Swinburne Test

 

This test is performed in order to find out

• Constant losses
• Efficiency at any load
• The Swinburne test is no load test in order to find out constant losses of the DC shunt motor or DC compound motor. 
• As it is no load test, it is not applicable for DC series motor. 
• The rated supply voltage is given to the DC machine. The rated speed of the DC machine is set by field rheostat.

Let

DC supply voltage = V

No load input current = I_o

Field current = I_Sh

No load armature current I_ao = I_o - I_sh

The no load input power = V I_o

The no load input power supply the following losses

Iron losses in the core

Friction and windage losses and

No load armature copper losses = I_o²R_a

Where R_a is hot armature resistance ( hot armature resistance at any temperature is find out by R_t = R_o ( 1 + α_ot )

• Resistance at 55^o C   R₅₅ = R_o( 1 + 55α_o )
• Resistance at 30^o C   R₃₀ = R_o( 1 + 30α_o )
• Resistance at 55^o C = R₅₅ /R₃₀  = ( 1 + 55α_o ) / ( 1 + 30α_o )

Where α₀ is temperature co – efficient of resistance = 1 / 234.5 at 0^oC

Constant Losses

• The constant losses can find out by subtracting no load armature copper losses from no load input power. 
• The constant losses is constant from no load to full load therefore the DC machine efficiency is find out at any load condition.

Constant losses W_c

= No load input power – No load armature copper losses

=VI_o – I_o²R_a

Efficiency : DC Motor

Input power = VI ( where I is full load current )

Output power = Input power – ( Armature copper losses + Field copper losses + Constant losses )

= VI – ( I_a²R_a +W_c )

Efficiency of DC motor = ( Output power / Input power ) × 100 %

Efficiency : Generator

Output power = VI

Input power = Output power + Losses

 = VI +I_a²R_a + W_c

Efficiency of DC Generator = ( Output power / Input power ) × 100 %

Observation table

Supply voltage	No load input current	Field current	Speed	Hot armature resistance
V	Io	ISH	N	Ra

 

Calculation

Motor efficiency at full load current

Input power = VI

                    = __________

Output power = Input power + armature copper losses + constant losses

W₀ = W_i + I_a²R_a + W_c

       = __________

DC Motor efficiency = ( W₀ / W_i ) × 100%

DC Generator at full load efficiency

Output power = VI

                       = __________

Input power = VI + I_a²R_a + W_c

                    = __________

DC Generator efficiency = ( W₀ / W_i ) × 100%

                                        = __________

Observation

Whether the efficiency of DC machine as DC motor and DC Generator is same?

Conclusion

The efficiency of the DC Machine can find out at any load condition.

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