermally equivalent S1 duty in order to choose an S1-type motor DC motor is used as an example, but the same concepts can be used for AC motors as well Lumped capacity thermal model can also be used for modelling the temperature rise of other devices

6 / 26 Outline Introduction Elementary Heat Transfer Model Motor Selection Based on the Average Temperature Rise

7 / 26 Elementary Heat Transfer Model Spatially uniform temperature T [K] Internal heat generation p d [W] Thermal capacitance C th = c p m [J/K] where c p [J/K/kg] is the specific heat capacity and m is the mass of the body Thermal resistance R th = 1 α th S [K/W] where α th [W/K/m 2 ] is the heat transfer coefficient and S is the surface area Ambient temperature T a Temperature T Power loss p d Thermal capacitance C th Thermal resistance R th of a surface

8 / 26 Heat Transfer Mechanisms Temperature differences in a system cause heat transfer Heat transfer coefficient consists of three terms α th = α cd + α cv + α em Conduction Between objects that are in physical contact, α cd Convection Due to free or forced fluid motion (typically air or liquid), α cv Radiation Electromagnetic radiation, α em

Lumped Thermal Capacity Model Temperature rise θ = T T a Generated heat = removed heat + stored heat Differential equation p d dt = 1 R th θdt + C th dθ p d = 1 dθ θ + C th R th dt Corresponding transfer function θ(s) p d (s) = R th 1 + τ th s, τ th = R th C th p d C th R th θ Thermal model i C u R Analogous electrical circuit 9 / 26

1 / 26 Step Response Temperature rise in steady state is θ = R th P d Response to the loss step can be solved from the differential equation ( ) θ(t) = θ + (θ θ ) 1 e t/τ th where τ th = R th C th is the time constant and θ = θ() θ θ θ θ.63θ θ θ P d.37θ τ th t τ th t

11 / 26 About Thermal Models in Electric Motor Drives Electric motors Temperature rise of winding insulations is the limiting factor: increase of 1 K in the operating temperature halves the lifetime Thermal time constants increase with the motor size (from a few minutes to hours in typical industrial motors) Thermal resistance may depend on the operating speed More detailed models are needed for accurate temperature-rise analysis Power converters Similar lumped capacity thermal model can be used Thermal time constants are short (typically seconds or tens of seconds)

12 / 26 Outline Introduction Elementary Heat Transfer Model Motor Selection Based on the Average Temperature Rise

13 / 26 Losses in Electric Motors Power losses = heat generation Resistive losses in conductors (aka copper losses) Core losses in the magnetic circuit (aka iron losses) Mechanical losses Additional losses (all other losses) We will consider only the resistive losses in the following Resistive losses are proportional to the square of the current p d = R a i 2 a Current i a = T M /k f depends on the torque and the flux factor Resistances depend on temperature (.4%/K)

14 / 26 Rated Values Refer to Continuous Duty Rated (catalogue) values of motors refer to continuous (S1) duty type Motor works at the constant load T N and speed ω N Constant resistive losses P dn = R a IN 2 = R atn 2/k fn Thermal equilibrium is reached with temperature rise θ N = R th P dn

15 / 26 Motor Selection for Periodic Duty S1-type motors are selected for periodic duty as well Resistive losses p d = R a i 2 a vary as a function of time Maximum temperature rise θ max should be roughly θ N Too low temperature rise leads to a big and expensive motor Too high temperature rise shortens the motor lifetime

16 / 26 Average Temperature Rise and Average Losses Average temperature rise over the period t c θ av = 1 tc θ dt t c Average losses over the period t c p d,av = 1 tc p d dt = 1 tc R a ia 2 dt = R a Ia,rms 2 t c t c depend on the rms current I a,rms = 1 tc ia 2 dt t c

Example: Temperature Rise Inside a Motor i a /I max 1 t c = 5 min I a,rms /I max Thermal time constant τ th = 15 min Cycle length t c = 5 min τ th much longer than t c θ max θ av in steady state Motor selection could be made based on I a,rms.5 θ/θ max 1.5 θ av /θ max 2 4 6 t (min) 2 4 6 t (min) 17 / 26

Example: Temperature Rise Inside a Power Converter i a /I max t c = 5 min 1 Thermal time constant τ th = 2 s Cycle length t c = 5 min τ th much shorter than t c θ max 4θ av Converter selection should be based on I max.5 θ/θ max 1.5 I a,rms /I max 5 1 15 t (min) θ av /θ max 5 1 15 t (min) 18 / 26

Steady-State Average Temperature Rise θ max θ av holds in steady state, if the thermal time constant τ th is much longer than the cycle length t c Steady-state temperature rise in S1 duty θ N = R th P dn = R th R a I 2 N Steady-state average temperature rise in periodic duty θ av = R th P d,av = R th R a I 2 a,rms θ av < θ N leads to the criterion I N > I a,rms (or equivalently T N > k fn I a,rms ) 19 / 26

2 / 26 Selection Criterion in the Full-Field Region Motor is assumed to operate only in the full-field region Current is proportional to the torque k f = k fn and ω M ω N i a = T M k fn Rms current becomes I a,rms = 1 1 tc T k fn t M 2 dt = 1 T M,rms c k fn I N > I a,rms can be expressed as T N > T M,rms in the full-field region

21 / 26 Selection Criterion in the Field-Weakening Region Motor is assumed to operate only in the field-weakening region k f = k fnω N ω M and ω M > ω N Current is proportional to the mechanical power i a = T M k f = T Mω M k fn ω N = p M k fn ω N Rms current becomes I a,rms = 1 1 tc p k fn ω N t M 2 dt = 1 P M,rms c k fn ω N I N > I a,rms can be expressed as T N > P M,rms /ω N in the field-weakening region

22 / 26 Combined Selection Criterion Including Both Regions If both regions are used in the period, the previous criteria can be combined Motor is assumed to operate in the full-field region at t =... t1 in the field-weakening region at t = t1... t c Effective torque becomes 1 t1 T M,ef = TM 2 t dt + 1 1 tc c t }{{} c ω 2 pm 2 dt N t }{{ 1 } ω M ω N ω M >ω N I N > I a,rms can be expressed as T N > T M,ef

23 / 26 Example: What is the Effective Torque of the Motor? Speed, torque, and power profiles of the motor are known Nominal speed of the motor is ω N = 25 rad/s ω M (rad/s) 2 15 1 5 T M (Nm) 2 1 1 2 p M (kw) 4 2 2 4 2 5 2 1 t (s) t (s) t (s)

24 / 26 Example: What is the Effective Torque of the Motor? Speed, torque, and power profiles of the motor are the same as in the previous example Nominal speed of the motor is ω N = 15 rad/s ω M (rad/s) 2 15 1 5 T M (Nm) 2 1 1 2 p M (kw) 4 2 2 4 2 5 ω N 2 1 t (s) t (s) t (s)

Example: Temperature Rise in a Selected Motor i a /I N 2 t c = 5 min I a,rms /I N Thermal time constant τ th = 15 min Cycle length t c = 5 min I a =.5I N for 4 min I a = 2I N for 1 min Rms current I a,rms = I N Maximum temperature rise is θ max 1.1θ N in this case 1 θ/θ N 1.5 θ av /θ N 2 4 6 t (min) 2 4 6 t (min) 25 / 26

26 / 26 About Motor Selection Some margin may be needed in practice, e.g., T N > 1.15T M,ef Previous selection criteria take in the account the thermal balance only Maximum torque and maximum speed of motors are limited Remember to check that the motor is able to produce the required torque at all speeds