Electrical Machine Design

Chapter 6
Thermal Design and Cooling Systems

Bachelors Program in Electrical Engineering
Electrical Machine Design   •   Bachelor

Thermal Design and Cooling Systems

Introduction

Thermal design determines whether an electrical machine can deliver its specified torque, power, efficiency, and service life without exceeding the allowable temperatures of its insulation, bearings, magnets, conductors, or lubricants. Electromagnetic losses become heat, and that heat must be stored temporarily or transferred to the surroundings.

Temperature is also coupled to electromagnetic performance. Copper resistance increases with temperature, permanent magnets may lose flux, semiconductor-fed machines may experience additional harmonic loss, and excessive thermal cycling can fatigue insulation and joints. Thermal design is therefore an integral part of electromagnetic, mechanical, and reliability design rather than a final verification step.

Learning Objectives

After completing this chapter, students should be able to:

  • identify the principal heat sources and heat-transfer paths in electrical machines;
  • formulate steady-state and transient lumped-parameter thermal networks;
  • estimate average temperatures and local hot-spot temperatures;
  • apply conduction, natural-convection, radiation, forced-air, and liquid-cooling equations;
  • size cooling systems using energy balances and thermal resistances;
  • coordinate insulation class, duty cycle, derating, sensors, and protection;
  • recognize the limitations of analytical models and the role of computational methods.

1. Heat Generation in Electrical Machines

The total machine loss is the heat-generation rate under thermally steady operating conditions:

Ploss=PCu+PFe+Pmech+Pstray+Paux

Here, PCu is conductor loss, PFe is core loss, Pmech includes bearing and windage losses, Pstray denotes additional load-dependent losses, and Paux represents auxiliary losses such as fans or pumps. All terms are measured in watts.

For a three-phase winding with phase resistance Rph and rms phase current Iph,

PCu=3Iph2Rph

The resistance must be evaluated at operating temperature. For copper over the usual machine temperature range,

R(T)=R(T0)[1+α0(T-T0)]

where α0≈0.00393 K-1 at T0=20°C. This linear relation is an engineering approximation; tabulated resistivity should be used where higher accuracy is required.

Core loss is commonly separated into hysteresis, classical eddy-current, and excess-loss components. Its spatial distribution depends on local flux-density waveform and frequency. Converter harmonics can increase both core and rotor losses beyond sinusoidal-supply predictions.

Important thermal constraints may include:

  • winding-insulation temperature;
  • permanent-magnet irreversible demagnetization temperature;
  • bearing and lubricant limits;
  • rotor-cage and retaining-sleeve temperature;
  • allowable coolant and enclosure temperatures;
  • limits imposed by nearby power electronics.

2. Heat-Transfer Mechanisms

2.1 Conduction

For one-dimensional steady conduction through a plane layer,

Q=\frac{kA(T1-T2)}{L}=\frac{T1-T2}{Rcond}
Rcond=\frac{L}{kA}

Here, Q is heat-transfer rate in watts, k is thermal conductivity in W·m⁻¹·K⁻¹, A is area in m², L is path length in metres, and Rcond is thermal resistance in K/W. The formula assumes constant conductivity, uniform area, and predominantly one-dimensional heat flow.

Contact interfaces introduce additional resistance because nominally touching surfaces make contact only at microscopic asperities. Slot liners, impregnation resin, air voids, oxide layers, and mechanical contact pressure strongly influence winding-to-core resistance.

For radial conduction through a cylindrical layer of length l,

Rcyl=\frac{\ln(r2/r1)}{2πkl}

This expression applies to axisymmetric radial conduction with no internal heat generation in the layer.

2.2 Convection

Convective heat transfer is represented by Newton's cooling law:

Q=hA(Ts-Tf)

where h is the convection coefficient in W·m⁻²·K⁻¹, Ts is surface temperature, and Tf is the appropriate bulk-fluid temperature. Thus,

Rconv=\frac{1}{hA}

The coefficient h is not a fluid property alone. It depends on geometry, orientation, flow velocity, surface condition, and whether the flow is laminar or turbulent.

2.3 Radiation

Radiative exchange between a small grey surface and large isothermal surroundings is approximated by

Qrad=εσA(Ts4-Tsur4)

Temperatures must be in kelvins. Here, ε is surface emissivity and σ=5.670×10-8 W·m-2·K-4 is the Stefan–Boltzmann constant. The expression assumes that the surroundings have a view factor close to unity. More complex enclosures require view factors and surface-to-surface radiation analysis.

