Retardation test of DC SHUNT MACHINE:

Retardation test (or) running 🏃 down test of dc machine
This method is used to find the constant losses(friction + windage + iron losses)

This is also a noload test. We can calculate the efficiency of a dc machine by knowing the armature amd shunt copper losses at a given load.

Consider a DC SHUNT MOTOR running at noload.
If the supply to the
                                   1.Armature is cut off 📴 and the supply to the field remains same then the KINETIC ENERGY ⚡ is used to overcome (friction + Windage +Field copper ) no load losses
                                   2.Armature as well as field to the field is cut off then the KINETIC ENERGY ⚡ is Used to overcome only the friction losses  + Windage losses.  (their will be no field copper loss because in second case their is no supply to the field)

How to conduct it:
It is very easy to conduct it we have to remove the supply and calculate the rate at which the speed falls.

Losses can be calculated by using formula (derivation):

     Since kinetic energy ⚡ is used to overcome losses:

                  W=(d(Iw^2)/2dt)
where     w(omega)= 2πN/60  in rad/sec
         
On differentiation we get
                   W= 2Iw (dw/dt)
Now substitute w= 2πN/60.
We get
                W= (((2π)÷(60))^2)(IN(dN/dt))

Where
          I=Moment of inertia of armature.
          N=speed of armature in rpm.

Accordingly  if the supply to the armature is cut off and supply to field is not cut off then 'W' indicates friction +windage+Field copper losses.

If supply to both armature as well as the field is Cut Off Then 'W' indicates only Friction + Windage Losses.

To find I:
                But it is difficult to find the moment of inertia of Armature i.e 'I' . It can be easily found by conducting another experiment

Firstly,the retardation test is conducted and dN/dt1 is noted

Secondly,the shaft of the motor is keyed to the flywheel 🎡 of known moment of inertia I1 and the time for the fall of same speed is noted i.e dN/dt2

Since in both cases same motor is used so losses are equal

Losses in case1:
W= (((2π)÷(60))^2)(IN(dN/dt1))

Losses in case2:
W=(((2π)÷(60))^2)((I+I1)N(dN/dt2))

N=speed in rpm
I=Moment of inertia of armature.
I1=Moment of inertia of flywheel
Equating above two equations we get I value if I1 value is known

Hence by retardation test we can calculate losses.