Versions Compared

Key

  • This line was added.
  • This line was removed.
  • Formatting was changed.

To ensure our motor controllers are always below their rated temperature of 70C (See data sheet to confirm) we must design a heat sink to keep the motor controllers at least 10C cooler. To do this we will integrate a cold plate into the skin of the aerobody. Since the vehicle will also be racing in very warm climates it is not unreasonable to experience ambient temperatures of 45C outside the car. We must therefore design a heat-sink to function with a 15C temperature delta (Δt = 60C-45C). This page details the required heat flow and thermal conductivity to ensure the motor controllers are kept at a functional temperature. 

Motor Controller Efficiency

The Power loss of the Wave-sculptor 22 motor controllers controller is given by the following equations equation where:

Ploss = Req*Io2 + (α*Io+β)*Vbus + Cfeq*V2bus


DescriptionValue
VbusBus (Battery) Voltage of the controller129.6V Nominal (100V - 151.2V)
IoOuput current of the controllerPmax/Vbus (Roughly, This will actually be off by ~1.2x)
ReqEquivalent resistance of the entire controller0.0108 Ohms
βConstant component of the switching loss (per unit of bus voltage)0.018153
αLinear component of the switching loss (per unit of bus voltage)0.003345
CfeqEquivalent capacitance*frequency of the controller0.00015625


Using an approximated value of Pmax for the entire car as 9kW, including mechanical and non-controller electrical losses, the per motor controller peak power is 4.5kW. Meaning the approximated output current will vary between 29.8A - 45A  (Nominal 34.7A) 

From this we can plot the efficiency of the motor controllers versus the voltage and the current using

...

the attached graph.

...

 https://www.desmos.com/calculator/kwklxrs8g3

From this plot we can clearly see as power increases the optimal voltage also increases. This optimal voltage range lies in our battery bus range and as such should be relatively optimized to minimize motor controller power losses

...

.

Using the above equation with a peak power of 4500W and a voltage of 80V (lowest reasonable bus voltage) we find the maximum power loss to be 50W. We will assume all of this energy is transferred into heat and as such this will be our required heat flow.


Heat-Sink Design

Since the cold plate will be directly attached to the exterior we need to find the minimum thermal conductivity of the cold plate as well as the minimum air speed to dissipate dissipate the heat through forced convection.

Heat Conduction

First we must find the maximum allowable thickness of the cold plate as this will impact if all power created in the motor controller can be dissipated through the plate. The simplified equation of heat flow through a material is governed by the following conduction equation:

Q = (k*A/T)*Δt

Where k is the thermal conductivity of the material stackup in question, A is the contact area, and T is the thickness.


DescriptionValue
AContact area0.029 m^2
TStack up thicknessTBD
kThermal Conductivity of the stackup, this value is approximated but should be recalculated with a correct connection200 W/m-K


Since A, Q, and Δt are known we can find the minimum value of k/T. 

k/T = Q/(A*Δt) = 50/(0.029*15) = 115 W/K

Since k is the series addition of the conductivity of the cold plate and the connection between the cold plate and the heat source we can approximate it as the conduction of aluminum and apply a reasonable safety factor once complete (For reference this connection should be made with a thermal interface material, TIM, to ensure maximum heat conduction across the joint). Therefore the maximum plate thickness is:

T = k/115 = 200/115 = 1.74m

This value is comically high and as such we do not have to worry if the stack up is exclusively aluminum.



To ensure the motor controllers will never over temp we will build safety factors into our design. One of them is to approximate all heat transfer from the controllers as being lost to forced air convection. The following details the minimum air speed required for this to occur. This number will then be given to the strategy team to ensure this speed is always kept while at maximum power draw.

Rate of convection heat transfer is governed by Newtons law of cooling, Qconv = h*A*(Δt) bb From this equation we will assume all waste heat from the motor controller will be lost through convection and so we can set Qconv = 50W. The area of the cold plate is taken from the dimensions of the motor controller and is 0.02925 m^2. We can now easily find the required value of the heat transfer coefficient, h.

h = Qconv/(A*(Δt) = 50/(15*0.02925) = 113.96


To



Reynolds number (Re): ratio of inertia forces to viscous forces in the fluid

Re = Vδ/v


Prandtl Number (Pr): a non-dimensional measure of relative thickness of the velocity and thermal boundary layer 

Pr = μCp/k 

ρ

Nusselt number (Nu): a non‐dimensional heat transfer coefficient representing the enhancement of heat transfer through a fluid as a result of
convection relative to conduction across the same fluid layer

Nu = h*δ/k 

For a flate plate this becomes → Nu = C*Re*Prn where C, m, and n are constants.

Since the flow will be turbulent at this point C = 0.037, m = 0.8, and n = 1/3



DescriptionValue to 
ρDensity1.109 kg/m3
vKinematic Viscosity, ρ/μ 1.750*10-5 m2/s
μDynamic Viscosity1.941*10-5 kg/m-s
kThermal Conductivity0.0269 W/m-K
h

δCharacteristic Length, for our purposes this will be LL = 0.117
CpSpecific heat capacity 1007 J/Kg-K
VvelocityTBD
LLength of the plate along the direction of flow0.117m
PrPrandtl Number0.7241
Area normal to the flow0.029 m^2

*Values in the above table are for air at 45C using the following database https://www.engineersedge.com/physics/viscosity_of_air_dynamic_and_kinematic_14483.htm




For more detailed information see the linked data sheet:

Google drive docs
urlhttps://drive.google.com/file/d/1XMdPXyh2Yv-zCBCM8KNiVlM04B9diN-6/view?usp=sharing