Radiation may be linearized as

Qrad=hradA(Ts-Tsur)
hrad=εσ(Ts+Tsur)(Ts2+Tsur2)

with absolute temperatures. Radiation and convection act in parallel when both transfer heat from the same surface.

3. Lumped-Parameter Thermal Networks

A lumped-parameter thermal network, or LPTN, replaces a distributed machine by thermal nodes connected by resistances. Each node is assigned a uniform temperature and may possess thermal capacitance.

The thermal–electrical analogy is:

Thermal quantityElectrical analogue
Heat-transfer rate, WCurrent, A
Temperature difference, KVoltage, V
Thermal resistance, K/WElectrical resistance, Ω
Thermal capacitance, J/KElectrical capacitance, F
Heat generation, WCurrent source, A

Thermal capacitance is

Cth=mcp

where m is mass in kg and cp is specific heat capacity in J·kg⁻¹·K⁻¹. A lumped node is credible only when internal temperature gradients within that node are small relative to gradients between nodes. The Biot number, Bi=hLc/k, is one useful indicator; lumped treatment is often acceptable when Bi<0.1, although a complete machine network may require finer subdivision.

3.1 Two-Node Winding–Frame Model

Let θw=Tw-Ta and θf=Tf-Ta be winding and frame temperature rises above ambient. The winding is connected only to the frame through Rwf, while the frame transfers heat to ambient through Rfa.

Two-node winding-frame thermal network thermal reference, θ = 0 P_w enters node Winding θ_w R_wf Frame θ_f P_f R_fa Ambient C_w C_f
Correct two-node winding-frame thermal network
Figure: Correct two-node winding-frame thermal network

Figure: Two-node thermal network with heat inputs entering their nodes, winding-to-frame resistance, frame-to-ambient resistance, and connected thermal capacitances.

Energy conservation gives

Cw\frac{dθw}{dt}=Pw-\frac{θw-θf}{Rwf}
Cf\frac{dθf}{dt}=Pf+\frac{θw-θf}{Rwf}-\frac{θf}{Rfa}

At steady state, the derivatives vanish:

θf=(Pw+Pf)Rfa
θw=θf+PwRwf

These equations are valid for the topology shown. An additional direct winding-to-ambient path would require another resistance and corresponding heat-flow terms.

Worked Example 1: Steady Temperatures

A machine has Pw=900 W, Pf=500 W, Rwf=0.035 K/W, and Rfa=0.045 K/W. For an ambient temperature of 40°C,

θf=(900+500)(0.045)=63.0 K
θw=63.0+(900)(0.035)=94.5 K

Therefore, Tf=103.0°C and Tw=134.5°C. A separately estimated local hot-spot allowance must be added if the winding node represents an average temperature.

3.2 First-Order Transient Model

For a single node with constant loss P, resistance Rth, capacitance Cth, and initial rise θ0,

θ(t)=PRth+(θ0-PRth)e-t/τ
τ=RthCth

This solution assumes constant parameters, ambient temperature, and loss during the interval. Piecewise-constant duty cycles can be analysed by applying it successively to each operating interval.

Worked Example 2: Periodic Duty Cycle

A one-node machine has Rth=0.060 K/W, Cth=20,000 J/K, and therefore τ=1,200 s. It dissipates 2,000 W for 10 minutes, followed by zero loss for 5 minutes.

The heating and cooling factors are

a=e-600/1200=0.6065, \qquad b=e-300/1200=0.7788

The on-state equilibrium rise is PRth=120 K. At periodic equilibrium,

θmax=120+(θmin-120)a, \qquad θmin=bθmax

Thus,

θmax=\frac{120(1-a)}{1-ab}=89.5 K

and θmin=69.7 K. Using average loss alone would neglect the peak temperature and may produce an unsafe design.

4. Natural Cooling and Radiation

Natural cooling requires no fan or pump, offering high reliability and low acoustic noise. Its thermal performance is sensitive to enclosure orientation, surface area, fin spacing, altitude, and obstruction.

For natural convection from a vertical plate, calculate

RaL=\frac{gβ(Ts-T_∞)L3}{να}

where g is gravitational acceleration, β is volumetric thermal-expansion coefficient, L is vertical characteristic length, ν is kinematic viscosity, and α is thermal diffusivity. Fluid properties are evaluated approximately at the film temperature (Ts+T_∞)/2.

