Thermal Model of a MotorPublished on 24/2/2012 & updated on 30/7/2018
At time ‘t’, let the motor has following parametersp1 = Heat developed, Joules/sec or watts p2 = Heat dissipated to the cooling medium, watts - W = Weight of the active parts of the machine. h = Specific heat, Joules per Kg per oC. A = Cooling Surface, m2 d = Co-efficient of heat transfer, Joules/Sec/m2/oC θ = Mean temperature rise oC
Now, if time dt, let the temperature rise of the machine be dθ, Therefore, heat absorbed in the machine = (Heat generated inside the machine – Heat dissipated to the surrounding cooling medicine) Where, dθ = p1dt - p2dt................(i) Since, p2 = θdA................(ii) Substituting (ii) in (i), we get Here, C is called the thermal capacity of the machine in watts/oC and D is the heat dissipation constant in watts/oC.
When we acquire the first order differential equation of the equation - We obtain the value of K by putting t = 0 in equation (iii) and get the solution as
So, from the above equation we can find out the rise in temperature inside a working machine, which is very near to being accurate and if we plot a graph for the variation of temperature risk with time during heating and cooling and thus the thermal modeling of a motor gets completed.
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