## 06/12/2020

### Swinburne Test

This test is performed in order to find out

• Constant losses
• 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 = Io

Field current = ISh

No load armature current Iao = Io - Ish

The no load input power = V Io

The no load input power supply the following losses

Iron losses in the core

Friction and windage losses and

No load armature copper losses = Io2Ra

Where Ra is hot armature resistance ( hot armature resistance at any temperature is find out by Rt = Ro ( 1 + αot )

• Resistance at 55o C   R55 = Ro( 1 + 55αo )
• Resistance at 30o C   R30 = Ro( 1 + 30αo )
• Resistance at 55o C = R55 /R30  = ( 1 + 55αo ) / ( 1 + 30αo )

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

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 Wc

=VIo – Io2Ra

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 – ( Ia2Ra +Wc )

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

Efficiency : Generator

Output power = VI

Input power = Output power + Losses

= VI +Ia2Ra + Wc

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 = Wi + Ia2Ra + Wc

= __________

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

DC Generator at full load efficiency

Output power = VI

= __________

Input power = VI + Ia2Ra + Wc

= __________

DC Generator efficiency = ( W0 / Wi ) × 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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