A widely used all-range correlation for an isothermal vertical plate is

NuL=\left[0.825+\frac{0.387RaL1/6}{[1+(0.492/Pr)9/16]8/27}\right]2
h=\frac{kNuL}{L}

The correlation is commonly applied for approximately 10-1≤RaL≤1012. It should not be transferred uncritically to enclosed passages or closely spaced fins.

Worked Example 3: Natural-Convection and Radiation Sizing

A painted vertical enclosure is limited to 80°C in 40°C ambient. Its exposed area is 0.80 m², vertical length is 0.60 m, and emissivity is 0.85. At the 60°C film temperature, take k=0.0287 W·m-1·K-1, Pr=0.71, ν=1.90×10-5 m2/s, α=2.68×10-5 m2/s, and β=1/333 K-1.

The Rayleigh number is

RaL=\frac{(9.81)(1/333)(40)(0.60)3}{(1.90×10-5)(2.68×10-5)}≈5.00×108

The correlation gives NuL≈99.4, so

h=\frac{(0.0287)(99.4)}{0.60}=4.75 W·m-2·K-1

Natural-convection heat removal is

Qconv=(4.75)(0.80)(40)=152 W

Radiative heat removal is

Qrad=(0.85)(5.670×10-8)(0.80)(353.154-313.154)≈229 W

Total removal is approximately 381 W. If 500 W must be rejected at the same temperatures, an approximate required area is

Areq=\frac{500}{h(80-40)+εσ(353.154-313.154)}≈1.05 m2

This area scaling is preliminary because changing the geometry may alter h, view factors, and fin efficiency. A prototype or validated computational model should confirm the design.

5. Forced-Air Cooling

Forced airflow increases the convection coefficient but adds fan power, noise, filters, ducts, and maintenance requirements. An overall air-energy balance gives

\dot ma=\frac{Q}{cp,a(Tout-Tin)}

and

\dot Va=\frac{\dot ma}{ρa}

where \dot ma is air mass flow in kg/s, \dot Va is volumetric flow in m³/s, and cp,a is specific heat in J·kg⁻¹·K⁻¹. This balance determines the minimum bulk airflow, not the local heat-transfer coefficient. Bypass flow and maldistribution require design margin.

Worked Example 4: Airflow Requirement

A machine rejects 3.2 kW to air. The allowed air-temperature increase is 15 K. Taking cp=1007 J·kg-1·K-1 and ρ=1.13 kg/m3,

\dot ma=\frac{3200}{(1007)(15)}=0.212 kg/s
\dot Va=\frac{0.212}{1.13}=0.188 m3/s

The fan must provide this flow at the actual system pressure rise, not at free-delivery conditions. Fan selection therefore uses the intersection of the fan curve and the duct-system curve.

6. Liquid and Direct Cooling

Liquid coolants have much greater volumetric heat capacity than air. Water jackets, cold plates, oil circuits, and glycol mixtures permit compact, high-power-density machines.

The coolant energy balance is

\dot mc=\frac{Q}{cp,cΔTc}

Channel-to-coolant resistance may be approximated as Rconv=1/(hA). The complete path also includes conduction through housings, contact interfaces, fouling layers, and spreading resistance.

Pump hydraulic power is

Phyd=Δp\dot V

The required electrical input is Ppump=Phyd/ηp, where ηp is overall pump efficiency.

Worked Example 5: Direct Liquid-Cooling Path

A stator cooling jacket must remove 6.0 kW using a water–glycol mixture with cp=3600 J·kg-1·K-1. The permissible coolant rise is 8 K.

\dot mc=\frac{6000}{(3600)(8)}=0.208 kg/s

If the channel coefficient is 1,800 W·m⁻²·K⁻¹ and wetted area is 0.45 m²,

Rconv=\frac{1}{(1800)(0.45)}=0.001235 K/W

The wall-to-mean-coolant temperature difference is approximately

ΔTwall-fluid=(6000)(0.001235)=7.41 K

With uniform heating, the mean coolant is about 4 K above its inlet temperature. The mean channel wall is therefore approximately 11.4 K above inlet temperature before adding solid conduction and interface temperature drops.

Direct cooling places coolant very close to, or in contact with, the heat-generating region. Examples include hollow conductors, dielectric-oil spray cooling, and cooling channels embedded near stator slots. Such systems reduce thermal resistance but introduce sealing, corrosion, erosion, dielectric compatibility, and pressure-integrity requirements.

7. Hot Spots and Electrothermal Coupling

A node temperature is generally a spatial average. Local hot spots may occur in end windings, slot centres, tooth tips, rotor bars, magnets, or regions of poor impregnation. A practical estimate is

Thot=Tnode+ΔTlocal

The local increment should come from detailed conduction analysis, finite-element or CFD results, embedded sensors, or calibrated prototype tests—not an arbitrary universal allowance.

Because winding loss depends on temperature, electrothermal calculations should be iterative:

  1. estimate temperature;
  2. calculate temperature-dependent resistance and loss;
  3. solve the thermal model;
  4. update temperature and repeat until convergence.

Strong cooling changes, saturation, and converter harmonics may also alter loss distribution. Under-relaxation can improve numerical stability in tightly coupled simulations.

8. Thermal Class, Duty, and Derating

IEC 60085 thermal classes are associated with insulation-system thermal endurance. Common class temperatures include 105, 120, 130, 155, 180, 200, and 220°C. The class temperature is not automatically the permissible measured winding temperature: ambient allowance, test method, hot-spot margin, contamination, mechanical stress, and required life must also be considered.

Machine duty classifications distinguish continuous operation from short-time, intermittent, starting, braking, and variable-load duties. Thermal time constants determine whether a short overload reaches steady temperature.

For an unchanged cooling system and approximately constant thermal resistance, allowable loss may be estimated from

Pallow=\frac{Tlimit-Ta}{Rth}

This is a steady-state relation. It is not valid without modification when Rth changes with speed, fan operation, orientation, or coolant conditions.

If copper loss dominates and resistance is treated as constant over a narrow range,

Iallow≈Ir\sqrt{\frac{Pallow}{Pr}}

This square-root rule is a special approximation, not a general machine derating law. Core, mechanical, harmonic, and temperature-dependent losses must be included where significant.

Altitude reduces air density and usually weakens air cooling. High ambient temperature reduces available temperature rise. Contaminated filters, blocked fins, low coolant flow, or reduced shaft speed in self-ventilated machines also require derating.

9. Thermal Protection and Monitoring

Common protective devices include resistance temperature detectors, thermistors, thermocouples, bearing-temperature sensors, coolant-flow switches, and thermal relays. Winding temperature can also be estimated from measured resistance:

T2=T0+\frac{R2/R1-1}{α0}

where R1 is resistance at known temperature T0 and R2 is the later resistance, assuming the linear resistance model.

Protection should normally provide graded responses: warning, controlled derating, shutdown, and—where safety requires—independent emergency trip. Sensor placement, failure detection, calibration, electromagnetic compatibility, and diagnostic coverage are essential. A sensor far from the hot spot may provide false confidence.

10. Illustrative Case: Cooling a Traction Motor

Background

An illustrative 75 kW permanent-magnet traction motor uses a stator water–glycol jacket. Continuous total loss is 4.8 kW, including 3.2 kW winding loss. Coolant enters at 51°C with mass flow 0.17 kg/s and cp=3600 J·kg-1·K-1. The calibrated winding-to-mean-coolant resistance is 0.018 K/W, and the local winding hot-spot increment is 8 K.

Problem

The designer must verify continuous temperature, a 60 s overload, and pump demand. The design winding limit is 140°C.

Analysis and Solution

The coolant rise is

ΔTc=\frac{4800}{(0.17)(3600)}=7.84 K

The mean coolant temperature is approximately 51+7.84/2=54.9°C. Continuous average winding temperature is

Tw=54.9+(3200)(0.018)=112.5°C

The predicted hot spot is 112.5+8=120.5°C, leaving about 19.5 K design margin.

During a 60 s peak, winding loss increases by 3.0 kW above its continuous value. Taking an effective winding capacitance of 12,000 J/K, the incremental time constant is

τ=(0.018)(12000)=216 s

The additional winding rise after 60 s is

Δθ=3000(0.018)(1-e-60/216)=13.1 K

The estimated peak hot spot is therefore 133.6°C, below the 140°C design limit.

If circuit pressure drop is 45 kPa and coolant density is 1,040 kg/m³, then \dot V=0.17/1040=1.63×10-4 m3/s. Hydraulic power is about 7.35 W. At 45% overall pump efficiency, electrical pump input is approximately 16.3 W.

Outcome

The design satisfies the stated thermal limit under the model assumptions. Validation should include worst-case inlet temperature, aged coolant, production tolerances, reduced flow, magnet temperature, sensor error, and repeated overload cycles.

Learning Points

  • Mean coolant temperature, not inlet temperature alone, governs the average thermal path.
  • Short overload capability depends on both thermal resistance and capacitance.
  • A low pump-power value does not eliminate requirements for pressure integrity and flow monitoring.
  • Calibrated hot-spot increments remain necessary even when average winding temperature is acceptable.

11. Modelling Limitations and Contemporary Practice

Lumped networks are computationally efficient and suitable for design iteration and real-time estimation, but their parameters vary with speed, temperature, flow regime, impregnation quality, and manufacturing tolerance. Computational fluid dynamics resolves airflow and coolant distribution, while thermal finite-element analysis resolves conduction and local gradients. These methods still require reliable losses, boundary conditions, and experimental validation.

Reduced-order models derived from CFD or test data are increasingly embedded in digital twins and motor controllers. Machine-learning estimators can infer inaccessible temperatures from current, speed, coolant, and sensor histories. Such models should be constrained by physical principles and validated outside their training conditions; extrapolation without uncertainty assessment is unsafe.

Standards concerning insulation systems, machine ratings, ingress protection, hazardous locations, coolant handling, and functional safety must be identified early. Environmentally responsible design also considers coolant toxicity, leakage, fan and pump energy, repairability, material life, and end-of-life disposal.

Design Best Practices

  • Develop the thermal network concurrently with electromagnetic design.
  • Map each major loss to its actual physical location.
  • Include radiation in natural-cooled designs when surface temperature and emissivity make it significant.
  • Use temperature-dependent material and fluid properties where necessary.
  • Analyse normal, overload, stalled, blocked-cooling, and sensor-failure conditions.
  • Verify fan and pump performance against system resistance curves.
  • Calibrate uncertain contact and convection resistances using tests.
  • Place sensors near critical regions and design fail-safe protective action.
  • Document assumptions, margins, model validity, and uncertainty.

Chapter Summary

Electrical-machine thermal design converts distributed losses into temperature predictions through conduction, convection, radiation, and thermal storage models. Lumped thermal networks provide efficient steady-state and transient analysis when their topology, parameters, and node definitions reflect the physical machine. Natural cooling requires combined convection and radiation analysis; forced-air and liquid systems require both energy balances and local heat-transfer assessment. Direct cooling can produce high power density but creates fluid compatibility and integrity challenges.

Safe design coordinates average and hot-spot temperatures with insulation class, component limits, ambient conditions, duty cycle, derating, and protection. Analytical models should be validated using finite-element analysis, CFD, and testing where spatial detail or uncertain boundary conditions are important.

Key Terms

Thermal resistance: Temperature difference divided by heat-transfer rate, measured in K/W.

Thermal capacitance: Heat-storage capacity of a body, equal to mass times specific heat.

Lumped-parameter thermal network: A discrete model of temperatures, heat sources, resistances, and capacitances.

Hot spot: Local region with the highest relevant component temperature.

Conduction: Heat transfer through a material caused by a temperature gradient.

Convection: Heat transfer between a surface and a moving fluid.

Natural convection: Buoyancy-driven fluid motion without a fan or pump.

Forced convection: Fluid motion created by mechanical equipment or imposed flow.

Radiation: Heat transfer by electromagnetic emission and absorption.

Emissivity: Ratio of a surface's radiative emission to that of an ideal blackbody.

Rayleigh number: Dimensionless measure governing buoyancy-driven convection.

Nusselt number: Dimensionless surface heat-transfer coefficient.

Thermal time constant: Product of thermal resistance and capacitance in a first-order model.

Film temperature: Representative mean temperature used to evaluate convection properties.

Derating: Reduction of allowable output to satisfy operational constraints.

Insulation thermal class: Classification associated with insulation-system thermal endurance.

Direct cooling: Cooling that places fluid very near to or in contact with the heat source.

Electrothermal coupling: Interaction between electrical losses and temperature.

Duty cycle: Sequence and duration of load, rest, starting, or braking conditions.

Review Questions

Short Answer Questions

  1. Identify the principal categories of loss in an electrical machine.
  2. Define thermal resistance and thermal capacitance, including their units.
  3. Why must absolute temperature be used in the radiation equation?
  4. What physical information does a machine's thermal time constant provide?
  5. Why may an average winding temperature fail to protect insulation adequately?

Long Answer Questions

  1. Derive the governing equations of the two-node winding–frame thermal network and explain every term.
  2. Compare natural-air, forced-air, jacket-liquid, and direct cooling in terms of performance, reliability, and design risk.
  3. Explain how insulation class, ambient temperature, hot-spot allowance, and expected life should be coordinated.
  4. Discuss the sources of uncertainty in an electrical-machine thermal model and suitable validation methods.
  5. Explain how duty cycle and temperature-dependent copper resistance produce electrothermal coupling.

Analytical / Application Questions

  1. A 1.5 kW heat source is connected to ambient through 0.055 K/W and has thermal capacitance 18 kJ/K. Determine its steady temperature rise and its rise after 12 minutes from ambient.
  2. Size the minimum coolant mass flow required to remove 8 kW when coolant specific heat is 3,700 J·kg⁻¹·K⁻¹ and its allowable rise is 10 K. Discuss why the resulting flow alone does not determine winding temperature.
  3. Construct a three-node thermal network for winding, core, and housing. State the required resistances, heat inputs, capacitances, boundary conditions, and measurements needed for calibration.

Multiple Choice Questions with Answers

Question 1

In the two-node winding–frame network, the winding is connected only to the frame through Rwf, and the frame is connected to ambient through Rfa. What is the steady frame temperature rise?

A. PwRwf

B. [[(P_w+P_f)R_{fa}]]

C. PfRwf

D. [[(P_w-P_f)R_{fa}]]

Correct Answer: B. [[(P_w+P_f)R_{fa}]]

Question 2

Which unit is correct for thermal resistance?

A. W/K

B. J/K

C. K/W

D. W·K

Correct Answer: C. K/W

Question 3

A body has mass 8 kg and specific heat 500 J·kg⁻¹·K⁻¹. What is its thermal capacitance?

A. 62.5 J/K

B. 4,000 J/K

C. 8,500 J/K

D. 40,000 J/K

Correct Answer: B. 4,000 J/K

Question 4

A cooling system removes 4,000 W using a fluid with cp=2,000 J·kg-1·K-1 and a 10 K temperature rise. What mass flow is required?

A. 0.020 kg/s

B. 0.100 kg/s

C. 0.200 kg/s

D. 2.000 kg/s

Correct Answer: C. 0.200 kg/s

Question 5

Why must temperatures in the Stefan–Boltzmann radiation equation be expressed in kelvins?

A. Radiation depends on the fourth power of absolute temperature.

B. Kelvins make emissivity equal to unity.

C. Convection coefficients are defined only in kelvins.

D. Celsius cannot express temperature differences.

Correct Answer: A. Radiation depends on the fourth power of absolute temperature.

Question 6

A first-order thermal model has Rth=0.08 K/W and Cth=15,000 J/K. What is its thermal time constant?

A. 120 s

B. 1,200 s

C. 12,000 s

D. 187,500 s

Correct Answer: B. 1,200 s

Question 7

Which change most directly increases the radiative heat transfer from an enclosure at a fixed surface and surroundings temperature?

A. Reducing surface emissivity

B. Increasing surface emissivity

C. Reducing specific heat

D. Increasing thermal capacitance

Correct Answer: B. Increasing surface emissivity

Question 8

Under which condition is the approximation Iallow≈Ir\sqrt{Pallow/Pr} most defensible?

A. Copper loss dominates and resistance is approximately constant over the range.

B. Mechanical loss dominates and varies linearly with current.

C. Cooling resistance changes sharply with speed.

D. Core loss is independent of voltage and frequency.

Correct Answer: A. Copper loss dominates and resistance is approximately constant over the range.

Question 9

A natural-cooled surface has h=5 W·m-2·K-1, area 1.2 m², and a surface-to-air difference of 30 K. What is its convective heat rejection?

A. 25 W

B. 150 W

C. 180 W

D. 360 W

Correct Answer: C. 180 W

Question 10

What is the principal purpose of adding thermal capacitance to a lumped thermal network?

A. To represent transient heat storage

B. To represent steady radiation only

C. To eliminate temperature gradients

D. To calculate electrical power factor

Correct Answer: A. To represent transient heat storage

Electrical Machine Design
Chapter 6 of 15